JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 08

Answer: (a) Benzoyl chloride, \(\ce{C6H5COCl}\)

Explanation:
1. Reactivity towards nucleophilic acyl substitution depends on how good a leaving group is and how electron deficient the carbonyl carbon is.
2. Chloride is the best leaving group and has no resonance donation, so the acid chloride is the most reactive.
3. The order is acid chloride > anhydride > ester > amide.

Teacher's Note:
1. Memorise the order of reactivity of acid derivatives: acid chloride, anhydride, ester, amide.

Answer: (b) \(r_2 > r_1\)

Explanation:
1. Both reactions are nucleophilic acyl substitution with ethoxy as the nucleophile.
2. The ethoxide ion is a much stronger nucleophile than neutral ethanol.
3. So the reaction with ethoxide is faster, and \(r_2 > r_1\).

Teacher's Note:
1. A negatively charged nucleophile is stronger than its neutral conjugate acid.

Answer: (b) The product is known as salol.

Explanation:
1. The product is methyl salicylate, which is oil of wintergreen, so statement a is correct.
2. Salol is phenyl salicylate, so statement b is incorrect.
3. The characteristic wintergreen smell is used as a test for methanol, so c is correct.

Teacher's Note:
1. Oil of wintergreen is methyl salicylate and salol is phenyl salicylate.

Answer: (d) both a and c

Explanation:
1. A benzoyl group is introduced, so it is benzoylation.
2. Benzoylation of an alcohol or phenol with benzoyl chloride in the presence of aqueous alkali is called the Schotten-Baumann reaction.
3. So both a and c are correct.

Teacher's Note:
1. Acetylation introduces the acetyl group, while benzoylation introduces the benzoyl group.

Answer: (b) benzoic acid

Explanation:
1. Bromobenzene and magnesium give phenylmagnesium bromide (X).
2. The Grignard reagent adds to \(\ce{CO2}\) to give a magnesium carboxylate.
3. Acidic hydrolysis gives benzoic acid.

Teacher's Note:
1. Grignard reagent plus carbon dioxide gives a carboxylic acid with one more carbon.

Answer: (a) acetic acid

Explanation:
1. Carboxylic acids form hydrogen-bonded dimers, giving the strongest intermolecular attraction.
2. Ethanol has hydrogen bonding but no dimer, while acetaldehyde and dimethyl ether have only dipole interactions.

Teacher's Note:
1. Boiling point order here: acid > alcohol > aldehyde > ether (of similar mass).

Answer: (a) P.C.C.

Explanation:
1. Pyridinium chlorochromate stops the oxidation of a primary alcohol at the aldehyde.
2. Aqueous \(\ce{KMnO4}\) and \(\ce{CrO3/H2O}\) are strong oxidants and carry the oxidation through to ethanoic acid.

Teacher's Note:
1. PCC is the standard reagent to stop at the aldehyde stage.

Answer: (d) \(\ce{CH3NH2}\)

Explanation:
1. Acetic acid and ammonia give ammonium acetate (P).
2. Heating gives acetamide (Q).
3. Hofmann bromamide degradation with \(\ce{NaOBr}\) removes one carbon, giving methylamine.

Teacher's Note:
1. Hofmann bromamide reaction gives a primary amine with one carbon less than the amide.

Answer: (c) \(\ce{CH3CO^{18}OC2H5 + H2O}\)

Explanation:
1. In esterification the acid loses \(-\text{OH}\) and the alcohol loses only the hydrogen (acyl-oxygen cleavage).
2. So the alcohol oxygen stays in the ester and the water formed is unlabelled.
3. The products are \(\ce{CH3CO^{18}OC2H5}\) and \(\ce{H2O}\).

Teacher's Note:
1. Acid gives OH and alcohol gives H, so the alcohol oxygen is retained in the ester.

Answer: (c) formic acid

Explanation:
1. Formic acid has an aldehyde group as well as a carboxyl group.
2. It reduces Tollens' reagent to silver and is itself oxidised to carbonic acid.
3. Benzoic and acetic acids cannot be oxidised further.

Teacher's Note:
1. Formic acid is the only carboxylic acid that gives Tollens' and Fehling's tests.