JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 04

Answer: (b) cuprous oxide

Explanation:
1. The aldehyde is oxidised to acetate, and Cu(II) in the tartrate complex is reduced to Cu(I).
2. The red precipitate is cuprous oxide, \( \ce{Cu2O} \).

Teacher's Note:
1. Silver mirror is for Tollens' reagent; red \( \ce{Cu2O} \) is for Fehling's.

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

Explanation:
1. Acetone is a ketone and, like other ketones, does not reduce Tollens' reagent.
2. It does give all the other addition-elimination and addition products.

Teacher's Note:
1. Only aldehydes (and a few reducing compounds) give a silver mirror.

Answer: (b) Grignard reagent

Explanation:
1. Both carbonyl compounds add Grignard reagents to give alcohols, so there is no visible difference.
2. Fehling, Tollens and Schiff's tests are positive only for the aldehyde.

Teacher's Note:
1. Choose a reagent that reacts with both in the same way.

Answer: (d) \( \ce{CH3-C#N} \)

Explanation:
1. \( \ce{CH3MgI} \) adds to methyl cyanide to give an imine salt.
2. Hydrolysis gives acetone.

Teacher's Note:
1. Nitrile + RMgX then H3O+ gives ketone.

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

Explanation:
1. Hydrolysis of the gem-dichloride gives acetophenone (A).
2. HCN addition gives the cyanohydrin, which has a new chiral centre formed by attack from either face with equal probability.
3. The product is a racemic mixture.

Teacher's Note:
1. Planar carbonyl carbon is attacked from both sides equally.

Answer: (a) benzoyl chloride + diphenylcadmium

Explanation:
1. Organocadmium compounds react with acid chlorides to give ketones and stop at that stage.
2. So PhCOCl + Ph2Cd gives Ph-CO-Ph.

Teacher's Note:
1. Gattermann-Koch gives benzaldehyde instead.

Answer: (c) \( \ce{CH3-C6H4-CH2-CO-Ph} \), a hindered aryl ketone

Explanation:
1. Bisulphite adducts form with aldehydes and methyl ketones with little steric hindrance.
2. A ketone with two large groups (aryl and benzyl-aryl) is too hindered.

Teacher's Note:
1. Large groups on both sides of C=O block the reaction.

Answer: (d) Friedel-Craft reaction

Explanation:
1. Friedel-Crafts acylation of arenes gives aryl ketones.
2. Rosenmund and Etard give aldehydes.

Teacher's Note:
1. Cannizzaro gives an alcohol and an acid.

Answer: (b) hydrolysis of sodium bisulphite adducts

Explanation:
1. Aldehydes and many ketones form crystalline bisulphite adducts which can be filtered and then hydrolysed to give the pure carbonyl compound.

Teacher's Note:
1. Adducts separate impurities that do not react.

Answer: (c) nucleophilic addition

Explanation:
1. The carbonyl carbon is electron-poor and is attacked by nucleophiles.
2. The reaction is a nucleophilic addition.

Teacher's Note:
1. C=O is polarised: C partly positive, O partly negative.

Answer: (c) Tollen's reagent

Explanation:
1. Only acetaldehyde reduces Tollens' reagent to give silver.
2. Both give iodoform and both give a DNP precipitate.

Teacher's Note:
1. Tollens' test is for aldehydes only.

Answer: (d) iodine and alkali

Explanation:
1. Acetaldehyde has a \( \ce{CH3-CO-} \) group and gives yellow iodoform.
2. Formaldehyde does not.

Teacher's Note:
1. Formaldehyde and acetaldehyde both reduce Fehling's and Tollens'.

Answer: (c) \( \ce{I2} \) and NaOH, then \( \ce{H+} \)

Explanation:
1. Iodoform reaction converts the methyl ketone into the carboxylate (loss of \( \ce{CHI3} \)) without touching the double bond.
2. Acidification gives the acid.

Teacher's Note:
1. KMnO4 would attack the double bond.

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

Explanation:
1. HCHO has no alpha hydrogen, so it undergoes Cannizzaro reaction.
2. One molecule is reduced to methanol and the other oxidised to sodium formate.

Teacher's Note:
1. Cannizzaro: disproportionation of aldehydes without alpha-H.

Answer: (c) \( \ce{HCHO} \)

Explanation:
1. Aldol condensation needs an alpha hydrogen.
2. Formaldehyde has none.

Teacher's Note:
1. No alpha-H means no enolate.

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

Explanation:
1. Aldehydes with no alpha-H cannot form enolates, so in strong alkali they disproportionate.

Teacher's Note:
1. Examples: HCHO, benzaldehyde.

Answer: (b) benzyl alcohol and sodium formate

Explanation:
1. This is a crossed Cannizzaro reaction.
2. Formaldehyde is the stronger hydride donor and is oxidised to formate; benzaldehyde is reduced to benzyl alcohol.

Teacher's Note:
1. In crossed Cannizzaro, HCHO is always oxidised.

Answer: (b) acetaldehyde

Explanation:
1. Aldol condensation of two molecules of acetaldehyde gives 3-hydroxybutanal, which on heating loses water to give crotonaldehyde.

Teacher's Note:
1. Crotonaldehyde = but-2-enal.

Answer: (c) \( \ce{R'-CH=N-R} \)

Explanation:
1. Primary amines condense with aldehydes to give imines (Schiff bases) with loss of water.

Teacher's Note:
1. This is a nucleophilic addition-elimination.

Answer: (c) 2,4-dinitrophenylhydrazine

Explanation:
1. Brady's reagent is a solution of 2,4-dinitrophenylhydrazine, which gives yellow-orange precipitates with aldehydes and ketones.

Teacher's Note:
1. Used as a test for the C=O group.

Answer: (d) \( \ce{CH3COCH3} \)

Explanation:
1. Forming an oxime shows a carbonyl group; no silver mirror shows it is not an aldehyde.
2. So it is acetone.

Teacher's Note:
1. C3H6O has isomers propanal and propanone.

Answer: (a) Fehling's solution

Explanation:
1. Aldehydes reduce Fehling's solution to red \( \ce{Cu2O} \) but ketones do not.

Teacher's Note:
1. Aromatic aldehydes do not reduce Fehling's but do reduce Tollens'.

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

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

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

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

Explanation:
1. Both contain the carbonyl group.
2. Aldehydes undergo nucleophilic addition while acid derivatives give nucleophilic acyl substitution; the reason is true but does not explain why the group is present.

Teacher's Note:
1. Addition versus substitution.