CBSE Class 12 Chemistry Chapter 7 Alcohols Phenols and Ethers Competency Based Questions Set 01

Competency Based Questions (MCQs)

Question 1. The correct order of boiling point of primary (1\(^\circ\)), secondary (2\(^\circ\)) and tertiary (3\(^\circ\)) alcohols is
(a) 1\(^\circ\) > 2\(^\circ\) > 3\(^\circ\)
(b) 3\(^\circ\) > 2\(^\circ\) > 1\(^\circ\)
(c) 2\(^\circ\) > 1\(^\circ\) > 3\(^\circ\)
(d) 2\(^\circ\) > 3\(^\circ\) > 1\(^\circ\)
Answer: (a) 1\(^\circ\) > 2\(^\circ\) > 3\(^\circ\)
In simple words: Straight-chain alcohols have more contact area and stronger attractions between molecules, so they require more heat to boil than branched ones.

Exam Tip: Remember that increased branching decreases surface area and weakens van der Waals forces, which always lowers the boiling point.

Question 2. \( \text{C}_2\text{H}_5\text{OH} + \text{SOCl}_2 \xrightarrow{\text{Pyridine}} \text{C}_2\text{H}_5\text{Cl} + \text{SO}_2 + \text{HCl} \) is
(a) Williamson’s reaction.
(b) Hofmann’s synthesis.
(c) Mendis reaction.
(d) Darzen’s reaction.
Answer: (d) Darzen’s reaction.
In simple words: The reaction of an alcohol with thionyl chloride in the presence of pyridine to form an alkyl chloride is called Darzen's reaction.

Exam Tip: Darzen's method is highly preferred for preparing alkyl chlorides because the side products (\( \text{SO}_2 \) and \( \text{HCl} \)) are escapable gases, leaving pure alkyl halide.

Question 3. The major organic product in the reaction, \( \text{CH}_3 — \text{O} — \text{CH(CH}_3)_2 + \text{HI} \rightarrow \) product: is/are
(a) \( \text{CH}_3\text{I} + \text{(CH}_3)_2\text{CHOH} \)
(b) \( \text{CH}_3\text{OH} + \text{(CH}_3)_2\text{CHI} \)
(c) \( \text{ICH}_2\text{OCH(CH}_3)_2 \)
(d) \( \text{CH}_3 - \text{O} - \text{CI(CH}_3)_2 \)
Answer: (a) \( \text{CH}_3\text{I} + \text{(CH}_3)_2\text{CHOH} \)
In simple words: The reaction proceeds via an \( \text{S}_{\text{N}}2 \) mechanism where the iodide ion attacks the smaller methyl group, leaving the larger isopropyl group as an alcohol.

Exam Tip: For unsymmetrical ethers with primary or secondary alkyl groups, nucleophilic attack by \( \text{I}^- \) always occurs at the smaller alkyl group.

Question 4. Which compound is predominantly formed when phenol is allowed to react with bromine in aqueous medium?
(a) Picric acid
(b) O-Bromophenol
(c) 2, 4, 6-Tribromophenol
(d) p-Bromophenol
Answer: (c) 2, 4, 6-Tribromophenol
In simple words: When mixed with bromine water, phenol reacts very quickly to form a white precipitate of 2,4,6-tribromophenol.

Exam Tip: Water highly polarizes phenol to form phenoxide ion, activating the ring strongly and resulting in trisubstitution. In non-polar solvents like \( \text{CS}_2 \), only monosubstitution occurs.

Question 5. Phenols are more acidic than alcohols because
(a) Phenoxide ion is stabilised by resonance
(b) Phenols are more soluble in polar solvents
(c) Phenoxide ion does not exhibit resonance
(d) Alcohols do not lose H atoms at all
Answer: (a) Phenoxide ion is stabilised by resonance
In simple words: Phenols easily release a hydrogen proton because the resulting phenoxide ion is stable, as its negative charge spreads throughout the aromatic ring.

Exam Tip: In alcohols, the alkoxide ion has no resonance stabilization, and the electron-donating alkyl group destabilizes it further, making alcohols much weaker acids than phenols.

Question 6. The compound B is formed in the sequence of the reaction given below: \( \text{C}_6\text{H}_5\text{OH} + \text{NaOH} + \text{CCl}_4 \xrightarrow{\text{Heat}} \text{A} \xrightarrow{\text{HCl}} \text{B} \) The compound B is
(a) Salicylaldehyde
(b) Benzoic acid
(c) Salicylic acid
(d) Cinnamic acid
Answer: (c) Salicylic acid
In simple words: Heating phenol with sodium hydroxide and carbon tetrachloride, followed by acid treatment, yields salicylic acid.

Exam Tip: Remember that using \( \text{CHCl}_3 \) with phenol gives salicylaldehyde, while using \( \text{CCl}_4 \) leads to salicylic acid.

Question 7. Which of the following reagents cannot be used to distinguish between phenol and benzyl alcohol?
(a) FeCl\(_3\)
(b) Litmus soln
(c) Br\(_2\)/CCl\(_4\)
(d) All of the options
Answer: (c) Br\(_2\)/CCl\(_4\)
In simple words: Neither phenol nor benzyl alcohol reacts noticeably with bromine in carbon tetrachloride, so this reagent cannot tell them apart.

Exam Tip: Phenol gives a characteristic violet color with neutral \( \text{FeCl}_3 \) and turns blue litmus red, whereas benzyl alcohol (an aliphatic alcohol) does not show these reactions.

Question 8. Identify Z in the series: \( \text{C}_3\text{H}_7\text{OH} \xrightarrow[160^\circ\text{-}180^\circ\text{C}]{\text{Conc. H}_2\text{SO}_4} \text{X} \xrightarrow{\text{Br}_2} \text{Y} \xrightarrow{\text{Excess of alc. KOH}} \text{Z} \)
(a) \( \text{CH}_3 - \text{CH(NH}_2) - \text{CH}_2(\text{NH}_2) \)
(b) \( \text{CH}_3 - \text{CH(OH)} - \text{CH}_2(\text{OH}) \)
(c) \( \text{CH}_3 - \text{C(OH)} = \text{CH}_2 \)
(d) \( \text{CH}_3 - \text{C} \equiv \text{CH} \)
Answer: (d) \( \text{CH}_3 - \text{C} \equiv \text{CH} \)
In simple words: Dehydrating propanol creates propene. Adding bromine yields 1,2-dibromopropane, which then loses two HBr units with strong base to yield propyne.

Exam Tip: Dehydrohalogenation of vicinal dihalides with excess strong base like alcoholic KOH always results in the formation of an alkyne.

Question 9. 1-propanol and 2-propanol can be best distinguished by
(a) Oxidation with KMnO\(_4\) followed by reaction with Fehling solution.
(b) Oxidation with acidic dichromate followed by reaction with Fehling solution.
(c) Oxidation by heating with copper followed by reaction with Fehling solution.
(d) Oxidation with concentrated H\(_2\)SO\(_4\) followed by reaction with Fehling solution.
Answer: (c) Oxidation by heating with copper followed by reaction with Fehling solution.
In simple words: Heating primary propanol with hot copper produces propanal, which reacts with Fehling's solution, while secondary propanol produces acetone, which does not react.

Exam Tip: Dehydrogenation of primary alcohols using hot copper yields aldehydes (Fehling positive), while secondary alcohols yield ketones (Fehling negative).

Question 10. \( \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{NaOH/CCl}_4} \text{X} \xrightarrow{\text{Zn dust/Heat}} \text{Y} \xrightarrow{\text{Na/Sodalime}} \text{Z} \) In the above sequence Z is
(a) Toluene
(b) Cresol
(c) Benzene
(d) Benzol
Answer: (c) Benzene
In simple words: Phenol turns into salicylic acid, then zinc dust removes its OH group to make benzoic acid, and finally sodalime removes the carboxyl group to form benzene.

Exam Tip: Sodalime decarboxylation (\( \text{NaOH} + \text{CaO} \)) of benzoic acid salts is a classic method to eliminate the carboxyl group and yield benzene.

Assertion-Reason Questions

Question. Assertion: Boiling points of alcohols are lower than hydrocarbons.
Reason: Among isomeric alcohols, boiling point decreases in the order: 1\(^\circ\) > 2\(^\circ\) > 3\(^\circ\).
Answer: (d) Assertion is wrong statement but reason is correct statement.
Alcohols participate in intermolecular hydrogen bonding, making their boiling points higher than those of comparable hydrocarbons. Within isomeric alcohols, branching decreases surface area, so the boiling point decreases in the order: 1\(^\circ\) > 2\(^\circ\) > 3\(^\circ\).
In simple words: Alcohols actually boil at higher temperatures than hydrocarbons because of hydrogen bonds. Branching in alcohols lowers their boiling points.

Exam Tip: Always look for intermolecular hydrogen bonding when comparing the boiling points of alcohols with other organic compounds of similar molecular mass.

Question. Assertion: Solubility of alcohols decreases with increase in size of alkyl/aryl groups.
Reason: Alcohols form H-bonding with water to show soluble nature.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
The capability to interact via hydrogen bonding falls as the hydrophobic hydrocarbon chain or aryl group grows larger.
In simple words: Both statements are true. Larger carbon chains do not mix well with water, which reduces the alcohol's overall solubility despite the OH group's hydrogen bonding.

Exam Tip: Remember that the alkyl group is hydrophobic (water-repelling), so larger alkyl groups shield the hydrophilic hydroxyl group, reducing water solubility.

Question. Assertion: tert-Butyl alcohol undergoes acid catalysed dehydration readily than propanol.
Reason: 3\(^\circ\) Alcohols do not give Victor-Meyer’s test.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Alcohols producing a more stable carbocation undergo dehydration much more easily. Since tert-butyl alcohol yields a highly stable tertiary carbocation, it reacts much faster than propanol.
In simple words: Tertiary butyl alcohol dehydrates faster because it forms a very stable intermediate carbocation. That it does not show a color in the Victor-Meyer test is true, but unrelated.

Exam Tip: The rate of acid-catalyzed dehydration of alcohols follows the stability order of carbocations: \( 3^\circ > 2^\circ > 1^\circ \).

Question. Assertion: o-Nitrophenol is more volatile than p-nitrophenol.
Reason: Intramolecular hydrogen bonding is present in o-nitrophenol while intermolecular H-bonding is in p-nitrophenol.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Intramolecular hydrogen bonding in o-nitrophenol prevents association between different molecules, whereas intermolecular hydrogen bonding in p-nitrophenol binds molecules strongly together, raising its boiling point.
In simple words: o-Nitrophenol has internal hydrogen bonds, so its molecules do not stick to each other and vaporize easily. p-Nitrophenol molecules link together tightly, making them less volatile.

Exam Tip: This difference in volatility due to intramolecular versus intermolecular hydrogen bonding allows these isomers to be separated by steam distillation.

Question. Assertion: Phenol is less acidic than p-nitrophenol.
Reason: Phenolate ion is more stable than p-nitrophenol ion.
Answer: (c) Assertion is correct statement but reason is wrong statement.
Phenol has a lower acidity than p-nitrophenol because the strongly electron-withdrawing nitro group stabilizes the p-nitrophenoxide ion much more than the unsubstituted phenoxide ion.
In simple words: The assertion is true, but the reason is false. The p-nitrophenoxide ion is actually more stable than the phenoxide ion due to the electron-pulling nitro group.

Exam Tip: Electron-withdrawing groups like nitro (\( \text{-NO}_2 \)) stabilize the phenoxide ion and increase acidity, especially when present at ortho and para positions.

Question. Assertion: With Br\(_2\)—H\(_2\)O, phenol gives 2,4,6-tribromophenol but with Br\(_2\)—CS\(_2\), it gives 4-bromophenol as the major product.
Reason: In water, ionisation of phenol is enhanced but in CS\(_2\), it is greatly suppressed.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Highly polar water promotes the ionization of phenol into the highly reactive phenoxide ion, leading to trisubstitution. In non-polar carbon disulfide, phenol remains mostly un-ionized, which limits reaction to monosubstitution.
In simple words: In water, phenol splits into ions easily, making it react very fast to add three bromines. In carbon disulfide, it stays neutral and only adds one bromine.

Exam Tip: Use aqueous bromine for full trisubstitution of phenol, and bromine in carbon disulfide at low temperatures for selective para-monosubstitution.

Question. Assertion: Phenol is more acidic than ethanol.
Reason: Phenoxide ion is resonance stabilised.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Phenol shows higher acidity than ethanol because the resulting phenoxide ion is highly stabilized by resonance. In contrast, the ethoxide ion is destabilized by the electron-donating ethyl group.
In simple words: Phenol easily releases its proton because the remaining phenoxide ion is stable due to resonance, whereas ethanol's ion has no stabilization.

Exam Tip: Resonance stabilization of the conjugate base always increases the acidity of the parent acid.

Question. Assertion: Phenol decomposes NaHCO\(_3\) solution to evolve CO\(_2\) gas.
Reason: Picric acid is 2, 4, 6-trinitrophenol.
Answer: (d) Assertion is wrong statement but reason is correct statement.
Phenol acts as a weaker acid than carbonic acid and therefore cannot decompose sodium bicarbonate to release carbon dioxide gas. Picric acid is indeed 2,4,6-trinitrophenol.
In simple words: Phenol is too weak an acid to react with sodium bicarbonate to produce carbon dioxide. However, picric acid is indeed the common name for 2,4,6-trinitrophenol.

Exam Tip: While phenol is too weak to react with \( \text{NaHCO}_3 \), highly acidic phenols like picric acid (due to three strong electron-withdrawing nitro groups) will easily decompose bicarbonate to evolve \( \text{CO}_2 \).

Question. Assertion: Reimer–Tiemann reaction of phenol with CHCl\(_3\) in NaOH at 340 K gives salicylaldehyde as the major product.
Reason: The reaction occurs through intermediate formation of \( ^+\text{CHCl}_2 \).
Answer: (c) Assertion is correct statement but reason is wrong statement.
The Reimer-Tiemann reaction of phenol with chloroform in sodium hydroxide yields salicylaldehyde. The active intermediate electrophile in this process is neutral dichlorocarbene (\( :\text{CCl}_2 \)), not a carbocation.
In simple words: The assertion is true because using chloroform gives salicylaldehyde. The reason is false because the actual intermediate is neutral dichlorocarbene, not a positively charged ion.

Exam Tip: Be sure to write the intermediate for the Reimer-Tiemann reaction as neutral dichlorocarbene (\( :\text{CCl}_2 \)), which acts as an electrophile.

Question. Assertion: Primary and secondary alcohols can be distinguished by Victor-Meyer’s test.
Reason: Primary alcohols form nitrolic acid which dissolves NaOH to form blood red colouration but secondary alcohols form pseudonitroles which give blue colouration with NaOH.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Primary and secondary alcohols are effectively distinguished via the Victor-Meyer test. Primary alcohols yield nitrolic acid, which reacts with base to give a blood-red solution, while secondary alcohols produce pseudonitroles that yield a distinct blue color with base.
In simple words: This chemical test clearly tells the alcohols apart because primary alcohols turn blood-red when treated with base, while secondary alcohols turn blue.

Exam Tip: Remember the color sequence for Victor-Meyer's test as RGB: Red for Primary (\( 1^\circ \)), Blue for Secondary (\( 2^\circ \)), and Colorless (Gray/No reaction) for Tertiary (\( 3^\circ \)).

Case Based Questions

Case Study 1
A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff’s reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60. Answer the following questions by choosing the most appropriate options:

Question (i) The compound (X) is an
(a) acid
(b) aldehyde
(c) alcohol
(d) ether
Answer: (d) ether
In simple words: Compound X does not react with sodium metal or Schiff's reagent, ruling out alcohols, acids, and aldehydes. This means it must be an ether.

Exam Tip: Ethers are chemically inert compared to alcohols and aldehydes, and do not react with active metals like sodium.

Question (ii) The IUPAC name of the acid formed is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
Answer: (b) ethanoic acid
In simple words: The carboxylic acid with an equivalent mass of 60 is ethanoic acid (acetic acid), which has the formula \( \text{CH}_3\text{COOH} \).

Exam Tip: Calculate equivalent weight of monocarboxylic acids using \( M = n \times E \). For \( E = 60 \), \( M = 60 \), which corresponds to \( \text{CH}_3\text{COOH} \).

Question (iii) Compound (Y) is
(a) ethyl iodide
(b) methyl iodide
(c) propyl iodide
(d) mixture of (a) and (b)
Answer: (a) ethyl iodide
In simple words: Refluxing diethyl ether with excess hydroiodic acid yields only ethyl iodide as the organic product.

Exam Tip: Symmetrical ethers react with excess \( \text{HI} \) to form two molecules of the same alkyl iodide.

Question (iv) Compound (Z) is
(a) methanol
(b) ethanol
(c) propanol
(d) butanol
Answer: (b) ethanol
In simple words: Hydrolyzing ethyl iodide yields ethanol, which can then be oxidized to form ethanoic acid.

Exam Tip: Primary alkyl halides readily undergo nucleophilic substitution with aqueous \( \text{KOH} \) to produce the corresponding primary alcohols.

Question (v) Compound (X) on treatment with excess of Cl\(_2\) in the presence of light gives
(a) \(\alpha\)-chlorodiethyl ether
(b) \(\alpha'\)-dichlorodiethyl ether
(c) perchlorodiethyl ether
(d) none of the options
Answer: (c) perchlorodiethyl ether
In simple words: In bright light, excess chlorine replaces every single hydrogen atom in diethyl ether with chlorine, forming perchlorodiethyl ether.

Exam Tip: Remember that chlorination of ethers in the dark yields monochloro or dichloro derivatives, whereas chlorination in light leads to complete substitution (perchlorination).

Case Study 2
Both alcohols and phenols are acidic in nature, but phenols are more acidic than alcohols. Acidic strength of alcohols mainly depends upon the inductive effect. Acidic strength of phenols depends upon a combination of both inductive effect and resonance effects of the substituent and its position on the benzene ring. Electron withdrawing groups increase the acidic strength of phenols whereas electron donating groups decrease the acidic strength of phenols. Phenol is weaker acid than carboxylic acid. Answer the following questions by choosing the most appropriate options:

Question (i) Phenols are highly acidic as compared to alcohols due to
(a) the higher molecular mass of phenols
(b) the stronger hydrogen bonds in phenols
(c) alkoxide ion is a strong conjugate base
(d) phenoxide ion is resonance stabilised.
Answer: (d) phenoxide ion is resonance stabilised.
In simple words: The phenoxide ion is highly stable because the negative charge is distributed over the aromatic ring, making phenols much more acidic than alcohols.

Exam Tip: While alkoxide ions have no resonance, phenoxide ions benefit from resonance stabilization, which strongly favors proton release.

Question (ii) The compound that does not liberate CO\(_2\) on treatment with aqueous sodium bicarbonate solution is
(a) benzoic acid
(b) benzenesulphonic acid
(c) salicylic acid
(d) carbolic acid
Answer: (d) carbolic acid
In simple words: Carbolic acid (phenol) is a very weak acid, even weaker than carbonic acid, so it cannot decompose sodium bicarbonate to release carbon dioxide gas.

Exam Tip: Stronger acids than carbonic acid (like carboxylic acids and sulphonic acids) react with sodium bicarbonate to release \( \text{CO}_2 \) gas, while simple phenols do not.

Case Study 3
An organic compound (A) having molecular formula C\(_6\)H\(_6\)O gives a characteristic colour with aqueous FeCl\(_3\) solution. (A) on treatment with CO\(_2\) and NaOH at 400 K under pressure gives (B), which on acidification gives a compound (C). The compound (C) reacts with acetyl chloride to give (D) which is a popular pain killer. Answer the following questions by choosing the most appropriate options:

Question (i) Compound (A) is
(a) 2-hexanol
(b) dimethyl ether
(c) phenol
(d) 2-methyl pentanol
Answer: (c) phenol
In simple words: Phenol has the formula C\(_6\)H\(_6\)O and gives a characteristic violet color when tested with ferric chloride.

Exam Tip: Phenols always give a characteristic violet/colored complex with neutral \( \text{FeCl}_3 \), a test not shown by aliphatic alcohols.

Question (ii) Compound (C) is
(a) salicylic acid
(b) salicyladehyde
(c) benzoic acid
(d) benzaldehyde
Answer: (a) salicylic acid
In simple words: Phenol reacts with sodium hydroxide and carbon dioxide to yield sodium salicylate, which turns into salicylic acid upon acidification.

Exam Tip: The carboxylation of phenol to salicylic acid is a key industrial reaction known as the Kolbe-Schmitt reaction.

Question (iii) Number of carbon atoms in compound (D) is
(a) 7
(b) 6
(c) 8
(d) 9
Answer: (d) 9
In simple words: Compound D is aspirin (acetylsalicylic acid), which contains exactly nine carbon atoms.

Exam Tip: Aspirin consists of a benzene ring (6 carbons), a carboxyl group (1 carbon), and an acetyl group (2 carbons), totaling 9 carbon atoms.

Question (iv) The conversion of compound (A) to (C) is known as
(a) Reimer-Tiemann reaction
(b) Kolbe’s reaction
(c) Smarts reaction
(d) Girgnard reaction
Answer: (b) Kolbe’s reaction
In simple words: Converting phenol to salicylic acid using carbon dioxide and base under high pressure is called Kolbe's reaction.

Exam Tip: Remember that Kolbe's reaction introduces a carboxyl group (\( \text{-COOH} \)) using \( \text{CO}_2 \), whereas Reimer-Tiemann introduces a formyl group (\( \text{-CHO} \)) using \( \text{CHCl}_3 \).

Question (v) Compound (A) on heating with compound (C) in presence of POCl\(_3\) gives a compound (D) which is used
(a) in perfumery as a flavouring agent
(b) as an antipyretic
(c) as an analgesic
(d) as an intestinal antiseptic
Answer: (d) as an intestinal antiseptic
In simple words: Heating phenol with salicylic acid yields phenyl salicylate (commonly known as salol), which is used as an intestinal antiseptic.

Exam Tip: Salol is not absorbed in the stomach but hydrolyzes in the alkaline medium of the intestine to release phenol and salicylic acid, acting as an effective antiseptic there.

Case Study 4
Reimer-Tiemann reaction introduces an aldehyde group, on aromatic ring of phenol, ortho to the hydroxyl group. This is a general method for the synthesis of substituted salicylaldehydes. Answer the following questions by choosing the most appropriate options:

Question (i) The electrophile in this reaction [A] is
(a) :CHCl
(b) \( ^+ \text{CHCl}_2 \)
(c) :CCl\(_2\)
(d) CCl\(_3\)
Answer: (c) :CCl\(_2\)
In simple words: The electrophile that attacks the phenol ring is neutral dichlorocarbene, which is generated from chloroform and base.

Exam Tip: Dichlorocarbene (\( :\text{CCl}_2 \)) contains a divalent carbon with six valence electrons, making it highly electrophilic.

Question (ii) Reimer-Tiemann reaction is an example of
(a) nucleophilic substitution reaction
(b) electrophilic substitution reaction
(c) nucleophilic addition reaction
(d) electrophilic addition reaction
Answer: (b) electrophilic substitution reaction
In simple words: This reaction is an electrophilic substitution because the electron-deficient dichlorocarbene attacks the electron-rich aromatic ring.

Exam Tip: Since the aromatic ring acts as a nucleophile, the reagent attacking it must be an electrophile, classifying the mechanism as an electrophilic aromatic substitution.

Question (iii) Which of the following reagents is used in the given reaction in step I?
(a) aq. NaOH + CH\(_3\)Cl
(b) aq. NaOH + CH\(_2\)Cl\(_2\)
(c) aq. NaOH + CHCl\(_3\)
(d) aq. NaOH + CCl\(_4\)
Answer: (c) aq. NaOH + CHCl\(_3\)
In simple words: Chloroform along with aqueous sodium hydroxide is used in the first step to create the active electrophile.

Exam Tip: Always make sure to distinguish between \( \text{CHCl}_3 \) (which forms formyl group) and \( \text{CCl}_4 \) (which forms carboxyl group) when choosing Reimer-Tiemann reagents.

Question (iv) When phenol reacts with chloroform in presence of KOH, the product formed is
(a) salicylic acid
(b) salicyladehyde
(c) both (a) and (b)
(d) none of the options
Answer: (b) salicyladehyde
In simple words: The reaction of phenol with chloroform and potassium hydroxide introduces a formyl group to form salicylaldehyde.

Exam Tip: Salicylaldehyde (2-hydroxybenzaldehyde) is the major product obtained when potassium or sodium hydroxide is used with chloroform.

Case Study 5
Dehydration of alcohols can lead to the formation of either alkenes or ethers. This dehydration can be carried out either with protonic acids such as conc. H\(_2\)SO\(_4\), H\(_3\)PO\(_4\) or catalysts such as anhydrous ZnCl\(_2\) or Al\(_2\)O\(_3\). When primary alcohols are heated with conc. H\(_2\)SO\(_4\) at 433-443 K, they undergo intramolecular dehydration to form alkenes. Secondary and tertiary alcohols undergo dehydration under milder conditions. The ease of dehydration of alcohols follows the order: 3\(^\circ\) > 2\(^\circ\) > 1\(^\circ\). The dehydration of alcohols always occurs in accordance with the Saytzeff’s rule. Primary alcohols when heated with protic acid at 413 K, give dialkyl ether. Answer the following questions by choosing the most appropriate options:

Question (i) Which one of the following alcohols undergoes acid-catalysed dehydration to alkenes most readily?
(a) (CH\(_3\))\intercal_2\)CHCH\(_2\)OH
(b) (CH\(_3\))\(_3\)COH
(c) CH\(_3\)CHOHCH\(_3\)
(d) CH\(_3\)CH\(_2\)CH\(_2\)OH
Answer: (b) (CH\(_3\))\(_3\)COH
In simple words: Tertiary butyl alcohol forms a highly stable tertiary carbocation intermediate, making it dehydrate much faster than primary or secondary alcohols.

Exam Tip: The ease of acid-catalyzed dehydration of alcohols always follows the carbocation stability order: tertiary (\( 3^\circ \)) > secondary (\( 2^\circ \)) > primary (\( 1^\circ \)).

Question (ii) Dehydration of alcohol is an example of which type of reaction?
(a) Substitution
(b) Elimination
(c) Addition
(d) Rearrangement
Answer: (b) Elimination
In simple words: Dehydration involves the removal of a water molecule from adjacent carbon atoms, which makes it an elimination reaction.

Exam Tip: Acid-catalyzed dehydration is typically a \( \beta \)-elimination reaction (specifically E1 for tertiary and secondary, E2 for primary alcohols).

Question (iii) The alcohol which does not give a stable compound on dehydration is
(a) ethyl alcohol
(b) methyl alcohol
(c) n-propyl alcohol
(d) n-butyl alcohol
Answer: (b) methyl alcohol
In simple words: Methyl alcohol only has one carbon, so it cannot form a stable double-bonded alkene upon losing water.

Exam Tip: Dehydration of methanol would yield a highly unstable singlet carbene or methylene species (\( :\text{CH}_2 \)), which is why it does not readily dehydrate.