Samacheer Kalvi Class 12 Chemistry Solutions Chapter 11 Hydroxyl Compounds and Ethers

Part - I Text Book Evaluation

I. Choose the correct answer

 

Question 1. blue colour in victormayer's test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X?
(a) CH3 CH (OH) CH2CH3
(b) CH3 – CH(OH) – CH3
(c) CH3 – C (OH) (CH3)2
(d) CH3 – CH2 – CH (OH) – CH2 – CH3
Answer: (a) CH3 CH (OH) CH2CH3
In simple words: When an alcohol reacts with metallic sodium, it produces hydrogen gas. By measuring the amount of hydrogen, we can find the alcohol's molar mass and then identify its structure. The alcohol that fits the data is CH3CH(OH)CH2CH3, which is a secondary alcohol.

🎯 Exam Tip: Always remember that the reaction 2R-OH + 2Na → 2RONa + H2 shows that 2 moles of alcohol produce 1 mole of hydrogen gas. This ratio is crucial for calculations.

 

Question 2. Which of the following compounds on reaction with methyl magnesium bromide will give tertiary alcohol.
(b) propanoic acid
(c) methyl propanoate
(d) acetaldehyde
Answer: (c) methyl propanoate
Solution:
When methyl propanoate reacts with methyl magnesium bromide, it first forms an intermediate, which on further reaction and hydrolysis gives a tertiary alcohol. This reaction involves two additions of the Grignard reagent to the ester.
\[ \text{CH}_3\text{MgBr} + \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{OCH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C}(\text{OMgBr})(\text{CH}_3)\text{OCH}_3 \]
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OMgBr})(\text{CH}_3)\text{OCH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{CH}_3 + \text{Mg}(\text{OCH}_3)\text{Br} \]
\[ \text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{CH}_3 + \text{CH}_3\text{MgBr} \rightarrow \text{CH}_3\text{CH}_2\text{C}(\text{OMgBr})(\text{CH}_3)_2 \]
\[ \text{CH}_3\text{CH}_2\text{C}(\text{OMgBr})(\text{CH}_3)_2 \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{C}(\text{OH})(\text{CH}_3)_2 \]
In simple words: Esters react with two equivalents of Grignard reagent, followed by hydrolysis, to produce tertiary alcohols. Methyl propanoate gives a tertiary alcohol through this process.

🎯 Exam Tip: Remember that esters react twice with Grignard reagents to yield tertiary alcohols, while aldehydes (except formaldehyde) yield secondary alcohols, and ketones yield tertiary alcohols.

 

Question 3. The X is
(a) \( \text{H}_3\text{C}\text{C(CH}_2\text{)} \text{CH}_2\text{OH} \)
(b) \( \text{H}_3\text{C}\text{CH}_2\text{CH(OH)}\text{CH}_2\text{CH}_3 \)
(c) \( \text{H}_2\text{C=C(CH}_2\text{)} \text{CH}_2\text{OH} \)
(d) None of these
Answer: (a) \( \text{H}_3\text{C}\text{C(CH}_2\text{)} \text{CH}_2\text{OH} \)
Solution: hydro boration – Anti markownikoff product i.e \( \text{CH}_3 – \text{CH}_2 – \text{CH} – \text{CH}_2 – \text{CH}_2 – \text{OH} \) The reaction proceeds via hydroboration, which is an anti-Markownikoff addition, meaning the -OH group adds to the less substituted carbon atom of the alkene. This leads to the formation of primary alcohols.
In simple words: This reaction uses borane and hydrogen peroxide, which is a special way to add water to an alkene. It makes the alcohol attach to the less crowded carbon, which is the opposite of the usual rule.

🎯 Exam Tip: Hydroboration-oxidation is a key reaction for synthesizing anti-Markownikoff products, specifically primary alcohols from terminal alkenes.

 

Question 4. In the reaction sequence, \( \text{CH}_2=\text{CH}_2 \xrightarrow{\text{HOCl}} \text{A} \xrightarrow{\text{X}} \text{Ethane} - 1,2-\text{diol} \). A and X respectively are
(a) Chloroethane and NaOH
(b) ethanol and H2SO4
(c) 2 – chloroethan – 1 – ol and NaHCO3
(d) ethanol and H2O
Answer: (c) 2 – chloroethan – 1 – ol and NaHCO3
Solution: When ethene reacts with hypochlorous acid (\( \text{HOCl} \)), it forms 2-chloroethan-1-ol (A). This compound (A) then reacts with sodium bicarbonate (\( \text{NaHCO}_3 \)) to produce ethane-1,2-diol. The bicarbonate removes the chloride and replaces it with a hydroxyl group, giving the diol.
\[ \text{CH}_2 = \text{CH}_2 \xrightarrow{\text{HOCl}} \text{CH}_2(\text{OH})-\text{CH}_2\text{Cl} \]
\( \implies \) Compound A is 2-chloroethan-1-ol.
\[ \text{CH}_2(\text{OH})-\text{CH}_2\text{Cl} \xrightarrow{\text{NaHCO}_3} \text{CH}_2(\text{OH})-\text{CH}_2(\text{OH}) + \text{NaCl} + \text{CO}_2 \]
\( \implies \) X is \( \text{NaHCO}_3 \).
In simple words: Ethene first gets a chlorine and an OH group, then the chlorine is replaced by another OH using sodium bicarbonate to make a two-OH alcohol called ethane-1,2-diol.

🎯 Exam Tip: Recognize that \( \text{HOCl} \) adds an -OH and a -Cl to an alkene, and subsequent treatment with a base like \( \text{NaHCO}_3 \) or \( \text{NaOH} \) can replace the -Cl with another -OH, forming a diol.

 

Question 5. Which one of the following is the strongest acid
(a) 2 – nitrophenol
(b) 4 – chlorophenol
(c) 4 – nitrophenol
(d) 3 – nitrophenol
Answer: (c) 4 – nitrophenol
In simple words: A compound is a stronger acid if it can release its hydrogen ion easily. Nitro groups help make phenols stronger acids, especially when they are at the para position, because they pull electrons and stabilize the negative charge left behind.

🎯 Exam Tip: Electron-withdrawing groups (like nitro) increase acidity, especially when positioned ortho or para to the -OH group, due to resonance stabilization of the phenoxide ion.

 

Question 6. on treatment with Con. H2SO4, predominately gives
(a) \( \text{CH}_2 \) cyclohexene
(b) \( \text{CH}_3 \) methylcyclohexene
(c) \( \text{CH}_3 \) methylcyclohexene (different position)
(d) \( \text{CH}_3 \) methylcyclohexene (different position)
Answer: (b) \( \text{CH}_3 \) methylcyclohexene
Solution: This reaction is an elimination (dehydration) of alcohol following Saytzeff's rule. Saytzeff's rule states that in an elimination reaction, the most stable (most substituted) alkene is formed as the major product. The starting material is a cyclohexanol with a methyl group. After protonation and loss of water, a carbocation is formed. Rearrangement (methyl shift) can occur to form a more stable tertiary carbocation, followed by the removal of a hydrogen to form the most substituted alkene. The product (b) is 1-methylcyclohexene.
In simple words: When a cyclohexanol is heated with concentrated sulfuric acid, it loses water to form an alkene. The main product will be the alkene that has the most alkyl groups attached to its double bond, following a rule called Saytzeff's rule.

🎯 Exam Tip: For dehydration reactions, always consider carbocation rearrangements to form more stable intermediates before applying Saytzeff's rule to determine the major alkene product.

 

Question 7. Carbolic acid is
(a) Phenol
(b) Picric acid
(c) benzoic acid
(d) phenylacetic acid
Answer: (a) Phenol
In simple words: Phenol is a chemical compound where a hydroxyl group is directly attached to a benzene ring. It is also known as carbolic acid.

🎯 Exam Tip: Many common organic compounds have both a systematic IUPAC name and a widely used common name. Knowing these common names, like "carbolic acid" for phenol, is important.

 

Question 8. Which one of the following will react with phenol to give salicyladehyde after hydrolysis
(a) Dichloro methane
(b) trichioroethane
(c) trichloro methane
(d) CO2
Answer: (c) trichloro methane (Riemer Tiemann reaction)
In simple words: The Riemer Tiemann reaction uses trichloro methane with a strong base to add an aldehyde group to phenol, creating salicylaldehyde. This reaction is very specific for making ortho-substituted products.

🎯 Exam Tip: The Riemer Tiemann reaction is a characteristic reaction of phenols used to introduce an aldehyde group at the ortho position, making salicylaldehyde a key product.

 

Question 9. \( \text{(CH}_3\text{)}_3 - \text{C-CH(OH)} \text{CH}_3 \xrightarrow{\text{Con H}_2\text{SO}_4} \text{X (major product)} \)
(a) \( \text{(CH}_3\text{)}_3 \text{CCH} = \text{CH}_2 \)
(b) \( \text{(CH}_3\text{)}_2\text{C} = \text{C(CH}_3\text{)}_2 \)
(c) \( \text{CH}_2 = \text{C(CH}_3\text{)}\text{CH}_2 – \text{CH}_2 - \text{CH}_3 \)
Answer: (b) \( \text{(CH}_3\text{)}_2\text{C} = \text{C(CH}_3\text{)}_2 \)
Solution: The dehydration of 3,3-dimethyl-2-butanol with concentrated \( \text{H}_2\text{SO}_4 \) follows Saytzeff's rule. The initial carbocation forms at the secondary carbon. However, a methyl shift occurs to form a more stable tertiary carbocation, which then eliminates a proton to form the most substituted alkene. The product formed is 2,3-dimethylbut-2-ene.
In simple words: When this alcohol loses water in the presence of sulfuric acid, it rearranges to form the most stable alkene. This stable alkene has its double bond surrounded by many other carbon groups.

🎯 Exam Tip: In dehydration reactions of alcohols, always check for possible carbocation rearrangements (hydride or alkyl shifts) that can lead to a more stable carbocation, before the final elimination step.

 

Question 10. The correct IUPAC name of the compound, \( \text{H}_3\text{C-CH(Cl)-CH(CH}_3\text{)-CH(CH}_3\text{)-CH}_2\text{OH} \)
(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol
(b) 2, 3 – dimethyl – 4 – chloropentan – 1 – ol
(c) 2, 3, 4 – trimethyl – 4 – chiorobutan – 1 – ol
(d) 4 – chioro – 2, 3, 4 – trimethyl pentan – 1 – ol
Answer: (a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol
In simple words: To name this compound, first find the longest carbon chain that includes the -OH group. Then, number the chain starting from the end closest to the -OH group. Finally, name the chlorine and methyl groups based on their positions.

🎯 Exam Tip: When naming alcohols with multiple substituents, the -OH group takes precedence in numbering the main chain, ensuring it receives the lowest possible number.

 

Question 11. Assertion: Phenol is more acidic than ethanol Reason: Phenoxide ion is resonance stabilized
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer: (a) if both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Phenol is a stronger acid than ethanol because the ion formed after phenol loses its hydrogen is very stable due to electron sharing across the ring. Ethanol's ion is not stable this way.

🎯 Exam Tip: Resonance stabilization of the conjugate base is a key factor in determining the acidity of organic compounds. A more stable conjugate base means a stronger acid.

 

Question 12. In the reaction \( \text{Ethanol} \xrightarrow{\text{PCl}_5} \text{X} \xrightarrow{\text{alc.KOH}} \text{Y} \xrightarrow{\text{H}_2\text{SO}_4\text{/H}_2\text{O}} \text{Z} \). Z the product Z is
(a) ethane
(b) ethoxyethane
(c) ethylbisuiphite
(d) ethanol
Answer: (d) ethanol
Solution: The reaction sequence is as follows:
Ethanol reacts with \( \text{PCl}_5 \) to form ethyl chloride (X):
\[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{CH}_2\text{Cl} \]
Ethyl chloride (X) reacts with alcoholic \( \text{KOH} \) to undergo elimination, forming ethene (Y):
\[ \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{alc.KOH}} \text{CH}_2=\text{CH}_2 \]
Ethene (Y) undergoes hydration with \( \text{H}_2\text{SO}_4\text{/H}_2\text{O} \) (hydrolysis) to reform ethanol (Z):
\[ \text{CH}_2=\text{CH}_2 \xrightarrow{\text{H}_2\text{SO}_4\text{/H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} \]
In simple words: Ethanol first turns into ethyl chloride, then into ethene. Finally, ethene reacts with water in the presence of acid to turn back into ethanol.

🎯 Exam Tip: This sequence demonstrates common transformations in organic chemistry: alcohol to alkyl halide, alkyl halide to alkene (elimination), and alkene back to alcohol (hydration).

 

Question 13. The reaction \( \text{Phenol} \xrightarrow{\text{NaH}} \text{Sodium phenoxide} \xrightarrow{\text{CH}_3\text{I}} \text{Anisole} \) can be classified as
(a) dehydration
(b) Williamson ether synthesis
(c) Williamson ether synthesis
(d) dehydrogenation of alcohol
Answer: (c) Williamson ether synthesis
Solution: The Williamson ether synthesis is a reaction where an alkoxide or phenoxide ion reacts with a primary alkyl halide to form an ether. In this case, phenol reacts with sodium hydride (NaH) to form sodium phenoxide, which then reacts with methyl iodide (CH3I) to produce anisole (methoxybenzene). This is a classic example of an \( \text{S}_{\text{N}}^2 \) reaction mechanism.
In simple words: This chemical reaction makes an ether from an alcohol and an alkyl halide. It starts by making a salt from the alcohol, then that salt reacts with a methyl compound to form the ether.

🎯 Exam Tip: The Williamson ether synthesis is highly versatile for creating both symmetrical and unsymmetrical ethers, but it works best with primary alkyl halides to avoid elimination side reactions.

 

Question 14. Isoprophylbcnzene on air oxidation in the presence of dilute acid gives
(a) C6H5COOH
(b) C6H5COCH3
(c) C6H5COC6H5
(d) C6H5 – OH
Answer: (d) C6H5 – OH (phenol)
In simple words: When isopropylbenzene is exposed to air and dilute acid, it undergoes a reaction that produces phenol. This process is a common industrial method for making phenol.

🎯 Exam Tip: This reaction is known as the cumene process, a commercially important method for synthesizing phenol and acetone from benzene and propene.

 

Question 15. Assertion: Phenol is more reactive than benzene towards electrophilic substitution reaction Reason: In the case of phenol. the intermediate arenium ion is more stabilized by resonance.
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer: (a) if both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Phenol reacts more easily than benzene with other chemicals because the -OH group helps stabilize the temporary charged molecule that forms during the reaction. This extra stability makes the reaction happen faster.

🎯 Exam Tip: The hydroxyl group (-OH) in phenol is an activating group and an ortho-para director, enhancing the reactivity of the benzene ring towards electrophilic substitution due to its electron-donating resonance effect.

 

Question 16. HO CH2 CH2 – OH on heating with periodic acid gives ...........
(a) methanoic acid
(b) Glyoxal
(c) methanol
(d) CO2
Answer: (c) methanol
In simple words: When 1,2-ethanediol (which is \( \text{HOCH}_2\text{CH}_2\text{OH} \)) is heated with periodic acid, it breaks apart to form two molecules of formaldehyde. Formaldehyde can then be oxidized to formic acid or reduced to methanol depending on further conditions. In this specific question, the answer points to methanol.

🎯 Exam Tip: Vicinal diols (like 1,2-ethanediol) undergo oxidative cleavage when treated with periodic acid (\( \text{HIO}_4 \)), breaking the carbon-carbon bond and forming aldehydes or ketones, depending on the substitution pattern of the diol.

 

Question 17. Which of the following compound can be used as artireeze in automobile radiators?
(a) methanol
(b) ethanol
(c) Neopentyl alcohol
(d) ethan -1, 2-diol
Answer: (d) ethan -1, 2-diol
In simple words: Ethane-1,2-diol, also known as ethylene glycol, is often used in car radiators to stop the water from freezing in cold weather. It helps the car engine work smoothly.

🎯 Exam Tip: Ethane-1,2-diol (ethylene glycol) is a highly effective antifreeze agent because it can form hydrogen bonds with water, disrupting the formation of ice crystals and lowering the freezing point of the mixture.

 

Question 18. The reaction \( \text{2-chlorocyclohexanol} \xrightarrow{\text{i) NaOH ii) CH}_2\text{I}_2} \text{1,2-epoxycyclohexane} \) is an example of
(a) Wurtz reaction
(b) cyclic reaction
(c) Williamson reaction
(d) Kolbe reactions
Answer: (c) Williamson reaction
In simple words: This reaction uses a chemical called NaOH and then another chemical with iodine to make a ring-shaped ether. This is a special way to make ethers, called the Williamson reaction, which forms a new oxygen-carbon bond inside a molecule.

🎯 Exam Tip: The intramolecular Williamson ether synthesis is particularly useful for forming cyclic ethers, where an alkoxide reacts with an alkyl halide within the same molecule.

 

Question 19. One mole of an organic compound (A) with the formula C3H8O reacts completely with two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound (A) is
(a) propan - 2 – ol
(b) propan- 1- ol
(c) ethoxy ethane
(d) methoxy ethane
Answer: (d) methoxy ethane
Solution: The organic compound (A) is methoxy ethane (\( \text{CH}_3\text{CH}_2\text{OCH}_3 \)).
When \( \text{CH}_3\text{CH}_2\text{OCH}_3 \) reacts with excess HI, it undergoes cleavage to form ethyl iodide (\( \text{CH}_3\text{CH}_2\text{I} \)) and methyl iodide (\( \text{CH}_3\text{I} \)).
\[ \text{CH}_3\text{CH}_2\text{OCH}_3 + \text{Excess HI} \rightarrow \text{CH}_3\text{CH}_2\text{I (X)} + \text{CH}_3\text{I (Y)} \]
When methyl iodide (Y) is boiled with aqueous alkali, it forms methanol (Z):
\[ \text{CH}_3\text{I} + \text{aqueous NaOH} \rightarrow \text{CH}_3\text{OH (Z)} + \text{NaI} \]
Methanol (Z) answers the iodoform test. Although typically methyl ketones and secondary alcohols give a positive iodoform test, primary alcohols like methanol can sometimes react under forcing conditions to produce the iodoform. However, in this context, the iodoform test result is the key to identifying the products and the original compound.
In simple words: The starting material is methoxy ethane. When it reacts with HI, it breaks into two iodide compounds. One of these, methyl iodide, when treated with alkali, becomes methanol. Methanol can give a positive iodoform test.

🎯 Exam Tip: For ether cleavage with HI, the reaction generally follows \( \text{S}_{\text{N}}^2 \) mechanism, yielding alkyl halides. The iodoform test helps identify compounds with \( \text{CH}_3\text{CO-} \) or \( \text{CH}_3\text{CH(OH)-} \) groups. Methanol can be oxidized to formaldehyde, which can then be further oxidized to give iodoform under specific conditions.

 

Question 20. Among the following ethers which one will produce methyl alcohol on treatment with hot HI?
(a) \( \text{(H}_3\text{C)}_3\text{-C-O-CH}_3 \)
(b) \( \text{(CH}_3\text{)}_2\text{-CH-CH}_2\text{-O-CH}_3 \)
(c) \( \text{CH}_3\text{-(CH}_2\text{)}_3\text{-O-CH}_3 \)
(d) \( \text{CH}_3\text{-CH}_2\text{-CH-O - CH}_3 \) with \( \text{CH}_3 \) attached to the CH
Answer: (a) \( \text{(H}_3\text{C)}_3\text{-C-O-CH}_3 \)
In simple words: When an ether reacts with hot HI, it breaks apart into an alcohol and an iodide. The ether that has a tertiary carbon group and a methyl group will form methyl alcohol because the HI will attack the tertiary carbon to make an iodide and release the methyl group as methyl alcohol.

🎯 Exam Tip: In the cleavage of unsymmetrical ethers with HI, if one of the alkyl groups is tertiary, the reaction follows an \( \text{S}_{\text{N}}^1 \) mechanism, leading to the formation of tertiary alkyl halide and primary alcohol (or methyl alcohol if the other group is methyl).

 

Question 21. Williamson synthesis of preparing dimethyl ether is a / an
(a) SN¹ reactions
(b) SN² reaction
(c) electrophilic addition
(d) electrophilic substitution
Answer: (b) SN² reaction
In simple words: The Williamson ether synthesis is a chemical reaction where an alkoxide ion attacks an alkyl halide. This attack happens from the back, which is typical for a type of reaction called \( \text{S}_{\text{N}}^2 \).

🎯 Exam Tip: The Williamson ether synthesis specifically relies on the \( \text{S}_{\text{N}}^2 \) mechanism, which means it works best with primary alkyl halides and strong nucleophiles (alkoxides or phenoxides).

 

Question 22. On reacting with neutral ferric chloride, phenol gives
(a) red colour
(b) violet colour
(c) dark green colour
(d) no colouration
Answer: (b) violet colour
In simple words: Phenol has a special group that reacts with neutral ferric chloride solution to produce a distinct violet color. This test is used to identify phenols.

🎯 Exam Tip: The ferric chloride test is a classic qualitative test for phenols. The characteristic color (often violet, but can vary) is due to the formation of a colored complex between the phenol and the \( \text{Fe}^{3+} \) ions.

II. Short Answer

 

Question 1. Identify the product (s) is/are formed when 1 – methoxy propane is heated with excess HI. Name the mechanism involved in the reaction.
Answer: When 1-methoxy propane is heated with excess HI, it cleaves to form methyl iodide (\( \text{CH}_3\text{I} \)) and 1-iodopropane (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{I} \)). The reaction is a nucleophilic substitution, specifically an \( \text{S}_{\text{N}}^2 \) mechanism.
\[ \text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{I} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \]
Then the alcohol reacts further with excess HI:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{I} + \text{H}_2\text{O} \]
The final products are methyl iodide and 1-iodopropane. This reaction involves the protonation of the ether oxygen, followed by an \( \text{S}_{\text{N}}^2 \) attack by the iodide ion on the less sterically hindered methyl carbon or primary propyl carbon. Since excess HI is used, the alcohol formed (1-propanol) reacts further to form 1-iodopropane. This is a crucial concept to remember for ether reactions.
In simple words: When 1-methoxy propane is heated with lots of HI, it breaks apart. It makes methyl iodide and 1-iodopropane. The reaction happens by an \( \text{S}_{\text{N}}^2 \) pathway.

🎯 Exam Tip: For ether cleavage with excess HI, both parts of the ether typically convert to alkyl iodides. Remember that \( \text{S}_{\text{N}}^2 \) mechanism leads to attack on the less hindered carbon.

 

Question 2. Draw the major product formed when 1 – ethoxyprop – 1 – ene is heated with one equivalent of HI
Answer: When 1-ethoxyprop-1-ene is heated with one equivalent of HI, the protonation of the double bond or the ether oxygen occurs. However, addition of HI to the alkene part usually follows Markovnikov's rule, and the ether linkage also undergoes cleavage. Given one equivalent, the major product is formed by the cleavage of the ether bond.
\[ \text{CH}_3\text{CH=CHOC}_2\text{H}_5 + \text{HI} \rightarrow \text{CH}_3\text{CH=CHOH} + \text{C}_2\text{H}_5\text{I} \]
The enol (\( \text{CH}_3\text{CH=CHOH} \)) then tautomerizes to propanal.
Alternatively, if the double bond is protonated first:
\[ \text{CH}_3\text{CH=CHOC}_2\text{H}_5 + \text{HI} \rightarrow \text{CH}_3\text{CH(I)-CH}_2\text{OC}_2\text{H}_5 \]
However, the structure in the source solution points to the formation of 1-iodopropane-1-ol. If the ether cleaves preferentially, then one possible set of products is ethanol and propenyl iodide. But the provided solution structure implies the addition across the double bond and subsequent ether cleavage, or a specific regioselectivity. Given the complexity and potential for tautomerization, we will represent the product as shown in the source, which implies an initial protonation of the ether oxygen followed by cleavage. The resulting product is 1-iodoprop-1-en-1-ol after initial protonation and then cleavage.
\[ \text{1-ethoxyprop-1-ene} \xrightarrow{\text{HI}} \text{Prop-1-en-1-oL (enol)} + \text{Iodoethane} \]
In simple words: When 1-ethoxyprop-1-ene reacts with HI, the ether bond breaks. This produces an alcohol with a double bond (an enol) and iodoethane. The enol formed is 1-prop-1-en-1-ol, which might then change into an aldehyde.

🎯 Exam Tip: Reactions of unsaturated ethers with HI can be complex, involving both ether cleavage and addition across the double bond. Pay close attention to regioselectivity (Markovnikov vs. anti-Markovnikov) and potential tautomerization of enols to aldehydes or ketones.

 

Question 3. Suggest a suitable reagent to prepare secondary alcohol with an identical groups using a Grignard reagent.
Answer: To prepare a secondary alcohol with identical alkyl groups using a Grignard reagent, we can react an aldehyde (other than formaldehyde) with the appropriate Grignard reagent, followed by hydrolysis. Specifically, if we want identical groups, we can use ethanal (acetaldehyde) with methyl magnesium bromide. Ethanal has one alkyl group, and the Grignard reagent adds another identical group.
\[ \text{CH}_3\text{MgBr} + \text{CH}_3\text{CHO (Ethanal)} \rightarrow \text{CH}_3\text{CH(OMgBr)CH}_3 \]
\[ \text{CH}_3\text{CH(OMgBr)CH}_3 \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH(OH)CH}_3 \text{ (2-propanol)} + \text{Mg(OH)Br} \]
The product is 2-propanol, a secondary alcohol with two identical methyl groups attached to the carbon bearing the hydroxyl group, in addition to the hydrogen. This is a very common way to prepare secondary alcohols.
In simple words: To make a secondary alcohol that has two same groups, you can use a Grignard reagent and ethanal. For example, methyl magnesium bromide reacting with ethanal will make 2-propanol, which is a secondary alcohol.

🎯 Exam Tip: Grignard reagents are powerful tools for forming new carbon-carbon bonds. Formaldehyde yields primary alcohols, other aldehydes yield secondary alcohols, and ketones yield tertiary alcohols upon reaction and hydrolysis.

 

Question 4. What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis
Answer: When methyl benzoate is treated with two moles of ethyl magnesium bromide, followed by acid hydrolysis, a tertiary alcohol with two ethyl groups attached to the carbonyl carbon will be formed. Methyl benzoate is an ester, and esters react with two equivalents of Grignard reagent.
First, one mole of Grignard reagent adds to the carbonyl, leading to the elimination of the methoxide group, forming a ketone (acetophenone in this case, then with another ethyl group).
\[ \text{C}_6\text{H}_5\text{COOCH}_3 + \text{C}_2\text{H}_5\text{MgBr} \rightarrow \text{C}_6\text{H}_5\text{COC}_2\text{H}_5 + \text{Mg(OCH}_3\text{)Br} \]
Then, the second mole of Grignard reagent adds to the newly formed ketone:
\[ \text{C}_6\text{H}_5\text{COC}_2\text{H}_5 + \text{C}_2\text{H}_5\text{MgBr} \rightarrow \text{C}_6\text{H}_5\text{C(OMgBr)(C}_2\text{H}_5\text{)}_2 \]
Finally, acid hydrolysis converts the alkoxide into the tertiary alcohol:
\[ \text{C}_6\text{H}_5\text{C(OMgBr)(C}_2\text{H}_5\text{)}_2 \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{C(OH)(C}_2\text{H}_5\text{)}_2 \]
The major product is 3-phenyl-3-pentanol, which is a tertiary alcohol.
In simple words: Methyl benzoate reacts with two parts of ethyl magnesium bromide. This makes a new chemical that, after adding acid and water, becomes a tertiary alcohol called 3-phenyl-3-pentanol.

🎯 Exam Tip: Esters react with two equivalents of Grignard reagents. The first equivalent forms a ketone, and the second equivalent then adds to the ketone to form a tertiary alcohol after hydrolysis. Always specify the final product's IUPAC name.

 

Question 5. Predict the major product, when 2-methyl but – 2 – ene is converted into alcohol in each of the following methods.
1. Acid-catalyzed hydration
2. Hydroboration
3. Hydroxylation using bayers reagent
Answer:
(i) Acid-catalyzed hydration (\( \text{H}_2\text{O/H}^+ \)) follows Markovnikov's rule, adding the -OH to the more substituted carbon of the alkene. For 2-methylbut-2-ene, this yields 2-methylbutan-2-ol (a tertiary alcohol).
\[ \text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3 \xrightarrow{\text{H}_2\text{O/H}^+} \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_2\text{-CH}_3 \]
(ii) Hydroboration-oxidation (\( \text{B}_2\text{H}_6 \), then \( \text{H}_2\text{O}_2/\text{OH}^- \)) follows anti-Markovnikov's rule, adding the -OH to the less substituted carbon. For 2-methylbut-2-ene, this yields 3-methylbutan-2-ol (a secondary alcohol). This method adds water across the double bond in an anti-Markovnikov fashion, and the stereochemistry is syn (addition to the same face).
\[ \text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3 \xrightarrow{\text{1. B}_2\text{H}_6 \text{ 2. H}_2\text{O}_2\text{/OH}^-} \text{CH}_3\text{-CH(CH}_3\text{)-CH(OH)-CH}_3 \]
(iii) Hydroxylation using Baeyer's reagent (alkaline \( \text{KMnO}_4 \)) results in syn-dihydroxylation, adding two -OH groups to the same face of the double bond. For 2-methylbut-2-ene, this yields 2-methylbutan-2,3-diol.
\[ \text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3 \xrightarrow{\text{alk.KMnO}_4 \text{ H}_2\text{O/[O]}} \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH(OH)-CH}_3 \]
In simple words: Different ways to add water to 2-methyl but-2-ene give different alcohols. Acidic water adds the OH where there are more carbons (Markovnikov). Hydroboration adds the OH where there are fewer carbons (anti-Markovnikov). Baeyer's reagent adds two OH groups at the same time to the double bond.

🎯 Exam Tip: Remember the regioselectivity and stereoselectivity of common alkene hydration reactions: acid-catalyzed hydration (Markovnikov), hydroboration-oxidation (anti-Markovnikov, syn addition), and Baeyer's reagent (syn dihydroxylation).

 

Question 6. Arrange the following in the increasing order of their boiling point and give a reason for your ordering
1. Butan – 2 – ol, Butan – 1 – SI, 2 – methylpropan – 2 – ol
2. Propan – 1 – ol, propan – 1, 2, 3 – triol, propan 1, 3 – diol, propan – 2 – ol
Answer:
1. Boiling points generally increase as the molecular mass increases because the Van der Waals forces of attraction become stronger. Among alcohols with the same number of atoms (isomeric alcohols), secondary (2°) alcohols have lower boiling points than primary (1°) alcohols. This difference is due to steric hindrance, which reduces the effectiveness of hydrogen bonding. For example, butan-2-ol has a lower boiling point than butan-1-ol. The increasing order of boiling points is: 2-methylpropan-2-ol \( < \) butan-2-ol \( < \) butan-1-ol.
2. Similarly, 2°-alcohols have lower boiling points than 1°-alcohols because hydrogen bonding is less extensive in secondary alcohols. Therefore, propan-1-ol has a higher boiling point than propan-2-ol. The boiling point also increases as the number of hydroxyl (hydrogen) groups in the molecule increases. The overall increasing order of boiling points is: propan-2-ol \( < \) propan-1-ol \( < \) propan-1,3-diol \( < \) propan-1,2,3-triol.
In simple words: Boiling points go up with molecular size. Primary alcohols generally boil hotter than secondary ones because of better hydrogen bonding. More -OH groups also mean higher boiling points.

🎯 Exam Tip: When comparing boiling points of isomers, always consider the strength and extent of hydrogen bonding first, followed by molecular shape (branching) and Van der Waals forces.

 

Question 7. Can we use nucleophiles such as \( \text{NH}_3 \), \( \text{CH}_3\text{O}^- \) for the Nucleophilic substitution of alcohols
Answer: To answer if \( \text{NH}_3 \) and \( \text{CH}_3\text{O}^- \) can be used as nucleophiles for the nucleophilic substitution of alcohols, we need to understand their nucleophilicity. The ability of a nucleophile to donate electrons to a positive center is called nucleophilicity.
1. The increasing order of nucleophilicity is generally observed as: \( \text{NH}_3 < \text{OH}^- < \text{CH}_3\text{O}^- \).
2. A higher electron density on an atom usually leads to increased nucleophilicity because there are more electrons available for donation.
3. Negatively charged species are almost always stronger nucleophiles compared to neutral species, as they have an excess of electrons.
4. An alkoxide ion, \( \text{RO}^- \), has an alkyl group attached, which can donate electrons through inductive effect and also allows for greater polarizability of its electron cloud. This makes the oxygen's lone pair electrons more easily available in \( \text{RO}^- \) than in \( \text{OH}^- \). Therefore, \( \text{CH}_3\text{O}^- \) is a better nucleophile for the nucleophilic substitution of alcohols. However, \( \text{NH}_3 \) cannot act as a suitable nucleophile for the nucleophilic substitution of alcohols in most cases. A methoxide ion is a powerful nucleophile due to its charge and electron-donating methyl group.
In simple words: Yes, \( \text{CH}_3\text{O}^- \) can be used because it's a strong nucleophile with extra electrons available. But \( \text{NH}_3 \) usually can't be used for alcohol substitution because it's not strong enough for these specific reactions. Charged molecules with alkyl groups are better at giving away electrons.

🎯 Exam Tip: Remember that nucleophilicity is affected by charge (anions are stronger than neutrals), electronegativity (less electronegative atoms are stronger nucleophiles within a period), and steric hindrance (bulkier groups hinder nucleophilicity).

 

Question 8. Is it possible to oxidise t – butyl alcohol using acidified dichromate to form a carbonyl compound.
Answer: Tertiary (3°) alcohols typically do not undergo oxidation reactions under normal conditions. This is because the carbon atom holding the hydroxyl group (\( \text{-OH} \)) has no hydrogen atoms directly attached to it, which are needed for typical oxidation. However, at elevated temperatures and with strong oxidizing agents like acidified dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4 \)), it is possible to oxidize t-butyl alcohol. In such conditions, the carbon-carbon bonds break (cleavage of C-C bonds), leading to a mixture of carboxylic acids and ketones, each containing a lesser number of carbon atoms than the original alcohol.
The oxidation process likely occurs via an initial dehydration of the alcohol under acidic conditions to form an alkene, followed by the oxidation of the alkene. For instance, t-butyl alcohol (2-methylpropan-2-ol) dehydrates to 2-methylprop-1-ene, which then undergoes oxidative cleavage. This is an unusual route but demonstrates that oxidation is possible under harsh conditions. \[ \underset{\text{t-butyl alcohol}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-OH}} \quad \xrightarrow{\text{H}^+/\text{K}_2\text{Cr}_2\text{O}_7 \text{ (strong oxidation)}} \quad \underset{\text{Acetone}}{\text{CH}_3\text{-C(O)-CH}_3} \quad + \quad \underset{\text{Formic acid}}{\text{HCOOH}} \] The formic acid can further oxidize to carbon dioxide and water.
In simple words: Normally, t-butyl alcohol is hard to oxidize. But if you heat it a lot with a strong chemical like acidified dichromate, its carbon bonds can break. This forms smaller molecules like acetone and formic acid, not a simple carbonyl compound directly from the alcohol group.

🎯 Exam Tip: Remember that primary alcohols oxidize to aldehydes and then carboxylic acids, secondary alcohols oxidize to ketones, but tertiary alcohols resist oxidation under mild conditions because they lack a hydrogen atom on the hydroxyl-bearing carbon.

 

Question 9. What happens when 1 – phenyl ethanol is treated with acidified \( \text{KMnO}_4 \).
Answer: When 1-phenyl ethanol, which is a secondary alcohol, is treated with acidified potassium permanganate (\( \text{KMnO}_4 \)), it undergoes an oxidation reaction. Acidified \( \text{KMnO}_4 \) is a strong oxidizing agent.
During this reaction, the secondary alcohol group (\( \text{-CH(OH)-} \)) is oxidized to a ketone group (\( \text{-C(O)-} \)). Therefore, 1-phenyl ethanol is converted into acetophenone. This is a common and effective method to prepare ketones from secondary alcohols. \[ \underset{\text{1-phenyl ethanol}}{\text{C}_6\text{H}_5\text{-}\underset{\text{OH}}{|}\text{CH}\text{-CH}_3} \quad \xrightarrow{\text{KMnO}_4 / \text{H}^+ \text{ [O]}} \quad \underset{\text{Acetophenone}}{\text{C}_6\text{H}_5\text{-C(O)-CH}_3} \quad + \quad \text{H}_2\text{O} \]
In simple words: When 1-phenyl ethanol meets strong acidified \( \text{KMnO}_4 \), it changes into acetophenone. This happens because the alcohol group is oxidized to a ketone group.

🎯 Exam Tip: Always remember that strong oxidizing agents like acidified \( \text{KMnO}_4 \) convert secondary alcohols to ketones. For primary alcohols, they would typically proceed to carboxylic acids.

 

Question 10. Write the mechanism of acid catalysed dehydration of ethanol to give ethene.
Answer: The acid-catalyzed dehydration of ethanol to produce ethene typically proceeds via an \( \text{E1} \) (Elimination, unimolecular) mechanism, especially at higher temperatures (around 170°C for primary alcohols, though the mechanism might have E2 character at lower temperatures or for primary alcohols). Here's a common mechanism:
**Step 1: Protonation of Alcohol** The oxygen atom in ethanol has lone pairs of electrons. It acts as a Lewis base and gets protonated by the acid (e.g., \( \text{H}_2\text{SO}_4 \)) to form a protonated alcohol, an alkyloxonium ion. This step converts the poor leaving group (\( \text{-OH} \)) into a good leaving group (\( \text{-OH}_2^+ \)). \[ \underset{\text{Ethanol}}{\text{CH}_3\text{-CH}_2\text{-OH}} \quad + \quad \text{H}^+ \quad \longrightarrow \quad \underset{\text{Protonated ethanol}}{\text{CH}_3\text{-CH}_2\text{-}\overset{+}{\text{O}}\text{H}_2} \]
**Step 2: Loss of Water to Form Carbocation** The protonated ethanol then loses a molecule of water. This step is unimolecular and is the rate-determining step. A primary carbocation is formed. \[ \underset{\text{Protonated ethanol}}{\text{CH}_3\text{-CH}_2\text{-}\overset{+}{\text{O}}\text{H}_2} \quad \longrightarrow \quad \underset{\text{Ethyl carbocation}}{\text{CH}_3\text{-}\overset{+}{\text{CH}}_2} \quad + \quad \text{H}_2\text{O} \]
**Step 3: Deprotonation to Form Alkene** The carbocation is highly reactive. An adjacent carbon's hydrogen is removed by a base (often another alcohol molecule or water acting as a base), leading to the formation of a double bond and regenerating the acid catalyst. \[ \underset{\text{Ethyl carbocation}}{\text{CH}_3\text{-}\overset{+}{\text{CH}}_2} \quad \longrightarrow \quad \underset{\text{Ethene}}{\text{CH}_2\text{=CH}_2} \quad + \quad \text{H}^+ \] The regenerated proton then continues the catalytic cycle.
In simple words: First, the alcohol gets an extra hydrogen from the acid. Then, a water molecule leaves, making a positively charged carbon (carbocation). Finally, another hydrogen leaves, forming a double bond and making ethene. The acid helps the reaction keep going.

🎯 Exam Tip: When writing dehydration mechanisms, clearly show the protonation, water loss (carbocation formation), and deprotonation steps. For primary alcohols, some textbooks simplify to E2 as the carbocation is unstable.

 

Question 11. How is phenol prepared form
1. chloro benzene
2. isopropyl benzene
Answer: Phenol can be prepared from chlorobenzene and isopropyl benzene through specific chemical processes.
**1. From Chlorobenzene (Dow's Process):** Chlorobenzene is converted to phenol using Dow's process, an industrial method. In this process, chlorobenzene is first heated with a concentrated aqueous solution of sodium hydroxide (\( \text{NaOH} \)) at a high temperature (around 633 K) and pressure (about 300 bar). This severe condition promotes nucleophilic substitution, producing sodium phenoxide. The sodium phenoxide is then acidified with dilute hydrochloric acid (\( \text{HCl} \)) to liberate phenol. \[ \underset{\text{Chlorobenzene}}{\text{C}_6\text{H}_5\text{Cl}} \quad \xrightarrow{\text{+ 2 NaOH, 633 K, 300 bar}} \quad \underset{\text{Sodium phenoxide}}{\text{C}_6\text{H}_5\text{ONa}} \quad \xrightarrow{\text{+ HCl}} \quad \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad + \quad \text{NaCl} \]
**2. From Isopropyl Benzene (Cumene Process):** Phenol is commercially produced from isopropyl benzene, also known as cumene. This process involves two main steps. First, cumene is oxidized by air (molecular oxygen) in the presence of a catalyst to form cumene hydroperoxide. This hydroperoxide is an unstable intermediate. In the second step, cumene hydroperoxide is treated with dilute acid (such as \( \text{H}_2\text{SO}_4 \)), which causes it to cleave and decompose into phenol and acetone. This is a very efficient industrial synthesis for both phenol and acetone. \[ \underset{\text{Cumene (Isopropylbenzene)}}{\text{C}_6\text{H}_5\text{CH(CH}_3\text{)}_2} \quad \xrightarrow{\text{Air, O}_2} \quad \underset{\text{Cumene hydroperoxide}}{\text{C}_6\text{H}_5\text{-}\underset{\text{OOH}}{|}\text{C(CH}_3\text{)}_2} \quad \xrightarrow{\text{H}^+/\text{H}_2\text{SO}_4} \quad \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad + \quad \underset{\text{Acetone}}{\text{CH}_3\text{COCH}_3} \]
In simple words: You can make phenol from chlorobenzene by reacting it with strong lye (NaOH) at high heat and pressure, then adding acid. Or, you can start with isopropyl benzene, bubble air through it, and then add a little acid to get phenol and acetone.

🎯 Exam Tip: For industrial preparations, it's crucial to mention the specific reagents, conditions (temperature, pressure, catalyst), and key intermediates. Remember Dow's process for chlorobenzene and the cumene hydroperoxide route for isopropyl benzene.

 

Question 12. Explain Kolbe's reaction
Answer: Kolbe's reaction, also known as Kolbe-Schmitt reaction, is an important method for synthesizing salicylic acid, a key precursor for aspirin. The reaction involves the carboxylation of sodium phenoxide.
The process begins by converting phenol into its more reactive form, sodium phenoxide, by treating it with a base like sodium hydroxide (\( \text{NaOH} \)). Sodium phenoxide is significantly more reactive than phenol towards electrophilic substitution reactions, particularly with carbon dioxide (\( \text{CO}_2 \)).
The sodium phenoxide is then heated under pressure with carbon dioxide at a moderate temperature (around 400 K) and pressure (4-7 bar). The \( \text{CO}_2 \) acts as a weak electrophile and substitutes on the aromatic ring, preferentially at the ortho position to the phenoxide group. This step forms sodium salicylate. Finally, acid hydrolysis of sodium salicylate yields salicylic acid. The stability of the phenoxide intermediate greatly assists this reaction. \[ \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad \xrightarrow{\text{NaOH}} \quad \underset{\text{Sodium phenoxide}}{\text{C}_6\text{H}_5\text{ONa}} \quad \xrightarrow{\text{+ CO}_2, \text{ 400 K, 4-7 bar}} \quad \underset{\text{Sodium salicylate}}{\text{C}_6\text{H}_4(\text{OH})\text{COONa}} \quad \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \quad \underset{\text{Salicylic acid}}{\text{C}_6\text{H}_4(\text{OH})\text{COOH}} \]
In simple words: Kolbe's reaction makes salicylic acid from phenol. First, phenol becomes sodium phenoxide. Then, this reacts with carbon dioxide under heat and pressure. Finally, adding acid turns it into salicylic acid. Sodium phenoxide is a better starting material because it reacts more easily.

🎯 Exam Tip: For Kolbe's reaction, remember the key steps: phenol to sodium phenoxide, carboxylation with \( \text{CO}_2 \) under specific conditions, and subsequent acidification. The ortho-directing nature of the phenoxide is critical for salicylic acid formation.

 

Question 13. Writes the chemical equation for Williamson synthesis of 2 – ethoxy – 2 – methyl pentane starting from ethanol and 2 – methyl pentan – 2 – ol
Answer: Williamson ether synthesis is a reaction used to form ethers by reacting an alkoxide with a primary alkyl halide. To synthesize 2-ethoxy-2-methylpentane, we need to choose the starting materials carefully to avoid undesirable elimination reactions. We should use a tertiary alkoxide and a primary alkyl halide, as this combination typically favors the substitution reaction over elimination.
**Step 1: Preparation of the Tertiary Alkoxide** We start with 2-methylpentan-2-ol, a tertiary alcohol. It is converted into its sodium alkoxide by reacting it with sodium metal or a strong base like sodium hydride. \[ \underset{\text{2-methylpentan-2-ol}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{\overset{\text{OH}}{|}\text{C}}\text{-CH}_2\text{-CH}_2\text{-CH}_3} \quad + \quad \text{Na} \quad \longrightarrow \quad \underset{\text{Sodium 2-methylpentan-2-oxide}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{\overset{\text{ONa}}{|}\text{C}}\text{-CH}_2\text{-CH}_2\text{-CH}_3} \quad + \quad \frac{1}{2}\text{H}_2 \]
**Step 2: Preparation of the Primary Alkyl Halide** Next, we prepare iodoethane, a primary alkyl halide, from ethanol. Ethanol reacts with hydroiodic acid (\( \text{HI} \)) to replace the hydroxyl group with an iodine atom. \[ \underset{\text{Ethanol}}{\text{CH}_3\text{-CH}_2\text{-OH}} \quad + \quad \text{HI} \quad \longrightarrow \quad \underset{\text{Iodoethane}}{\text{CH}_3\text{-CH}_2\text{-I}} \quad + \quad \text{H}_2\text{O} \]
**Step 3: Williamson Ether Synthesis** Finally, the sodium 2-methylpentan-2-oxide (the tertiary alkoxide) reacts with iodoethane (the primary alkyl halide). The alkoxide acts as a nucleophile, attacking the carbon bearing the halogen, leading to the formation of the ether and sodium iodide (\( \text{NaI} \)). This specific combination works well to maximize the desired ether product. \[ \underset{\text{Sodium 2-methylpentan-2-oxide}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{\overset{\text{ONa}}{|}\text{C}}\text{-CH}_2\text{-CH}_2\text{-CH}_3} \quad + \quad \underset{\text{Iodoethane}}{\text{CH}_3\text{-CH}_2\text{-I}} \quad \longrightarrow \quad \underset{\text{2-ethoxy-2-methylpentane}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{\overset{\text{OCH}_2\text{CH}_3}{|}\text{C}}\text{-CH}_2\text{-CH}_2\text{-CH}_3} \quad + \quad \text{NaI} \]
In simple words: To make 2-ethoxy-2-methylpentane, first turn 2-methylpentan-2-ol into its sodium salt (alkoxide). Then, turn ethanol into iodoethane. Mix these two together, and they will react to form the desired ether. This method prefers using a bulky alkoxide with a simple halide to get the ether.

🎯 Exam Tip: For Williamson synthesis, always use a primary alkyl halide and a bulky alkoxide (if one component is tertiary) to promote SN2 reaction and avoid E2 elimination, which can be a competing side reaction.

 

Question 14. Write the structure of the aldehyde, carboxylic acid and ester that yield 4 – methylpent – 2 – en – 1 – ol.
Answer: To obtain 4-methylpent-2-en-1-ol, which is a primary allylic alcohol, the corresponding aldehyde, carboxylic acid, and ester would have the following structures:
**1. Aldehyde:** The aldehyde that would yield 4-methylpent-2-en-1-ol upon reduction is 4-methylpent-2-enal. The aldehyde group (\( \text{-CHO} \)) would be reduced to a primary alcohol group (\( \text{-CH}_2\text{OH} \)). \[ \underset{\text{4-methylpent-2-enal}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{|}\text{CH}\text{-CH=CH-CHO}} \]
**2. Carboxylic Acid:** The carboxylic acid that would yield 4-methylpent-2-en-1-ol upon strong reduction is 4-methylpent-2-enoic acid. The carboxylic acid group (\( \text{-COOH} \)) would be reduced to a primary alcohol group (\( \text{-CH}_2\text{OH} \)). \[ \underset{\text{4-methylpent-2-enoic acid}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{|}\text{CH}\text{-CH=CH-COOH}} \]
**3. Ester:** The ester that would yield 4-methylpent-2-en-1-ol upon reduction is methyl 4-methylpent-2-enoate. The ester group (\( \text{-COOR} \)) would be reduced to a primary alcohol group (\( \text{-CH}_2\text{OH} \)) and an alcohol (methanol in this case). \[ \underset{\text{Methyl 4-methylpent-2-enoate}}{\text{CH}_3\text{-}\underset{\text{CH}_3}{|}\text{CH}\text{-CH=CH-COOCH}_3} \]
In simple words: If you want to make 4-methylpent-2-en-1-ol, you could reduce an aldehyde called 4-methylpent-2-enal, a carboxylic acid called 4-methylpent-2-enoic acid, or an ester called methyl 4-methylpent-2-enoate. All these have the same carbon skeleton, and their functional groups get turned into alcohol.

🎯 Exam Tip: When working backward from an alcohol, remember that primary alcohols can be formed by reducing aldehydes, carboxylic acids, or esters. Ensure the carbon chain and position of double bonds (if any) are maintained correctly.

 

Question 15. What is metamerism? Give the structure and IUPAC name of metamers of 2 – methoxy propane
Answer: **Metamerism:** Metamerism is a special type of structural isomerism where compounds have the same molecular formula and the same functional group. However, they differ in the nature of the alkyl groups attached to the polyvalent functional group. This means the carbon chain distribution around the functional group is different. Ethers, ketones, and secondary amines are common examples of compounds that exhibit metamerism.
**Metamers of 2-methoxypropane:** 2-methoxypropane has the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \). The metamers will also have this same formula and an ether (\( \text{-O-} \)) functional group, but with different alkyl groups surrounding the oxygen atom.
1. **2-Methoxypropane:** This is the given compound. The oxygen atom is bonded to a methyl group and an isopropyl group. \[ \text{CH}_3\text{-O-CH(CH}_3\text{)}_2 \]
2. **1-Methoxypropane (Methyl propyl ether):** In this isomer, the oxygen atom is bonded to a methyl group and a n-propyl group. The alkyl chain on one side of the oxygen is different from 2-methoxypropane. \[ \text{CH}_3\text{-O-CH}_2\text{-CH}_2\text{-CH}_3 \]
3. **Ethoxyethane (Diethyl ether):** In this isomer, the oxygen atom is bonded to two ethyl groups. The distribution of carbon atoms around the oxygen is different from both 2-methoxypropane and 1-methoxypropane. \[ \text{CH}_3\text{-CH}_2\text{-O-CH}_2\text{-CH}_3 \] These three compounds are metamers because they all have the same molecular formula (\( \text{C}_4\text{H}_{10}\text{O} \)) and the same ether functional group, but the alkyl groups attached to the oxygen atom are arranged differently.
In simple words: Metamerism is when molecules have the same chemical formula and the same main group (like an ether), but the carbon chains on either side of that main group are different. For 2-methoxypropane, other metamers include 1-methoxypropane and ethoxyethane, which just means the carbons are arranged differently around the oxygen.

🎯 Exam Tip: When identifying metamers, always ensure the molecular formula and functional group are identical. The key difference lies solely in the varying length or branching of alkyl chains directly attached to the polyvalent functional group's central atom.

 

Question 16. How are the following conversions effected
1. benzyl chloride to benzyl alcohol
2. benzyl alcohol to benzoic acid
Answer: Here's how to achieve the requested conversions:
**1. Benzyl Chloride to Benzyl Alcohol:** This conversion is typically carried out by a nucleophilic substitution reaction. Benzyl chloride reacts with an aqueous solution of a strong base, such as sodium hydroxide (\( \text{NaOH} \)) or potassium hydroxide (\( \text{KOH} \)). The chloride atom, which is a good leaving group, is replaced by the hydroxyl group (\( \text{-OH} \)), yielding benzyl alcohol. \[ \underset{\text{Benzyl chloride}}{\text{C}_6\text{H}_5\text{CH}_2\text{Cl}} \quad \xrightarrow{\text{Aq. NaOH}} \quad \underset{\text{Benzyl alcohol}}{\text{C}_6\text{H}_5\text{CH}_2\text{OH}} \quad + \quad \text{NaCl} \]
**2. Benzyl Alcohol to Benzoic Acid:** This conversion requires an oxidation reaction. Benzyl alcohol, a primary alcohol, can be oxidized to benzoic acid using strong oxidizing agents. A common reagent for this purpose is acidified potassium dichromate (\( \text{Na}_2\text{Cr}_2\text{O}_7/\text{H}^+ \)) or potassium permanganate (\( \text{KMnO}_4/\text{H}^+ \)). The reaction typically proceeds in two steps: first, the alcohol oxidizes to an aldehyde (benzaldehyde), which then further oxidizes to the carboxylic acid (benzoic acid). \[ \underset{\text{Benzyl alcohol}}{\text{C}_6\text{H}_5\text{CH}_2\text{OH}} \quad \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7/\text{H}^+ \text{ [O]}} \quad \underset{\text{Benzaldehyde}}{\text{C}_6\text{H}_5\text{CHO}} \quad \xrightarrow{\text{further [O]}} \quad \underset{\text{Benzoic acid}}{\text{C}_6\text{H}_5\text{COOH}} \]
In simple words: To change benzyl chloride to benzyl alcohol, react it with a water-based solution of sodium hydroxide. To change benzyl alcohol to benzoic acid, treat it with an acid like acidified dichromate, which makes it lose hydrogens and gain oxygen.

🎯 Exam Tip: When converting between functional groups, identify the type of reaction (e.g., substitution, oxidation, reduction). For primary alcohols, strong oxidation leads directly to carboxylic acids, while milder oxidation can stop at aldehydes.

 

Question 17. Complete the following reactions
(i) \( \text{CH}_3\text{-CH}_2\text{-OH} \quad \xrightarrow{\text{PBr}_3} \quad \text{A} \quad \xrightarrow{\text{aq.NaOH}} \quad \text{B} \quad \xrightarrow{\text{Na}} \quad \text{C} \)
(ii) \( \text{C}_6\text{H}_5\text{-OH} \quad \xrightarrow{\text{Zn dust}} \quad \text{A} \quad \xrightarrow{\text{CH}_3\text{Cl, Anhy. AlCl}_3} \quad \text{B} \quad \xrightarrow{\text{KMnO}_4, \text{H}^+} \quad \text{C} \)
(iii) \( \text{Anisole} \quad \xrightarrow{\text{t-butylchloride, AlCl}_3} \quad \text{A} \quad \xrightarrow{\text{Cl}_2/\text{FeCl}_3} \quad \text{B} \quad \xrightarrow{\text{HBr}} \quad \text{C} \)
(iv) \( \underset{\text{1-(1-methylcyclohexyl)enol}}{\text{Cyclic alcohol with methyl and OH}} \quad \xrightarrow{\text{H}^+, \text{-H}_2\text{O}} \quad \text{A} \quad \xrightarrow{\text{i) O}_3, \text{ ii) H}_2\text{O}} \quad \text{B} \)
Answer: Let's complete each of the given reaction sequences:
(i) **Ethanol to Sodium Ethoxide:** Ethanol reacts with phosphorus tribromide (\( \text{PBr}_3 \)) to form bromoethane (A). This is a substitution reaction. Bromoethane then reacts with aqueous sodium hydroxide (\( \text{aq. NaOH} \)) via nucleophilic substitution to regenerate ethanol (B). Finally, ethanol reacts with sodium metal (\( \text{Na} \)) to form sodium ethoxide (C). \[ \underset{\text{Ethanol}}{\text{CH}_3\text{-CH}_2\text{-OH}} \quad \xrightarrow{\text{PBr}_3} \quad \underset{\text{Bromoethane (A)}}{\text{CH}_3\text{-CH}_2\text{-Br}} \quad \xrightarrow{\text{aq. NaOH}} \quad \underset{\text{Ethanol (B)}}{\text{CH}_3\text{-CH}_2\text{-OH}} \quad \xrightarrow{\text{Na}} \quad \underset{\text{Sodium Ethoxide (C)}}{\text{CH}_3\text{-CH}_2\text{-ONa}} \]
(ii) **Phenol to Benzoic Acid (Friedel-Crafts Alkylation and Oxidation):** Phenol is reduced by zinc dust (\( \text{Zn dust} \)) to produce benzene (A). Benzene (A) undergoes Friedel-Crafts alkylation with chloromethane (\( \text{CH}_3\text{Cl} \)) in the presence of anhydrous aluminum chloride (\( \text{Anhy. AlCl}_3 \)) to form toluene (B). Toluene (B) is then oxidized by potassium permanganate (\( \text{KMnO}_4 \)) under acidic conditions (\( \text{H}^+ \)) to form benzoic acid (C). The alkyl group on the benzene ring is oxidized to a carboxylic acid group. \[ \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad \xrightarrow{\text{Zn dust}} \quad \underset{\text{Benzene (A)}}{\text{C}_6\text{H}_6} \quad \xrightarrow{\text{CH}_3\text{Cl, Anhy. AlCl}_3} \quad \underset{\text{Toluene (B)}}{\text{C}_6\text{H}_5\text{CH}_3} \quad \xrightarrow{\text{KMnO}_4, \text{H}^+} \quad \underset{\text{Benzoic Acid (C)}}{\text{C}_6\text{H}_5\text{COOH}} \]
(iii) **Anisole Reactions (Electrophilic Substitution and Cleavage):** Anisole undergoes Friedel-Crafts alkylation with t-butyl chloride in the presence of \( \text{AlCl}_3 \) to form 4-t-butyl anisole (A). Then, 4-t-butyl anisole undergoes chlorination with \( \text{Cl}_2/\text{FeCl}_3 \) (electrophilic substitution) predominantly at the ortho or para position relative to the activating methoxy group, but given the existing t-butyl group, this leads to 4-t-butyl-2-chloroanisole (B). Finally, this compound undergoes ether cleavage with \( \text{HBr} \) to yield 4-t-butyl-2-chlorophenol and bromomethane (C). \[ \underset{\text{Anisole}}{\text{C}_6\text{H}_5\text{OCH}_3} \quad \xrightarrow{\text{t-butylchloride, AlCl}_3} \quad \underset{\text{4-t-butyl anisole (A)}}{\text{C}_6\text{H}_4(\text{OCH}_3)\text{C(CH}_3\text{)}_3} \] \[ \underset{\text{4-t-butyl anisole (A)}}{\text{C}_6\text{H}_4(\text{OCH}_3)\text{C(CH}_3\text{)}_3} \quad \xrightarrow{\text{Cl}_2/\text{FeCl}_3} \quad \underset{\text{4-t-butyl-2-chloroanisole (B)}}{\text{C}_6\text{H}_3(\text{OCH}_3)(\text{Cl})\text{C(CH}_3\text{)}_3} \] \[ \underset{\text{4-t-butyl-2-chloroanisole (B)}}{\text{C}_6\text{H}_3(\text{OCH}_3)(\text{Cl})\text{C(CH}_3\text{)}_3} \quad \xrightarrow{\text{HBr}} \quad \underset{\text{4-t-butyl-2-chlorophenol}}{\text{C}_6\text{H}_3(\text{OH})(\text{Cl})\text{C(CH}_3\text{)}_3} \quad + \quad \underset{\text{Bromomethane (C)}}{\text{CH}_3\text{Br}} \]
(iv) **Cyclic Enol to Alkene and then Ozonolysis Products:** A cyclic alcohol with a double bond (1-(1-methylcyclohexyl)enol) undergoes acid-catalyzed dehydration (\( \text{H}^+, \text{-H}_2\text{O} \)) to form an alkene, 1-methylcyclohexene (A). This alkene (A) then undergoes ozonolysis, where it reacts with ozone (\( \text{O}_3 \)) followed by reductive workup (e.g., with \( \text{H}_2\text{O} \)). Ozonolysis cleaves the carbon-carbon double bond, leading to the formation of an open-chain compound with both aldehyde and ketone functional groups. This product is 6-oxoheptanal (B). \[ \underset{\text{1-(1-methylcyclohexyl)enol}}{\text{C}_6\text{H}_9(\text{CH}_3)\text{OH}} \quad \xrightarrow{\text{H}^+, \text{-H}_2\text{O}} \quad \underset{\text{1-methylcyclohexene (A)}}{\text{C}_6\text{H}_9\text{CH}_3} \] \[ \underset{\text{1-methylcyclohexene (A)}}{\text{C}_6\text{H}_9\text{CH}_3} \quad \xrightarrow{\text{i) O}_3, \text{ ii) H}_2\text{O}} \quad \underset{\text{6-oxoheptanal (B)}}{\text{CHO-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-C(O)CH}_3} \] The actual structure from the diagram for B would be an open chain with methyl ketone and aldehyde end. \[ \underset{\text{6-oxoheptanal (B)}}{\text{CH}_3\text{-C(O)-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}} \]
In simple words: These are step-by-step changes. (i) Ethanol becomes bromoethane, then back to ethanol, then sodium ethoxide. (ii) Phenol turns into benzene, then toluene, then benzoic acid. (iii) Anisole gets a t-butyl group, then a chlorine, then splits into a chlorophenol and bromomethane. (iv) A cyclic alcohol loses water to become an alkene, then ozone breaks this alkene into a molecule with both an aldehyde and a ketone group.

🎯 Exam Tip: When completing reactions, always identify the reagent's function (e.g., oxidizing agent, reducing agent, electrophile, nucleophile) and the type of reaction (e.g., substitution, elimination, addition). Pay close attention to conditions like "anhy. AlCl3" for Friedel-Crafts and "H+" for acid catalysis.

 

Question 18. O.44g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 \( \text{cm}^3 \) of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.
Answer: We need to identify a monohydric alcohol (one \( \text{-OH} \) group) that reacts with methyl magnesium iodide (\( \text{CH}_3\text{MgI} \)) to produce methane, and when oxidized by PCC (Pyridinium Chlorochromate), forms a carbonyl compound that gives a silver mirror test (Tollens' test for aldehydes).
**1. Calculate moles of methane released:** At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22400 \( \text{cm}^3 \). Given methane volume = 112 \( \text{cm}^3 \). Moles of \( \text{CH}_4 = \frac{112 \text{ cm}^3}{22400 \text{ cm}^3/\text{mol}} = 0.005 \text{ mol} \).
**2. Determine moles of alcohol:** The reaction between an alcohol and a Grignard reagent to produce methane is: \( \text{R-OH} + \text{CH}_3\text{MgI} \rightarrow \text{CH}_4 + \text{R-OMgI} \) This shows a 1:1 molar ratio. Therefore, moles of alcohol = moles of methane = 0.005 mol.
**3. Calculate the molar mass of the alcohol:** Given mass of alcohol = 0.44 g. Molar mass \( (M) = \frac{\text{Mass (W)}}{\text{Moles (n)}} = \frac{0.44 \text{ g}}{0.005 \text{ mol}} = 88 \text{ g/mol} \).
**4. Determine the molecular formula (\( \text{C}_n\text{H}_{2n+1}\text{OH} \)):** The general formula for a monohydric alcohol is \( \text{C}_n\text{H}_{2n+1}\text{OH} \). Its molar mass is \( (12 \times n) + (1 \times (2n+1)) + (16 \times 1) + (1 \times 1) = 88 \). \( 12n + 2n + 1 + 16 + 1 = 88 \) \( 14n + 18 = 88 \) \( 14n = 88 - 18 \) \( 14n = 70 \) \( n = \frac{70}{14} = 5 \). So, the molecular formula of the alcohol is \( \text{C}_5\text{H}_{11}\text{OH} \).
**5. Identify the alcohol based on oxidation reaction:** The alcohol, upon oxidation with PCC, forms a carbonyl compound that gives a silver mirror test. The silver mirror test is characteristic of aldehydes. PCC is a mild oxidizing agent that converts primary alcohols to aldehydes and secondary alcohols to ketones. Since an aldehyde is formed, the alcohol must be a primary alcohol. A primary alcohol with 5 carbons is 1-pentanol. \[ \underset{\text{1-pentanol}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}} \quad \xrightarrow{\text{PCC}} \quad \underset{\text{Pentanal}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}} \] Pentanal, being an aldehyde, would give a positive silver mirror test. \[ \underset{\text{Pentanal}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}} \quad + \quad \text{2Ag}^+ \quad \longrightarrow \quad \underset{\text{Pentanoic acid}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-COOH}} \quad + \quad \text{2Ag(s)} \text{ (silver mirror)} \] The identified compound is **1-pentanol**.
In simple words: We found out the alcohol has 5 carbons and a molar mass of 88. Since it forms an aldehyde that passes the silver mirror test when oxidized by PCC, it must be a primary alcohol. So, the alcohol is 1-pentanol.

🎯 Exam Tip: For problems involving Grignard reagents and gas liberation, remember that active hydrogens (from -OH, -COOH, -NH2) react with Grignard reagents to produce alkanes in a 1:1 molar ratio. Also, recall the specific products of PCC oxidation for different types of alcohols.

 

Question 19. Complete the following reactions
(i) \( \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad \xrightarrow{\text{C}_6\text{H}_5\text{COCl}, \text{ OH}^-} \quad \text{A} \quad \xrightarrow{\text{Nitration}} \quad \text{B (major product)} \)
(ii) \( \underset{\text{1-phenyl-3-methylbutan-2-ol}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-CH(OH)-CH(CH}_3\text{)_2}} \quad \xrightarrow{\text{Conc. H}_2\text{SO}_4} \quad \text{X (major product)} \)
Answer: Let's complete the following reactions:
(i) **Phenol reaction with Benzoyl chloride and Nitration:** Phenol reacts with benzoyl chloride (\( \text{C}_6\text{H}_5\text{COCl} \)) in the presence of a base (\( \text{OH}^- \)) to form phenyl benzoate (A). This is a Schotten-Baumann reaction (esterification). Phenyl benzoate then undergoes nitration, which introduces a nitro group (\( \text{-NO}_2 \)) onto the benzene ring. The major product (B) will be the para-nitrophenyl benzoate, as the ester group is ortho/para directing. \[ \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad + \quad \underset{\text{Benzoyl chloride}}{\text{C}_6\text{H}_5\text{COCl}} \quad \xrightarrow{\text{OH}^-} \quad \underset{\text{Phenyl benzoate (A)}}{\text{C}_6\text{H}_5\text{COOC}_6\text{H}_5} \] \[ \underset{\text{Phenyl benzoate (A)}}{\text{C}_6\text{H}_5\text{COOC}_6\text{H}_5} \quad \xrightarrow{\text{Nitration}} \quad \underset{\text{p-Nitrophenyl benzoate (B)}}{\text{C}_6\text{H}_5\text{COOC}_6\text{H}_4\text{-NO}_2 \text{ (Major product)}} \] (The source diagram for (i) also shows o-nitrophenyl benzoate as a minor product).
(ii) **Dehydration of 1-phenyl-3-methylbutan-2-ol:** 1-phenyl-3-methylbutan-2-ol is a secondary alcohol. When treated with concentrated sulfuric acid (\( \text{Conc. H}_2\text{SO}_4 \)), it undergoes acid-catalyzed dehydration. This reaction proceeds via a carbocation intermediate, which can undergo rearrangements to form more stable carbocations. According to Saytzeff's rule, the major product (X) will be the most substituted and stable alkene. The reaction involves protonation of the alcohol, loss of water to form a secondary carbocation, followed by a hydride shift or methyl shift to form a more stable tertiary carbocation, and then elimination of a proton to yield the alkene. For 1-phenyl-3-methylbutan-2-ol: \[ \underset{\text{1-phenyl-3-methylbutan-2-ol}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-CH(OH)-CH(CH}_3\text{)_2}} \quad \xrightarrow{\text{Conc. H}_2\text{SO}_4 \text{ (-H}_2\text{O)}} \quad \underset{\text{1-phenyl-3-methylbut-1-ene (X, Major product)}}{\text{C}_6\text{H}_5\text{-CH=C(CH}_3\text{)_2}} \] (The source diagram shows a complex rearrangement to give \( \text{C}_6\text{H}_5\text{-CH}_2\text{-C(CH}_3\text{)=CH}_2 \) as major product and \( \text{C}_6\text{H}_5\text{-CH}_2\text{-CH=C(CH}_3\text{)_2} \) as minor. This implies a 1,2-phenyl shift or 1,2-hydride shift from the secondary carbocation and then elimination. Following the diagram's depiction, the major product involves a methyl shift leading to a tertiary carbocation followed by elimination.) Let's follow the major product shown in the source diagram as: \[ \underset{\text{1-phenyl-3-methylbutan-2-ol}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-CH(OH)-CH(CH}_3\text{)_2}} \quad \xrightarrow{\text{Conc. H}_2\text{SO}_4 \text{ (-H}_2\text{O)}} \quad \underset{\text{1-phenyl-2,3-dimethylbut-2-ene (X, Major product)}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-C(CH}_3\text{)=CH(CH}_3\text{)}} \] The source diagram is inconsistent with itself and the text. Given the rule, I will present the most plausible major product from the typical dehydration of 1-phenyl-3-methylbutan-2-ol, which is 1-phenyl-3-methylbut-1-ene (more substituted alkene via rearrangement). However, if I strictly follow the "major product" structure shown as an output (which is \( \text{C}_6\text{H}_5\text{-CH}_2\text{-C = C(CH}_3\text{)_2} \)), the mechanism would involve an alkyl shift after carbocation formation. This structure is 1-phenyl-2,3-dimethylbut-2-ene. Let's use the main product mentioned by the text rule: "According to Saytzeff s rule...". This would imply the most substituted alkene. Using the structures from the diagram that explicitly shows the major product: \[ \underset{\text{1-phenyl-3-methylbutan-2-ol}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-CH(OH)-CH(CH}_3\text{)_2}} \quad \xrightarrow{\text{Conc. H}_2\text{SO}_4} \quad \underset{\text{1-phenyl-2,3-dimethylbut-2-ene (Major, X)}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-C(CH}_3\text{)=CH(CH}_3\text{)}} \quad + \quad \underset{\text{1-phenyl-3-methylbut-1-ene (Minor)}}{\text{C}_6\text{H}_5\text{-CH}_2\text{-CH=C(CH}_3\text{)_2}} \] The text "According to Saytzeff s rule..." generally means forming the most substituted alkene, which often involves carbocation rearrangement.
In simple words: For (i), phenol first combines with benzoyl chloride to make phenyl benzoate, then adding a nitro group creates p-nitrophenyl benzoate. For (ii), when 1-phenyl-3-methylbutan-2-ol is heated with strong sulfuric acid, it loses water and rearranges to form the most stable alkene, 1-phenyl-2,3-dimethylbut-2-ene, as the main product.

🎯 Exam Tip: For dehydration reactions, always consider the possibility of carbocation rearrangements (hydride or alkyl shifts) if they can lead to a more stable carbocation, as this will influence the major alkene product according to Saytzeff's rule.

 

Question 20. Phenol is distilled with Zn dust gives (A) followed by Friedel – crafts alkylation with propyl chloride to give a compound B, B on oxidation gives (C). Identify A,B and C.
Answer: Let's identify compounds A, B, and C based on the reaction sequence provided:
**1. Phenol to Benzene (A):** When phenol is distilled with zinc dust (\( \text{Zn dust} \)), the hydroxyl group (\( \text{-OH} \)) is removed, and the benzene ring is reduced to form benzene (A). This is a classic reduction of phenol. \[ \underset{\text{Phenol}}{\text{C}_6\text{H}_5\text{OH}} \quad \xrightarrow{\text{Zn dust}} \quad \underset{\text{Benzene (A)}}{\text{C}_6\text{H}_6} \]
**2. Benzene to n-Propyl Benzene (B):** Benzene (A) then undergoes Friedel-Crafts alkylation. It reacts with propyl chloride (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \)) in the presence of anhydrous aluminum chloride (\( \text{Anhy. AlCl}_3 \)) as a catalyst. This reaction introduces an n-propyl group to the benzene ring, forming n-propyl benzene (B). \[ \underset{\text{Benzene (A)}}{\text{C}_6\text{H}_6} \quad + \quad \underset{\text{Propyl chloride}}{\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}} \quad \xrightarrow{\text{Anhy. AlCl}_3} \quad \underset{\text{n-Propyl benzene (B)}}{\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_3} \]
**3. n-Propyl Benzene to Benzoic Acid (C):** Finally, n-propyl benzene (B) is oxidized using potassium permanganate (\( \text{KMnO}_4 \)). In the presence of a strong oxidizing agent, any alkyl group attached to an aromatic ring that has at least one benzylic hydrogen (a hydrogen atom on the carbon directly attached to the ring) will be oxidized to a carboxylic acid group. For n-propyl benzene, the entire propyl chain is oxidized to a carboxyl group, forming benzoic acid (C). \[ \underset{\text{n-Propyl benzene (B)}}{\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}_3} \quad \xrightarrow{\text{KMnO}_4} \quad \underset{\text{Benzoic Acid (C)}}{\text{C}_6\text{H}_5\text{COOH}} \] Therefore:
**A is Benzene**
**B is n-Propyl benzene**
**C is Benzoic acid**
In simple words: Phenol turns into benzene (A) when mixed with zinc. Then benzene reacts with propyl chloride to become n-propyl benzene (B). Finally, n-propyl benzene is oxidized to form benzoic acid (C). This happens because any carbon chain connected to a benzene ring that has a hydrogen on the first carbon can be fully oxidized.

🎯 Exam Tip: When an alkylbenzene is oxidized by a strong agent like \( \text{KMnO}_4 \), the alkyl chain attached to the benzene ring is completely oxidized to a carboxyl group, regardless of the length of the alkyl chain, as long as there's a benzylic hydrogen.

 

Question 21. Identify A, B, C, D and write the complete equation.
Answer: Let's identify the products A, B, C, and D in the given reaction sequence, which involves Grignard reactions and subsequent transformations.
**Step 1: Reaction of \( \text{CH}_3\text{MgBr} \) with Cyclohexanone (to form intermediate A, then B upon hydrolysis):** Methyl magnesium bromide (\( \text{CH}_3\text{MgBr} \)), a Grignard reagent, reacts with cyclohexanone. The methyl group adds to the carbonyl carbon, and the magnesium bromide portion attaches to the oxygen, forming an intermediate magnesium alkoxide. This alkoxide is product A. Upon acidic hydrolysis (\( \text{H}_3\text{O}^+ \)), the alkoxide yields a tertiary alcohol. \[ \underset{\text{Methyl magnesium bromide}}{\text{CH}_3\text{MgBr}} \quad + \quad \underset{\text{Cyclohexanone}}{\text{C}_6\text{H}_{10}\text{O}} \quad \longrightarrow \quad \underset{\text{1-methylcyclohexylmagnesium bromide (A)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{OMgBr}} \] \[ \underset{\text{1-methylcyclohexylmagnesium bromide (A)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{OMgBr}} \quad \xrightarrow{\text{H}_3\text{O}^+} \quad \underset{\text{1-methylcyclohexanol (B)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{OH}} \quad + \quad \text{Mg(OH)Br} \]
**Step 2: Reaction of 1-methylcyclohexanol (B) with HBr (to form C):** The tertiary alcohol, 1-methylcyclohexanol (B), reacts with hydrobromic acid (\( \text{HBr} \)) to form a bromoalkane. This is a nucleophilic substitution reaction, where the \( \text{-OH} \) group is replaced by \( \text{-Br} \). \[ \underset{\text{1-methylcyclohexanol (B)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{OH}} \quad + \quad \text{HBr} \quad \longrightarrow \quad \underset{\text{1-bromo-1-methylcyclohexane (C)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{Br}} \quad + \quad \text{H}_2\text{O} \]
**Step 3: Reaction of (C) to form a Grignard reagent, then with Formaldehyde (to form D):** The bromoalkane (C) reacts with magnesium metal in ether to form a new Grignard reagent, 1-methylcyclohexylmagnesium bromide. This Grignard reagent then reacts with formaldehyde (\( \text{HCHO} \)), followed by acidic hydrolysis (\( \text{H}_3\text{O}^+ \)), to produce a primary alcohol (D). \[ \underset{\text{1-bromo-1-methylcyclohexane (C)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{Br}} \quad \xrightarrow{\text{Mg/Ether}} \quad \underset{\text{1-methylcyclohexylmagnesium bromide}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{MgBr}} \] \[ \underset{\text{1-methylcyclohexylmagnesium bromide}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{MgBr}} \quad + \quad \underset{\text{Formaldehyde}}{\text{HCHO}} \quad \xrightarrow{\text{i) Addition, ii) H}_3\text{O}^+} \quad \underset{\text{(1-methylcyclohexyl)methanol (D)}}{\text{C}_6\text{H}_{10}(\text{CH}_3)\text{CH}_2\text{OH}} \] Therefore:
**A is 1-methylcyclohexylmagnesium bromide** (intermediate alkoxide)
**B is 1-methylcyclohexanol**
**C is 1-bromo-1-methylcyclohexane**
**D is (1-methylcyclohexyl)methanol**
In simple words: First, methyl magnesium bromide reacts with cyclohexanone to give an intermediate (A), which then becomes 1-methylcyclohexanol (B) after water is added. Next, 1-methylcyclohexanol reacts with HBr to form 1-bromo-1-methylcyclohexane (C). Finally, 1-bromo-1-methylcyclohexane is turned back into a Grignard reagent, which then reacts with formaldehyde and water to produce (1-methylcyclohexyl)methanol (D).

🎯 Exam Tip: Grignard reactions are versatile! Remember that formaldehyde gives primary alcohols, aldehydes give secondary alcohols, and ketones give tertiary alcohols upon reaction with Grignard reagents followed by hydrolysis. Also, know how to convert alcohols to halides and vice versa.

 

Question 22. What will be the product for the following reaction
\( \underset{\text{Acetyl chloride}}{\text{CH}_3\text{-C(O)-Cl}} \quad \xrightarrow{\text{i) CH}_3\text{MgBr, ii) H}_3\text{O}^+} \quad \text{X} \quad \xrightarrow{\text{acidic K}_2\text{Cr}_2\text{O}_7} \quad \text{A} \)
Answer: Let's determine the products for the given reaction sequence.
**1. Formation of Product X:** Acetyl chloride (\( \text{CH}_3\text{C(O)Cl} \)) reacts with methyl magnesium bromide (\( \text{CH}_3\text{MgBr} \)), a Grignard reagent. The Grignard reagent adds to the carbonyl carbon, and the chloride leaves. If one equivalent of Grignard reagent is added, a ketone is formed. If excess Grignard reagent is used, the ketone further reacts to form a tertiary alkoxide, which on hydrolysis yields a tertiary alcohol. The question implies the formation of Acetone as X. \[ \underset{\text{Acetyl chloride}}{\text{CH}_3\text{-C(O)-Cl}} \quad + \quad \underset{\text{Methyl magnesium bromide}}{\text{CH}_3\text{MgBr}} \quad \longrightarrow \quad \text{CH}_3\text{-C(O)-CH}_3 \quad + \quad \text{MgClBr} \] (The source diagram shows \( \text{CH}_3\text{C(O)Cl} \) reacts with \( \text{CH}_3\text{MgBr} \) to form \( \text{CH}_3\text{C(OMgBr)CH}_3 \), which then forms \( \text{CH}_3\text{C(O)CH}_3 \) upon \( \text{H}_3\text{O}^+ \) in the text. Let's trace it consistently.) \[ \underset{\text{Acetyl chloride}}{\text{CH}_3\text{-C(O)-Cl}} \quad \xrightarrow{\text{i) CH}_3\text{MgBr}} \quad \underset{\text{Intermediate}}{\text{CH}_3\text{-}\underset{\text{Cl}}{\overset{\text{OMgBr}}{|}\text{C}}\text{-CH}_3} \quad \xrightarrow{\text{ii) H}_3\text{O}^+} \quad \underset{\text{Acetone (X)}}{\text{CH}_3\text{-C(O)-CH}_3} \] So, **X is Acetone**.
**2. Formation of Product A:** Acetone (X), which is a ketone, is then oxidized under strong acidic conditions with potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+ \)). Ketones are generally resistant to oxidation compared to aldehydes, but under harsh conditions, they undergo carbon-carbon bond cleavage. For acetone, this oxidative cleavage produces carboxylic acids with fewer carbon atoms. \[ \underset{\text{Acetone (X)}}{\text{CH}_3\text{-C(O)-CH}_3} \quad \xrightarrow{\text{Acidic K}_2\text{Cr}_2\text{O}_7} \quad \underset{\text{Acetic acid (A)}}{\text{CH}_3\text{COOH}} \] So, **A is Acetic acid**.
In simple words: First, acetyl chloride reacts with methyl magnesium bromide and water to make acetone (X). Then, acetone is oxidized by acidified dichromate to form acetic acid (A).

🎯 Exam Tip: Remember that acyl halides react with Grignard reagents to form ketones, and ketones can be further oxidized to carboxylic acids under harsh conditions with C-C bond cleavage. Acetic acid is a common product from acetone oxidation.

 

Question 23. How will you convert acetylene into n – butyl alcohol.
Answer: To convert acetylene (\( \text{CH}\equiv\text{CH} \)) into n-butyl alcohol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \)), a series of reactions are required. This involves lengthening the carbon chain and converting a triple bond into a primary alcohol.
**1. Hydration of Acetylene to Acetaldehyde:** Acetylene is first hydrated using mercuric sulfate (\( \text{HgSO}_4 \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)) as catalysts. This reaction adds water across the triple bond to form acetaldehyde. \[ \underset{\text{Acetylene}}{\text{CH}\equiv\text{CH}} \quad \xrightarrow{\text{H}_2\text{O}/( \text{H}_2\text{SO}_4, \text{ HgSO}_4 )} \quad \underset{\text{Acetaldehyde}}{\text{CH}_3\text{CHO}} \]
**2. Aldol Condensation of Acetaldehyde:** Two molecules of acetaldehyde undergo aldol condensation in the presence of a dilute base (like \( \text{dil NaOH} \)). This forms 3-hydroxybutanal (an aldol). \[ \underset{\text{Acetaldehyde}}{\text{CH}_3\text{CHO}} \quad \xrightarrow{\text{dil NaOH}} \quad \underset{\text{3-hydroxybutanal (Aldol)}}{\text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}} \]
**3. Dehydration of Aldol to Crotonaldehyde:** The 3-hydroxybutanal then undergoes dehydration (loss of a water molecule) upon heating or in the presence of acid to form an \( \alpha,\beta \)-unsaturated aldehyde, crotonaldehyde. \[ \underset{\text{3-hydroxybutanal}}{\text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}} \quad \xrightarrow{\text{- H}_2\text{O}} \quad \underset{\text{Crotonaldehyde}}{\text{CH}_3\text{-CH=CH-CHO}} \]
**4. Reduction of Crotonaldehyde to n-Butyl Alcohol:** Finally, crotonaldehyde is completely reduced using a strong reducing agent like hydrogen gas with a nickel catalyst (\( \text{H}_2/\text{Ni} \)). This reduces both the carbon-carbon double bond and the aldehyde group to a primary alcohol, resulting in n-butyl alcohol. \[ \underset{\text{Crotonaldehyde}}{\text{CH}_3\text{-CH=CH-CHO}} \quad \xrightarrow{\text{H}_2/\text{Ni or [H]}} \quad \underset{\text{n-Butyl alcohol}}{\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}} \]
In simple words: First, acetylene gets water added to it to make acetaldehyde. Then, two of these acetaldehyde molecules join up, lose water, and form crotonaldehyde. Finally, adding hydrogen to crotonaldehyde makes it fully saturated and converts the aldehyde part into an alcohol, giving n-butyl alcohol.

🎯 Exam Tip: This synthesis involves chain lengthening (aldol condensation) and functional group transformations. Remember the sequence: hydration of alkyne to aldehyde, aldol condensation, dehydration, and complete reduction (hydrogenation) of both the double bond and aldehyde.

 

Question 24. Predict the product A, B, X and Y in the following sequence of reaction
\( \underset{\text{butan-2-ol}}{\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3} \quad \xrightarrow{\text{SOCl}_2} \quad \text{A} \quad \xrightarrow{\text{Mg/ether}} \quad \text{B} \)
\( \underset{\text{butan-2-ol}}{\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3} \quad \xrightarrow{\text{Cu/573 K}} \quad \text{X} \quad \xrightarrow{\text{B (Grignard reagent)}, \text{ H}_3\text{O}^+} \quad \text{Y} \)
Answer: Let's identify the products A, B, X, and Y in the given reaction sequence starting from butan-2-ol.
**1. Formation of A and B:** * **Butan-2-ol to A:** Butan-2-ol, a secondary alcohol, reacts with thionyl chloride (\( \text{SOCl}_2 \)). This converts the hydroxyl group into a chlorine atom via nucleophilic substitution, forming 2-chlorobutane (A). \[ \underset{\text{Butan-2-ol}}{\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3} \quad \xrightarrow{\text{SOCl}_2} \quad \underset{\text{2-chlorobutane (A)}}{\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3} \] * **A to B:** 2-chlorobutane (A) then reacts with magnesium metal (\( \text{Mg} \)) in dry ether. This reaction inserts magnesium into the carbon-halogen bond, forming a Grignard reagent, sec-butyl magnesium chloride (B). \[ \underset{\text{2-chlorobutane (A)}}{\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3} \quad \xrightarrow{\text{Mg/ether}} \quad \underset{\text{sec-butyl magnesium chloride (B)}}{\text{CH}_3\text{CH}_2\text{CH(MgCl)CH}_3} \]
**2. Formation of X and Y:** * **Butan-2-ol to X:** When butan-2-ol is passed over heated copper at 573 K, it undergoes dehydrogenation (oxidation). This converts the secondary alcohol into a ketone, 2-butanone (X). \[ \underset{\text{Butan-2-ol}}{\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3} \quad \xrightarrow{\text{Cu/573 K}} \quad \underset{\text{2-butanone (X)}}{\text{CH}_3\text{CH}_2\text{C(O)CH}_3} \] * **X with B to Y:** 2-butanone (X) reacts with the Grignard reagent, sec-butyl magnesium chloride (B). The sec-butyl group from the Grignard reagent adds to the carbonyl carbon of 2-butanone, forming a tertiary alkoxide intermediate. Subsequent acidic hydrolysis (\( \text{H}_3\text{O}^+ \)) converts this alkoxide into a tertiary alcohol, 3,4-dimethyl-3-hexanol (Y). \[ \underset{\text{2-butanone (X)}}{\text{CH}_3\text{CH}_2\text{C(O)CH}_3} \quad + \quad \underset{\text{sec-butyl magnesium chloride (B)}}{\text{CH}_3\text{CH}_2\text{CH(MgCl)CH}_3} \quad \longrightarrow \quad \text{intermediate alkoxide} \] \[ \text{intermediate alkoxide} \quad \xrightarrow{\text{H}_3\text{O}^+} \quad \underset{\text{3,4-dimethyl-3-hexanol (Y)}}{\text{CH}_3\text{CH}_2\text{-}\underset{\text{OH}}{\overset{\text{CH}_3}{|}}\text{C}\text{-}\underset{\text{CH}_3}{|}\text{CH}\text{-CH}_2\text{CH}_3} \] Therefore:
**A is 2-chlorobutane**
**B is sec-butyl magnesium chloride**
**X is 2-butanone**
**Y is 3,4-dimethyl-3-hexanol**
In simple words: Butan-2-ol becomes 2-chlorobutane (A) with \( \text{SOCl}_2 \), then 2-chlorobutane makes a Grignard reagent (B) with magnesium. Separately, butan-2-ol changes to 2-butanone (X) when heated with copper. Finally, 2-butanone (X) reacts with the Grignard reagent (B) and water to form a complex alcohol, 3,4-dimethyl-3-hexanol (Y).

🎯 Exam Tip: This problem combines several key reactions: alcohol to alkyl halide, Grignard reagent formation, alcohol oxidation (dehydrogenation), and Grignard addition to a ketone. Ensure you can identify the product types and name them correctly.

 

Question 25. 3,3 – dimethyl butane – 2 – ol on treatment with conc. H2SO4 to give tetramethyl ethylene as a major product. Suggest suitable mechanisms.
Answer: When 3,3-dimethylbutan-2-ol is treated with concentrated sulfuric acid (\( \text{Conc. H}_2\text{SO}_4 \)), it undergoes acid-catalyzed dehydration. This reaction proceeds via a carbocation intermediate, which often involves rearrangements to form a more stable carbocation. According to Saytzeff's rule, the major product is the most substituted alkene. In this case, tetramethylethylene (2,3-dimethylbut-2-ene) is the major product.
**Mechanism:**
**Step 1: Protonation of the Alcohol** The hydroxyl group (\( \text{-OH} \)) of 3,3-dimethylbutan-2-ol accepts a proton from the sulfuric acid, forming a protonated alcohol (an alkyloxonium ion). This transforms the poor leaving group (\( \text{-OH} \)) into a good leaving group (\( \text{-OH}_2^+ \)). \[ \underset{\text{3,3-dimethylbutan-2-ol}}{\text{CH}_3\text{-}\underset{\text{OH}}{|}\text{CH}\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3} \quad + \quad \text{H}^+ \quad \longrightarrow \quad \text{CH}_3\text{-}\underset{\text{OH}_2^+}{|}\text{CH}\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3 \]
**Step 2: Formation of Secondary Carbocation** The protonated alcohol then loses a water molecule, leading to the formation of a secondary carbocation. This is the rate-determining step. \[ \text{CH}_3\text{-}\underset{\text{OH}_2^+}{|}\text{CH}\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3 \quad \xrightarrow{\text{-H}_2\text{O}} \quad \underset{\text{Secondary carbocation}}{\text{CH}_3\text{-}\overset{+}{C}\text{H}\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3} \]
**Step 3: Carbocation Rearrangement (1,2-Methyl Shift)** The secondary carbocation is less stable than a tertiary one. A 1,2-methyl shift occurs, where one of the methyl groups from the adjacent carbon (which is a tertiary carbon) migrates to the positively charged carbon. This results in the formation of a more stable tertiary carbocation. \[ \underset{\text{Secondary carbocation}}{\text{CH}_3\text{-}\overset{+}{C}\text{H}\text{-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3} \quad \xrightarrow{\text{1,2-Methyl Shift}} \quad \underset{\text{Tertiary carbocation}}{\text{CH}_3\text{-CH}_2\text{-}\underset{\text{CH}_3}{\overset{+}{\text{C}}}\text{-CH}_3} \]
**Step 4: Elimination of a Proton to Form Alkenes** The tertiary carbocation then loses a proton from an adjacent carbon to form an alkene. According to Saytzeff's rule, the major product is the most substituted alkene. * **Major Product (Tetramethylethylene):** Proton removal from the \( \alpha \)-carbon's hydrogen adjacent to the tertiary carbocation, forming a highly substituted double bond. \[ \underset{\text{Tertiary carbocation}}{\text{CH}_3\text{-CH}_2\text{-}\underset{\text{CH}_3}{\overset{+}{\text{C}}}\text{-CH}_3} \quad \xrightarrow{\text{-H}^+} \quad \underset{\text{2,3-dimethylbut-2-ene (Major)}}{\text{CH}_3\text{-CH=}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3} \] * **Minor Product (3,3-dimethylbut-1-ene):** Proton removal from the \( \beta \)-carbon's hydrogen, leading to a less substituted alkene. \[ \underset{\text{Tertiary carbocation}}{\text{CH}_3\text{-CH}_2\text{-}\underset{\text{CH}_3}{\overset{+}{\text{C}}}\text{-CH}_3} \quad \xrightarrow{\text{-H}^+} \quad \underset{\text{3,3-dimethylbut-1-ene (Minor)}}{\text{CH}_2\text{=CH-}\underset{\text{CH}_3}{\overset{\text{CH}_3}{|}}\text{C}\text{-CH}_3} \]
In simple words: First, the alcohol gets an extra hydrogen. Then it loses water, leaving a positive spot (carbocation). This positive spot then shifts to a more stable place by moving a methyl group. Finally, a hydrogen leaves from a nearby carbon, forming a double bond. The main product is tetramethylethylene because it's the most stable alkene that can be formed.

🎯 Exam Tip: For acid-catalyzed dehydration of alcohols, especially secondary or tertiary ones with branching, always anticipate carbocation rearrangements (hydride or alkyl shifts) if they lead to a more stable carbocation. The most substituted alkene is generally the major product (Saytzeff's rule).

III. Evaluate yourself

 

Question 1. Classify the following alcohols as 1º, 2º, and 3º and give their IUPAC Names.
Answer: Let's classify each alcohol as primary (1°), secondary (2°), or tertiary (3°) and provide their IUPAC names. The classification depends on the number of carbon atoms directly attached to the carbon bearing the hydroxyl group (\( \text{-OH} \)).

In simple words: To classify alcohols, look at the carbon that holds the -OH group. If that carbon is connected to one other carbon, it's primary (1°). If it's connected to two other carbons, it's secondary (2°). If it's connected to three other carbons, it's tertiary (3°). Then, name them using IUPAC rules by finding the longest chain with the -OH group and numbering carbons correctly.

🎯 Exam Tip: For IUPAC naming, find the longest continuous carbon chain containing the hydroxyl group. Number the chain to give the hydroxyl group the lowest possible number. List substituents alphabetically with their positions. For classification, count carbons directly attached to the C-OH carbon.

Question 2. Write all the possible isomers of alcohol having the molecular formula \( \text{C}_{15}\text{H}_{12}\text{O} \) and their IUPAC names.
Answer: Eight different alcohol structures are possible with the molecular formula \( \text{C}_{15}\text{H}_{12}\text{O} \). These are called isomers because they have the same formula but different arrangements of atoms. Finding all possible isomers helps in understanding how complex molecules can be formed and named.
In simple words: For the chemical formula \( \text{C}_{15}\text{H}_{12}\text{O} \), there are eight different alcohol shapes we can draw, each with its own special name.

🎯 Exam Tip: When asked to write isomers, always ensure that each isomer has the exact same molecular formula but different structural arrangements, and provide the correct IUPAC name for each to avoid losing marks.

 

Question 3. Suggest a suitable carbonyl compound for the preparation of pent – 2 – en – 1 ol using LiAlH4.
Answer: To make pent-2-en-1-ol using \( \text{LiAlH}_4 \), we should start with 2-pentenal. Lithium aluminum hydride (LiAlH4) is a strong reducing agent that can convert aldehydes into primary alcohols without affecting any carbon-carbon double bonds present in the molecule.
\( \text{CH}_3\text{-CH}_2\text{-CH} = \text{CH}-\text{CHO} \)
(2 - pentenal)
\( \frac{\text{LiAlH}_4}{\text{ }} \)
\( \implies \text{CH}_3\text{-CH}_2\text{-CH} = \text{CH}-\text{CH}_2\text{-OH} \)
(Pent-2-en-1-ol)
In simple words: We can make pent-2-en-1-ol by using the chemical 2-pentenal and a reducing agent called lithium aluminum hydride. This changes the aldehyde part to an alcohol, but keeps the double bond as it is.

🎯 Exam Tip: Remember that \( \text{LiAlH}_4 \) is a selective reducing agent; it reduces carbonyl groups (like aldehydes and ketones) but generally leaves carbon-carbon double or triple bonds untouched, which is key for synthesizing unsaturated alcohols.

 

Question 4. 2 – methylpropan – 2 – ene \( \text{H}_2\text{SO}_4/\text{H}_2\text{O} \) ?
Answer: When 2-methylpropan-2-ene reacts with water in the presence of sulfuric acid, it undergoes hydration following Markovnikov's rule. This means the hydrogen atom from water adds to the carbon atom of the double bond that already has more hydrogen atoms, and the hydroxyl group adds to the carbon with fewer hydrogen atoms, leading to the formation of a tertiary alcohol.
\[ \text{CH}_3\text{-C(CH}_3\text{)} = \text{CH}_2 \frac{\text{H}_2\text{SO}_4}{\text{H}_2\text{O}} \implies \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_3 \] (2-methyl-2-propanol)
In simple words: When 2-methylpropene mixes with water and acid, it forms 2-methyl-2-propanol. The -OH group from water attaches to the carbon that has fewer hydrogen atoms.

🎯 Exam Tip: Always remember Markovnikov's rule for the addition of unsymmetrical reagents (like H2O, HX) to unsymmetrical alkenes: the hydrogen adds to the carbon with more hydrogens, and the other part adds to the carbon with fewer hydrogens.

 

Question 5. How will you prepare the following using a Grignard reagent?
1. t – butyl alcohol
2. allyl alcohol
Answer:
(i) To prepare t-butyl alcohol using a Grignard reagent, we can react acetone with methylmagnesium bromide. Acetone is a ketone, and the addition of a Grignard reagent to a ketone followed by hydrolysis produces a tertiary alcohol. The methyl group from the Grignard reagent adds to the carbonyl carbon, and the magnesium-halide part is removed during hydrolysis.
\[ \text{CH}_3\text{-C}(=\text{O})\text{-CH}_3 + \text{CH}_3\text{MgBr} \implies \text{CH}_3\text{-C(OMgBr)(CH}_3\text{)-CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_3 + \text{Mg(OH)Br} \] (Acetone) (Methylmagnesium bromide) (t-butyl alcohol)
(ii) To prepare allyl alcohol using a Grignard reagent, we can react formaldehyde with vinylmagnesium bromide. Formaldehyde is an aldehyde, and the addition of a Grignard reagent to an aldehyde followed by hydrolysis produces a primary alcohol. The vinyl group from the Grignard reagent adds to the carbonyl carbon, and hydrolysis removes the magnesium-halide part.
\[ \text{H-C}(=\text{O})\text{-H} + \text{CH}_2 = \text{CH-MgBr} \implies \text{H-C(OMgBr)(H)-CH}=\text{CH}_2 \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{H-C(OH)(H)-CH}=\text{CH}_2 + \text{Mg(OH)Br} \] (Formaldehyde) (Vinylmagnesium bromide) (Allyl alcohol)
In simple words: We can make t-butyl alcohol by mixing acetone with a methyl Grignard reagent. To make allyl alcohol, we mix formaldehyde with a vinyl Grignard reagent. Both reactions need water and acid at the end.

🎯 Exam Tip: Remember that Grignard reagents react with formaldehyde to give primary alcohols, with other aldehydes to give secondary alcohols, and with ketones to give tertiary alcohols.

 

Question 6. Identify the product (s) in the following reactions. Write their IUPAC names and mention the mechanism involved in the reactions.
I cyclopentanol \( \xrightarrow{\text{H}_2\text{SO}_4} \) ?
iii) neopentyl alcohol \( \xrightarrow{\text{PCl}_5} \) ?
ii) butan-1-ol \( \xrightarrow{\text{NaBr}/\text{H}_2\text{SO}_4} \) ?
Answer:
(i) Cyclopentanol undergoes dehydration in the presence of concentrated sulfuric acid to form cyclopentene. This reaction follows the E1 mechanism, which involves the formation of a carbocation intermediate.
\[ \text{Cyclopentanol} \xrightarrow{\text{Conc H}_2\text{SO}_4} \text{Cyclopentene} + \text{H}_2\text{O} \] Mechanism: E1 mechanism (Dehydration)
(ii) Butan-1-ol reacts with sodium bromide in the presence of sulfuric acid to form 1-bromobutane. This is a nucleophilic substitution reaction, specifically an \( \text{S}_{\text{N}}2 \) mechanism, as it involves a primary alcohol.
\[ \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} \xrightarrow{\text{NaBr}/\text{H}_2\text{SO}_4} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-Br} + \text{NaHSO}_4 + \text{H}_2\text{O} \] (1-butanol) (1-bromobutane)
Mechanism: \( \text{S}_{\text{N}}2 \) mechanism
(iii) Neopentyl alcohol (2,2-dimethylpropan-1-ol) reacts with \( \text{PCl}_5 \) to form 1-chloro-2,2-dimethylpropane. This is a nucleophilic substitution reaction. Since neopentyl alcohol is a primary alcohol, this reaction typically follows an \( \text{S}_{\text{N}}2 \) mechanism.
\[ \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-CH}_2\text{-OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-CH}_2\text{-Cl} + \text{POCl}_3 + \text{HCl} \] (Neopentyl alcohol) (1-chloro-2,2-dimethylpropane)
Mechanism: \( \text{S}_{\text{N}}2 \) mechanism
In simple words: (i) Cyclopentanol loses water with strong acid to make cyclopentene. (ii) Butan-1-ol turns into 1-bromobutane when reacted with sodium bromide and acid. (iii) Neopentyl alcohol turns into 1-chloro-2,2-dimethylpropane when reacted with \( \text{PCl}_5 \). These are all common ways to change alcohols into other compounds.

🎯 Exam Tip: Always identify the type of alcohol (primary, secondary, tertiary) as it dictates the most likely reaction mechanism (E1, E2, SN1, SN2) and the products formed in dehydration and substitution reactions.

 

Question 7. What is the major product obtained when 2, 3 – dimethyl pentan – 3 – ol is heated in the presence of H2SO4
Answer: When 2,3-dimethylpentan-3-ol is heated with sulfuric acid, it undergoes dehydration. According to Saytzeff's rule, the major product will be the most substituted alkene. The hydroxyl group removes with a hydrogen from an adjacent carbon, and the most stable alkene (with more alkyl groups attached to the double bond carbons) is preferred. This leads to the formation of 3-ethyl-2,3-dimethylbut-1-ene as the major product.
\[ \text{CH}_3\text{-CH(CH}_3\text{)-C(OH)(CH}_3\text{)-CH}_2\text{-CH}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{-CH(CH}_3\text{)-C}(=\text{CH}_2\text{)-CH}_2\text{-CH}_3 \text{ (Major Product)} + \text{H}_2\text{O} \] (2,3-dimethylpentan-3-ol) (3-ethyl-2,3-dimethylbut-1-ene)
In simple words: When 2,3-dimethylpentan-3-ol is heated with acid, it loses water and forms a new molecule with a double bond. The main product is the one where the double bond has the most other carbon groups attached, making it very stable.

🎯 Exam Tip: Saytzeff's rule is crucial for predicting the major product in elimination reactions: the hydrogen atom is removed from the \( \beta \)-carbon that has the fewest hydrogen atoms, leading to the most stable (most substituted) alkene.

 

Question 8. Which of the following set of reactants will give 1 – methoxy – 4 – nitrobenzene.
(i) \( \text{O}_2\text{N-C}_6\text{H}_4\text{-Br} + \text{CH}_3\text{ONa} \)
(ii) \( \text{O}_2\text{N-C}_6\text{H}_4\text{-ONa} + \text{CH}_3\text{Br} \)
Answer: (ii) \( \text{O}_2\text{N-C}_6\text{H}_4\text{-ONa} + \text{CH}_3\text{Br} \)
In Williamson ether synthesis, an alkoxide reacts with an alkyl halide. For aromatic ethers like 1-methoxy-4-nitrobenzene, it is better to use sodium 4-nitrophenoxide and methyl bromide. This is because the bromine in bromonitrobenzene is strongly bonded to the benzene ring and cannot be easily replaced in a nucleophilic substitution. However, the phenoxide ion is an excellent nucleophile.
\[ \text{O}_2\text{N-C}_6\text{H}_4\text{-ONa} + \text{CH}_3\text{Br} \implies \text{O}_2\text{N-C}_6\text{H}_4\text{-OCH}_3 + \text{NaBr} \] (Sodium 4-nitrophenoxide) (Methyl bromide) (1-methoxy-4-nitrobenzene)
In simple words: To make 1-methoxy-4-nitrobenzene, we should use sodium 4-nitrophenoxide and methyl bromide. We cannot use bromonitrobenzene with sodium methoxide because the bromine on the ring is too strong to remove.

🎯 Exam Tip: For Williamson ether synthesis involving aromatic rings, always use an alkoxide (or phenoxide) and an alkyl halide where the halogen is on the alkyl group. If the halogen is directly attached to the benzene ring, it's generally unreactive for \( \text{S}_{\text{N}}2 \) reactions.

 

Question 9. What happens when m – cresol is treated with an acidic solution of sodium dichromate?
Answer: When m-cresol (3-methylphenol) is treated with an acidic solution of sodium dichromate, the benzylic carbon (the carbon directly attached to the aromatic ring that also has hydrogen atoms) will be oxidized. This reaction will convert the methyl group into a carboxylic acid group, forming 3-hydroxybenzoic acid. This is because strong oxidizing agents like dichromate can oxidize alkyl groups on a benzene ring if they have benzylic hydrogens.
\[ \text{OH-C}_6\text{H}_4\text{-CH}_3 \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{OH-C}_6\text{H}_4\text{-COOH} \] (m-cresol) (3-hydroxybenzoic acid)
In simple words: If you mix m-cresol with strong acid and sodium dichromate, the methyl part of the m-cresol changes into an acid group. This makes 3-hydroxybenzoic acid.

🎯 Exam Tip: Remember that alkyl groups attached to a benzene ring can be oxidized to carboxylic acids if they have at least one benzylic hydrogen. Strong oxidizing agents like acidic dichromate or permanganate are used for this transformation.

 

Question 10. When phenol is treated with propan – 2 – ol in the presence of HF, Friedel – Craft reaction takes place. Identify the products.
Answer: When phenol reacts with propan-2-ol in the presence of hydrogen fluoride (HF) as a catalyst, a Friedel-Crafts alkylation occurs. This reaction introduces an isopropyl group onto the phenol ring. Phenol is an ortho/para directing group, meaning the isopropyl group will attach at positions ortho and para to the hydroxyl group. Thus, the products are 2-isopropylphenol (ortho-product) and 4-isopropylphenol (para-product). The para-product is usually the major product due to less steric hindrance.
\[ \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{-CH(OH)-CH}_3 \xrightarrow{\text{HF}} \text{C}_6\text{H}_4\text{(OH)-CH(CH}_3\text{)}_2 \] (Phenol) (Propan-2-ol) (2-isopropylphenol and 4-isopropylphenol)
In simple words: Phenol reacts with propan-2-ol and HF acid to add an isopropyl group to the phenol ring. The two main things made are 2-isopropylphenol and 4-isopropylphenol, with the 4-isopropylphenol usually being the most produced.

🎯 Exam Tip: In Friedel-Crafts alkylation of phenols, the hydroxyl group is an activating and ortho/para-directing group. Always consider both ortho and para products, with the para product often being favored due to reduced steric hindrance.

 

Question 11. Given the IUPAC name for the following ethers and classify them as simple or mixed.
Answer: Ethers are compounds where an oxygen atom is connected to two alkyl or aryl groups. They can be classified as simple if both groups are identical, or mixed if the groups are different. Naming them correctly requires identifying the longest carbon chain and the smaller alkoxy group.
In simple words: We need to name these ethers using IUPAC rules and say if they are "simple" (same groups on both sides of oxygen) or "mixed" (different groups).

🎯 Exam Tip: To classify ethers, look at the two groups attached to the oxygen atom. If they are identical, it's a simple ether; if they are different, it's a mixed ether. For IUPAC naming, prioritize the longer carbon chain as the alkane and the smaller group as an alkoxy substituent.

 

Question 12. 1. Which of the following reaction will give 1 – methoxy – 4 – nitrobenzene.
1. 4 – nitro – 1 – bromobenzene + sodium methoxide.
2. 4 – nitrosodium phenoxide + bromomethane
Answer: The reaction that will give 1-methoxy-4-nitrobenzene is: 4-nitrosodium phenoxide + bromomethane.
This is an example of Williamson ether synthesis. In this reaction, the phenoxide ion acts as a nucleophile and attacks the methyl bromide, leading to the formation of the ether. Using 4-nitro-1-bromobenzene with sodium methoxide will not work easily because the bromine is directly attached to the benzene ring and is unreactive towards \( \text{S}_{\text{N}}2 \) displacement.
\[ \text{O}_2\text{N-C}_6\text{H}_4\text{-ONa} + \text{CH}_3\text{Br} \implies \text{O}_2\text{N-C}_6\text{H}_4\text{-OCH}_3 + \text{NaBr} \] (4-nitrosodium phenoxide) (Bromomethane) (1-methoxy-4-nitrobenzene)
In simple words: To make 1-methoxy-4-nitrobenzene, we should mix 4-nitrosodium phenoxide with bromomethane. The other choice won't work well because the bromine is too tightly held by the benzene ring.

🎯 Exam Tip: For aromatic ether synthesis, ensure the nucleophile is an aromatic alkoxide (phenoxide) and the electrophile is an alkyl halide. The reverse (aryl halide + alkyl alkoxide) typically fails due to the low reactivity of aryl halides in nucleophilic substitution.

 

Question 13. Arrange the following compounds in the increasing order of their acid strength. propan 1 – ol, 2, 4, 6 – trinitroptienol, 3 – nitrophenol, 3,5 – dinitrophenol, phenol, 4 – methyl phenol.
Answer: The acidity of alcohols and phenols depends on the stability of their conjugate base (alkoxide or phenoxide ion). Phenols are generally more acidic than alcohols because the phenoxide ion is stabilized by resonance, unlike the alkoxide ion from alcohols. Electron-withdrawing groups increase acidity, while electron-donating groups decrease it. The effect of electron-withdrawing groups is more pronounced at ortho and para positions due to direct resonance interaction.
Here's the increasing order of acid strength:
Propan-1-ol \( < \) 4-methylphenol \( < \) phenol \( < \) 3-nitrophenol \( < \) 3,5-dinitrophenol \( < \) 2,4,6-trinitrophenol.
Propan-1-ol is the least acidic as its conjugate base is not resonance stabilized. 4-methylphenol has an electron-donating methyl group, making it less acidic than phenol. Phenol is more acidic than alcohols due to resonance stabilization. Nitrophenols are more acidic than phenol because the nitro group is an electron-withdrawing group that stabilizes the phenoxide ion. The more nitro groups, or the better their position (ortho/para), the stronger the acid. 2,4,6-trinitrophenol (picric acid) is the most acidic due to three strong electron-withdrawing nitro groups at ortho and para positions, providing maximum resonance stabilization.
In simple words: Alcohols are weakest acids. Phenols are stronger because they can spread out negative charge. Adding groups that pull electrons makes phenols even stronger acids, especially if these groups are in the right places. Adding groups that push electrons makes phenols weaker acids. So, trinitrophenol is the strongest, and propan-1-ol is the weakest.

🎯 Exam Tip: To compare acid strengths of alcohols and phenols, remember: 1. Phenols \( > \) Alcohols. 2. Electron-withdrawing groups increase phenolic acidity (strongest at ortho/para). 3. Electron-donating groups decrease phenolic acidity (weakest at ortho/para). 4. More electron-withdrawing groups generally mean higher acidity.

 

Question 14. 1 mole of HI is allowed to react with t – butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer: When 1 mole of HI reacts with t-butyl methyl ether, the reaction follows an \( \text{S}_{\text{N}}1 \) mechanism because the ether contains a tertiary alkyl group. The first step involves protonation of the ether's oxygen. This forms an oxonium ion. Then, the C-O bond breaks to form a stable tertiary carbocation and methanol. Finally, the iodide ion attacks the tertiary carbocation to form t-butyl iodide. The products are t-butyl iodide and methyl alcohol.
\[ \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-O-CH}_3 + \text{H}^+ \implies \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-O}^+\text{H-CH}_3 \] (t-butyl methyl ether) (Protonated ether)
\( \implies \text{CH}_3\text{-C(CH}_3\text{)}_2^+ + \text{CH}_3\text{OH} \) (Tertiary carbocation + Methanol)
\( \implies \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-I} \) (t-butyl iodide)
\[ \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-O-CH}_3 \xrightarrow{\text{HI}} \text{CH}_3\text{-C(CH}_3\text{)}_2\text{-I} + \text{CH}_3\text{OH} \] (t-butyl methyl ether) (t-butyl iodide) (Methyl alcohol)
In simple words: When t-butyl methyl ether reacts with HI, it forms t-butyl iodide and methyl alcohol. This happens because the bond on the t-butyl side breaks easily, making a stable carbon ion that iodide then attacks.

🎯 Exam Tip: Remember that ethers with tertiary alkyl groups react via an \( \text{S}_{\text{N}}1 \) mechanism with HX, leading to the formation of a tertiary alkyl halide and an alcohol. Ethers with primary or secondary alkyl groups typically follow an \( \text{S}_{\text{N}}2 \) mechanism.

 

12th Chemistry Guide Hydroxy Compounds and Ethers Additional Questions and Answers

Part – II - Additional Questions

I. Choose the correct answer

 

Question 1. In ethanol the – OH group is attached to hybridised carbon
(a) sp
(b) \( \text{sp}^2 \)
(c) \( \text{sp}^3 \)
(d) \( \text{dsp}^2 \)
Answer: (c) \( \text{sp}^3 \)
In simple words: In an ethanol molecule, the carbon atom that holds the -OH group has four single bonds, which means it is \( \text{sp}^3 \) hybridized.

🎯 Exam Tip: To determine hybridization, count the number of sigma bonds and lone pairs around the atom. Four electron domains (e.g., four single bonds) mean \( \text{sp}^3 \) hybridization.

 

Question 2. In Vinyl alcohol the -OH group is attached to hvbridised carbon
(a) sp
(b) \( \text{sp}^2 \)
(c) \( \text{sp}^3 \)
(d) \( \text{dsp}^2 \)
Answer: (b) \( \text{sp}^2 \)
In simple words: In vinyl alcohol, the carbon atom connected to the -OH group is also part of a double bond. This means it is \( \text{sp}^2 \) hybridized.

🎯 Exam Tip: A carbon atom involved in a double bond (with three electron domains) is \( \text{sp}^2 \) hybridized, while a carbon in a single bond (with four electron domains) is \( \text{sp}^3 \) hybridized.

 

Question 3. Sorbitol is
(a) monohydric alcohol
(b) dihydric alcohol
(c) trihydric alcohol
(d) hexa hydric alcohol
Answer: (d) hexa hydric alcohol
In simple words: Sorbitol is a sugar alcohol that has six hydroxyl (-OH) groups in its structure, making it a hexa hydric alcohol.

🎯 Exam Tip: Remember that 'hydric' refers to the number of hydroxyl (-OH) groups in an alcohol. 'Mono' means one, 'di' means two, 'tri' means three, and 'hexa' means six.

 

Question 4. The IUPAC name of sorbitol is
(a) Ethan -1, 2 – diol
(b) Propan – 1, 2, 3 – triol
(c) Hexan – 1, 2, 3, 4, 5, 6 – hexol
(d) Ethenol
Answer: (c) Hexan – 1, 2, 3, 4, 5, 6 – hexol
In simple words: Sorbitol is officially named Hexan-1,2,3,4,5,6-hexol because it's a six-carbon chain with a hydroxyl group on each carbon.

🎯 Exam Tip: For polyols (alcohols with multiple -OH groups), the IUPAC name specifies the longest carbon chain as the alkane, followed by the suffix '-ol' with numbers indicating the positions of all hydroxyl groups.

 

Question 5. The IUPAC name of \( \text{CH}_2 = \text{CH – C(OH) – CH}_3 \)
(a) prop – 2 – en – 1 – ol
(b) 2 – methyl but – 3 – en – 2 ol
(c) 3 – methyl but – 1 – en – 3 – ol
(d) 1,1 – di methyl prop – 2 – en – 1 – ol
Answer: (b) 2 – methyl but – 3 – en – 2 ol
In simple words: The correct name for this molecule is 2-methylbut-3-en-2-ol. It's a four-carbon chain with a double bond at carbon 3 and an -OH group and a methyl group at carbon 2.

🎯 Exam Tip: When naming compounds with both a double bond and an alcohol group, prioritize the alcohol suffix '-ol' and ensure the longest continuous carbon chain containing both functional groups is identified. Number the chain to give the hydroxyl group the lowest possible number, then locate the double bond and any substituents.

 

Question 6. The C - O - H bond angle in methanol is
(a) 107°
(b) 109°.5°
(c) 108.9°
(d) 104°
Answer: (c) 108.9°
In simple words: The angle between the carbon, oxygen, and hydrogen atoms in methanol is around 108.9 degrees. This is close to the ideal tetrahedral angle but slightly smaller due to the lone pairs on the oxygen.

🎯 Exam Tip: While an \( \text{sp}^3 \) hybridized atom typically has 109.5° bond angles, the presence of lone pairs on the oxygen atom in alcohols causes slight repulsion, compressing the bond angle (e.g., C-O-H) to be slightly less than 109.5°.

 

Question 7. CH3-CH2-a + NaOH → CH3-CH2-OH + NaCl This reaction follows mechanism
(a) \( \text{S}_{\text{N}}1 \)
(b) \( \text{S}_{\text{N}}2 \)
(c) \( \text{E}^1 \)
(d) \( \text{E}^2 \)
Answer: (b) \( \text{S}_{\text{N}}2 \)
In simple words: This reaction, where an ethyl group connected to 'a' turns into ethanol using NaOH, happens through a \( \text{S}_{\text{N}}2 \) pathway. This means the new group comes in as the old group leaves, all in one smooth step.

🎯 Exam Tip: For primary alkyl halides (like ethyl halide), nucleophilic substitution with a strong base like NaOH typically follows an \( \text{S}_{\text{N}}2 \) mechanism because there's less steric hindrance for the nucleophile to attack from the back.

 

Question 8. Conversion of isobutyl chloride and isopropyl chloride into isobutyl alcohol and isopropyl alcohol respectively using dilute aquous NaOH follow the mechnisms
(a) \( \text{S}_{\text{N}}1 \) & \( \text{S}_{\text{N}}2 \)
(b) both \( \text{S}_{\text{N}}1 \)
(c) \( \text{S}_{\text{N}}2 \) & \( \text{S}_{\text{N}}1 \)
(d) both \( \text{S}_{\text{N}}2 \)
Answer: (c) \( \text{S}_{\text{N}}2 \) & \( \text{S}_{\text{N}}1 \)
In simple words: Changing isobutyl chloride to isobutyl alcohol uses a \( \text{S}_{\text{N}}2 \) path. Changing isopropyl chloride to isopropyl alcohol uses a \( \text{S}_{\text{N}}1 \) path. This is because isobutyl chloride is a primary halide and isopropyl chloride is a secondary halide.

🎯 Exam Tip: Primary alkyl halides typically undergo \( \text{S}_{\text{N}}2 \) reactions, while secondary alkyl halides can undergo both \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) depending on conditions (solvent, nucleophile strength). Here, dilute aqueous NaOH favors \( \text{S}_{\text{N}}2 \) for primary and \( \text{S}_{\text{N}}1 \) for secondary due to carbocation stability.

 

Question 9. Addition of water across the double bond of an unsymmetrical alkene follows
(a) Saytzeffs rule
(b) Markownikoff's rule
(c) anti Markownikoff's rule
(d) Popoff's rule
Answer: (b) Markownikoff's rule
In simple words: When water adds to a double bond that is not symmetrical, the hydrogen part of the water attaches to the carbon with more hydrogen atoms already there. This is known as Markovnikov's rule.

🎯 Exam Tip: Markovnikov's rule states that in the addition of an unsymmetrical reagent (like H-X or H-OH) to an unsymmetrical alkene, the hydrogen atom adds to the carbon atom of the double bond that has a greater number of hydrogen atoms already attached.

 

Question 10. Hydroboration of an alkene follows
(a) Saytzeff's rule
(b) Markownikoff's rule
(c) anti Markownikoff's rule
(d) Popoff's rule
Answer: (c) anti Markownikoff's rule
In simple words: Hydroboration is a reaction where hydrogen and boron add to a double bond in a way that is opposite to Markovnikov's rule. This helps to form alcohols with the -OH group on the less substituted carbon.

🎯 Exam Tip: Hydroboration-oxidation is a method to convert alkenes to alcohols in an anti-Markovnikov fashion, meaning the hydroxyl group attaches to the less substituted carbon atom of the original double bond.

 

Question 11. Addition of water to 2-methyl propene in presence of cone, sulphuric acid gives
(a) 2-propanol
(b) 2 – methyl – 2 – propanol
(c) 2 – methyl – 1 – propanol
(d) 2 – butanol
Answer: (b) 2 – methyl – 2 – propanol
When 2-methyl propene reacts with water in the presence of concentrated sulfuric acid, it undergoes hydration following Markovnikov's rule. The hydroxyl group adds to the more substituted carbon, resulting in a tertiary alcohol, 2-methyl-2-propanol. This is a common method for synthesizing alcohols from alkenes.
\[ \text{CH}_3\text{-C}(=\text{CH}_2)\text{-CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}_2\text{SO}_4} \text{CH}_3\text{-C(OH)(CH}_3\text{)-CH}_3 \] (2-methyl propene) (2-methyl-2-propanol)
In simple words: When water is added to 2-methyl propene with sulfuric acid, the -OH group joins the carbon with fewer hydrogen atoms. This creates 2-methyl-2-propanol.

🎯 Exam Tip: For hydration of alkenes under acidic conditions, always apply Markovnikov's rule: the hydrogen adds to the less substituted carbon and the hydroxyl group to the more substituted carbon, typically forming the more stable carbocation intermediate.

 

Question 12. 2 – methyl propene reacts with diborane followed by H2O2 in presence of NaOH gives
(a) 2 – proponol
(b) 2 – methyl – 2 – propanol
(c) 2 – methyl – 1 – propanol
(d) propanol
Answer: (c) 2 – methyl – 1 – propanol
In simple words: When 2-methyl propene reacts with diborane and then with hydrogen peroxide and sodium hydroxide, it forms 2-methyl-1-propanol. This reaction follows the anti-Markovnikov rule.

🎯 Exam Tip: Hydroboration-oxidation (reaction with diborane followed by H2O2/NaOH) is a powerful method for synthesizing alcohols from alkenes in an anti-Markovnikov manner, meaning the hydroxyl group attaches to the less substituted carbon, and the reaction is syn-addition.

 

Question 13. To prepare a primary alcohol a Grignard reagent must be reacted with
(a) HCHO
(b) RCHO
(c) RCOR
(d) RNH2
Answer: (a) HCHO
In simple words: When you want to make a primary alcohol using a Grignard reagent, you need to react it with formaldehyde (HCHO). Formaldehyde is the simplest aldehyde.

🎯 Exam Tip: Remember that formaldehyde is the only aldehyde that produces primary alcohols with Grignard reagents, as it has two hydrogen atoms on the carbonyl carbon.

 

Question 14. To prepare a secondary alcohol a Grignard reagent must be reacted with
(a) HCHO
(b) RCHO
(c) RCOR
(d) RNH2
Answer: (b) RCHO
In simple words: To create a secondary alcohol using a Grignard reagent, you must react it with an aldehyde other than formaldehyde, like RCHO. Aldehydes (RCHO) have one alkyl group.

🎯 Exam Tip: Distinguish between HCHO (formaldehyde) for primary alcohols and other RCHO (aldehydes) for secondary alcohols when using Grignard reagents.

 

Question 15. To prepare a tertiary alcohol a Grignard reagent must be reacted with
(a) HCHO
(b) RCHO
(c) RCOR
(d) RNH2
Answer: (c) RCOR
In simple words: If you want to make a tertiary alcohol using a Grignard reagent, you need to react it with a ketone, which has the general formula RCOR. Ketones have two alkyl groups.

🎯 Exam Tip: Ketones (RCOR) are used to form tertiary alcohols, while aldehydes (RCHO and HCHO) form primary and secondary alcohols, respectively.

 

Question 16. Phenyl magnesium bromide reacts with formaldehyde followed by hydrolysis gives
(a) phenol
(b) Benzyl alcohol
(c) Benzaldehyde
(d) Benzoic acid
Answer: (b) Benzyl alcohol
\( \text{C}_{6}\text{H}_{5}\text{MgBr} + \text{HCHO} \rightarrow \text{C}_{6}\text{H}_{5}\text{CH}_{2}\text{OMgBr} \)
\( \implies \) \( \text{C}_{6}\text{H}_{5}\text{CH}_{2}\text{OMgBr} + \text{H}_{3}\text{O}^{+} \rightarrow \text{C}_{6}\text{H}_{5}\text{CH}_{2}\text{OH} + \text{Mg(OH)Br} \)
In simple words: Phenyl magnesium bromide reacts with formaldehyde and then water to make benzyl alcohol. This reaction creates a primary alcohol.

🎯 Exam Tip: Formaldehyde (HCHO) always yields primary alcohols when reacted with Grignard reagents, like phenyl magnesium bromide.

 

Question 17. Name the product obtained when 1 mole of methyl magnesium bromide reacts with ethyl methanoate
(a) propan-2-ol
(b) propan-1-ol
(c) ethanal
Answer: (a) propan-2-ol
In simple words: Ethyl methanoate reacts with methyl magnesium bromide in two steps, and after adding water, it forms propan-2-ol. Esters generally require two moles of Grignard reagent for this conversion.

🎯 Exam Tip: Esters generally react with two equivalents of Grignard reagents to produce alcohols. Formate esters (like ethyl methanoate) produce secondary alcohols, while other esters produce tertiary alcohols.

 

Question 18. The reducing agent X is
\( \text{CH}_{3}\text{COCH}_{2}\text{CH}_{2}\text{COOH} \xrightarrow{\text{X} / \text{H}_{3}\text{O}} \text{CH}_{3}\text{-CHOH-CH}_{2}\text{CH}_{2}\text{COOH} \)
(a) Raney Ni
(b) NaBH4
(c) LiA/H4
(d) Na - Hg
Answer: (b) NaBH4
In simple words: The reducing agent X is sodium borohydride (\( \text{NaBH}_{4} \)) because it changes only the ketone part of the molecule into an alcohol, leaving the acid part untouched. This is a selective reduction.

🎯 Exam Tip: Remember that \( \text{NaBH}_{4} \) is a milder reducing agent that can selectively reduce ketones and aldehydes, whereas \( \text{LiAlH}_{4} \) is a stronger agent that reduces carboxylic acids, esters, aldehydes, and ketones.

 

Question 19. The best reducing agent to prepare unsaturated alcohols from unsaturated carbonyl compounds is
(a) Raney Ni
(b) NaBH4
(c) LiA/H4
(d) both (b) & (c)
Answer: (d) both (b) & (c)
In simple words: To make unsaturated alcohols from unsaturated carbonyl compounds, both \( \text{NaBH}_{4} \) and \( \text{LiAlH}_{4} \) are good because they only reduce the carbonyl group, leaving the double bond alone. These are powerful yet selective reducing agents.

🎯 Exam Tip: For unsaturated carbonyl compounds, select a reducing agent that does not reduce the carbon-carbon double bond, like \( \text{NaBH}_{4} \) or \( \text{LiAlH}_{4} \).

 

Question 20. Ethene is converted into ethylene glycol by treating with
(a) acidified potassium dichromate
(b) alkaline potassium permanganate
Answer: (b) alkaline potassium permanganate
In simple words: To turn ethene into ethylene glycol, you treat it with alkaline potassium permanganate. This is also called Baeyer's test.

🎯 Exam Tip: Alkaline potassium permanganate (Baeyer's reagent) is used for the syn-dihydroxylation of alkenes, adding two -OH groups to the same side of the double bond.

 

Question 21. Cold dilute alkaline potassium permanganate is called as
(a) Fenton's reagent
(b) Tollen's reagent
(c) Baeyer's reagent
(d) Schiff's reagent
Answer: (c) Baeyer's reagent
In simple words: Cold, dilute alkaline potassium permanganate is known as Baeyer's reagent. This reagent helps to test if a chemical compound has double or triple bonds.

🎯 Exam Tip: Baeyer's reagent is a key test for unsaturation; its purple color disappears or a brown precipitate forms when it reacts with an alkene or alkyne.

 

Question 22. Glyceryl palmitate on alkaline hydrolysis gives
(a) glycerol and soap
(b) glycerol and sodium palmitate
(c) glycerol and sodium hexadentate
(d) all the options
Answer: (d) all the options
In simple words: When glyceryl palmitate is broken down with alkali and water, it makes glycerol and soap (which is sodium palmitate). So, all options correctly describe the products.

🎯 Exam Tip: Alkaline hydrolysis of fats like glyceryl palmitate is called saponification, yielding glycerol and a soap (the sodium salt of the fatty acid).

 

Question 23. The preparation of glycerol and soap from oils and fats is known as
(a) esterification
(b) saponification
(c) hydroboration
Answer: (b) saponification
In simple words: Making glycerol and soap from oils and fats is called saponification. This is how soap is made.

🎯 Exam Tip: Saponification is the industrial process for producing soap by hydrolyzing triglycerides with a strong base.

 

Question 24. Lucas reagent is
(a) dil HCl & ZnCl2
(b) Cone. HCl & anhydrous ZnCl2
(c) dil H2SO4 & AlCl3
(d) Conc. H2SO4 & anhydrous AlCl3
Answer: (b) Cone. HCl & anhydrous ZnCl2
In simple words: Lucas reagent is a mix of concentrated hydrochloric acid and dry zinc chloride. It's used to tell different types of alcohols apart.

🎯 Exam Tip: The Lucas test uses this reagent to classify alcohols; tertiary alcohols react fastest, followed by secondary, and primary alcohols react very slowly or not at all at room temperature.

 

Question 25. Which among the following will give immediate turbidity with Lucas reagent?
(a) propan-2-ol
(b) propan-1-ol
(c) 2-methyl propan-2-ol
(d) 2-methyl propan-1-ol
Answer: (c) 2-methyl propan-2-ol
In simple words: 2-methyl propan-2-ol will cause immediate cloudiness with Lucas reagent because it is a tertiary alcohol. Tertiary alcohols react quickly.

🎯 Exam Tip: Tertiary alcohols show immediate turbidity with Lucas reagent due to the rapid formation of a stable carbocation and insoluble alkyl chloride.

 

Question 26. In Victor Meyer's test, the alcohol which gives blue colour is
(a) propan-2-ol
(b) propan-1-ol
Answer: (a) propan-2-ol
In simple words: A secondary alcohol, like propan-2-ol, gives a blue color in Victor Meyer's test. Primary alcohols turn red, and tertiary alcohols stay colorless.

🎯 Exam Tip: Remember the color code for Victor Meyer's test: Red for primary (1°), Blue for secondary (2°), and Colorless for tertiary (3°) alcohols.

 

Question 27. In the given equation \( \text{C}_{6}\text{H}_{5}\text{OH} \xrightarrow{\text{H}_{2}\text{SO}_{4}, 110^{\circ}\text{C}} \text{[X]} \xrightarrow{\text{(i) NaOH fusion (ii) H}^{+}} \text{[Y]} \) is
(a) OH
(b) OH SO3H
(c) OH
(d) OH
Answer: (c) OH
\( \text{C}_{6}\text{H}_{5}\text{OH} \xrightarrow{\text{H}_{2}\text{SO}_{4}, 110^{\circ}\text{C}} \text{HO-C}_{6}\text{H}_{4}\text{-SO}_{3}\text{H} \) (p-hydroxybenzenesulfonic acid, X)
\( \implies \) \( \text{HO-C}_{6}\text{H}_{4}\text{-SO}_{3}\text{H} \xrightarrow{\text{(i) NaOH fusion (ii) H}^{+}} \text{C}_{6}\text{H}_{5}\text{OH} \) (Phenol, Y)
In simple words: Phenol reacts with sulfuric acid to make a sulfonic acid (X). Then, by reacting X with sodium hydroxide and acid, you get phenol (Y) back. This is a way to make phenol.

🎯 Exam Tip: NaOH fusion of sulfonic acid derivatives is a common method for preparing phenols.

 

Question 28. Among butyl alcohols which one have the lowest boiling point?
(a) n-butyl alcohol
(b) isobutyl alcohol
(c) sec-butyl alcohol
(d) ter - butyl alcohol
Answer: (d) ter - butyl alcohol
In simple words: Tertiary butyl alcohol has the lowest boiling point because it has the most branches in its structure, which reduces how strongly molecules stick together. More branching means less surface area for contact.

🎯 Exam Tip: Boiling points of isomeric alcohols decrease with increased branching due to reduced surface area for van der Waals interactions, making tertiary alcohols have the lowest boiling points.

 

Question 29. Which among the alcohol has a higher boiling point?
(a) n-butyl alcohol
(b) n-propyl alcohol
(c) ethyl alcohol
(d) methyl alcohol
Answer: (a) n-butyl alcohol
In simple words: N-butyl alcohol has the highest boiling point because it is the largest molecule among the choices, meaning its molecules stick together more strongly. Longer chains allow for more interactions.

🎯 Exam Tip: Boiling points increase with molecular weight (longer carbon chain) within a homologous series due to stronger van der Waals forces.

 

Question 30. Which among the following is less soluble in water?
(a) n-butyl alcohol
(b) n-propyl alcohol
(c) ethyl alcohol
(d) methyl alcohol
Answer: (a) n-butyl alcohol
In simple words: N-butyl alcohol is least soluble in water because its carbon chain is the longest, making it more oily and less able to mix with water. The non-polar part becomes dominant.

🎯 Exam Tip: As the hydrocarbon part of an alcohol gets bigger, its water solubility decreases because the non-polar part becomes more dominant.

 

Question 31. Conversion of alcohols into alkyl halide is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer: (b) Nucleophilic substitution
In simple words: Turning alcohols into alkyl halides is a nucleophilic substitution reaction, meaning one part (the -OH group) is swapped for another (a halogen). This happens when a nucleophile attacks.

🎯 Exam Tip: Recognize substitution reactions where one group is directly replaced by another, often involving a nucleophile attacking an electrophilic carbon.

 

Question 32. Conversion of 2 - methyl - 1 - propanol into 2 - methyl - 1 - bromopropane is reaction
(a) SN2
(b) SN¹
(c) E2
(d) E¹
Answer: (a) SN2
In simple words: Changing 2-methyl-1-propanol into 2-methyl-1-bromopropane happens through an \( \text{S}_{\text{N}}2 \) reaction. This is a one-step reaction.

🎯 Exam Tip: Primary alcohols usually undergo \( \text{S}_{\text{N}}2 \) reactions for conversion to alkyl halides, especially with strong nucleophiles and good leaving groups.

 

Question 33. Primary alcohols undergo dehydration by mechanism
(a) SN2
(b) SN¹
(c) E2
(d) E¹
Answer: (d) E¹
In simple words: Primary alcohols can lose water (dehydrate) through an \( \text{E}^{1} \) mechanism, where a carbocation is formed first, and then an alkene. This usually needs strong acid and heat.

🎯 Exam Tip: While tertiary alcohols readily undergo \( \text{E}^{1} \) dehydration due to stable carbocations, primary alcohols usually prefer \( \text{E}^{2} \) elimination or require harsh conditions for \( \text{E}^{1} \) due to the instability of primary carbocations.

 

Question 34. 2 - methyl - 2 - propanol when reacted with cone. H2SO4 gives 2 - methyl propene. This reaction follows mechanism
(a) SN2
(b) SN¹
(c) E2
(d) E¹
Answer: (d) E¹
In simple words: 2-methyl-2-propanol, a tertiary alcohol, loses water to form 2-methyl propene through an \( \text{E}^{1} \) mechanism. This happens easily because of the stable carbocation formed.

🎯 Exam Tip: Tertiary alcohols favor \( \text{E}^{1} \) dehydration due to the stability of the intermediate tertiary carbocation, leading to rapid formation of alkenes.

 

Question 35. During intramolecular dehydration of 3, 3 - dimethyl - 2 - butanol the major product obtained is
(a) 2, 3 - dimethyl but - 1 - ene
(b) 2, 3 - dimethyl but - 2 - ene
(c) 3, 3 - dimethyl but - 1 - ene
(d) 3, 3 - dimethyl but - 2 - ene
Answer: (b) 2, 3 - dimethyl but - 2 - ene
In simple words: When 3,3-dimethyl-2-butanol loses water, the main product is 2,3-dimethylbut-2-ene, following Saytzeff's rule for the most stable alkene. The most stable alkene has more groups attached to the double bond.

🎯 Exam Tip: For dehydration reactions leading to multiple alkene products, Saytzeff's rule predicts the most substituted (most stable) alkene as the major product.

 

Question 36. The major product obtained when phenol reacts with con.H2SO4 at 280 K is
(a) Salicyclic acid
(b) picric acid
(c) o-phenol sulphonic acid
(d) p-phenol sulphonic acid
Answer: (c) o-phenol sulphonic acid
In simple words: When phenol reacts with concentrated sulfuric acid at 280 K, it mostly forms o-phenol sulfonic acid. This happens because the reaction is fast at this position.

🎯 Exam Tip: Remember that the sulfonation of phenol is temperature-dependent: low temperature favors ortho-substitution (kinetic control), while high temperature favors para-substitution (thermodynamic control).

 

Question 37. During intramolecular dehydration of alcohols say tzeff's rule favours the formation of
(a) unstable alkenes
(b) less substituted alkenes
(c) more substituted alkenes
(d) none of the options
Answer: (c) more substituted alkenes
In simple words: Saytzeff's rule says that during alcohol dehydration, the main product will be the alkene with more branches (more substituted). More branches make the alkene more stable.

🎯 Exam Tip: Saytzeff's rule is crucial for predicting the major product in elimination reactions where multiple alkene isomers can be formed, always favoring the most stable (highly substituted) alkene.

 

Question 39. In Swern oxidation of alcohols into aldehydes/ketones the oxidising agent used is
(a) Pyridinium chlorochromate (PCC)
(b) Dimethyl sulfoxide (DMSO)
(c) Alkaline potassium permanganate (Baeyer's reagent)
(d) Ferrous sulphate / H2O2 (Fenton's reagent)
Answer: (b) Dimethyl sulfoxide (DMSO)
In simple words: Dimethyl sulfoxide (DMSO) is the main oxidizing agent used in Swern oxidation to turn alcohols into aldehydes or ketones. It's a gentle way to do this.

🎯 Exam Tip: Swern oxidation is favored for its mild conditions and selectivity, preventing further oxidation of aldehydes to carboxylic acids.

 

Question 40. To detoxify the alcohol produced in animals by the fermentaion of food, the oxidising agent used is
(a) ADH
(b) ADP
(c) NAD
(d) ATP
Answer: (c) NAD
In simple words: To clean up alcohol in animals, NAD+ is used as the oxidizing agent. It helps break down the alcohol.

🎯 Exam Tip: NAD+ is a vital coenzyme in many metabolic oxidation-reduction reactions, including alcohol detoxification, by acting as an electron acceptor.

 

Question 41. The catalyst which catalyses the oxidation of toxic alcohols into non-toxic aldehyde in animals is
(a) ADH
(b) ADP
(c) NAD
(d) ATP
Answer: (a) ADH
In simple words: The enzyme ADH (alcohol dehydrogenase) helps animals break down toxic alcohols into safer aldehydes. This is the first step in cleaning the body from alcohol.

🎯 Exam Tip: Enzymes are biological catalysts; ADH specifically catalyzes the oxidation of alcohols in biological systems.

 

Question 42. When ethan-1, 2-diol is heated with anhydrous ZnCl2 under pressure it gives
(a) Ethanol
(b) Ethanal
(c) Ethanoic acid
(d) Ethene
Answer: (b) Ethanal
In simple words: Heating ethan-1,2-diol with dry zinc chloride under pressure turns it into ethanal. This involves losing a water molecule and rearranging.

🎯 Exam Tip: Anhydrous \( \text{ZnCl}_{2} \) is a Lewis acid often used to promote dehydration and rearrangement reactions of alcohols and diols.

 

Question 43. Glycerol can be oxidised to meso oxalic acid by
(a) dil HNO3
(b) HIO4
(c) Bismuth nitrate
(d) Fenton's reagent
Answer: (c) Bismuth nitrate
In simple words: Glycerol can be changed into meso oxalic acid using bismuth nitrate as the oxidizing agent. This chemical helps in a gentle oxidation process.

🎯 Exam Tip: Different oxidizing agents yield different products from glycerol; knowing the specificity of each reagent is key.

 

Question 44. The alcohol used in the manufacture of dynamite is
(a) methanol
(b) ethanol
(c) ethv lene glycol
(d) glycerol
Answer: (d) glycerol
In simple words: Glycerol is the alcohol used to make nitroglycerine, which is then used in dynamite. It is a main part of this explosive.

🎯 Exam Tip: Glycerol's structure with three hydroxyl groups allows for the formation of trinitrate, an explosive compound like nitroglycerine.

 

Question 45. The correct order of acidic nature of alcohols is
(a) Ethanol < propan - 2 - ol < 2 - methyl propan - 2- ol
(b) Propan - 2 - ol < Ethanol < 2 - methyl propan - 2 - ol
(c) 2 - methyl propan - 2 - ol < Ethanol < propan - 2 - ol
(d) 2 - methyl propan - 2 - ol
Answer: (c) 2 - methyl propan - 2 - ol < propan - 2 - ol < Ethanol
In simple words: The stronger an alcohol is as an acid, the less branched it usually is. So, tertiary alcohols (like 2-methyl propan-2-ol) are the least acidic, and primary alcohols (like ethanol) are the most acidic.

🎯 Exam Tip: Remember that electron-donating alkyl groups decrease alcohol acidity, so primary alcohols are more acidic than secondary, which are more acidic than tertiary alcohols.

 

Question 46. 1, 3 - dihydroxy benzene is commonly known as
(a) Cresol
(b) Catechol
(c) Resorcinol
(d) Quinol
Answer: (c) Resorcinol
In simple words: 1,3-dihydroxybenzene is commonly known as resorcinol. It is a chemical that has two -OH groups on a benzene ring at specific spots.

🎯 Exam Tip: Remember the common names for dihydroxybenzenes: catechol (1,2-), resorcinol (1,3-), and hydroquinone/quinol (1,4-).

 

Question 47. Which of the following is the correct systematic name for 1,2,3-trihydroxy benzene?
(a) 1, 2, 3 - trihydroxy benzene
(h) 1, 2, 4 - trihydroxy beneze
(c) 1, 3, 5 - trihydroxv benzene
(d) 1, 4, 5 - tri hydroxy benzene
Answer: (a) 1, 2, 3 - trihydroxy benzene
In simple words: The correct name for the chemical with three -OH groups on a benzene ring at positions 1, 2, and 3 is 1,2,3-trihydroxy benzene. This is also called pyrogallol.

🎯 Exam Tip: IUPAC nomenclature for substituted benzenes involves numbering the carbon atoms to give the substituents the lowest possible set of numbers.

 

Question 48. Oricinol is
(a) 1, 2 - dihydroxy benzene
(h) 3 - methyl phenol :
(c) 3, 5 - dihydroxy toluene
(d) 3, 5 - dimethyl toluene
Answer: (c) 3, 5 - dihydroxy toluene
In simple words: Orcinol is another name for 3,5-dihydroxytoluene. It is a natural compound with two -OH groups and a methyl group.

🎯 Exam Tip: Familiarize yourself with common names for substituted phenols and benzenes, as they are often used interchangeably with systematic IUPAC names.

 

Question 49. Which of the following reaction will give ether?
(a) \( \text{C}_{2}\text{H}_{5}\text{ONa} + \text{C}_{2}\text{H}_{5}\text{Br} \rightarrow \)
(b) \( \text{CH}_{3}\text{CH}_{2}\text{OH} \xrightarrow{\text{Con. H}_{2}\text{SO}_{4}, 413 \text{ K}} \)
(c) \( \text{CH}_{3}\text{CH}_{2}\text{OH} + \text{CH}_{2}\text{N}_{2} \xrightarrow{\text{HBF}_{4}} \)
(d) All of the options
Answer: (d) All of the options
In simple words: All three reactions - Williamson ether synthesis, alcohol dehydration with acid, and alcohol methylation with diazomethane - can make ethers. These are all valid ways to produce them.

🎯 Exam Tip: Be familiar with the different synthetic routes to ethers, including Williamson synthesis (alkoxide + alkyl halide), acid-catalyzed dehydration of alcohols (especially primary alcohols at specific temperatures), and methylation with diazomethane.

 

Question 50. Phenol on oxidation with acidified K2Cr2O7gives
(a) 1,4- benzoquinone
(b) 1,4- benzoquinone
(c) cyclohexanol
(d) cumene
Answer: (b) 1,4- benzoquinone
In simple words: Phenol, when mixed with an acidic potassium dichromate, changes into 1,4-benzoquinone. This strong chemical reaction turns the phenol into a different type of compound.

🎯 Exam Tip: Remember that phenols are easily oxidized, and strong oxidizing agents often convert them to quinones.

 

Question 51. Which among the following is a simple ether?
(a) 1 - methoxy propane
(b) 2 - methoxy propane
(c) methoxy methane
(d) methoxy benzene
Answer: (c) methoxy methane
In simple words: Methoxy methane is a simple ether because it has the same group (methyl) on both sides of the oxygen atom. Simple ethers have identical groups around the oxygen.

🎯 Exam Tip: Simple ethers have identical alkyl or aryl groups attached to the ether oxygen, while mixed ethers have different groups.

 

Question 52. The oxygen atom of ether is hybridised
(a) sp
(b) sp²
(c) sp³
(d) dsp²
Answer: (c) sp³
In simple words: The oxygen atom in an ether is \( \text{sp}^{3} \) hybridised, meaning it uses four hybrid orbitals for its two bonds and two lone pairs. This gives it a bent shape.

🎯 Exam Tip: The hybridization of central atoms in organic molecules can be determined by counting the number of sigma bonds and lone pairs.

 

Question 53. Williamson's synthesis is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer: (b) Nucleophilic substitution
In simple words: Williamson's synthesis is a nucleophilic substitution reaction where one group replaces another. A negatively charged group (alkoxide) attacks and replaces a leaving group.

🎯 Exam Tip: Williamson ether synthesis typically follows an \( \text{S}_{\text{N}}2 \) mechanism, making primary alkyl halides the preferred electrophiles to avoid elimination side reactions.

 

Question 54. Which among the following is more reactive towards ethers?
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (d) HI
In simple words: Hydrogen iodide (HI) is the strongest acid among the hydrogen halides. This makes it the most reactive when breaking open ether molecules. Its strong acidity helps speed up the reaction.

🎯 Exam Tip: Remember that stronger acids generally lead to more reactive species in chemical reactions, especially in bond cleavage processes like with ethers.

 

Question 55. The products obtained when methoxy ethane reacted with one mole of HI are
(a) methanol & iodoethane
(b) iodomethane & ethanol
(c) iodomethane & jodo ethane
(d) Methanol & ethanol
Answer: (b) iodomethane & ethanol
In simple words: When methoxy ethane reacts with one mole of HI, the HI will break the ether bond. This gives iodomethane (methyl iodide) and ethanol.

🎯 Exam Tip: For unsymmetrical ethers reacting with HI, the reaction follows an \( S_N^2 \) mechanism for primary and secondary ethers, leading to the formation of the alkyl halide with the smaller alkyl group.

 

Question 56. Which is used as a precursor to the synthesis of perfumes and insecticide
(a) Phenol
Answer: (a) Phenol
In simple words: Phenol is a starting material used to make many chemicals, including those found in perfumes and insecticides. It's a key building block in these industries.

🎯 Exam Tip: Phenol's versatility in chemical reactions makes it a crucial intermediate in the synthesis of a wide range of organic compounds.

 

Question 57. According to Lewis concept of acids and bases an ether is
(a) acidic
(b) basic
(c) neutral
(d) amphoteric
Answer: (b) basic
In simple words: Ethers have oxygen atoms with lone pairs of electrons. These lone pairs can be given to other molecules, making ethers act as Lewis bases.

🎯 Exam Tip: A Lewis base is any substance that can donate a pair of non-bonding electrons. The oxygen in ethers makes them good electron pair donors.

 

Question 58. \( A \xrightarrow{K_2Cr_2O_7 / dil H_2SO_4} B \xrightarrow{CH_3MgI / H_2O} \underset{CH_3}{|} \ CH_3-\underset{C}{|}-OH \)
The reactant A is
(a) \( CH_3COCH_3 \)
(b) \( C_2H_5OH \)
(c) \( CH_3-CH(OH)CH_3 \)
(d) \( CH_3CHO \)
Answer: (c) \( CH_3-CH(OH)CH_3 \)
In simple words: The reaction starts with a secondary alcohol. It gets oxidized to a ketone (B). Then, reacting with a Grignard reagent, it forms a tertiary alcohol. The reactant A is a secondary alcohol.

🎯 Exam Tip: Recognize the sequence: secondary alcohol \( \rightarrow \) ketone \( \rightarrow \) tertiary alcohol via Grignard reaction. This helps identify the starting material.

 

Question 59. In the reaction Ethanol \( \xrightarrow{PCl_5} X \xrightarrow{alc.KOH} Y \xrightarrow{alkaline KMnO_4} Z \) the product Z is
(a) Propan-1,3-diol
(b) Ethan -1,2-diol
(c) Propan -1,2,3- Triol
(d) Ethanal
Answer: (b) Ethan -1,2-diol
In simple words: Ethanol first reacts with \( PCl_5 \) to form chloroethane (X). Then, chloroethane reacts with alcoholic KOH to form ethene (Y). Finally, ethene reacts with alkaline \( KMnO_4 \) (Baeyer's reagent) to form ethane-1,2-diol (Z). This is a simple way to make a diol from an alcohol.

🎯 Exam Tip: Knowing the products of common reactions like halogenation, dehydrohalogenation, and Baeyer's test is key to solving multi-step synthesis problems.

 

Question 60. Which of the following has lowest boiling point?
(a) phenol
(b) O-nitro phenol
(c) m-nitro phenol
(d) p-nitro phenol
Answer: (b) O-nitro phenol
In simple words: O-nitrophenol has an intramolecular hydrogen bond, meaning the bond is formed within the same molecule. This makes it less likely to form bonds with other molecules, so it has a lower boiling point and can evaporate easily. Other nitrophenols form bonds between different molecules, which raises their boiling points.

🎯 Exam Tip: Intramolecular hydrogen bonding reduces intermolecular forces, leading to lower boiling points and higher volatility compared to isomers with intermolecular hydrogen bonding.

II. Assertion and Reason

 

Question 1. Assertion : Tertiary alcohols undergo dehydration more readily than primary alcohol. Reason: Tertiary alcohols are less acidic than primary alcohol.
(a) if both assertion and reason are true and reason is the correct explanation of assertion
(b) if both assertion and reason are true but reason is not the correct explanation of assertion
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer: (b) if both assertion and reason are true but reason is not the correct explanation of assertion
In simple words: Both statements are true. Tertiary alcohols lose water (dehydrate) more easily than primary ones because they form a more stable carbocation intermediate. Tertiary alcohols are also less acidic than primary alcohols because of the electron-donating alkyl groups. However, the reason for easier dehydration is the stability of the carbocation, not their acidity.

🎯 Exam Tip: For assertion-reason questions, first check if both statements are true individually, then evaluate if the reason directly explains the assertion.

 

Question 2. Assertion (A): Alcohols can act as Bronsted acids Reason (R) : It is due to the presence of unshared electron pairs on oxygen which make them proton acceptors.
(a) Both (A) and (R) are correct, R explain A.
(b) Both (A) and (R) are correct, R does not explain A.
(c) (A) is correct but R is correct.
(d) (A) is wrong but R is correct.
Answer: (d) (A) is wrong but R is correct
In simple words: Alcohols have a hydrogen atom that can be removed, so they can act as Brønsted acids. However, their oxygen also has lone pairs of electrons, which means they can accept protons and act as Brønsted bases. So, alcohols can be both Brønsted acids and bases, not just acids. The reason given is correct for their basic nature.

🎯 Exam Tip: Remember that species capable of both donating and accepting protons (like alcohols) are amphoteric, acting as Brønsted acids or bases depending on the reaction environment.

 

Question 3. Assertion: Alcohols are more acidic than aliphatic alcohols. Reason (R) : Alkyl substituted phenols show a decreased acidity due to the electron releasing +I effect of alkyl group.
(a) Both (A) and (R) are correct, R explain A.
(b) Both (A) and (R) are correct, R does not explain A.
(c) (A) is correct but R is correct.
(d) (A) is wrong but R is correct.
Answer: (b) Both (A) and (R) are correct, R does not explain A.
In simple words: The first statement is true; phenols are more acidic because the phenoxide ion they form is stable due to resonance. The second statement is also true; alkyl groups release electrons, making phenols less acidic when they are substituted. However, the reason does not explain why *phenols* in general are more acidic than *aliphatic alcohols*. It explains a specific effect on phenol acidity.

🎯 Exam Tip: For comparing acidity, always consider the stability of the conjugate base formed after losing a proton. Resonance stabilization greatly increases acidity.

 

Question 4. Assertion (A): Orthonitro phenol is slightly soluble in water and more volatile whereas p – nitro phenol is more soluble in water and less volatile. Reason (R) : Orthonitro phenol forms intramolecular hydrogen bonding and p-nitro phenol forms intermolecular hydrogen bonding.
(a) Both (A) and (R) are correct, R explains A.
(b) Both (A) and (R) are correct, R does not explain A.
(c) (A) is correct but R is correct.
(d) (A) is wrong but R is correct.
Answer: (a) Both (A) and (R) are correct, R explains A.
In simple words: Both statements are correct, and the reason explains the assertion well. O-nitrophenol has hydrogen bonding within its own molecule, which makes it less attracted to water and more likely to evaporate. P-nitrophenol forms hydrogen bonds with other p-nitrophenol molecules, making it more soluble in water and harder to evaporate.

🎯 Exam Tip: Understand the difference between intramolecular (within molecule) and intermolecular (between molecules) hydrogen bonding, as it significantly impacts physical properties like solubility and boiling point.

III. Pick Out The Correct Statements

 

Question 1.
(i) The structure of – OH group attached to a \( sp^3 \)- hybridised carbon in alcohol is similar to the structure of – OH group attached to a hydrogen in water, ie., 'V' shaped.
(ii) Due to lone pair – lone pair repulsion, the – C – O – H bond angle in methanol is reduced to 108.9° from the regular tetrahedral bond angle of 109.5°.
(iii) With Grignard reagent formaldehyde gives primary alcohol and other aldehydes give secondary alcohols.
(iv) Reaction of Grignard reagent with aldehydes and ketones to form alcohols is an example for nucleophilic substitution reaction.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (i) & (iv)
Answer: (c) (i) & (iii)
In simple words: Statements (i) and (iii) are correct. Alcohols have an oxygen atom bonded to a \( sp^3 \) carbon and hydrogen, similar to water, giving a bent or 'V' shape. Formaldehyde reacting with a Grignard reagent always gives a primary alcohol, while other aldehydes give secondary alcohols. Statement (ii) mentions a bond angle reduction to 108.9°, not 104°. Statement (iv) is incorrect because the Grignard reaction is a nucleophilic *addition*, not substitution.

🎯 Exam Tip: Know the specific products formed when Grignard reagents react with different carbonyl compounds (formaldehyde, other aldehydes, ketones) to produce primary, secondary, or tertiary alcohols.

 

Question 2.
(i) Lower alcohols are waxy solids and the higher members are colourless liquids.
(ii) Due to the presence of intermolecular hydrogen bonding alcohols have higher boiling point than the corresponding alkanes, aldehydes, ethers etc.
(iii) Due to the formation of intermolecular hydrogen bonding with water lower alcohols are highly soluble in water.
(iv) Among isomeric alcohols primary alcohols have lower boiling point and tertiary alcohols have higher boiling points.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Statements (ii) and (iii) are correct. Alcohols have strong intermolecular hydrogen bonds, which means they need more energy to boil, giving them higher boiling points than similar compounds without these bonds. Also, small alcohols can form hydrogen bonds with water molecules, making them dissolve well in water. Statement (i) is incorrect because lower alcohols are liquids, not waxy solids. Statement (iv) is incorrect because primary alcohols typically have higher boiling points than their isomeric tertiary alcohols due to less steric hindrance for hydrogen bonding.

🎯 Exam Tip: Hydrogen bonding is a key factor affecting the physical properties of alcohols, leading to higher boiling points and increased solubility in water for smaller molecules.

 

Question 3.
(i) In Swern Oxidation DMSO, oxalyl chloride and triethylamine are used to convert alcohols into aldehydes / ketones.
(ii) Vapours of primary alcohol passed over heated copper at 573 K undergoes dehydration..
(iii) Methanol reacts with ethanoic acid in presence of an acid to form ethylethanoate.
(iv) Vapours of tertiary alcohols react with heated copper at 573 K to give alkenes.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Statements (i) and (iv) are correct. Swern oxidation uses DMSO, oxalyl chloride, and triethylamine to turn alcohols into aldehydes or ketones. Tertiary alcohols, when heated over copper at 573 K, undergo dehydration to form alkenes. Statement (ii) is incorrect because primary alcohols undergo *dehydrogenation* (form aldehydes) under these conditions, not dehydration. Statement (iii) is incorrect because methanol and ethanoic acid form *methyl* ethanoate, not ethyl ethanoate.

🎯 Exam Tip: Distinguish between dehydrogenation (removal of \( H_2 \)) for primary/secondary alcohols and dehydration (removal of \( H_2O \)) for tertiary alcohols when heated over copper.

 

Question 4.
(i) Glycerol contains two primary alcoholic group and one secondary alcoholic group.
(ii) Oxidation of glycerol with cone. \( HNO_3 \) gives mainly formic acid.
(iii) Glycerol is used in making printing inks, and stamp pad ink.
(iv) Glycerol is used as a preservative for biological specimens
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (i) & (iii)
(d) (i) & (iv)
Answer: (c) (i) & (iii)
In simple words: Statements (i) and (iii) are correct. Glycerol's structure indeed has two primary and one secondary alcohol groups. It is also used in printing inks and stamp pads. Statement (ii) is incorrect; concentrated \( HNO_3 \) mainly oxidizes glycerol to glyceric acid. Statement (iv) is incorrect as ethanol, not glycerol, is commonly used as a preservative for biological specimens.

🎯 Exam Tip: Knowing the functional groups in a molecule like glycerol and its common industrial uses can help you identify correct statements. Always verify specific reaction products.

IV. Pick Out The Incorrect Statements

 

Question 1.
(i) The overall reaction of hydroboration is hydration of an alkene.
(ii) Hydroboration reaction occurs according to Markownikoff's rule.
(iii) Raney Ni does not reduce the Carbon – Carbon double bond present in the Carbonyl compound to form unsaturated alcohols.
(iv) When two or more functional groups are present in a molecule sodium borohydride to used as a reducing agent to reduce the more reactive group
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Statements (ii) and (iii) are incorrect. Hydroboration follows anti-Markownikoff's rule, meaning the hydrogen adds to the less substituted carbon. Also, Raney Ni *can* reduce the carbon-carbon double bond in carbonyl compounds, not just the carbonyl group itself. So, to make unsaturated alcohols, other reagents like \( LiAlH_4 \) or \( NaBH_4 \) are preferred.

🎯 Exam Tip: Remember that hydroboration is an anti-Markownikoff addition, and selective reduction of a carbonyl group without affecting a C=C double bond requires specific reducing agents like \( NaBH_4 \).

 

Question 2.
(i) Ethanol forms a turbidity with Lucas reagent within 10 minutes.
(ii) 2 – Methyl propan 2 – ol forms immediate turbidity with Lucas reagent due to the formation of insoluble 2 – chloro – 2 – methyl propane.
(iii) Ehanol forms red colour in Victor Meyer's test.
(iv) 2 – 2 – dimethyl propan – 1 – ol gives a colourless solution in victor Meyer's test,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (d) (i) & (iv)
In simple words: Statements (i) and (iv) are incorrect. Ethanol (a primary alcohol) does *not* form turbidity with Lucas reagent in 10 minutes; it reacts very slowly or needs heating. Also, 2,2-dimethylpropan-1-ol (a primary alcohol) gives a *red* color in Victor Meyer's test, not a colorless solution. Statement (ii) is correct; tertiary alcohols react immediately with Lucas reagent. Statement (iii) is also correct; ethanol (a primary alcohol) gives a red color in Victor Meyer's test.

🎯 Exam Tip: Lucas test and Victor Meyer's test are important for distinguishing between primary, secondary, and tertiary alcohols based on reactivity and color changes. Memorize the characteristic results for each type of alcohol.

 

Question 3.
(i) Alcohols undergo nucleophilic substitution reaction with hydrohaloacids to form alkyl halides.
(ii) Alkyl halide formation from primary alcohols follow E2 mechanism.
(iii) Alkyl halide formation from tertiary alcohols follow \( S_N^2 \) mechanism.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (b) (ii) & (iii)
In simple words: Statements (ii) and (iii) are incorrect. Alkyl halide formation from primary alcohols primarily follows an \( S_N^2 \) mechanism, not E2. Alkyl halide formation from tertiary alcohols follows an \( S_N^1 \) mechanism, not \( S_N^2 \). Statement (i) is correct; alcohols react with hydrohaloacids via nucleophilic substitution to produce alkyl halides.

🎯 Exam Tip: Understand the preferred mechanisms ( \( S_N^1 \), \( S_N^2 \), E1, E2) for primary, secondary, and tertiary alcohols and alkyl halides, as it's crucial for predicting reaction pathways and products.

 

Question 4.
(i) In phenol the carbon bearing the – OH group is \( sp^2 \) hybridised.
(ii) Unlike alcohols phenol reacts with sodium hydroxide to form sodium phenoxide.
(iii) In substituted phenols, the electron with drawing groups decreases the acidic nature of phenol.
(iv) Alkyl substituted phenols show increased acidity due to electron releasing +1 effect of alkyl group.
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer: (c) (iii) & (iv)
In simple words: Statements (iii) and (iv) are incorrect. Electron-withdrawing groups actually *increase* the acidic nature of phenols, making the -OH proton easier to remove. Conversely, alkyl groups are electron-releasing, which *decreases* the acidity of phenols, not increases it. Statements (i) and (ii) are correct: the carbon attached to the -OH in phenol is \( sp^2 \) hybridized, and phenols react with \( NaOH \) to form phenoxides.

🎯 Exam Tip: Remember that electron-withdrawing groups enhance acidity by stabilizing the conjugate base, while electron-donating groups destabilize it, thereby reducing acidity.

V. Match The Following

 

Question 1. Oxidising agent Oxidation product of glycerol

Answer:
i) d glyceric acid & tartronic acid
ii) e glyceric acid
iii) a meso oxalic acid
iv) f glycerose
v) b formaldehyde & formic acid
vi) c oxalic acid
In simple words: Different oxidizing agents change glycerol into different products. Dilute nitric acid gives glyceric and tartronic acids. Concentrated nitric acid produces glyceric acid. Bismuth nitrate yields meso oxalic acid. Fenton's reagent gives glycerose. Periodic acid produces formaldehyde and formic acid. Acidified potassium permanganate produces oxalic acid.

🎯 Exam Tip: For matching questions involving reagents and products, focus on the functional group transformations and the strength of the oxidizing agent to predict the outcome.

 

Question 2. Example Type of Phenol

Answer:
i) d Substituted phenol
ii) c Dihydric phenol
iii) a Monohydric phenol
iv) b Trihydric phenol
In simple words: This matching exercise helps classify phenols based on their structure and the number of hydroxyl groups they contain. Orcinol is a substituted phenol. Catechol has two hydroxyl groups, making it a dihydric phenol. Cresol has one hydroxyl group, making it a monohydric phenol. Phloroglucinol has three hydroxyl groups, classifying it as a trihydric phenol.

🎯 Exam Tip: Classify phenols by counting the number of hydroxyl (-OH) groups attached directly to the benzene ring. Common names often indicate their substitution pattern.

 

Question 3. Compound Uses

Answer:
i) c Carbolic soaps
ii) b Perfumery
iii) a Antifreeze
In simple words: Phenol is used in carbolic soaps because of its antiseptic properties. Glycerol is a common ingredient in perfumery and cosmetics due to its moisturizing qualities. Glycol, specifically ethylene glycol, is widely used as an antifreeze in car engines to prevent freezing and boiling.

🎯 Exam Tip: Familiarize yourself with the practical applications of common organic compounds, as these are often tested in factual recall questions.

 

Question 4. Type of Alcohol Example

Answer:
i) e Neo pentyl alcohol
ii) d Isopropyl alcohol
iii) f 2-phenyl propan-2-ol
iv) c Ethylene glycol
v) a Glycerol
vi) b Sorbitol
In simple words: This matching helps link alcohol types to their specific examples. Neo pentyl alcohol is a primary alcohol. Isopropyl alcohol is a secondary alcohol. 2-phenyl propan-2-ol is a tertiary alcohol. Ethylene glycol is a dihydric alcohol (has two -OH groups). Glycerol is a trihydric alcohol (three -OH groups). Sorbitol is a polyhydric alcohol (many -OH groups).

🎯 Exam Tip: To classify alcohols, look at the carbon atom directly bonded to the hydroxyl group: if it's attached to one other carbon, it's primary; two carbons, secondary; three carbons, tertiary. Polyhydric alcohols have multiple hydroxyl groups.

VI. Two Marks Questions

 

Question 1. Illustrate Markownikoff's rule in the addition of water to an alkene
Answer: Markownikoff's rule states that when a protic acid (like \( H_2O \) in the presence of an acid catalyst, acting as \( H^+\) and \( OH^- \)) adds to an unsymmetrical alkene, the hydrogen atom of the protic acid attaches to the carbon atom of the double bond that already has more hydrogen atoms. The negative part of the acid (like \( OH^- \)) attaches to the carbon atom with fewer hydrogen atoms. This ensures the formation of the more stable carbocation intermediate.
\( CH_3-CH=CH_2 + H_2O \xrightarrow{Conc.H_2SO_4} CH_3-CH(OH)-CH_3 \)
Propene \( \rightarrow \) Propan-2-ol
In simple words: When water adds to a double bond in a molecule that is not symmetrical, the hydrogen part of the water goes to the carbon that already has more hydrogens. The oxygen and hydrogen part (OH) goes to the carbon with fewer hydrogens. This is how the most common product is formed.

🎯 Exam Tip: Always remember "Hydrogen goes rich" or "the rich get richer" when applying Markownikoff's rule to predict the major product of addition reactions to unsymmetrical alkenes.

 

Question 2. Write a note on hydroboration reaction
Answer: The hydroboration-oxidation reaction is a two-step process to convert alkenes into alcohols, following an anti-Markownikoff addition pattern. In the first step, diborane \( (BH_3)_2 \) reacts with an alkene to form an intermediate trialkyl borane. In the second step, this trialkyl borane is treated with hydrogen peroxide \( (H_2O_2) \) in the presence of sodium hydroxide \( (NaOH) \), which converts it into an alcohol. The overall reaction is a hydration of an alkene. This method allows for the formation of less substituted alcohols from alkenes.
In simple words: Hydroboration is a way to change a double bond into an alcohol. First, a molecule called diborane adds to the double bond. Then, with hydrogen peroxide and a base, it turns into an alcohol. This is special because the alcohol forms on the carbon that normally would get the hydrogen if it followed the usual rule.

🎯 Exam Tip: Hydroboration-oxidation is crucial for synthesizing anti-Markownikoff alcohols from alkenes, which is often difficult to achieve with direct acid-catalyzed hydration.

 

Question 3. Convert crotonaldehyde into crotylalcohol
Answer: Lithium aluminum hydride (\( \text{LiAlH}_4 \)) is a powerful reducing agent that can convert aldehydes and ketones into alcohols. However, it does not affect carbon-carbon double bonds. This makes it ideal for reducing crotonaldehyde (which has a carbon-carbon double bond) to crotyl alcohol without changing the double bond. The reaction involves adding hydrogen across the carbonyl group.
\( \text{CH}_3-\text{CH}=\text{CH}-\text{CHO} \xrightarrow{\text{(i) LiAlH}_4\text{, (ii) H}_2\text{O}} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2\text{OH} \)
In simple words: Crotonaldehyde can be changed into crotyl alcohol using a special chemical called lithium aluminum hydride. This chemical adds hydrogen to the aldehyde part but leaves the double bond alone.

🎯 Exam Tip: Remember that \( \text{LiAlH}_4 \) is a selective reducing agent; it reduces carbonyl groups but typically leaves carbon-carbon double bonds untouched, unlike hydrogen gas with nickel catalyst, which would reduce both.

 

Question 4. An organic compound (A) – C3H8O3 used as sweetening agent, which on oxidation with Fenton's reagent gives a mixture of compounds B and C. Identify A, B & C. Write possible reactions.
Answer:
(A) is Glycerol (\( \text{C}_3\text{H}_8\text{O}_3 \)). Glycerol is a sweet-tasting, non-toxic compound often used as a sweetening agent. It's a triol, meaning it has three hydroxyl (-OH) groups. Its chemical structure helps it act as a humectant and sweetener.
When glycerol is oxidized with Fenton's reagent (\( \text{FeSO}_4 + \text{H}_2\text{O}_2 \)), it forms a mixture of glyceraldehyde (B) and dihydroxyacetone (C). These are both three-carbon compounds, one an aldehyde and the other a ketone.

The reactions are:
\[ \text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH} \xrightarrow{\text{[O]}} \text{CH}_2\text{OH}-\text{CHOH}-\text{CHO} \text{ (Glyceraldehyde, B)} \]
\[ \text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH} \xrightarrow{\text{[O]}} \text{CH}_2\text{OH}-\text{CO}-\text{CH}_2\text{OH} \text{ (Dihydroxyacetone, C)} \]
In simple words: The sweet compound (A) is glycerol. When glycerol is mixed with Fenton's reagent, it changes into two new compounds: glyceraldehyde (B) and dihydroxyacetone (C). Glycerol has three alcohol parts, and the reagent oxidizes some of these parts.

🎯 Exam Tip: Glycerol is a polyol, and its oxidation products vary depending on the oxidizing agent and conditions. Fenton's reagent is a non-selective oxidant.

 

Question 5. What is saponification?
Answer: Saponification is a chemical process where oils or fats react with a strong alkali (like sodium hydroxide or potassium hydroxide) to produce glycerol and a salt of a fatty acid, which is commonly known as soap. This reaction is essentially the hydrolysis of an ester in the presence of a base. It's an ancient process used to make soap from animal fats or vegetable oils.

\[ \text{CH}_2\text{O}-\text{CO}-\text{R} \\ \text{CH O}-\text{CO}-\text{R} \\ \text{CH}_2\text{O}-\text{CO}-\text{R} \quad \text{ (Fat/Oil)} \quad + \quad 3\text{NaOH} \quad \xrightarrow{\Delta} \quad \text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH} \quad \text{ (Glycerol)} \quad + \quad 3\text{RCOONa} \quad \text{ (Soap)} \]
In simple words: Saponification is how soap is made. It happens when fats or oils are boiled with a strong base, creating soap and glycerol.

🎯 Exam Tip: Remember that saponification is a specific type of hydrolysis reaction – base-catalyzed ester hydrolysis – and it always yields soap and glycerol.

 

Question 6. What is Saytzeff's rule
Answer: Saytzeff's rule (also spelled Zaitsev's rule) is an empirical rule in organic chemistry that predicts the major product in an elimination reaction (like dehydration of alcohols). The rule states that in an elimination reaction, the alkene product that is formed in the greatest amount is the one with the most substituted double bond (i.e., the alkene with the greatest number of alkyl groups attached to the double bond carbons). This more substituted alkene is generally more stable. This rule is often observed when multiple alkene products are possible.
In simple words: Saytzeff's rule helps predict which alkene will form most often when a molecule loses atoms. It says the alkene with more carbon groups attached to its double bond will be the main product because it's usually more stable.

🎯 Exam Tip: Understand that Saytzeff's rule applies to elimination reactions (E1 and E2) and helps identify the most stable (major) alkene product. Conversely, Hofmann's rule predicts the least substituted alkene as the major product under specific conditions (e.g., with bulky bases).

 

Question 7. Convert propan -1 – ol to propanal.
Answer: To convert propan-1-ol (a primary alcohol) to propanal (an aldehyde), a mild oxidizing agent is needed that can stop the oxidation at the aldehyde stage and prevent further oxidation to a carboxylic acid. Pyridinium chlorochromate (PCC) is an excellent reagent for this transformation. PCC is a complex of chromium trioxide, pyridine, and hydrochloric acid. It oxidizes primary alcohols to aldehydes and secondary alcohols to ketones without over-oxidizing.

\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH} \quad \xrightarrow{\text{PCC}} \quad \text{CH}_3-\text{CH}_2-\text{CHO} \]
In simple words: To change propan-1-ol into propanal, we use a gentle oxidizer called PCC. This chemical helps to remove hydrogen and turn the alcohol into an aldehyde without changing it too much.

🎯 Exam Tip: Using strong oxidizing agents like potassium permanganate or chromic acid would typically convert primary alcohols directly to carboxylic acids, so PCC is crucial for selective aldehyde formation.

 

Question 8. Write a note on Swern Oxidation
Answer: Swern oxidation is a mild and highly selective method for oxidizing primary alcohols to aldehydes and secondary alcohols to ketones. This reaction uses dimethyl sulfoxide (DMSO) as the oxygen source, oxalyl chloride \( \text{(COCl)}_2 \) (or another activating agent like trifluoroacetic anhydride) as the activator, and a tertiary amine (like triethylamine) as the base. The reaction proceeds at very low temperatures (typically -60°C to -78°C), which helps prevent over-oxidation and side reactions. It is widely used in organic synthesis because it tolerates a wide range of functional groups and produces water-soluble by-products that are easy to remove.

\[ \text{CH}_3-\text{CHOH}-\text{CH}_3 \quad + \quad \text{H}_3\text{C-S-CH}_3 \quad + \quad \text{Cl-C-C-Cl} \quad \xrightarrow{\text{(i) DMSO, Oxalyl chloride (ii) (C}_2\text{H}_5)_3\text{N}} \quad \text{CH}_3-\text{CO}-\text{CH}_3 \quad + \quad (\text{CH}_3)_2\text{S} \quad + \quad \text{CO}_2 \quad + \quad \text{CO} \quad + \quad 2\text{HCl} \]
In simple words: Swern oxidation is a gentle way to turn alcohols into aldehydes or ketones. It uses specific chemicals like DMSO and oxalyl chloride at very cold temperatures to avoid making unwanted by-products.

🎯 Exam Tip: Note the key reagents (DMSO, oxalyl chloride, triethylamine) and the low temperature requirement, which are characteristic of Swern oxidation and ensure selective oxidation.

 

Question 9. Write note on biological oxidation that occurs in animals.
Answer: In animals, the primary biological oxidation of alcohol happens when the body processes food or consumes alcoholic drinks. When alcohol is ingested, it is mainly broken down in the liver. An enzyme called alcohol dehydrogenase (ADH) plays a crucial role in this process. ADH converts alcohol (ethanol) into acetaldehyde, which is itself toxic. Acetaldehyde is then further oxidized into acetic acid by another enzyme called aldehyde dehydrogenase (ALDH), and acetic acid can eventually be broken down into carbon dioxide and water for energy. Nicotinamide adenine dinucleotide (NAD+) acts as an important coenzyme in these oxidation reactions, accepting hydrogen atoms. This entire process helps the body remove harmful alcohol.

\[ \text{CH}_3\text{CH}_2\text{OH} \quad + \quad \text{NAD}^+ \quad \xrightarrow{\text{ADH}} \quad \text{CH}_3\text{CHO} \quad + \quad \text{NADH} \quad + \quad \text{H}^+ \]
In simple words: Animals break down alcohol in their bodies through a process called biological oxidation. An enzyme called ADH in the liver changes alcohol into a substance called acetaldehyde, which is then changed again to be removed.

🎯 Exam Tip: Focus on the role of alcohol dehydrogenase (ADH) and NAD+ in converting ethanol to acetaldehyde, and the further detoxification to acetic acid. This pathway is vital for alcohol metabolism.

 

Question 10. What happens when ethylene glycol is reacted with PI3?
Answer: When ethylene glycol (ethane-1,2-diol) reacts with phosphorus triiodide \( \text{(PI}_3) \), the hydroxyl (-OH) groups are replaced by iodine atoms, leading to the formation of 1,2-diiodoethane. This is a common method for converting alcohols into alkyl halides. The phosphorus triiodide acts as a halogenating agent. The 1,2-diiodoethane product is relatively unstable and can easily undergo further reactions, such as elimination to form ethene, especially if heated.

\[ \text{CH}_2\text{OH}-\text{CH}_2\text{OH} \quad + \quad \text{PI}_3 \quad \longrightarrow \quad \text{CH}_2\text{I}-\text{CH}_2\text{I} \]
\[ \text{CH}_2\text{I}-\text{CH}_2\text{I} \quad \xrightarrow{\text{-I}_2} \quad \text{CH}_2=\text{CH}_2 \text{ (Ethene)} \]
In simple words: When ethylene glycol reacts with \( \text{PI}_3 \), its -OH parts are replaced by iodine, making a compound called 1,2-diiodoethane. This compound can then easily change into ethene.

🎯 Exam Tip: Remember that \( \text{PI}_3 \) (and \( \text{PCl}_3 \), \( \text{PBr}_3 \)) are common reagents used to convert alcohols to alkyl halides by replacing the hydroxyl group with a halogen atom.

 

Question 11. What happens when ethylene glycol is heated with conc. HNO3 and conc. H2SO4?
Answer: When ethylene glycol is treated with a mixture of concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) at a low temperature, it undergoes nitration. This reaction replaces the hydrogen atoms of the hydroxyl groups with nitro groups \( \text{( -NO}_2) \), forming 1,2-dinitroxyethane, also known as ethylene glycol dinitrate (EGDN) or dinitroglycol. Concentrated sulfuric acid acts as a dehydrating agent and catalyst, helping to form the nitronium ion \( \text{(NO}_2^+) \) needed for nitration. This product is a powerful explosive, similar to nitroglycerine.

\[ \text{CH}_2\text{OH}-\text{CH}_2\text{OH} \quad + \quad 2\text{HNO}_3 \quad \xrightarrow{\text{Conc H}_2\text{SO}_4} \quad \text{CH}_2\text{O-NO}_2-\text{CH}_2\text{O-NO}_2 \quad + \quad 2\text{H}_2\text{O} \]
In simple words: When ethylene glycol reacts with strong nitric acid and sulfuric acid, it turns into dinitroglycol. This new substance is very explosive.

🎯 Exam Tip: This is a nitration reaction of an alcohol, specifically a polyol. The sulfuric acid helps remove water, driving the reaction to completion, and forms the nitronium ion. The product is an explosive due to the presence of nitrate ester groups.

 

Question 12. How is nitroglycerine prepared
Answer: Nitroglycerine, also known as glyceryl trinitrate (GTN), is prepared by the nitration of glycerol (propane-1,2,3-triol). This is achieved by carefully reacting glycerol with a mixture of concentrated nitric acid (\( \text{HNO}_3 \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) at a controlled low temperature. The sulfuric acid acts as a catalyst and dehydrating agent, facilitating the formation of the nitronium ion (\( \text{NO}_2^+ \)) and absorbing the water produced, which helps prevent hydrolysis of the nitrate ester. Nitroglycerine is a highly explosive liquid and is used in dynamite and as a vasodilator in medicine.

\[ \text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH} \quad + \quad 3\text{HONO}_2 \quad \xrightarrow{\text{Conc H}_2\text{SO}_4\text{, -3H}_2\text{O}} \quad \text{CH}_2\text{O-NO}_2-\text{CHO-NO}_2-\text{CH}_2\text{O-NO}_2 \quad \text{ (Nitroglycerine)} \]
In simple words: Nitroglycerine is made by mixing glycerol with strong nitric acid and sulfuric acid in a controlled way. The reaction creates a powerful explosive liquid.

🎯 Exam Tip: Highlight the use of glycerol and the mixed acid mixture (\( \text{HNO}_3/\text{H}_2\text{SO}_4 \)) for nitration. Emphasize the explosive nature of the product and its dual uses (explosive and medicinal).

 

Question 13. Write about the dehydration of glycerol
Answer: Dehydration of glycerol (propane-1,2,3-triol) typically occurs when it is heated with strong dehydrating agents such as concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)), potassium hydrogen sulfate (\( \text{KHSO}_4 \)), or phosphorus pentoxide (\( \text{P}_2\text{O}_5 \)). Under these conditions, glycerol loses two molecules of water to form acrolein (prop-2-enal). Acrolein is an unsaturated aldehyde with a distinct pungent smell. This reaction is often used as a test for glycerol. The mechanism involves initial protonation of a hydroxyl group, followed by elimination of water and subsequent tautomerization.

\[ \text{CH}_2\text{OH}-\text{CHOH}-\text{CH}_2\text{OH} \quad \xrightarrow{\text{KHSO}_4\text{, heat}} \quad \text{CH}_2=\text{CH}-\text{CHO} \quad + \quad 2\text{H}_2\text{O} \]
In simple words: When glycerol is heated with substances that remove water, like \( \text{KHSO}_4 \), it loses water and turns into a smelly compound called acrolein.

🎯 Exam Tip: Recognize that dehydration of glycerol to acrolein is a characteristic reaction and serves as a qualitative test for the presence of glycerol due to acrolein's distinctive odor.

 

Question 14. Convert aniline into phenol
Answer: To convert aniline into phenol, a two-step process called diazotization followed by hydrolysis is used. First, aniline reacts with nitrous acid (\( \text{NaNO}_2/\text{HCl} \)) at a low temperature (0-5°C or 273-278 K) to form an unstable intermediate called benzenediazonium chloride. In the second step, this benzenediazonium chloride solution is heated with water. The diazonium group is replaced by a hydroxyl (-OH) group, releasing nitrogen gas and hydrochloric acid, thereby yielding phenol. This is a common synthetic route in organic chemistry.

\[ \text{C}_6\text{H}_5\text{NH}_2 \quad + \quad \text{NaNO}_2 \quad + \quad 2\text{HCl} \quad \xrightarrow{273-278 \text{ K}} \quad \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \quad + \quad \text{NaCl} \quad + \quad 2\text{H}_2\text{O} \]
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \quad + \quad \text{H}_2\text{O} \quad \xrightarrow{\text{heat}} \quad \text{C}_6\text{H}_5\text{OH} \quad + \quad \text{N}_2 \quad + \quad \text{HCl} \]
In simple words: Aniline can be changed into phenol by a two-step reaction. First, aniline reacts to form a diazonium salt, and then this salt is heated with water to make phenol.

🎯 Exam Tip: Remember the critical low-temperature condition for diazotization (0-5°C) to prevent decomposition of the diazonium salt, which is highly unstable at higher temperatures.

 

Question 15. What is Schotten – Baumann reaction
Answer: The Schotten-Baumann reaction is a method for the acylation of alcohols or amines, typically involving benzoylation. In this reaction, an acyl chloride (most commonly benzoyl chloride) reacts with an alcohol or amine in the presence of an aqueous base (like sodium hydroxide). The base neutralizes the hydrochloric acid produced during the reaction, which shifts the equilibrium towards product formation and also activates the phenol or amine for nucleophilic attack. This reaction is often used to prepare esters or amides. The classic example is the benzoylation of phenol to form phenyl benzoate.

\[ \text{C}_6\text{H}_5\text{OH} \quad + \quad \text{C}_6\text{H}_5\text{COCl} \quad \xrightarrow{\text{NaOH/Pyridine}} \quad \text{C}_6\text{H}_5\text{COOC}_6\text{H}_5 \quad + \quad \text{HCl} \]
In simple words: The Schotten-Baumann reaction is a way to add a benzoyl group to alcohols or amines. It uses benzoyl chloride and a base to create new compounds like esters or amides.

🎯 Exam Tip: Key features include the use of benzoyl chloride, an aqueous base, and often an organic solvent (like pyridine) to ensure the reaction proceeds efficiently, as the product is often insoluble in water.

 

Question 16. Write a note on williamson ether synthesis?
Answer: Williamson ether synthesis is a widely used organic reaction for preparing symmetrical and asymmetrical ethers. It involves the reaction of an alkoxide (a salt of an alcohol, typically sodium alkoxide) with a primary alkyl halide or tosylate. The alkoxide acts as a nucleophile, attacking the carbon atom bearing the halogen (or tosylate) in an \( \text{S}_N2 \) reaction, displacing the leaving group and forming an ether. For optimal yields and to avoid elimination side reactions, it is best to use a primary alkyl halide and a simple alkoxide. This method is effective for creating a wide variety of ethers.

\[ \text{RO}^- \text{Na}^+ \quad + \quad \text{R'X} \quad \longrightarrow \quad \text{ROR'} \quad + \quad \text{NaX} \]
Example: Methyl iodide reacts with sodium phenoxide to form anisole.
\[ \text{C}_6\text{H}_5\text{ONa} \quad + \quad \text{CH}_3\text{I} \quad \longrightarrow \quad \text{C}_6\text{H}_5\text{OCH}_3 \quad + \quad \text{NaI} \]
In simple words: Williamson ether synthesis is a chemical way to make ethers. It combines an alcohol salt (alkoxide) with a primary alkyl halide. This method is good for making many types of ethers.

🎯 Exam Tip: To avoid elimination side reactions (which produce alkenes instead of ethers), always choose a primary alkyl halide when dealing with secondary or tertiary alkoxides.

 

Question 17. Convert phenol into (i) 1, 4 – benzoquionone (ii) cyclo hexanol
Answer:
(i) Conversion of phenol to 1,4-benzoquinone (Oxidation):
Phenol can be oxidized to 1,4-benzoquinone using strong oxidizing agents such as acidic potassium dichromate \( \text{(K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4) \). This reaction involves the removal of hydrogen atoms and the formation of two carbonyl groups on the benzene ring, resulting in the quinone structure. The quinone is a conjugated diketone.

\[ \text{C}_6\text{H}_5\text{OH} \quad \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4} \quad \text{1,4-benzoquinone} \]
(ii) Conversion of phenol to cyclohexanol (Reduction):
Phenol can be reduced to cyclohexanol by catalytic hydrogenation. This process involves reacting phenol with hydrogen gas in the presence of a metal catalyst, such as nickel (Ni), at elevated temperatures and pressures (typically around 160°C). The benzene ring undergoes complete hydrogenation, converting all double bonds to single bonds and forming a cyclic alcohol.

\[ \text{C}_6\text{H}_5\text{OH} \quad + \quad 3\text{H}_2 \quad \xrightarrow{\text{Ni, 160°C}} \quad \text{C}_6\text{H}_{11}\text{OH} \text{ (Cyclohexanol)} \]
In simple words: Phenol can become 1,4-benzoquinone through oxidation, which adds oxygen and changes its structure. It can also become cyclohexanol through reduction, where hydrogen is added to remove the double bonds.

🎯 Exam Tip: Note the difference in reagents for oxidation (dichromate) and reduction (hydrogen with nickel catalyst). Oxidation typically involves oxygen addition or hydrogen removal, while reduction involves hydrogen addition or oxygen removal.

 

Question 18. How is picric acid prepared?
Answer: Picric acid, chemically known as 2,4,6-trinitrophenol, is prepared by the nitration of phenol. However, direct nitration of phenol with concentrated nitric acid can be dangerous and lead to uncontrolled reactions. Therefore, a two-step process is usually employed. First, phenol is treated with concentrated sulfuric acid to form phenol-2,4-disulfonic acid. This step protects the ring from over-oxidation and activates it for subsequent nitration. Then, this disulfonic acid is treated with concentrated nitric acid, which replaces the sulfonic acid groups and introduces three nitro groups, forming 2,4,6-trinitrophenol (picric acid). Picric acid is a strong acid and an explosive.

\[ \text{C}_6\text{H}_5\text{OH} \quad \xrightarrow{\text{Conc H}_2\text{SO}_4\text{, 298K}} \quad \text{Phenol-2,4-disulfonic acid} \quad \xrightarrow{\text{Conc HNO}_3} \quad \text{2,4,6-trinitrophenol} \]
In simple words: Picric acid is made from phenol. First, phenol is treated with sulfuric acid, and then with strong nitric acid. This adds three nitro groups to the phenol, creating picric acid, which is an explosive.

🎯 Exam Tip: Emphasize the indirect nitration method via sulfonation to prevent uncontrolled reactions and achieve the desired tri-substituted product. Direct nitration of phenol can be problematic.

 

Question 19. Differentiate phenol and ethanol
Answer: Phenol and ethanol, both containing hydroxyl groups, have distinct chemical properties that allow them to be differentiated. Phenol is an aromatic alcohol where the -OH group is directly attached to a benzene ring, making it weakly acidic due to resonance stabilization of the phenoxide ion. Ethanol is an aliphatic alcohol, lacking this aromatic stabilization. Their differences are evident in several reactions:

In simple words: Phenol and ethanol are different types of alcohols. Phenol reacts to make a red-orange dye and turns purple with ferric chloride, showing it's slightly acidic. Ethanol does not do these things.

🎯 Exam Tip: When differentiating between compounds, always provide at least two distinct chemical tests with their observable results. The ferric chloride test is a classic differentiator for phenols.

 

Question 20. Give four uses or diethyl ether.
Answer: Diethyl ether, commonly known as ether, is a highly volatile, flammable liquid with several important uses:
1. **Surgical Anaesthetic:** Historically, it was widely used as a general anaesthetic for surgery due to its ability to induce unconsciousness and muscle relaxation. However, it has largely been replaced by safer alternatives due to its flammability and side effects.
2. **Solvent:** It is an excellent non-polar solvent for organic reactions and extractions. Many organic compounds dissolve well in diethyl ether, making it valuable in laboratories and industrial processes for dissolving fats, oils, resins, and other non-polar substances.
3. **Starting Fluid:** Because of its high volatility and low ignition temperature, it is used as a starting fluid for diesel and gasoline engines, especially in cold weather. It helps engines to ignite more easily.
4. **Refrigerant:** Due to its low boiling point and high heat of vaporization, it can be used as a refrigerant, though it's less common now compared to modern refrigerants.
In simple words: Diethyl ether has many uses, like being a sleeping gas for operations, a chemical to dissolve things, a fluid to start engines easily, and even a cooling liquid.

🎯 Exam Tip: When listing uses, provide concise descriptions for each to show understanding beyond just naming them. Focus on properties that make it suitable for each application (e.g., volatility for starting fluid, solvent nature for dissolving).

 

Question 21. Write the mechanism of intermolecular dehydration of alcohols?
Answer: Intermolecular dehydration of alcohols, typically performed with concentrated sulfuric acid at around 413 K (140°C), leads to the formation of ethers. This reaction is an \( \text{S}_N2 \) type mechanism and involves two molecules of alcohol. The overall reaction removes a molecule of water between two alcohol molecules.

**Mechanism:**
**Step 1: Protonation of alcohol.** The alcohol's oxygen atom, with its lone pairs, attacks a proton from the acid catalyst. This forms a protonated alcohol (an oxonium ion), making the -OH group a better leaving group (as water).
\[ \text{CH}_3\text{CH}_2\text{OH} \quad + \quad \text{H}_2\text{SO}_4 \quad \longrightarrow \quad \text{CH}_3\text{CH}_2\text{O}^+\text{H}_2 \quad + \quad \text{HSO}_4^- \]

**Step 2: Nucleophilic attack by another alcohol molecule.** A second molecule of alcohol acts as a nucleophile, attacking the carbon atom bonded to the protonated hydroxyl group. This displaces a molecule of water (the good leaving group).
\[ \text{CH}_3\text{CH}_2\text{OH} \quad + \quad \text{CH}_3\text{CH}_2\text{O}^+\text{H}_2 \quad \longrightarrow \quad \text{CH}_3\text{CH}_2\text{O}^+\text{H}(\text{CH}_2\text{CH}_3) \quad + \quad \text{H}_2\text{O} \]

**Step 3: Deprotonation.** The positively charged intermediate oxonium ion loses a proton to a base (often \( \text{HSO}_4^- \) or another alcohol molecule), regenerating the acid catalyst and forming the neutral ether product.
\[ \text{CH}_3\text{CH}_2\text{O}^+\text{H}(\text{CH}_2\text{CH}_3) \quad \xrightarrow{\text{-H}^+} \quad \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \]
In simple words: When two alcohol molecules are heated with strong acid, they lose a water molecule between them and join together to form an ether. First, an alcohol gains a proton, then another alcohol attacks it, pushing out water, and finally, a proton is removed to make the ether.

🎯 Exam Tip: Distinguish intermolecular dehydration (forms ethers at lower temperatures, e.g., 413 K) from intramolecular dehydration (forms alkenes at higher temperatures, e.g., 443 K).

 

Question 22. Write the mechanism of nucleophilic substitution of ethers
Answer: Ethers can undergo nucleophilic substitution when reacted with strong acids, particularly hydrogen halides (HI, HBr, HCl). The mechanism depends on the type of ether (primary, secondary, tertiary) and the conditions. However, a common reaction involves the cleavage of the C-O bond. For ethers with a primary alkyl group, an \( \text{S}_N2 \) mechanism is common, while tertiary alkyl ethers tend to react via an \( \text{S}_N1 \) mechanism.

**Mechanism (for primary alkyl ethers, \( \text{S}_N2 \)-like):**
**Step 1: Protonation of the ether.** The oxygen atom of the ether, with its lone pairs, gets protonated by the strong acid, forming an oxonium ion. This makes the carbon-oxygen bond more susceptible to nucleophilic attack.
\[ \text{CH}_3-\text{O}-\text{CH}_2\text{CH}_3 \quad + \quad \text{HI} \quad \longrightarrow \quad \text{CH}_3-\text{O}^+\text{H}-\text{CH}_2\text{CH}_3 \quad + \quad \text{I}^- \]

**Step 2: Nucleophilic attack and displacement.** The halide ion (iodide in this case), being a good nucleophile, attacks the less sterically hindered carbon atom (the methyl carbon in this example) of the protonated ether. This causes the cleavage of the C-O bond and displacement of the alcohol molecule.
\[ \text{CH}_3-\text{O}^+\text{H}-\text{CH}_2\text{CH}_3 \quad + \quad \text{I}^- \quad \longrightarrow \quad \text{CH}_3\text{I} \quad + \quad \text{CH}_3\text{CH}_2\text{OH} \]
If excess HI is present, the alcohol formed can further react to form another alkyl halide:
\[ \text{CH}_3\text{CH}_2\text{OH} \quad + \quad \text{HI} \quad \longrightarrow \quad \text{CH}_3\text{CH}_2\text{I} \quad + \quad \text{H}_2\text{O} \]
In simple words: Ethers can be broken apart by strong acids. First, the ether grabs a proton. Then, a halide ion attacks the carbon with fewer groups around it, breaking the ether bond and forming an alkyl halide and an alcohol. If there is a lot of acid, the alcohol will also turn into an alkyl halide.

🎯 Exam Tip: Note that for mixed ethers, the halide ion preferentially attacks the less hindered carbon in an \( \text{S}_N2 \) pathway. If one of the alkyl groups is tertiary, the reaction follows an \( \text{S}_N1 \) mechanism, with the halide attacking the tertiary carbon to form a carbocation.

 

Question 23. What is autooxidation of ethers?
Answer: Autooxidation of ethers is a spontaneous, slow oxidation reaction that occurs when ethers are exposed to atmospheric oxygen, especially over long periods or in the presence of light. This process forms hydroperoxides and dialkyl peroxides. These peroxide compounds are highly unstable, non-volatile, and explosive, particularly when heated or concentrated. Therefore, ethers must be stored in dark, airtight containers, usually with an antioxidant, to prevent the formation of these hazardous peroxides. Before use, old ether samples are often tested for peroxides. The reaction involves a free radical chain mechanism where oxygen adds to the alpha-carbon position (the carbon next to the ether oxygen).

\[ \text{R-O-R'} \quad + \quad \text{O}_2 \quad \xrightarrow{\text{slow}} \quad \text{R-O-CH(R'')-O-O-H} \text{ (Hydroperoxide)} \]
\[ \text{R-O-R'} \quad + \quad \text{O}_2 \quad \xrightarrow{\text{slow}} \quad \text{R-O-CH(R'')-O-O-CH(R'')-O-R} \text{ (Dialkyl Peroxide)} \]
In simple words: Ethers slowly react with air (oxygen) over time, a process called autooxidation. This creates dangerous, explosive compounds called peroxides. Because of this, ethers need to be stored carefully to prevent accidents.

🎯 Exam Tip: The main danger of autooxidation is the formation of explosive peroxides. Always store ethers properly and test for peroxides before distilling them, as concentration can lead to explosions.

 

Question 24. Phenol \( \text{Zn dust} \xrightarrow{\text{A}} \text{CH}_3\text{Cl} \xrightarrow{\text{B}} \text{alk KMnO}_4 \xrightarrow{\text{C}} \) Identify A, B and C
Answer:
This reaction sequence describes the conversion of phenol to benzene (A), followed by Friedel-Crafts alkylation to toluene (B), and finally oxidation to benzoic acid (C).

**Step 1: Phenol to Benzene (A)**
When phenol is distilled with zinc dust, the zinc removes the hydroxyl group, leading to the formation of benzene. This is a reduction reaction.
\[ \text{C}_6\text{H}_5\text{OH} \quad + \quad \text{Zn} \quad \xrightarrow{\text{distillation}} \quad \text{C}_6\text{H}_6 \text{ (Benzene, A)} \quad + \quad \text{ZnO} \]

**Step 2: Benzene to Toluene (B)**
Benzene undergoes Friedel-Crafts alkylation with methyl chloride (\( \text{CH}_3\text{Cl} \)) in the presence of anhydrous aluminum chloride (\( \text{AlCl}_3 \)) as a Lewis acid catalyst. This reaction introduces a methyl group onto the benzene ring, forming toluene.
\[ \text{C}_6\text{H}_6 \text{ (A)} \quad + \quad \text{CH}_3\text{Cl} \quad \xrightarrow{\text{Anhy. AlCl}_3} \quad \text{C}_6\text{H}_5\text{CH}_3 \text{ (Toluene, B)} \quad + \quad \text{HCl} \]

**Step 3: Toluene to Benzoic Acid (C)**
Toluene, when treated with an alkaline potassium permanganate solution (\( \text{alk. KMnO}_4 \)), undergoes oxidation. The methyl group attached to the benzene ring (benzylic carbon) is oxidized to a carboxylic acid group, forming benzoic acid. This is a common method for oxidizing alkyl side chains on aromatic rings.
\[ \text{C}_6\text{H}_5\text{CH}_3 \text{ (B)} \quad \xrightarrow{\text{alk. KMnO}_4} \quad \text{C}_6\text{H}_5\text{COOH} \text{ (Benzoic acid, C)} \]

Therefore:
A = Benzene
B = Toluene
C = Benzoic acid
In simple words: First, phenol turns into benzene (A) when reacted with zinc. Then, benzene reacts with methyl chloride to become toluene (B). Finally, toluene is oxidized with potassium permanganate to form benzoic acid (C).

🎯 Exam Tip: Remember the specific reagents and conditions for each step: Zn dust for reduction of phenol, Friedel-Crafts alkylation (alkyl halide + \( \text{AlCl}_3 \)) for introducing alkyl groups, and strong oxidants like \( \text{KMnO}_4 \) for oxidizing benzylic carbons to carboxylic acids.

 

Question 25. Complete the reaction
\[ \text{C}_6\text{H}_5-\text{O}-\text{CH}_2-\text{C}_6\text{H}_5 \quad + \quad \text{HI} \quad \longrightarrow \quad ? \]
Answer: This is an ether cleavage reaction with a hydrogen halide (HI). The reaction typically proceeds via \( \text{S}_N1 \) or \( \text{S}_N2 \) mechanisms, depending on the structure of the alkyl groups attached to the ether oxygen. In this case, we have benzyl phenyl ether. The C-O bond with the benzyl group is more reactive to cleavage. The products will be phenol and benzyl iodide.

\[ \text{C}_6\text{H}_5-\text{O}-\text{CH}_2-\text{C}_6\text{H}_5 \quad + \quad \text{HI} \quad \longrightarrow \quad \text{C}_6\text{H}_5\text{OH} \quad + \quad \text{C}_6\text{H}_5\text{CH}_2\text{I} \]
The reaction involves the protonation of the ether oxygen, followed by the iodide ion attacking the benzyl carbon, which is an \( \text{S}_N2 \) reaction. Benzyl is a good leaving group and also forms a relatively stable carbocation intermediate if the reaction was \( \text{S}_N1 \). The C-O bond of the phenyl group is very strong due to resonance and is not easily cleaved.
In simple words: When benzyl phenyl ether reacts with HI, it breaks apart. The products formed are phenol and benzyl iodide.

🎯 Exam Tip: In ether cleavage by HI, the bond between oxygen and the alkyl group that can form a more stable carbocation (e.g., tertiary, allylic, benzylic) or is less hindered (for \( \text{S}_N2 \)) is cleaved. The aryl-oxygen bond (like in phenol) is usually stable and not cleaved.

 

Question 1. Explain the structure of the functional group of alcohol
Answer: The functional group of an alcohol is the hydroxyl group, -OH, attached to a saturated carbon atom (an sp³ hybridized carbon). The structure of the C-O-H bond in alcohols is similar to the structure of H-O-H in water, but with one hydrogen replaced by an alkyl group. The oxygen atom is sp³ hybridized, and it has two lone pairs of electrons and forms two sigma bonds (one with carbon and one with hydrogen).

Due to the repulsion between the two lone pairs of electrons on the oxygen atom and the bonding pairs, the C-O-H bond angle in methanol is slightly less than the ideal tetrahedral angle of 109.5°. For example, in methanol \( \text{(CH}_3\text{OH)} \), the C-O-H bond angle is approximately 108.9°. This slight deviation from the perfect tetrahedral angle occurs because lone pairs occupy more space than bonding pairs, pushing the bonding pairs closer together. The C-O bond length is typically around 141 pm.


In simple words: The alcohol group has an oxygen atom linked to a carbon and a hydrogen. This oxygen has two pairs of unshared electrons. These unshared electrons push the other bonds a little, making the C-O-H angle slightly smaller than expected.

🎯 Exam Tip: Remember to mention the sp³ hybridization of the oxygen atom, the presence of lone pairs, and how lone pair-bond pair repulsion affects the bond angle, making it slightly less than the ideal tetrahedral angle.

 

Question 2. Convert acetaldehyde into crotyl alcohol
Answer: To convert acetaldehyde into crotyl alcohol, a series of reactions involving aldol condensation and reduction can be used. Crotyl alcohol has a four-carbon chain with a double bond. Acetaldehyde has two carbons, so we need to extend the carbon chain.

**Step 1: Aldol Condensation.** Acetaldehyde undergoes aldol condensation in the presence of a dilute base (like NaOH) to form 3-hydroxybutanal (aldol product). This reaction combines two acetaldehyde molecules.
\[ \text{CH}_3\text{CHO} \quad + \quad \text{CH}_3\text{CHO} \quad \xrightarrow{\text{dil NaOH}} \quad \text{CH}_3-\text{CHOH}-\text{CH}_2-\text{CHO} \]

**Step 2: Dehydration.** The 3-hydroxybutanal then dehydrates (loses a molecule of water) upon heating to form crotonaldehyde (but-2-enal). This creates the desired carbon-carbon double bond.
\[ \text{CH}_3-\text{CHOH}-\text{CH}_2-\text{CHO} \quad \xrightarrow{\text{-H}_2\text{O}} \quad \text{CH}_3-\text{CH}=\text{CH}-\text{CHO} \]

**Step 3: Reduction.** Finally, crotonaldehyde is selectively reduced to crotyl alcohol (but-2-en-1-ol) using a reducing agent like lithium aluminum hydride (\( \text{LiAlH}_4 \)) or catalytic hydrogenation with \( \text{H}_2 \). \( \text{LiAlH}_4 \) selectively reduces the carbonyl group without affecting the carbon-carbon double bond. If \( \text{H}_2/\text{Ni} \) is used, it should be controlled carefully to only reduce the aldehyde, or a stronger reducing agent like \( \text{LiAlH}_4 \) can be preferred.
\[ \text{CH}_3-\text{CH}=\text{CH}-\text{CHO} \quad \xrightarrow{\text{(i) LiAlH}_4\text{, (ii) H}_2\text{O}} \quad \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2\text{OH} \]
In simple words: To change acetaldehyde into crotyl alcohol, you first combine two acetaldehyde molecules in a special way to make a longer chain. Then, you remove water to create a double bond. Finally, you use a chemical like \( \text{LiAlH}_4 \) to turn the aldehyde part into an alcohol without touching the double bond.

🎯 Exam Tip: This conversion involves building a longer carbon chain (aldol condensation) and then selectively reducing the carbonyl group. Remember that \( \text{LiAlH}_4 \) reduces C=O but not C=C, which is crucial for forming unsaturated alcohols.

 

Question 3. Explain Lucas test of differentiating three types of alcohols
Answer: The Lucas test is a qualitative chemical test used to differentiate between primary (1°), secondary (2°), and tertiary (3°) alcohols. It involves reacting an alcohol with the Lucas reagent, which is a solution of anhydrous zinc chloride (\( \text{ZnCl}_2 \)) in concentrated hydrochloric acid (\( \text{HCl} \)). The reaction is a nucleophilic substitution where the hydroxyl group of the alcohol is replaced by a chlorine atom, forming an alkyl chloride. Alkyl chlorides are generally insoluble in the Lucas reagent, so their formation is observed as turbidity (cloudiness) in the solution.

**Explanation:**
* **Tertiary (3°) alcohols** react very quickly because they form stable tertiary carbocations, leading to immediate turbidity.
* **Secondary (2°) alcohols** react at a moderate rate, forming less stable secondary carbocations, and show turbidity within 5-10 minutes.
* **Primary (1°) alcohols** are the least reactive; they do not form stable carbocations and typically require heating to show turbidity, or may not react at all at room temperature.
In simple words: The Lucas test tells different kinds of alcohols apart by how fast they turn cloudy when mixed with Lucas reagent. Tertiary alcohols react fastest and turn cloudy right away. Secondary alcohols take a few minutes, and primary alcohols usually don't react at all unless heated.

🎯 Exam Tip: Remember the order of reactivity (3° > 2° > 1°) and relate it to the stability of the carbocation intermediate formed, which drives the \( \text{S}_N1 \)-like mechanism in this test.

 

Question 4. Write a note on catalytic dehydrogenation of three types of alcohols.
Answer: Catalytic dehydrogenation is a reaction where hydrogen gas is removed from an organic compound in the presence of a catalyst, typically heated copper metal at 573 K. This method is used to distinguish between primary, secondary, and tertiary alcohols based on their different reactions.

1. **Primary Alcohols:** When the vapors of a primary alcohol are passed over heated copper at 573 K, it undergoes dehydrogenation (removal of \( \text{H}_2 \)) to form an aldehyde.
\[ \text{CH}_3-\text{CH}_2-\text{OH} \quad \xrightarrow{\text{Cu, 573 K}} \quad \text{CH}_3-\text{CHO} \quad + \quad \text{H}_2 \]
2. **Secondary Alcohols:** Secondary alcohols, under the same conditions (heated copper at 573 K), also undergo dehydrogenation, but they form ketones.
\[ \text{CH}_3-\text{CHOH}-\text{CH}_3 \quad \xrightarrow{\text{Cu, 573 K}} \quad \text{CH}_3-\text{CO}-\text{CH}_3 \quad + \quad \text{H}_2 \]
3. **Tertiary Alcohols:** Tertiary alcohols do not undergo dehydrogenation under these conditions. Instead, they undergo dehydration (elimination of water) to form alkenes. This happens because the catalyst facilitates the removal of water rather than hydrogen from the highly substituted carbon.
\[ \text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)\text{CH}_3 \quad \xrightarrow{\text{Cu, 573 K}} \quad \text{CH}_3-\text{C}(=\text{CH}_2)\text{CH}_3 \quad + \quad \text{H}_2\text{O} \]
In simple words: When vapors of different alcohols pass over hot copper, they react differently. Primary alcohols lose hydrogen to become aldehydes. Secondary alcohols also lose hydrogen but become ketones. Tertiary alcohols, however, lose water to form alkenes instead.

🎯 Exam Tip: The key differentiator is that primary and secondary alcohols undergo dehydrogenation, while tertiary alcohols undergo dehydration. Remember the temperature (573 K) and catalyst (Cu) are specific for this test.

 

Question 5. Write about the oxidation of ethylene glycol
Answer: Ethylene glycol (ethane-1,2-diol) is a diol, meaning it has two hydroxyl (-OH) groups. Its oxidation products vary significantly depending on the strength of the oxidizing agent and the reaction conditions. Both hydroxyl groups can be oxidized sequentially or simultaneously. The primary alcohol groups can be oxidized to aldehydes, then to carboxylic acids.

**1. Oxidation with Mild Oxidizing Agents (e.g., dilute \( \text{HNO}_3 \), alkaline \( \text{KMnO}_4 \)):**
- Under mild conditions, one primary alcohol group can be oxidized to an aldehyde, forming glycolaldehyde.
\[ \text{CH}_2\text{OH}-\text{CH}_2\text{OH} \quad \xrightarrow{\text{[O]}} \quad \text{CHO}-\text{CH}_2\text{OH} \text{ (Glycolaldehyde)} \]
- Further oxidation of glycolaldehyde can lead to glycolic acid (by oxidizing the aldehyde group) or glyoxal (by oxidizing the second alcohol group).
\[ \text{CHO}-\text{CH}_2\text{OH} \quad \xrightarrow{\text{[O]}} \quad \text{COOH}-\text{CH}_2\text{OH} \text{ (Glycolic acid)} \]
\[ \text{CHO}-\text{CH}_2\text{OH} \quad \xrightarrow{\text{[O]}} \quad \text{CHO}-\text{CHO} \text{ (Glyoxal)} \]

**2. Oxidation with Strong Oxidizing Agents (e.g., concentrated \( \text{HNO}_3 \), acidified \( \text{KMnO}_4 \)):**
- Strong oxidation can lead to the formation of glyoxalic acid (by oxidizing one alcohol and one aldehyde group to carboxylic acids) or oxalic acid (by oxidizing both alcohol groups completely to carboxylic acids).
\[ \text{CHO}-\text{COOH} \text{ (Glyoxalic acid)} \quad \xrightarrow{\text{[O]}} \quad \text{COOH}-\text{COOH} \text{ (Oxalic acid)} \]
- If both alcohol groups are oxidized to carboxylic acids, the final product is oxalic acid.

**3. Oxidation with Periodic Acid (\( \text{HIO}_4 \)):**
- Periodic acid is a selective oxidant for vicinal diols (1,2-diols). It cleaves the C-C bond between the two hydroxyl-bearing carbons, forming formaldehyde. This is often called Malaprade reaction.
\[ \text{CH}_2\text{OH}-\text{CH}_2\text{OH} \quad + \quad \text{HIO}_4 \quad \longrightarrow \quad 2\text{HCHO} \quad + \quad \text{HIO}_3 \]
In simple words: Ethylene glycol can be oxidized in many ways. Gentle oxidation can turn it into aldehyde or acid forms. Strong oxidation turns it into oxalic acid. A special chemical called periodic acid breaks it apart to form formaldehyde.

🎯 Exam Tip: The key takeaway is that ethylene glycol's oxidation products are highly dependent on the oxidant's strength and type. Be ready to list various products like glycolaldehyde, glyoxal, glycolic acid, glyoxalic acid, and oxalic acid, and especially the cleavage to formaldehyde by periodic acid.

 

Question 6. Why is C-O-C bond angle in ether slightly greater than the tetrahedral bond angle.
Answer: The C-O-C bond angle in ethers is typically slightly greater than the ideal tetrahedral bond angle of 109.5°, usually around 110-112°. This is due to the presence of bulky alkyl groups attached to the oxygen atom.

Here's a breakdown:
1. **sp³ Hybridization of Oxygen:** The oxygen atom in ethers is sp³ hybridized, similar to that in water and alcohols. It forms two sigma bonds (with two carbon atoms) and has two lone pairs of electrons.
2. **Steric Repulsion:** While the lone pairs exert some repulsion, the primary reason for the increased bond angle is the steric repulsion between the two relatively bulky alkyl groups attached to the oxygen. These alkyl groups (e.g., methyl, ethyl) are larger than hydrogen atoms and require more space.
3. **Minimizing Strain:** To minimize the steric strain caused by the close proximity of these alkyl groups, the C-O-C bond angle opens up slightly beyond the ideal tetrahedral angle. For example, in dimethyl ether (\( \text{CH}_3\text{OCH}_3 \)), the C-O-C bond angle is approximately 111.7°.
In contrast, in water, where only small hydrogen atoms are attached to oxygen, the H-O-H bond angle (104.5°) is smaller than the tetrahedral angle due to lone pair-lone pair repulsion.


In simple words: The C-O-C angle in ethers is a bit wider than a perfect tetrahedral angle. This happens because the carbon groups attached to the oxygen are quite big and push each other away, needing more space.

🎯 Exam Tip: The increased bond angle in ethers is primarily due to steric hindrance from bulky alkyl groups, contrasting with the smaller angle in water which is dominated by lone pair repulsion.

 

Question 7. For the preparation of mixed ether having primary and tertiary alkyl group, primary alkyl halide and alkoxide are used. Why?
Answer: Primary alkyl halides are more likely to react using the \( \mathrm{S}_{\mathrm{N}}^{2} \) mechanism. This means that for creating an ether with both primary and tertiary alkyl groups, we should use a primary alkyl halide and an alkoxide. This helps to achieve the desired product through a substitution reaction.
However, if a tertiary alkyl halide is used with a primary alkoxide, the reaction will mostly lead to elimination instead of substitution. This is because the bulky tertiary alkyl group makes it harder for the nucleophile to attack, favoring the removal of atoms to form a double bond (alkene).
In simple words: To make an ether with a primary and a bulky (tertiary) part, we use a primary alkyl halide and an alkoxide. This works well because primary halides prefer to swap parts (substitution). If we use a bulky alkyl halide instead, it will mostly lose atoms and form a double bond (elimination) rather than swapping parts.

🎯 Exam Tip: Remember the distinction between substitution (\( \mathrm{S}_{\mathrm{N}} \)) and elimination (E) reactions; for Williamson ether synthesis, \( \mathrm{S}_{\mathrm{N}}^{2} \) is preferred, especially with primary alkyl halides.

 

Question 8. Convert anisole into (i) Methoxy toluene (ii) methoxy acetophenone
Answer:
(i) Conversion of Anisole to Methoxy toluene (Friedel-Crafts alkylation):
\[ \begin{array}{l} \mathrm{OCH_3} \\ \quad \text{Anisole} \end{array} \qquad \xrightarrow{\mathrm{+ CH_3Cl \ (Anhyd. AlCl_3)}} \qquad \begin{array}{l} \mathrm{OCH_3} \\ \qquad\mathrm{CH_3} \\ \text{2-methoxy toluene(minor)} \end{array} \qquad + \qquad \begin{array}{l} \mathrm{OCH_3} \\ \qquad\qquad\mathrm{CH_3} \\ \text{4-methoxy toluene (major)} \end{array} \]
(ii) Conversion of Anisole to Methoxy acetophenone (Friedel-Crafts acylation):
\[ \begin{array}{l} \mathrm{OCH_3} \\ \quad \text{Anisole} \end{array} \qquad \xrightarrow{\mathrm{+ CH_3COCl \ (Anhyd. AlCl_3)}} \qquad \begin{array}{l} \mathrm{OCH_3} \\ \qquad\mathrm{COCH_3} \\ \text{2-methoxy acetophenone (minor)} \end{array} \qquad + \qquad \begin{array}{l} \mathrm{OCH_3} \\ \qquad\qquad\mathrm{COCH_3} \\ \text{4-methoxy Acetophenone (major)} \end{array} \]
In simple words: Anisole can be changed into methoxy toluene by adding \( \mathrm{CH_3Cl} \) with \( \mathrm{AlCl_3} \). It can also be changed into methoxy acetophenone by adding \( \mathrm{CH_3COCl} \) with \( \mathrm{AlCl_3} \). These are called Friedel-Crafts reactions, which add groups to the benzene ring.

🎯 Exam Tip: Always remember that Friedel-Crafts reactions (alkylation and acylation) on anisole will yield a mixture of ortho and para products, with the para product usually being the major one due to less steric hindrance.

 

Question 9. 375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. What is the formula of the alcohol?
Answer:
When an alcohol (R-OH) reacts with methyl magnesium iodide (Grignard reagent), it produces methane gas:
\( \mathrm{C_nH_{2n+1}-OH + CH_3MgI \rightarrow CH_4 + Mg(OH)I} \)
From the reaction, 1 mole of alcohol produces 1 mole of methane.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22400 \( \mathrm{cm^3} \) (or mL).
Given: Volume of methane gas liberated = 140 mL
Number of moles of methane = \( \frac{\text{Volume of methane}}{\text{Molar volume at STP}} = \frac{140 \text{ mL}}{22400 \text{ mL/mole}} = 0.00625 \) moles
Since 1 mole of alcohol produces 1 mole of methane, the number of moles of alcohol is also 0.00625 moles.
Given mass of alcohol = 375 mg = \( 375 \times 10^{-3} \) g = 0.375 g
Molar mass of alcohol (M) = \( \frac{\text{Mass}}{\text{Number of moles}} = \frac{0.375 \text{ g}}{0.00625 \text{ mol}} = 60 \text{ g/mol} \)
The general formula for a monohydric alcohol is \( \mathrm{C_nH_{2n+1}OH} \).
Molar mass of \( \mathrm{C_nH_{2n+1}OH} \) = \( \mathrm{n(12) + (2n+1)(1) + 1(16) + 1(1)} = 60 \)
\( \mathrm{12n + 2n + 1 + 16 + 1 = 60} \)
\( \mathrm{14n + 18 = 60} \)
\( \mathrm{14n = 60 - 18} \)
\( \mathrm{14n = 42} \)
\( \mathrm{n = \frac{42}{14} = 3} \)
Therefore, the formula of the alcohol is \( \mathrm{C_3H_7OH} \). This calculation helps us determine the correct molecular structure.
In simple words: We know how much methane gas is made from a small amount of alcohol. This helps us figure out how many moles of alcohol there were. Then, by dividing the mass of the alcohol by its moles, we get its total weight. Knowing the general formula for alcohols, we can then find out how many carbon atoms it has, which is 3. So, the alcohol is \( \mathrm{C_3H_7OH} \).

🎯 Exam Tip: Always convert given mass to grams and volume to liters (or \( \mathrm{cm^3} \)) and use the molar volume at STP (22.4 L or 22400 \( \mathrm{cm^3} \)) for gas calculations in stoichiometry problems.

 

Question 10. How will you convert methanol into ethanol?
Answer: Methanol can be converted to ethanol through a series of reactions:
\[ \mathrm{CH_3OH \xrightarrow{PI_3} CH_3I \xrightarrow{KCN} CH_3CN \xrightarrow{H_2O/H^+} CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH} \]
Methanol is first converted to iodomethane using \( \mathrm{PI_3} \). Then, iodomethane reacts with potassium cyanide (KCN) to form ethanenitrile (or methyl cyanide). This nitrile is then hydrolyzed with acid (\( \mathrm{H_2O/H^+} \)) to produce ethanoic acid. Finally, ethanoic acid is reduced using lithium aluminium hydride (\( \mathrm{LiAlH_4} \)) to yield ethanol.
In simple words: To change methanol into ethanol, first turn methanol into an iodine compound, then into a cyanide compound. Next, convert the cyanide into an acid. Lastly, use a strong reducer to change the acid into ethanol. This process adds one carbon atom.

🎯 Exam Tip: Remember that KCN followed by hydrolysis adds a carbon atom to the chain, and \( \mathrm{LiAlH_4} \) is a powerful reducing agent used to convert carboxylic acids to primary alcohols.

 

Question 1. Convert methyl magnesium iodide into (i) ethanol (ii) propan – 2 – ol (iii) 2 – methyl propan – 2 – ol
Answer:
(i) Conversion of Methyl magnesium iodide to Ethanol:
\[ \mathrm{CH_3MgI + H-C=O \xrightarrow{Ether} H-C-OMgI \xrightarrow{H_3O^+} H-C-OH + Mg(OH)I} \]
\[ \mathrm{\qquad Formaldehyde \qquad \quad \quad \quad \quad \quad Ethanol} \]
(ii) Conversion of Methyl magnesium iodide to Propan-2-ol:
\[ \mathrm{CH_3MgI + CH_3-C=O \xrightarrow{Ether} CH_3-C-OMgI \xrightarrow{H_3O^+} CH_3-C-OH + Mg(OH)I} \]
\[ \mathrm{\qquad Acetaldehyde \qquad \quad \quad \quad \quad \quad Propan-2-ol} \]
(iii) Conversion of Methyl magnesium iodide to 2-methylpropan-2-ol:
\[ \mathrm{CH_3MgI + CH_3-C=O \xrightarrow{Ether} CH_3-C-OMgI \xrightarrow{H_3O^+} CH_3-C-OH + Mg(OH)I} \]
\[ \mathrm{\qquad Acetone \qquad \quad \quad \quad \quad \quad 2-methylpropan-2-ol} \]
In simple words: To make different alcohols from methyl magnesium iodide, you react it with different carbonyl compounds. Formaldehyde gives a primary alcohol (ethanol), acetaldehyde gives a secondary alcohol (propan-2-ol), and acetone gives a tertiary alcohol (2-methylpropan-2-ol). The Grignard reagent adds to the carbonyl, and then water is added to finish the reaction.

🎯 Exam Tip: Remember that Grignard reagents are powerful tools for synthesizing alcohols; the type of alcohol (primary, secondary, or tertiary) depends on the carbonyl compound used.

 

Question 2. Bring about the following conversion using Grignard reagents (i) Methanal into phenyl methanol (ii) Ethanal into butan – 2 – ol (iii) Propanone into 2 – methyl hexan – 2 -ol (iv) Ethyl methanoate into propan – 2 – ol (v) Formaldehyde into propan-1-ol
Answer:
(i) Conversion of Methanal to Phenyl methanol:
\[ \mathrm{C_6H_5MgBr + H-C=O \xrightarrow{Ether} C_6H_5-CH_2OMgBr \xrightarrow{H_3O^+} C_6H_5-CH_2OH + Mg(OH)Br} \]
\[ \mathrm{Phenyl \ magnesium \ bromide \quad Methanal \qquad \qquad \qquad \qquad Phenyl \ methanol} \]
(ii) Conversion of Ethanal to Butan-2-ol:
\[ \mathrm{CH_3CH_2MgBr + CH_3-C=O \xrightarrow{Ether} CH_3CH_2-CH(OMgBr)CH_3 \xrightarrow{H_3O^+} CH_3CH_2-CH(OH)CH_3 + Mg(OH)Br} \]
\[ \mathrm{Ethyl \ magnesium \ bromide \quad Ethanal \qquad \qquad \qquad \qquad Butan-2-ol} \]
(iii) Conversion of Propanone to 2-methylhexan-2-ol:
\[ \mathrm{CH_3(CH_2)_3MgBr + CH_3-C=O \xrightarrow{Ether} CH_3(CH_2)_3-C(OMgBr)(CH_3)_2 \xrightarrow{H_3O^+} CH_3(CH_2)_3-C(OH)(CH_3)_2 + Mg(OH)Br} \]
\[ \mathrm{Butyl \ magnesium \ bromide \quad Propanone \qquad \qquad \qquad \qquad 2-methylhexan-2-ol} \]
(iv) Conversion of Ethyl methanoate to Propan-2-ol:
\[ \mathrm{CH_3MgBr + H-C(=O)OC_2H_5 \xrightarrow{Ether} CH_3-C(OMgBr)(H)OC_2H_5 \xrightarrow{H_3O^+} CH_3-CH(OH)CH_3 + Mg(OH)Br + C_2H_5OH} \]
\[ \mathrm{Methyl \ magnesium \ bromide \quad Ethyl \ methanoate \qquad \qquad \qquad \qquad Propan-2-ol} \]
(v) Conversion of Formaldehyde to Propan-1-ol:
\[ \mathrm{C_2H_5MgBr + H-C=O \xrightarrow{Ether} C_2H_5-CH_2OMgBr \xrightarrow{H_2O/H^+} C_2H_5CH_2OH + Mg(OH)Br} \]
\[ \mathrm{Ethyl \ magnesium \ bromide \quad Formaldehyde \qquad \qquad \qquad \qquad Propan-1-ol} \]
In simple words: Grignard reagents are used to create various alcohols by reacting them with different aldehydes, ketones, or esters. The Grignard reagent adds to the carbonyl group, and then acid hydrolysis converts the intermediate into the final alcohol. For example, phenylmagnesium bromide with methanal gives phenyl methanol, while ethylmagnesium bromide with ethanal yields butan-2-ol.

🎯 Exam Tip: Pay close attention to the number of carbon atoms in the Grignard reagent and the carbonyl compound, as this determines the carbon skeleton of the final alcohol. Remember that esters reacting with Grignard reagents often require two moles of the Grignard reagent.

 

Question 3. An ether (A) \( \mathrm{C_5H_{12}O} \) when heated with excess of hot concentration HI, produced two alkyl halides, which on hydrolysis forms compound (B) and (C). Oxidation of (B) gives an acid (D) where as oxidation of (C) gives ketone (E). Identify A, B, C, D and E and write the chemical equation.
Answer:
Given: Ether (A) has formula \( \mathrm{C_5H_{12}O} \). It reacts with excess hot concentrated HI to give two alkyl halides. After hydrolysis, these halides form compounds (B) and (C). Oxidation of (B) yields an acid (D), and oxidation of (C) yields a ketone (E).
Based on the information:
1. The ether must be 2-ethoxy propane, as it's the only ether \( \mathrm{C_5H_{12}O} \) that would give products leading to an acid and a ketone upon oxidation after HI reaction and hydrolysis. \[ \mathrm{CH_3-CH_2-O-CH(CH_3)_2 \xrightarrow{Con. HI} CH_3-CH_2-I + CH_3-CH(I)CH_3} \]
\[ \mathrm{(A) \quad 2-ethoxy \ propane \qquad \quad \qquad \qquad (I) \quad \qquad \qquad \qquad (II)} \]
2. Hydrolysis of alkyl halides (I) and (II): \[ \mathrm{CH_3-CH_2-I \xrightarrow{Hydrolysis, \ aq \ KOH} CH_3-CH_2-OH} \]
\[ \mathrm{(I) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (B) \quad Ethanol} \]
\[ \mathrm{CH_3-CH(I)CH_3 \xrightarrow{Hydrolysis, \ aq \ KOH} CH_3-CH(OH)CH_3} \]
\[ \mathrm{(II) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (C) \quad 2-propanol} \]
3. Oxidation of (B) and (C):
Oxidation of (B) (Ethanol) to Acid (D): \[ \mathrm{CH_3-CH_2-OH \xrightarrow{acidified \ Na_2Cr_2O_7} CH_3-CHO \xrightarrow{acidified \ Na_2Cr_2O_7} CH_3-COOH} \]
\[ \mathrm{(B) \quad Ethanol \qquad \qquad \qquad \qquad \quad \quad \quad (D) \quad Ethanoic \ acid} \]
Oxidation of (C) (2-propanol) to Ketone (E): \[ \mathrm{CH_3-CH(OH)CH_3 \xrightarrow{acidified \ Na_2Cr_2O_7} CH_3-C(=O)CH_3} \]
\[ \mathrm{(C) \quad 2-propanol \qquad \qquad \qquad \qquad \quad \quad (E) \quad Propanone} \]
The final identification is as follows:

In simple words: We have an ether that breaks down into two different alkyl halides. When these halides are treated with water, they become alcohols. One alcohol turns into an acid when oxidized, and the other turns into a ketone. By tracing these reactions backward, we can find the original ether and all the intermediate compounds.

🎯 Exam Tip: When solving conversion problems, work backward from the final products to identify the intermediates and initial reactants. Remember the oxidation products of primary (acid) and secondary (ketone) alcohols.

 

Question 4. Explain \( \mathrm{E_1} \) mechanism involved in the dehydration of tertiary butyl alcohol
Answer: Tertiary alcohols readily undergo dehydration via the \( \mathrm{E_1} \) (Elimination, unimolecular) mechanism to form alkenes. This reaction involves the formation of a stable carbocation intermediate. The process happens in three main steps:

**Step 1: Protonation of alcohol**
The hydroxyl group of the alcohol gets protonated by the acid (\( \mathrm{H_2SO_4} \)). This converts the poor leaving group (-OH) into a good leaving group (water, \( \mathrm{H_2O} \)).
\[ \begin{array}{l} \mathrm{CH_3} \\ \mathrm{CH_3-C-O-H} \\ \mathrm{CH_3} \end{array} \xrightarrow{\mathrm{H-OSO_3H}} \begin{array}{l} \mathrm{CH_3} \\ \mathrm{CH_3-C-\overset{\oplus}{O}-H} \\ \mathrm{CH_3 \ H} \end{array} \]

**Step 2: Dissociation of oxonium ion to form a carbocation**
The protonated alcohol loses a water molecule, forming a stable tertiary carbocation. This is the slowest step and is therefore the rate-determining step in the \( \mathrm{E_1} \) mechanism.
\[ \begin{array}{l} \mathrm{H_3C} \\ \mathrm{CH_3-\overset{\oplus}{C}-\overset{\oplus}{O}-H} \\ \mathrm{CH_3 \ H} \end{array} \xrightarrow{\mathrm{Slow}} \begin{array}{l} \mathrm{H_3C} \\ \mathrm{CH_3-\overset{\oplus}{C}} \\ \mathrm{CH_3} \end{array} \mathrm{+ H_2O} \]

**Step 3: Deprotonation of carbocation to form an alkene**
A proton is removed from a carbon atom adjacent to the carbocation center by a base (often a water molecule or the conjugate base of the acid), leading to the formation of a double bond and an alkene. This final step restores the acid catalyst.
\[ \begin{array}{l} \mathrm{H_3C} \\ \mathrm{H_2C \quad H} \\ \mathrm{\quad \overset{\oplus}{C}} \\ \mathrm{CH_3} \end{array} \xrightarrow{\mathrm{fast}} \begin{array}{l} \mathrm{H_3C} \\ \mathrm{CH_3-C=CH_2} \\ \mathrm{\quad \quad CH_3} \end{array} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad 2-methylprop-1-ene} \]
This detailed breakdown shows how the hydroxyl group is removed and a double bond is formed through a carbocation intermediate. Tertiary alcohols are favored for this mechanism because their carbocations are more stable.
In simple words: \( \mathrm{E_1} \) dehydration of tertiary alcohols happens in three steps. First, the alcohol gets a hydrogen atom, making water a good leaving group. Second, the water leaves, forming a charged carbon (carbocation). Third, another hydrogen is removed, creating a double bond and forming an alkene. Tertiary alcohols do this easily because their charged carbons are very stable.

🎯 Exam Tip: Always identify the rate-determining step in a mechanism. For \( \mathrm{E_1} \) reactions, it's the formation of the carbocation, which explains why tertiary carbocations lead to faster reaction rates due to their stability.

 

Question 5. An organic compound \( \mathrm{C_2H_6O} \) (A) heated with Cone. \( \mathrm{H_2SO_4} \) at 443K to give and unsaturated hydrocarbon \( \mathrm{C_2H_4} \) (B), which on treatment with Baeyer's reagent to give compound \( \mathrm{C_2H_6O_2} \) (C). Which is used as antifreeze in automobile radiator. Compound (C) distilled with cone. \( \mathrm{H_2SO_4} \) to give cyclic compound \( \mathrm{C_4H_8O_2} \) (D). Compound (A) is heated with cone \( \mathrm{H_2SO_4} \) at 413K to give compound \( \mathrm{C_4H_{10}O} \) (E). Identify compounds (A) to (E) and write equations Compound (A) is Ethanol
Answer:
Compound (A) is Ethanol (\( \mathrm{C_2H_6O} \)).

1. When ethanol (A) is heated with concentrated \( \mathrm{H_2SO_4} \) at 443K, it undergoes dehydration to form an unsaturated hydrocarbon (B), which is ethene (\( \mathrm{C_2H_4} \)).
\[ \mathrm{CH_3CH_2OH \xrightarrow{Con. \ H_2SO_4, \ 443K} CH_2=CH_2 + H_2O} \]
\[ \mathrm{(A) \quad Ethanol \qquad \qquad \qquad \qquad (B) \quad Ethene} \]
2. Ethene (B) on treatment with Baeyer's reagent (cold alkaline \( \mathrm{KMnO_4} \)) undergoes hydroxylation to give compound (C), which is ethane-1,2-diol (ethylene glycol, \( \mathrm{C_2H_6O_2} \)). This compound is known for its use as an antifreeze.
\[ \mathrm{CH_2=CH_2 + H_2O \xrightarrow{Cold \ alkaline \ KMnO_4} CH_2(OH)-CH_2(OH)} \]
\[ \mathrm{(B) \quad Ethene \qquad \qquad \qquad \qquad \qquad \qquad (C) \quad Ethane-1,2-diol} \]
3. When ethane-1,2-diol (C) is distilled with concentrated \( \mathrm{H_2SO_4} \), it undergoes intramolecular dehydration to form a cyclic compound (D), which is 1,4-dioxane (\( \mathrm{C_4H_8O_2} \)).
\[ \mathrm{2 \ CH_2(OH)-CH_2(OH) \xrightarrow{Con. \ H_2SO_4, \ -2H_2O}} \]
\[ \mathrm{\quad \qquad \qquad Ethane-1,2-diol} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad \quad \quad O} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad CH_2 \quad CH_2} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad | \qquad \quad |} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad CH_2 \quad CH_2} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad \quad \quad O} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (D) \quad 1,4-dioxane} \]
4. When ethanol (A) is heated with concentrated \( \mathrm{H_2SO_4} \) at 413K (a lower temperature than 443K), it undergoes intermolecular dehydration to form an ether (E), which is diethyl ether (\( \mathrm{C_4H_{10}O} \)).
\[ \mathrm{2 \ CH_3CH_2OH \xrightarrow{Con. \ H_2SO_4, \ 413K} CH_3CH_2-O-CH_2CH_3 + H_2O} \]
\[ \mathrm{(A) \quad Ethanol \qquad \qquad \qquad \qquad \qquad \qquad (E) \quad Diethyl \ ether} \]
Summary of compounds:

In simple words: We start with ethanol (A). When heated strongly, it turns into ethene (B). Ethene reacts with Baeyer's reagent to make ethylene glycol (C), which is an antifreeze. Heating ethylene glycol makes a ring-shaped molecule called 1,4-dioxane (D). If ethanol is heated less strongly, it forms diethyl ether (E).

🎯 Exam Tip: Remember that the temperature and concentration of \( \mathrm{H_2SO_4} \) dictate the product of alcohol dehydration: higher temperature (443K) favors alkene formation, while lower temperature (413K) favors ether formation.

 

Question 6. Explain the oxidation of glycerol
Answer: Glycerol (\( \mathrm{CH_2OH-CHOH-CH_2OH} \)) is a trihydric alcohol, meaning it has three hydroxyl (-OH) groups. Its oxidation products vary significantly depending on the strength of the oxidizing agent and the reaction conditions. Here's how it generally works:

**Oxidation with Dilute \( \mathrm{HNO_3} \):**
Dilute nitric acid acts as a mild oxidizing agent, selectively oxidizing the primary alcohol groups to carboxylic acid groups, leading to glyceric acid and tartronic acid.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{i) Dilute } \mathrm{HNO_3} & \text{Glyceric acid & Tartronic acid} \\ \hline \end{array} \]

**Oxidation with Concentrated \( \mathrm{HNO_3} \):**
Concentrated nitric acid is a stronger oxidizing agent. It oxidizes all primary and secondary alcohol groups, forming glyceric acid as the main product.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{ii) Concentrated } \mathrm{HNO_3} & \text{Glyceric acid} \\ \hline \end{array} \]

**Oxidation with Bismuth Nitrate:**
Bismuth nitrate oxidizes glycerol to mesoxalic acid.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{iii) Bismuth nitrate} & \text{Mesoxalic acid} \\ \hline \end{array} \]

**Oxidation with Fenton's Reagent (\( \mathrm{FeSO_4 + H_2O_2} \)) or Bromine water (\( \mathrm{Br_2/H_2O} \)) or \( \mathrm{NaOBr} \)):**
These reagents oxidize glycerol to a mixture of glyceraldehyde and dihydroxyacetone, which are the initial products of mild oxidation. Glycerose is a general term for this mixture.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{iv) Fenton's reagent} & \text{Mixture of glyceraldehyde and dihydroxyacetone (glycerose)} \\ \hline \end{array} \]

**Oxidation with Periodic Acid (\( \mathrm{HIO_4} \)) or Lead Tetraacetate (LTA):**
These are specific reagents for vicinal diols. They cleave the C-C bonds, producing formaldehyde and formic acid.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{v) Periodic Acid } \mathrm{HIO_4} & \text{Formaldehyde & Formic acid} \\ \hline \end{array} \]

**Oxidation with Acidified \( \mathrm{KMnO_4} \):**
Acidified potassium permanganate is a very strong oxidizing agent that oxidizes glycerol completely to oxalic acid.
\[ \begin{array}{|l|l|} \hline \textbf{Oxidizing agent} & \textbf{Product formed} \\ \hline \text{vi) Acidified } \mathrm{KMnO_4} & \text{Oxalic acid} \\ \hline \end{array} \]
Here is a summary of the reaction scheme:
\[ \begin{array}{l} \mathrm{CH_2OH} \\ \mathrm{CHOH} \\ \mathrm{CH_2OH} \\ \text{Glycerol} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{CHO} \\ \mathrm{CHOH} \\ \mathrm{CH_2OH} \\ \text{Glyceraldehyde} \\ \text{(2,3-dihydroxy propanal)} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{COOH} \\ \mathrm{CHOH} \\ \mathrm{CH_2OH} \\ \text{Glyceric acid} \\ \text{(2,3-dihydroxy propanoic acid)} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{COOH} \\ \mathrm{CHOH} \\ \mathrm{COOH} \\ \text{Tartronic acid} \\ \text{(2-hydroxy propan-1,3-dioic acid)} \end{array} \]
\[ \begin{array}{l} \mathrm{CH_2OH} \\ \mathrm{CHOH} \\ \mathrm{CH_2OH} \\ \text{Glycerol} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{CH_2OH} \\ \mathrm{C=O} \\ \mathrm{CH_2OH} \\ \text{Dihydroxyacetone} \\ \text{(1,3-dihydroxy propan-2-one)} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{COOH} \\ \mathrm{C=O} \\ \mathrm{COOH} \\ \text{Mesoxalic acid} \\ \text{(2-Oxo propan-1,3-dioic acid)} \end{array} \xrightarrow{\mathrm{[O]}} \begin{array}{l} \mathrm{COOH} \\ \mathrm{COOH} \\ \text{Oxalic acid (ethane-1,2-dioic acid)} \end{array} \]
In simple words: Glycerol has three alcohol groups, so it can be oxidized in many ways. The final products depend on how strong the chemical used for oxidation is. Mild oxidizers might create aldehydes or ketones, while stronger ones can turn it into different acids like glyceric acid, tartronic acid, or even oxalic acid.

🎯 Exam Tip: When discussing the oxidation of polyols like glycerol, always specify the oxidizing agent and conditions, as they determine the selectivity and extent of oxidation, leading to different products.

 

Question 7. Write a note on (i) Riemer Tiemann reaction (ii) Phthalein reaction (iii) coupling reaction
Answer:
(i) **Riemer-Tiemann Reaction:**
This reaction is used to introduce a formyl (-CHO) group at the ortho-position of a phenol. Phenol reacts with chloroform (\( \mathrm{CHCl_3} \)) in the presence of an aqueous alkali, typically sodium hydroxide (NaOH). The reaction usually forms salicylaldehyde as the major product. The mechanism involves an intermediate formation of substituted benzal chloride.
\[ \mathrm{OH} \qquad \qquad \qquad \qquad \mathrm{O^-Na^+} \qquad \qquad \qquad \qquad \mathrm{O^-Na^+} \qquad \qquad \qquad \qquad \mathrm{OH} \]
\[ \quad \mathrm{Phenol} \qquad \xrightarrow{\mathrm{CHCl_3, \ aq.NaOH}} \qquad \mathrm{C_6H_5ONa} \xrightarrow{\mathrm{CHCl_2}} \qquad \mathrm{C_6H_5ONa} \xrightarrow{\mathrm{NaOH}} \qquad \mathrm{CHO} \xrightarrow{\mathrm{H^+}} \mathrm{CHO} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad Salicylaldehyde} \]

(ii) **Phthalein Reaction:**
In this reaction, phenol is heated with phthalic anhydride in the presence of concentrated sulfuric acid (\( \mathrm{H_2SO_4} \)). This condensation reaction forms a phthalein dye, such as phenolphthalein. This reaction is important for synthesizing pH indicators.
\[ \begin{array}{l} \mathrm{OH} \\ \quad \text{Phenol} \end{array} \qquad \qquad \begin{array}{l} \qquad \mathrm{O} \\ \quad \mathrm{C=O} \\ \mathrm{C=O} \\ \text{Phthalic anhydride} \end{array} \qquad \xrightarrow{\mathrm{Con. \ H_2SO_4}} \qquad \begin{array}{l} \mathrm{HO} \\ \mathrm{OH} \\ \text{Phenolphthalein} \end{array} \]

(iii) **Coupling Reaction:**
Phenols undergo coupling reactions with benzene diazonium chloride in an alkaline solution. This reaction forms an azo dye, which is characterized by an azo group (-N=N-). For example, phenol reacts to form p-hydroxy azobenzene, which is a red-orange dye. These dyes are widely used in the textile industry.
\[ \mathrm{N_2Cl + \qquad OH \xrightarrow{NaOH, \ 273-278K} N=N-OH} \]
\[ \mathrm{Benzene \ diazonium \ chloride \quad Phenol \qquad \qquad \qquad \qquad p-hydroxy \ azobenzene} \]
In simple words: The Riemer-Tiemann reaction puts a -CHO group on phenol. The Phthalein reaction makes dyes like phenolphthalein by mixing phenol with phthalic anhydride. Coupling reactions join phenol with diazonium salts to make colorful azo dyes.

🎯 Exam Tip: For each named reaction, clearly state the reactants, reagents, and the major product, highlighting the key functional group changes or introductions. Mentioning the type of reaction (e.g., electrophilic substitution, condensation) is also beneficial.

 

Question 8. Write any three methods of preparing ethers
Answer: Ethers can be prepared through several methods, each with its own advantages and specific applications. Here are three common methods:

(i) **Intermolecular Dehydration of Alcohols:**
This method involves heating alcohols with concentrated sulfuric acid (\( \mathrm{H_2SO_4} \)) at a specific temperature (usually around 413K). Two molecules of alcohol combine, eliminating a water molecule, to form an ether. This method is effective for symmetrical ethers derived from primary alcohols. For example, ethanol yields diethyl ether.
\[ \mathrm{2 \ CH_3CH_2OH \xrightarrow{Conc. \ H_2SO_4, \ 413K} CH_3CH_2-O-CH_2CH_3 + H_2O} \]

(ii) **Williamson Ether Synthesis:**
This is a very versatile method for preparing both symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide. The reaction proceeds via an \( \mathrm{S_N2} \) mechanism. Primary alkyl halides work best to minimize elimination side reactions. For example, sodium methoxide reacts with bromoethane to form methoxyethane.
\[ \mathrm{CH_3ONa + CH_3CH_2Br \rightarrow CH_3-O-CH_2CH_3 + NaBr} \]

(iii) **Methylation of Alcohol with Diazomethane:**
Methyl ethers can be prepared by treating an alcohol with diazomethane (\( \mathrm{CH_2N_2} \)) in the presence of a catalyst like fluoroboric acid (\( \mathrm{HBF_4} \)). This reaction selectively adds a methyl group to the alcohol's oxygen atom.
\[ \mathrm{CH_3CH_2OH + CH_2N_2 \xrightarrow{HBF_4, \ \Delta} CH_3CH_2-O-CH_3 + N_2} \]
In simple words: Ethers can be made in a few ways. One way is to heat alcohols with strong acid to remove water, joining two alcohol molecules. Another is the Williamson synthesis, where an alkyl halide reacts with an alkoxide. Lastly, you can add a methyl group to an alcohol using diazomethane.

🎯 Exam Tip: When describing ether synthesis, always specify the type of alcohol or alkyl halide (primary, secondary, tertiary) as it can influence the reaction mechanism and products (e.g., \( \mathrm{S_N2} \) vs. elimination in Williamson synthesis).

 

Question 9. How does diethylether react with the following (i) \( \mathrm{Cl_2/light} \) (ii) \( \mathrm{PCl_5} \) (iii) \( \mathrm{dilH_2SO_4} \) (iv) \( \mathrm{CH_3COCl/anhy.ZnCl_2} \)
Answer: Diethyl ether reacts differently depending on the specific reagent and conditions:

(i) **Reaction with \( \mathrm{Cl_2/light} \):**
Under the influence of light, diethyl ether undergoes free radical chlorination, where hydrogen atoms are replaced by chlorine atoms. This reaction can lead to multiple chlorination products, like perchloro diethyl ether, where all hydrogens are replaced by chlorine.
\[ \mathrm{CH_3CH_2-O-CH_2CH_3 \xrightarrow{10 \ Cl_2, \ light} CCl_3CCl_2-O-CCl_2CCl_3 + 10 \ HCl} \]
\[ \mathrm{Diethyl \ ether \qquad \qquad \qquad \qquad \qquad \qquad Perchloro \ diethyl \ ether} \]

(ii) **Reaction with \( \mathrm{PCl_5} \):**
Phosphorus pentachloride (\( \mathrm{PCl_5} \)) reacts with diethyl ether to cleave the ether bond, forming two molecules of ethyl chloride and phosphorus oxychloride.
\[ \mathrm{CH_3CH_2-O-CH_2CH_3 \xrightarrow{PCl_5} 2 \ CH_3CH_2-Cl + POCl_3} \]

(iii) **Reaction with Dilute \( \mathrm{H_2SO_4} \):**
Diethyl ether undergoes hydrolysis in the presence of dilute sulfuric acid, regenerating the original alcohol (ethanol). This reaction requires heating.
\[ \mathrm{CH_3CH_2-O-CH_2CH_3 \xrightarrow{dil \ H_2SO_4, \ H_2O} 2 \ CH_3CH_2-OH} \]

(iv) **Reaction with \( \mathrm{CH_3COCl/anhy.ZnCl_2} \):**
Diethyl ether reacts with acetyl chloride (\( \mathrm{CH_3COCl} \)) in the presence of anhydrous zinc chloride (\( \mathrm{ZnCl_2} \)) in a reaction called Friedel-Crafts acylation (when benzene is involved) or, in this context, an acylation that cleaves the ether, producing an alkyl halide and an ester. Here, it yields ethyl chloride and ethyl acetate.
\[ \mathrm{CH_3CH_2-O-CH_2CH_3 + CH_3COCl \xrightarrow{Anhy. \ ZnCl_2} CH_3CH_2-Cl + CH_3COOCH_2CH_3} \]
\[ \mathrm{\qquad Diethyl \ ether \qquad \qquad \qquad \qquad \qquad Ethyl \ chloride \quad Ethyl \ acetate} \]
In simple words: Diethyl ether reacts in several ways: with chlorine and light, all its hydrogen atoms can be swapped for chlorine. With \( \mathrm{PCl_5} \), it breaks into two ethyl chloride molecules. With dilute acid and water, it turns back into ethanol. And with acetyl chloride, it breaks into an ethyl chloride and an ester called ethyl acetate.

🎯 Exam Tip: For ether reactions, note whether the ether linkage is preserved (e.g., chlorination) or cleaved (e.g., with \( \mathrm{PCl_5} \), \( \mathrm{H_2SO_4} \), or \( \mathrm{CH_3COCl} \)), as this is a key indicator of the reaction type.

 

Question 10. Convert (i) methanol into tertiary butyl alcohol (ii) ethanol into 1- butanol
Answer:
(i) **Methanol into Tertiary Butyl Alcohol (2-methylpropan-2-ol):**
To convert methanol to tertiary butyl alcohol, we need to increase the carbon chain and introduce a tertiary carbon atom. This can be achieved using a Grignard reagent with a ketone, after converting methanol to an alkyl halide, then to a Grignard reagent itself, and finally reacting with a ketone. A more direct approach involves converting methanol to iodomethane, which is then used to form a Grignard reagent. This Grignard reagent reacts with acetone, followed by hydrolysis, to form tertiary butyl alcohol.
\[ \mathrm{CH_3OH \xrightarrow{PI_3} CH_3I \xrightarrow{Mg, \ Ether} CH_3MgI \xrightarrow{CH_3COCH_3} (CH_3)_3C-OMgI \xrightarrow{H_3O^+} (CH_3)_3C-OH} \]
\[ \mathrm{Methanol \qquad \qquad \qquad \qquad \qquad \qquad Acetone \qquad \qquad \qquad \qquad \quad 2-methylpropan-2-ol} \]

(ii) **Ethanol into 1-Butanol:**
This conversion requires increasing the carbon chain length. A common method involves converting ethanol to an alkyl halide, then to a Grignard reagent, and reacting it with formaldehyde, followed by hydrolysis.
Alternatively, ethanol can be converted to an alkyl halide, which then reacts with a cyanide, followed by reduction and a final reaction step. A simpler approach involves converting ethanol to an alkyl halide, then to an organometallic reagent, and reacting with a suitable carbonyl compound (like ethanal, then converting to another intermediate, or via the simpler Grignard route):
\[ \mathrm{CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{Mg, \ Ether} CH_3CH_2MgBr \xrightarrow{CH_3CH_2CHO} CH_3CH_2CH(OMgBr)CH_2CH_3 \xrightarrow{H_3O^+} CH_3CH_2CH(OH)CH_2CH_3} \]
\[ \mathrm{Ethanol \qquad \qquad \qquad \qquad \qquad \quad Propionaldehyde \qquad \qquad \qquad \qquad \quad 1-Butanol} \]
Alternatively, using formaldehyde to get a primary alcohol with increased carbon chain:
\[ \mathrm{CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{Mg, \ Ether} CH_3CH_2MgBr \xrightarrow{HCHO} CH_3CH_2CH_2OMgBr \xrightarrow{H_3O^+} CH_3CH_2CH_2OH} \]
\[ \mathrm{Ethanol \qquad \qquad \qquad \qquad \qquad \quad Formaldehyde \qquad \qquad \qquad \qquad \quad Propan-1-ol \ (not \ 1-Butanol)} \]
To get 1-Butanol (4 carbons), you need to react ethyl magnesium bromide with an aldehyde having two carbons (ethanal) and then react it with another Grignard, or use a Grignard with three carbons and react it with formaldehyde. The most direct method for 1-Butanol from ethanol is by converting ethanol to ethyl iodide, reacting with KCN, reducing the nitrile to primary amine, converting to diazonium salt, and then treating with \( \mathrm{H_2O} \).
However, a simpler Grignard pathway to lengthen the chain to 4 carbons: ethylmagnesium bromide reacts with an epoxide like ethylene oxide, followed by hydrolysis.
\[ \mathrm{CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{Mg, \ Ether} CH_3CH_2MgBr \xrightarrow{\quad \quad \quad \text{Ethylene oxide}} CH_3CH_2CH_2CH_2OMgBr \xrightarrow{H_3O^+} CH_3CH_2CH_2CH_2OH} \]
\[ \mathrm{Ethanol \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad Butan-1-ol} \]
In simple words: To change methanol into tertiary butyl alcohol, convert methanol into a Grignard reagent, then react it with acetone. To change ethanol into 1-butanol, first convert ethanol into an ethyl Grignard reagent. Then, react this Grignard reagent with ethylene oxide, which helps add two carbons to the chain. Finally, add water to get 1-butanol.

🎯 Exam Tip: Grignard reactions are excellent for carbon chain elongation. Remember that reacting a Grignard reagent with ketones yields tertiary alcohols, with aldehydes yields secondary alcohols (except formaldehyde giving primary), and with epoxides yields primary alcohols with increased carbon count.

 

Question 11. Complete the reaction (x) \( \mathrm{CH_3-CH=C-CH_3} \xrightarrow{(i) \ O_3 \ (ii) \ Zn/H_2O} \ ? \) (y) \( \mathrm{CH_3COCH_3 \xrightarrow{Mg-Hg, \ H_2O} \ ? \)
Answer:
(x) Reaction of 2-methylbut-2-ene with \( \mathrm{O_3} \) followed by \( \mathrm{Zn/H_2O} \) (Ozonolysis):
Ozonolysis of an alkene breaks the carbon-carbon double bond and replaces it with carbonyl groups. For 2-methylbut-2-ene, the products are acetone (a ketone) and acetaldehyde (an aldehyde).
\[ \mathrm{CH_3-CH=C-CH_3 \xrightarrow{(i) \ O_3} \quad CH_3-CH \quad CH_3 \quad \quad \quad \quad CH_3 \quad C=O \quad O} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \quad O \quad \quad O \quad \quad \quad \quad O} \]
\[ \mathrm{2-methylbut-2-ene \qquad \qquad \qquad \quad \xrightarrow{(ii) \ Zn/H_2O} \quad CH_3-CHO \quad + \quad CH_3-C(=O)CH_3} \]
\[ \mathrm{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad Ethanal \quad \quad \quad \quad Propanone} \]

(y) Reduction of acetone with \( \mathrm{Mg-Hg/H_2O} \) (Pinacol formation):
Ketones, upon reduction with magnesium amalgam (\( \mathrm{Mg-Hg} \)) in the presence of water, undergo reductive coupling to form pinacols, which are symmetrical diols.
\[ \mathrm{2 \ CH_3COCH_3 \xrightarrow{Mg-Hg, \ H_2O} \quad CH_3-C(OH)(CH_3)-C(OH)(CH_3)_2} \]
\[ \mathrm{Acetone \qquad \qquad \qquad \qquad \qquad \qquad 2,3-dimethylbutane-2,3-diol \ (Pinacol)} \]
In simple words: For reaction (x), breaking a double bond in an alkene using ozone and then water (ozonolysis) gives you an aldehyde (ethanal) and a ketone (propanone). For reaction (y), taking two acetone molecules and treating them with magnesium amalgam and water (reductive coupling) joins them together to form a special kind of alcohol with two -OH groups, called a pinacol.

🎯 Exam Tip: Ozonolysis is a powerful reaction for determining the position of a double bond in an alkene. Remember that reductive coupling of ketones forms symmetrical diols (pinacols).

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