RBSE Solutions Class 12 Chemistry Chapter 11 Organic Compounds with Functional Group C

RBSE Class 12 Chemistry Chapter 11 Text Book Questions

RBSE Class 12 Chemistry Chapter 11 Multiple Choice Questions

Question 1. Which of the following is not obtained in any condition by the reaction of ethanol and conc.H2SO4?
(a) CH3CHO
(b) CH3CH2HSO4
(c) C2H4
(d) CH3CH2OCH2CH3
Answer: (d) CH3CH2OCH2CH3
In simple words: When ethanol reacts with concentrated sulfuric acid, products like acetaldehyde, ethyl hydrogen sulfate, and ethene can form, but diethyl ether (CH3CH2OCH2CH3) is typically not produced under these conditions. Diethyl ether formation usually needs specific temperature control or a different approach.

🎯 Exam Tip: Remember that the reaction of ethanol with sulfuric acid is highly dependent on temperature. For example, at lower temperatures (around 140°C), it favors ether formation, while at higher temperatures (around 170°C), it leads to ethene.

Question 2. Given the reaction \( R-MgX + HCHO \xrightarrow [ Ether ]{ Dry } [P] \), here \( [P] \) is:
(a) RCH2OH
(b) \( CH_3-\underset{O}{\overset{||}{C}}-CH_3 \)
(c) CH3OH
(d) \( R-\underset{O}{\overset{||}{C}}-R \)
Answer: (a) RCH2OH
In simple words: When a Grignard reagent reacts with formaldehyde, it forms an intermediate adduct which, upon hydrolysis (adding \( H^+ H_2O \)), always yields a primary alcohol like RCH2OH. The R-group from the Grignard reagent attaches to the carbon atom of formaldehyde.

🎯 Exam Tip: Remember the general rule for Grignard reagents: formaldehyde gives primary alcohols, other aldehydes give secondary alcohols, and ketones give tertiary alcohols upon hydrolysis.

Question 4. The final product obtained by the reaction between alcohol and phosphorus pentachloride is
(a) Chloro alkene
(b) Dichloro alkene
(c) Chloroalkane
(d) Dichloro alkane
Answer: (c) Chloroalkane
In simple words: When an alcohol reacts with phosphorus pentachloride (PCl5), the hydroxyl group (-OH) of the alcohol is replaced by a chlorine atom. This process turns the alcohol into a chloroalkane, also known as an alkyl chloride.

🎯 Exam Tip: Know that PCl5, PCl3, and SOCl2 are common reagents used to convert alcohols into alkyl halides. PCl5 is particularly effective for replacing the -OH group with -Cl.

Question 5. Which of the following phenol is strongest?
(a) o - nitrophenol
(b) m - nitrophenol
(c) p - nitrophenol
(d) p - chlorophenol
Answer: (c) p - nitrophenol
In simple words: A phenol becomes more acidic when electron-withdrawing groups are attached to its ring. The nitro group (\( -NO_2 \)) is a strong electron-withdrawing group, and its effect is strongest when it's at the para position because it allows for better stabilization of the phenoxide ion through resonance.

🎯 Exam Tip: Electron-withdrawing groups (like \( -NO_2 \)) increase acidity by stabilizing the conjugate base (phenoxide ion) through resonance and inductive effects, with resonance being most effective at ortho and para positions.

Question 6. Victor Mayer test is not given by
(a) C2H5OH
(b) (CH3)3COH
(c) (CH3)2CHOH
(d) CH3CH2CH2OH
Answer: (b) (CH3)3COH
In simple words: The Victor-Meyer test is used to tell apart primary, secondary, and tertiary alcohols based on the color produced. Primary and secondary alcohols give different colors, but tertiary alcohols do not react in a way that shows a color, meaning they do not give this test.

🎯 Exam Tip: The Victor-Meyer test involves several steps and is crucial for distinguishing between 1°, 2°, and 3° alcohols based on the final color (blood red for 1°, blue for 2°, colorless for 3°).

Question 7. Which one of the following is the strongest acid?
(a) Phenol
(b) m - chlorophenol
(c) Benzyl alcohol
(d) p - chlorophenol
Answer: (d) p - chlorophenol
In simple words: Among the options, p-chlorophenol is the strongest acid because the chlorine atom is an electron-withdrawing group. This group helps to make the phenoxide ion more stable by pulling electrons away, especially when it's in the para position, which increases the acidity compared to regular phenol, meta-chlorophenol, or benzyl alcohol.

🎯 Exam Tip: Acidity of phenols is enhanced by electron-withdrawing groups and reduced by electron-donating groups. The position of the substituent (ortho, meta, para) is also critical due to resonance and inductive effects.

Question 9. Phenol + Chloroform + Base → Main product. Main product is
(a) Salicylaldehyde
(b) Formaldehyde
(c) ketone
(d) Acetaldehyde
Answer: (a) Salicylaldehyde
In simple words: This reaction is known as the Reimer-Tiemann reaction. In this process, phenol reacts with chloroform in the presence of a strong base to form salicylaldehyde as the main product. An aldehyde group is added to the ortho position of the phenol.

🎯 Exam Tip: The Reimer-Tiemann reaction is a classic example of electrophilic substitution on phenol, where the \( -CHO \) group is introduced preferentially at the ortho position due to hydrogen bonding with the hydroxyl group.

Question 10. The product formed when ether is passed on alumina at 653 K is
(a) Alkene
(b) Alkane
(c) Alcohol
(d) Phenol
Answer: (a) Alkene
In simple words: When an ether is heated to a high temperature (like 653 K) over an alumina catalyst, it undergoes dehydration. This reaction removes water molecules from the ether structure and converts it into an alkene.

🎯 Exam Tip: Remember that ethers can undergo various reactions. Dehydration over catalysts like alumina at high temperatures is a key method for converting ethers into alkenes.

RBSE Class 12 Chemistry Chapter 11 Very Short Answer Type Questions

Question 1. Write the general formula of alcohol.
Answer: The general formula for alcohols is \( C_nH_{2n+1}OH \), or simply \( R-OH \), where \( R \) represents an alkyl group. This formula shows that an alcohol has a hydroxyl group (-OH) attached to a carbon chain. For example, ethanol is \( C_2H_5OH \).
In simple words: The basic way to write an alcohol is \( R-OH \), where \( R \) is a carbon chain and \( OH \) is the special alcohol part. Another common way is \( C_nH_{2n+1}OH \).

🎯 Exam Tip: Understand that \( R \) represents any alkyl group, highlighting the functional group (-OH) which defines the alcohol class.

Question 3. Write the name of the product formed when Grignard's reagent reacts with formaldehyde.
Answer: When a Grignard reagent reacts with formaldehyde, a primary alcohol is formed upon subsequent hydrolysis. This reaction is a two-step process: first, the Grignard reagent adds to the carbonyl carbon of formaldehyde to form an adduct, and then this adduct is hydrolyzed to yield the primary alcohol.

In simple words: Grignard's reagent reacting with formaldehyde makes a primary alcohol.

🎯 Exam Tip: Always remember that formaldehyde is unique among aldehydes in that it reacts with Grignard reagents to specifically produce primary alcohols. Other aldehydes produce secondary alcohols, and ketones produce tertiary alcohols.

Question 4. Write the order of acidity for primary, secondary and tertiary alcohols.
Answer: The order of acidity for primary, secondary, and tertiary alcohols is: Primary (1°) > Secondary (2°) > Tertiary (3°). This means that primary alcohols are the most acidic, followed by secondary alcohols, while tertiary alcohols are the least acidic. The acidity of alcohols is influenced by the electron-donating effect of alkyl groups, which destabilizes the alkoxide ion, a key factor in determining acidity. Primary alcohols have fewer alkyl groups, making them more acidic.
In simple words: Primary alcohols are the most acidic. Then come secondary alcohols, and tertiary alcohols are the least acidic.

🎯 Exam Tip: Acidity generally decreases as the number of alkyl groups attached to the carbon bearing the -OH group increases, due to their electron-donating inductive effect destabilizing the conjugate base (alkoxide ion).

Question 5. Write Fries rearrangement.
Answer: Fries Rearrangement is a reaction where a phenyl ester undergoes rearrangement to form hydroxyaryl ketones. This happens when the phenyl ester is heated in nitrobenzene with an anhydrous Lewis acid like aluminium chloride (AlCl3). During the rearrangement, the acyl group migrates from the phenolic oxygen to either an ortho or para position on the benzene ring. This rearrangement is useful for creating specific substituted phenols.
In simple words: Fries rearrangement changes a phenyl ester into a special kind of ketone called a hydroxyaryl ketone. This needs heat and a strong acid like AlCl3, and the acyl part moves from the oxygen to the benzene ring.

🎯 Exam Tip: The Fries rearrangement is an important reaction for synthesizing ortho- and para-substituted hydroxyaryl ketones, with the product distribution often influenced by reaction conditions like temperature and solvent.

Question 6. What happens when phenol is kept open in the air?
Answer: When phenol is left open in the air, it slowly gets oxidized. This oxidation process causes phenol to turn into a pink-colored compound called p-benzoquinone. This color change is a common observation for phenol exposed to the environment, showing its susceptibility to air oxidation.

In simple words: If you leave phenol out in the air, it slowly changes color to pink because it gets oxidized and turns into p-benzoquinone.

🎯 Exam Tip: Phenols are easily oxidized due to the presence of the hydroxyl group directly attached to the benzene ring, making them prone to air oxidation and color changes.

Question 7. What is the effect of electron withdrawing groups on the acidity of phenol?
Answer: When an electron-withdrawing group, such as a nitro group (\( -NO_2 \)), is attached to the benzene ring of phenol, it increases the acidic strength of the phenol. This happens because these groups stabilize the phenoxide ion (the conjugate base formed after phenol loses a proton) by pulling electron density away from the oxygen. When electron-withdrawing groups are in the ortho or para positions, their effect is stronger because they can stabilize the negative charge more effectively through resonance, making the phenol more acidic.
In simple words: Electron-withdrawing groups make phenol more acidic. They pull electrons away, making the phenol's ion more stable, especially if they are at the ortho or para spots.

🎯 Exam Tip: Remember that electron-withdrawing groups (like \( -NO_2 \), \( -X \), \( -COOH \)) increase acidity, while electron-donating groups (like \( -CH_3 \), \( -OCH_3 \)) decrease acidity. The resonance effect is particularly strong at ortho and para positions.

Question 8. Write the general formula of ether.
Answer: The general formula for an ether is \( R-O-R' \), where \( R \) and \( R' \) represent alkyl or aryl groups. In simple ethers, both \( R \) and \( R' \) are the same, for example, diethyl ether (\( CH_3CH_2OCH_2CH_3 \)). In mixed ethers, \( R \) and \( R' \) are different groups, like ethyl methyl ether (\( CH_3OCH_2CH_3 \)). The oxygen atom forms a bridge between two hydrocarbon groups.
In simple words: The basic formula for an ether is \( R-O-R' \). \( R \) and \( R' \) are carbon chains. If they are the same, it's a simple ether; if they are different, it's a mixed ether.

🎯 Exam Tip: Ethers are characterized by the \( -O- \) functional group linking two organic substituents. Be able to distinguish between simple and mixed ethers based on the symmetry of the R groups.

RBSE Class 12 Chemistry Chapter 11 Short Answer Type Questions

Question 2. Write a method for preparation of primary alcohol from Grignard's reagent.
Answer:
(i) One method to prepare primary alcohols from Grignard reagents involves reacting the Grignard reagent with formaldehyde. The Grignard reagent acts as a nucleophile and attacks the carbonyl carbon of formaldehyde. This forms an addition product, an adduct, which upon hydrolysis (treatment with dilute acid) yields a primary alcohol. This process is highly specific for primary alcohol synthesis using formaldehyde.

(ii) Another way is to react a Grignard reagent with oxirane (ethylene oxide). The Grignard reagent opens the epoxide ring, forming an intermediate adduct. When this adduct is hydrolyzed, it gives a primary alcohol with two more carbon atoms than the original Grignard reagent. This reaction is useful for extending the carbon chain.
In simple words: You can make primary alcohols by mixing a Grignard reagent with formaldehyde or with oxirane. Both reactions first make an intermediate product, which then turns into a primary alcohol when you add water and acid.

🎯 Exam Tip: Remember that Grignard reagents are versatile and can be used to synthesize 1°, 2°, or 3° alcohols depending on the carbonyl compound used (formaldehyde for 1°, other aldehydes for 2°, ketones for 3°, and oxirane for 1° with chain extension).

Question 3. Alcohols are soluble in water whereas diethyl ether is not, explain with reason.
Answer: Alcohols are soluble in water because they can form strong intermolecular hydrogen bonds with water molecules. The hydroxyl group (-OH) in alcohols allows for both hydrogen bond donation and acceptance, leading to extensive hydrogen bonding between alcohol and water. Diethyl ether, on the other hand, can only act as a hydrogen bond acceptor (through its oxygen atom) and cannot donate hydrogen bonds. Its longer hydrocarbon chains also increase its non-polar character, further reducing its ability to mix with water. The presence of strong hydrogen bonding is crucial for water solubility.
In simple words: Alcohols dissolve in water because they can make strong connections called hydrogen bonds with water. Diethyl ether cannot make these strong connections as easily, and its larger non-polar part makes it not mix well with water.

🎯 Exam Tip: The ability to form hydrogen bonds is a primary factor determining water solubility for organic compounds. Molecules with more polar groups and shorter non-polar chains tend to be more soluble.

Question 4. Explain two commercial methods for the preparation of phenol.
Answer: Two commercial methods for preparing phenol are:
(i) From Cumene: Phenol is commonly produced from cumene (isopropyl benzene). Cumene is first made by reacting benzene with propene using an acid catalyst like \( H_3PO_4 \) (a Friedel-Crafts alkylation). Then, oxygen is bubbled through the cumene to form cumene hydroperoxide. Finally, this hydroperoxide is decomposed in the presence of an aqueous acid, yielding phenol and acetone. This method is economically important because it also produces acetone, a valuable co-product.

(ii) From Coaltar: Phenol can also be obtained from the middle oil fraction of coal tar, which is collected during coal tar distillation at a temperature range of 440 to 503 K. This fraction contains both phenol and naphthalene. After cooling, naphthalene crystals separate out. The remaining oily solution is then washed with dilute \( H_2SO_4 \) to remove any basic impurities. Phenol is subsequently extracted from this solution using aqueous sodium hydroxide (NaOH) to form soluble sodium phenate. Carbon dioxide (\( CO_2 \)) is then passed through the sodium phenate solution to regenerate phenol.

In simple words: Phenol is made commercially in two main ways. One way uses cumene, which is first made from benzene and then turned into phenol and acetone. The other way extracts phenol from coal tar, cleaning it up through several steps.

🎯 Exam Tip: The cumene process is highly valued for its co-production of acetone. The coal tar method highlights the importance of fractional distillation in separating complex mixtures.

Question 5. Phenols show less acidity in comparison to carboxylic acids. Explain with reason.
Answer: Phenols are less acidic than carboxylic acids because of the differing stability of their conjugate bases. In carboxylic acids, the negative charge of the carboxylate ion (\( RCOO^- \)) is equally shared (delocalized) over two electronegative oxygen atoms through resonance, making it very stable. In contrast, for phenols, the negative charge of the phenoxide ion (\( C_6H_5O^- \)) is delocalized over one oxygen atom and several carbon atoms of the benzene ring. This means the charge is not as effectively stabilized on the highly electronegative oxygen atom as it is in a carboxylate ion. The more stable the conjugate base, the stronger the acid. Therefore, carboxylic acids ionize to a greater extent, producing more \( H^+ \) ions, making them stronger acids than phenols.

In simple words: Carboxylic acids are stronger acids than phenols because their leftover ion (carboxylate) is more stable. This stability comes from sharing the negative charge evenly between two oxygen atoms, which doesn't happen as well in phenol's ion (phenoxide).

🎯 Exam Tip: The stability of the conjugate base is a key factor in determining acid strength. Greater delocalization of negative charge on more electronegative atoms leads to higher stability and stronger acidity.

Question 6. Write the following reactions:
1. Gattermann Reaction.
2. Reimer-Tiemann reaction
3. Duff reaction
Answer:
1. Gattermann Reaction:
When phenol reacts with a mixture of hydrogen cyanide (HCN) and hydrogen chloride (HCl) in the presence of a zinc chloride (\( ZnCl_2 \)) catalyst, it forms an aldimine as an intermediate. This aldimine then undergoes hydrolysis (reaction with water) to produce p-hydroxybenzaldehyde. This reaction is a method for introducing an aldehyde group onto an aromatic ring.

2. Reimer-Tiemann Reaction:
This reaction involves the treatment of phenol with chloroform (\( CHCl_3 \)) in the presence of strong bases such as sodium hydroxide (NaOH) or potassium hydroxide (KOH). In this process, an aldehyde (\( -CHO \)) group is introduced into the benzene ring, typically at the ortho position relative to the phenolic group. The main product formed is salicylaldehyde. This reaction is important for the synthesis of ortho-formylated phenols.

3. Duff Reaction:
The Duff reaction is used to synthesize ortho-formylated phenols by reacting phenol with hexamethylenetetramine (\( (CH_2)_6N_4 \)) and boric acid (\( H_3BO_3 \)) in the presence of glycerol. The reaction introduces an aldehyde group (\( -CHO \)) primarily at the ortho position, resulting in the formation of salicylaldehyde. It is a useful method for directing formylation to a specific position on the phenol ring.

In simple words: The Gattermann, Reimer-Tiemann, and Duff reactions are all ways to add an aldehyde group to a phenol. They use different starting materials and conditions, but the result is adding a -CHO group, usually to the position next to the existing -OH group.

🎯 Exam Tip: These three reactions are important named reactions for the formylation of phenols. Pay close attention to the specific reagents and conditions for each to distinguish between them.

Question 7. Explain the halogenation reactions of diethyl ether.
Answer: Halogenation reactions of diethyl ether involve the substitution of hydrogen atoms by halogen atoms (like chlorine or bromine).
(i) In the dark: When diethyl ether reacts with a halogen (\( X_2 \)) in the dark, the hydrogen atoms on the alpha-carbon (the carbon directly attached to the oxygen) are substituted by halogen atoms. This reaction is selective for the alpha-hydrogens due to the electron-donating effect of the oxygen atom which stabilizes the intermediate radicals. The product formed is an alpha, alpha'-dihaloether.
\( CH_3-CH-O-CH-CH_3 \)
\( \quad \quad | \quad \quad \quad | \)
\( \quad \quad X \quad \quad X \)
\( \alpha, \alpha'- \text{Dihaloether} \)
(ii) In the presence of sunlight: When diethyl ether is heated with chlorine in the presence of sunlight (UV light), a more extensive substitution occurs. All the hydrogen atoms in diethyl ether can be replaced by chlorine atoms, leading to the formation of a per-chloro-diethyl ether. This reaction is less selective and proceeds through a free-radical mechanism, which is typical for halogenation in the presence of light.
\( CH_3CH_2OCH_2CH_3 \xrightarrow {10Cl_2} CCl_3-CCl_2-O-CCl_2 + HCl \)
\( \quad \quad \quad \quad \quad \quad \text{Sunlight} \)
In simple words: Halogenation reactions add chlorine or bromine to diethyl ether. If done in the dark, only hydrogen atoms next to the oxygen get replaced. If done in sunlight, almost all hydrogen atoms can be replaced by the halogen.

🎯 Exam Tip: Distinguish between the halogenation of ethers in the dark (alpha-substitution) and in the presence of light (per-halogenation) based on the reaction mechanism and products formed.

RBSE Class 12 Chemistry Chapter 11 Long Answer Type Questions

Question 1. What does alcohol form by reaction with the following:
1. PCl3
2. SOCI2
Answer:
1. Reaction with PCl3:
When alcohols react with phosphorus trichloride (\( PCl_3 \)), the hydroxyl group (-OH) of the alcohol is replaced by a chlorine atom, leading to the formation of alkyl halides. In this reaction, three molecules of alcohol react with one molecule of \( PCl_3 \) to produce three molecules of alkyl chloride and one molecule of phosphorous acid (\( H_3PO_3 \)). This is a general method to convert alcohols into their corresponding chloroalkanes.
\( 3R-OH + PCl_3 \rightarrow 3R-Cl + H_3PO_3 \)
\( \text{Alcohol} \quad \quad \quad \quad \quad \text{Alkyl halide} \)
For example:
\( 3CH_3OH + PCl_3 \rightarrow 3CH_3-Cl + H_3PO_3 \)
\( \text{Methanol} \quad \quad \quad \quad \quad \text{Methyl chloride} \)
2. Reaction with SOCl2:
Alcohols react with thionyl chloride (\( SOCl_2 \)) in the presence of pyridine to form alkyl chlorides. This reaction is often preferred because the by-products, sulfur dioxide (\( SO_2 \)) and hydrogen chloride (\( HCl \)), are gaseous and escape easily, leaving behind a pure alkyl chloride. Pyridine is used to neutralize the \( HCl \) formed, driving the reaction to completion. This method is considered one of the best for preparing alkyl chlorides from alcohols due to the ease of product separation.
\( R-OH + SOCl_2 \xrightarrow {Pyridine} R-Cl + SO_2\uparrow + HCl\uparrow \)
\( \text{Alcohol} \quad \text{thionylchloride} \quad \quad \quad \quad \quad \text{Alkyl chloride} \)
For example:
\( CH_3CH_2-OH + SOCl_2 \xrightarrow {Pyridine} CH_3CH_2-Cl + SO_2\uparrow + HCl\uparrow \)
\( \text{Ethanol} \quad \quad \quad \quad \quad \quad \text{Chloroethane} \)
In simple words: Alcohols react with \( PCl_3 \) to make alkyl chlorides and phosphoric acid. They also react with \( SOCl_2 \) (often with pyridine) to make alkyl chlorides, but this time, the extra products are gases that float away, leaving a cleaner product.

🎯 Exam Tip: Remember that both \( PCl_3 \) and \( SOCl_2 \) are effective reagents for converting alcohols to alkyl halides, but \( SOCl_2 \) is often preferred in organic synthesis due to the gaseous nature of its by-products, which simplifies purification.

Question 3. Write the substitution reactions of diethyl ether.
Answer: Substitution reactions of diethyl ether primarily involve the halogenation of the alkyl groups. When diethyl ether is reacted with chlorine or bromine, hydrogen atoms present on the alpha-carbon (the carbon atoms directly attached to the oxygen) can be replaced by halogen atoms. This process leads to the formation of halogenated ethers. Depending on the reaction conditions, the substitution can be selective or extensive. For instance, in the presence of sunlight, a complete substitution of all hydrogen atoms by halogen atoms can occur, forming a per-chloro-diethyl ether. This highlights the reactivity of the C-H bonds in ethers towards free radical substitution.
\( CH_3-CH-O-CH-CH_3 \)
\( \quad \quad | \quad \quad \quad | \)
\( \quad \quad X \quad \quad X \)
\( \alpha, \alpha'- \text{Perchlorodiethyl ether} \)
(ii) Halogenation in presence of sunlight:
\( CH_3CH_2OCH_2CH_3 \xrightarrow {10Cl_2} CCl_3-CCl_2-O-CCl_2 + HCl \)
\( \quad \quad \quad \quad \quad \quad \text{Sunlight} \)
In simple words: Substitution reactions of diethyl ether involve replacing its hydrogen atoms with halogen atoms like chlorine or bromine. This can happen selectively at carbons next to oxygen, or more widely if there's sunlight, leading to heavily halogenated ethers.

🎯 Exam Tip: The halogenation of ethers demonstrates the free-radical nature of these substitution reactions, with different products formed depending on the presence or absence of light to initiate the reaction.