JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 38

Answer: (b) Acid chloride, R–CO–Cl

Explanation:
1. The rate depends on the leaving group ability of the group attached to the acyl carbon.
2. Cl⁻ is the best leaving group and chlorine donates only weakly to the carbonyl.
3. So the acid chloride hydrolyses fastest: acid chloride > anhydride > ester > amide.
4. Answer: (b)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (a) Phenylethanoic acid

Explanation:
1. HBr converts benzyl alcohol to benzyl bromide (A).
2. KCN gives benzyl cyanide, C₆H₅CH₂CN (B).
3. Acid hydrolysis of the nitrile gives phenylethanoic (phenylacetic) acid, C₆H₅CH₂COOH.
4. Answer: (a)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (b)

Explanation:
1. P₂O₅ dehydrates acetic acid to acetic anhydride (A).
2. PCl₅ converts it to acetyl chloride (B).
3. Friedel-Crafts acylation of benzene with CH₃COCl/AlCl₃ gives acetophenone (C).
4. Answer: (b)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (d) Iodoform : anaesthetic

Explanation:
1. Ethanoic acid is used as a solvent, esters of benzoic acid in perfumes and flavours, and sodium benzoate as a food preservative.
2. Iodoform is an antiseptic. Chloroform (not iodoform) was used as an anaesthetic.
3. Answer: (d)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (d) CH₃CH₂CH₃

Explanation:
1. Tollens' reagent oxidises butanal to butanoic acid (X).
2. NaHCO₃ gives sodium butanoate (Y).
3. Soda lime decarboxylation removes CO₂ and gives propane (Z).
4. Answer: (d)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (c) CH₃COCH₃

Explanation:
1. A Grignard reagent gives a primary alcohol with HCHO, a secondary alcohol with other aldehydes, and a tertiary alcohol with ketones.
2. Acetone gives 2-methylpropan-2-ol.
3. Answer: (c)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (a)

Explanation:
1. Cyclopentanone has alpha-hydrogens, so with dilute base it undergoes self aldol condensation.
2. The aldol, 2-(1-hydroxycyclopentyl)cyclopentanone, loses water on heating to give the conjugated 2-cyclopentylidenecyclopentanone.
3. Answer: (a)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (d)

Explanation:
1. PCC oxidises cyclohexanol to cyclohexanone (A).
2. Formaldehyde has no alpha-hydrogen and acts only as the acceptor in a crossed aldol reaction at the alpha-carbon of the ketone.
3. Heating with base dehydrates the aldol, so B is 2-methylenecyclohexanone.
4. Answer: (d)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (d) Both (a) and (b)

Explanation:
1. Baeyer's reagent hydroxylates the C=C bond to give propane-1,2-diol (X).
2. Periodic acid cleaves the C–C bond between the two OH-bearing carbons.
3. The products are CH₃CHO and HCHO.
4. Answer: (d)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (b)

Explanation:
1. KMnO₄ cleaves the ring to give adipic acid (X).
2. Ca(OH)₂ forms calcium adipate (Y).
3. Heating the calcium salt gives cyclopentanone (Z) by an intramolecular ketone formation.
4. Answer: (b)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (c)

Explanation:
1. PCl₃ gives propanoyl chloride (A).
2. Friedel-Crafts acylation gives propiophenone (B).
3. Wolff-Kishner reduction (hydrazine/base) converts C=O to CH₂, giving propylbenzene (C).
4. Answer: (c)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (b) 3

Explanation:
1. One mole of PCl₅ replaces C=O by CCl₂, giving 2,2-dichlorobutane (x = 1).
2. Two moles of NaNH₂ remove two HCl to form the alkyne (y = 2).
3. The internal alkyne does not use any more amide. x + y = 3.
4. Answer: (b)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (d) Pent-2-ene

Explanation:
1. Both products must be aldehydes (positive Fehling's test), so each carbon of the double bond carries an H.
2. Pent-2-ene gives CH₃CHO (Y) and CH₃CH₂CHO (Z).
3. Acetaldehyde gives the iodoform test, while propanal (Z) does not.
4. 2-Methylbut-2-ene gives acetone, which does not respond to Fehling's solution, and pent-1-ene gives formaldehyde and butanal, so they do not fit.
5. Answer: (d)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (a) Glyoxal + 2 HCHO

Explanation:
1. CH₂=CH–CH=CH₂ has two C=C bonds.
2. The terminal CH₂ groups give two molecules of HCHO, and the two internal CH carbons, which stay joined, give OHC–CHO (glyoxal).
3. Answer: (a)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (c) Propanone

Explanation:
1. Hydration of ethyne gives acetaldehyde (X).
2. CH₃MgX adds to give propan-2-ol (Y), a secondary alcohol.
3. Oxidation of a secondary alcohol gives a ketone: propanone (Z).
4. Answer: (c)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (a) X = acetophenone, Y = m-nitroacetophenone

Explanation:
1. Friedel-Crafts acylation with acetic anhydride gives acetophenone (X).
2. The acetyl group withdraws electrons and is meta directing, so nitration gives m-nitroacetophenone (Y).
3. Answer: (a)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (a)

Explanation:
1. Sodium methoxide causes an intramolecular Claisen (Dieckmann) condensation, giving the cyclic beta-keto ester, methyl 2-oxocyclopentanecarboxylate (A).
2. Hydrolysis gives the beta-keto acid (B).
3. Heating removes CO₂ to give cyclopentanone (C).
4. Answer: (a)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.

Answer: (b)

Explanation:
1. Electrolytic reduction of nitrobenzene in conc. H₂SO₄ gives p-aminophenol (A) (through phenylhydroxylamine and the Bamberger rearrangement).
2. One equivalent of acetic anhydride acetylates the more nucleophilic NH₂ group.
3. B is paracetamol (4-acetamidophenol), an analgesic and antipyretic.
4. Answer: (b)

Teacher's Note:
1. Follow the reaction sequence one step at a time, naming each intermediate, and note which step changes the carbon count.