JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 11

Answer: (b) cuprous oxide

Explanation:
1. Aldehydes reduce the Cu²⁺ complex of Fehling's solution to red Cu₂O: CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O + 3H₂O.

Teacher's Note:
1. The red precipitate is cuprous oxide, Cu₂O.
2. Benzaldehyde does not respond to Fehling's solution.

Answer: (c) a silver mirror with Tollen's reagent

Explanation:
1. All ketones add nucleophiles like NH₂OH, phenylhydrazine and NaHSO₃ (methyl ketones). Tollens' reagent is a mild oxidant that oxidises only aldehydes, so acetone gives no silver mirror.

Teacher's Note:
1. Silver mirror and Fehling's tests identify aldehydes.

Answer: (b) Grignard reagent

Explanation:
1. Fehling's, Tollens' and Schiff's reagents react only with the aldehyde. Both compounds react with Grignard reagents (to give alcohols), so it cannot distinguish them.

Teacher's Note:
1. Aldehydes give 2° alcohols, and ketones give 3° alcohols with Grignard reagents.

Answer: (c) butan-2-one

Explanation:
1. The silver salt with Br₂ gives bromoethane by the Hunsdiecker reaction (A).
2. Mg gives ethylmagnesium bromide (B).
3. A Grignard reagent adds to a nitrile; hydrolysis of the imine gives the ketone CH₃CH₂COCH₃ (butan-2-one).

Teacher's Note:
1. Nitriles with Grignard reagents give ketones.

Answer: (a) 50 percent d + 50 percent l (racemic mixture)

Explanation:
1. Hydrolysis of the gem-dichloride gives acetophenone (A). HCN addition to the planar carbonyl carbon occurs equally from both faces to give the racemic cyanohydrin (B), which is optically inactive.

Teacher's Note:
1. Attack on a planar carbonyl group is equally likely from either side.

Answer: (c) A is true, but R is false.

Explanation:
1. Formic acid (pKa 3.75) is stronger than benzoic acid (pKa 4.2). The assertion is true.
2. A stronger acid has a lower pKa, so the reason is false.

Teacher's Note:
1. Lower pKa means stronger acid.

Answer: (a) benzoyl chloride + diphenylcadmium

Explanation:
1. Acid chlorides react with dialkyl or diaryl cadmium to give ketones: 2C₆H₅COCl + (C₆H₅)₂Cd → 2C₆H₅COC₆H₅ + CdCl₂. Option 3 is the Gattermann–Koch type synthesis of benzaldehyde.

Teacher's Note:
1. Organocadmium reagents stop at the ketone stage.

Answer: (c) benzyl 4-methylphenyl ketone, C₆H₅COCH₂C₆H₄CH₃

Explanation:
1. NaHSO₃ adds to aldehydes and to unhindered methyl or cyclic ketones. A large diaryl-type ketone like C₆H₅COCH₂C₆H₄CH₃ is sterically hindered and does not react.

Teacher's Note:
1. Steric hindrance prevents bisulphite addition in bulky ketones.

Answer: (b) hydrolysis of their sodium bisulphite adducts

Explanation:
1. Aldehydes and methyl ketones form crystalline bisulphite adducts that separate from impurities; the carbonyl compound is regenerated with acid or alkali.

Teacher's Note:
1. Bisulphite adducts are water soluble and regenerate the carbonyl compound.

Answer: (c) nucleophilic addition

Explanation:
1. The carbonyl carbon is electron deficient (C^δ+=O^δ−) and is attacked by nucleophiles, giving addition reactions.

Teacher's Note:
1. Aldehydes are more reactive than ketones.

Answer: (c) Tollen's reagent

Explanation:
1. Both give a positive iodoform test and both form 2,4-DNP derivatives. Only acetaldehyde reduces Tollens' reagent.

Teacher's Note:
1. Tollens' reagent distinguishes aldehydes from ketones.

Answer: (d) iodine and alkali

Explanation:
1. Acetaldehyde has the CH₃CO– group and gives yellow iodoform with I₂/NaOH. Formaldehyde does not. Both reduce Fehling's and Tollens' reagents.

Teacher's Note:
1. Iodoform test is positive for CH₃CO– and CH₃CH(OH)– groups.

Answer: (c) a mixture of methanol and sodium formate

Explanation:
1. HCHO has no α-hydrogen, so in conc. NaOH it undergoes Cannizzaro reaction: 2HCHO + NaOH → CH₃OH + HCOONa.

Teacher's Note:
1. Cannizzaro is a disproportionation of an aldehyde without α-H.

Answer: (c) HCHO

Explanation:
1. Aldol condensation requires an α-hydrogen. HCHO has none.

Teacher's Note:
1. Check for α-hydrogen atoms.

Answer: (c) aldehydes not containing α-hydrogen atoms

Explanation:
1. Aldehydes without α-H (HCHO, C₆H₅CHO, (CH₃)₃CCHO) undergo disproportionation in strong alkali to an alcohol and a carboxylate.

Teacher's Note:
1. Those with α-H prefer aldol condensation.

Answer: (b) benzyl alcohol and sodium formate

Explanation:
1. Formaldehyde is the better hydride donor, so it is oxidised to sodium formate, while benzaldehyde is reduced to benzyl alcohol.

Teacher's Note:
1. In crossed Cannizzaro the more reactive aldehyde is oxidised.

Answer: (b) acetaldehyde

Explanation:
1. Two molecules of acetaldehyde undergo aldol addition to 3-hydroxybutanal, which loses water on heating to give crotonaldehyde.

Teacher's Note:
1. Aldol condensation followed by dehydration gives an α,β-unsaturated aldehyde.

Answer: (c) R′CH=N–R (a Schiff base)

Explanation:
1. RNH₂ + R′CHO → R′CH=NR + H₂O. The product is an imine (Schiff base).

Teacher's Note:
1. This is a nucleophilic addition–elimination (condensation) reaction.

Answer: (c) 2,4-dinitrophenylhydrazine

Explanation:
1. Brady's reagent is 2,4-dinitrophenylhydrazine in methanol and sulphuric acid. It gives orange or yellow precipitates with aldehydes and ketones.

Teacher's Note:
1. It is the test for the carbonyl group.

Answer: (d) CH₃COCH₃

Explanation:
1. It has a C=O group (oxime formation) but is not an aldehyde (no Tollens' reaction), so it is the ketone, propanone.

Teacher's Note:
1. Both propanal and propanone give oximes, but only propanal gives silver mirror.

Answer: (a) acidic reduction (Zn–Hg/conc. HCl)

Explanation:
1. Clemmensen reduction (Zn–Hg/HCl) converts C=O into CH₂. Wolff–Kishner reduction (NH₂NH₂/KOH) does the same in basic medium.

Teacher's Note:
1. Clemmensen is acidic and Wolff–Kishner is basic.

Answer: (b) aldehyde

Explanation:
1. Acetaldol (3-hydroxybutanal) loses water to give crotonaldehyde, an α,β-unsaturated aldehyde.

Teacher's Note:
1. Dehydration of aldol gives conjugated carbonyl compounds.

Answer: (a) NaHSO₃

Explanation:
1. Acetaldehyde forms a water-soluble bisulphite adduct, while acetophenone (a hindered aryl ketone) does not.

Teacher's Note:
1. Bisulphite adducts of aldehydes can be separated by filtration.

Answer: (d) PCC (pyridinium chlorochromate)

Explanation:
1. Strong oxidants like KMnO₄, K₂Cr₂O₇ and CrO₃ in aqueous acid oxidise primary alcohols further to acids. PCC in CH₂Cl₂ stops at the aldehyde.

Teacher's Note:
1. PCC is a mild oxidising agent.

Answer: (b) a β-hydroxy aldehyde or ketone

Explanation:
1. The enolate attacks the carbonyl carbon of a second molecule to give an aldol, a β-hydroxy aldehyde or ketone.

Teacher's Note:
1. Aldol means aldehyde plus alcohol.

Answer: (d) bicarbonate

Explanation:
1. Propanoic acid transfers H⁺ to HCO₃⁻, giving H₂CO₃, which decomposes to CO₂ and H₂O. The carbon in CO₂ comes from the bicarbonate ion.

Teacher's Note:
1. The acid only supplies the proton.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both are true. The amide is stabilised by delocalisation of the nitrogen lone pair with the carbonyl group, so it hydrolyses less easily, but this does not explain why both are derivatives of benzoic acid.

Teacher's Note:
1. Reactivity: acid chloride > anhydride > ester > amide.