JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 34

Answer: (b) 2-cyclohexylidenecyclohexan-1-one

Explanation:
1. PCC oxidises the secondary alcohol to cyclohexanone (A).
2. Dilute NaOH causes aldol addition of two molecules of cyclohexanone, and on heating water is lost to give the conjugated enone.
3. The C=C is between the alpha carbon of one ring and the carbonyl-derived carbon of the other, giving 2-cyclohexylidenecyclohexanone.

Teacher's Note:
1. Heating the aldol product gives the condensation product.
2. The double bond forms in conjugation with C=O.

Answer: (a) \( \ce{B2H6} \)/diglyme, then \( \ce{H2O2/NaOH} \), then PCC in \( \ce{CH2Cl2} \)

Explanation:
1. Hydroboration-oxidation gives the anti-Markovnikov alcohol \( \ce{C6H5CH2CH2OH} \).
2. PCC oxidises the primary alcohol only up to the aldehyde.

Teacher's Note:
1. Anti-Markovnikov hydration places OH on the terminal carbon.
2. PCC stops at the aldehyde stage.

Answer: (b) Gattermann aldehyde synthesis

Explanation:
1. HCN and HCl with \( \ce{ZnCl2} \) introduce the CHO group on a phenol or phenolic ether ring, which is the Gattermann reaction.
2. The Gattermann-Koch reaction uses CO and HCl for benzene and alkylbenzenes.

Teacher's Note:
1. Gattermann uses HCN, Gattermann-Koch uses CO.
2. Perkin reaction gives cinnamic acids.

Answer: (c) PCC in \( \ce{CH2Cl2} \)

Explanation:
1. The primary alcohol must stop at the aldehyde stage without affecting the double bond.
2. PCC in dichloromethane does this.
3. Dichromate would over-oxidise to the acid, and \( \ce{OsO4} \) would hydroxylate the double bond.

Teacher's Note:
1. PCC and Collins reagent give aldehydes from primary alcohols.
2. Strong oxidants give acids.

Answer: (c) cyclopentanone

Explanation:
1. Oxidation cleaves the ring to give hexanedioic (adipic) acid (A).
2. The calcium salt of adipic acid on heating loses calcium carbonate and cyclises to cyclopentanone (Ruzicka-type pyrolysis).

Teacher's Note:
1. Adipic acid gives a five-membered ring ketone.
2. Pimelic acid would give cyclohexanone.

Answer: (c) iodoform

Explanation:
1. Acetone has the \( \ce{CH3CO-} \) group, so it gives the iodoform reaction.
2. The product is yellow iodoform \( \ce{CHI3} \) together with sodium acetate.

Teacher's Note:
1. Methyl ketones give the iodoform test.
2. The trihalogenated ketone is cleaved by base.

Answer: (b) piperidine

Explanation:
1. The oxime of cyclopentanone undergoes the Beckmann rearrangement to the six-membered lactam, piperidin-2-one (B).
2. \( \ce{LiAlH4} \) reduces the amide C=O completely to CH2, giving piperidine.

Teacher's Note:
1. Beckmann rearrangement expands the ring by inserting N.
2. LiAlH4 converts amides to amines.

Answer: (c) a = \( \ce{Br2/H+} \), b = alc. KOH

Explanation:
1. Acid-catalysed bromination replaces the alpha hydrogen (here the ring carbon bearing CHO) by bromine.
2. Alcoholic KOH then eliminates HBr to give the conjugated alkene.

Teacher's Note:
1. Alcoholic KOH favours elimination, aqueous KOH favours substitution.
2. \( \ce{Br2/CCl4} \) would add to a C=C, not substitute.

Answer: (a) 4-bromomandelonitrile, \( \ce{4-BrC6H4CH(OH)CN} \)

Explanation:
1. Cyanide adds to the carbonyl carbon (nucleophilic addition) and protonation gives the cyanohydrin.
2. The aryl bromide is untouched.

Teacher's Note:
1. Cyanohydrins form with HCN or NaCN with acid.
2. The ring bromine is inert here.

Answer: (c) \( \ce{CH3COOH} \)

Explanation:
1. Heating ammonium acetate gives acetamide (X).
2. \( \ce{P2O5} \) dehydrates the amide to methyl cyanide (Y).
3. Acid hydrolysis of the nitrile gives acetic acid (Z).

Teacher's Note:
1. Amide dehydrates to nitrile.
2. Nitrile hydrolyses to acid.

Answer: (a) butanoyl chloride

Explanation:
1. Hofmann bromamide degradation removes one carbon as \( \ce{CO2} \).
2. Propylamine must come from butanamide, which comes from butanoyl chloride.

Teacher's Note:
1. Hofmann degradation shortens the chain by one carbon.
2. 2-Methylpropanamide would give isopropylamine.

Answer: (b) allyl 2-hydroxypropanoate

Explanation:
1. An acid and an alcohol with a mineral acid catalyst give an ester (Fischer esterification).
2. The free secondary OH of lactic acid is unaffected, so X is \( \ce{CH3CH(OH)COOCH2CH=CH2} \).

Teacher's Note:
1. Esterification joins the acid carbonyl to the alcohol oxygen.
2. The alcohol part keeps its double bond.

Answer: (d) \( \ce{CH3CHO} \) (\( \ce{H2SO4} \))

Explanation:
1. Acid chlorides, acetic acid with a mineral acid catalyst, and anhydrides all acylate the alcohol to give butyl acetate.
2. An aldehyde does not give an ester with an alcohol under acid conditions (it gives an acetal).

Teacher's Note:
1. Acyl derivatives esterify alcohols.
2. Aldehydes give acetals.

Answer: (c) acetaldehyde

Explanation:
1. The Cannizzaro reaction requires an aldehyde with no alpha hydrogen.
2. Acetaldehyde has alpha hydrogens and undergoes aldol reaction instead.
3. The others have no alpha hydrogen.

Teacher's Note:
1. No alpha-H means Cannizzaro.
2. Aldehydes with alpha-H give aldol.

Answer: (a) cyclopent-1-ene-1-carbaldehyde

Explanation:
1. An intramolecular aldol reaction occurs between the alpha carbon of one CHO and the other CHO group, forming a five-membered ring.
2. Dehydration gives the conjugated cyclopent-1-ene-1-carbaldehyde.

Teacher's Note:
1. Five- and six-membered rings form easily in intramolecular aldols.
2. Count the ring atoms formed.

Answer: (a) \( \ce{CH3CH2CH=N-NHCONH2} \)

Explanation:
1. B gives a silver mirror, so it is propanal, and A is propan-1-ol.
2. Propanal condenses with semicarbazide to give the semicarbazone \( \ce{CH3CH2CH=NNHCONH2} \), with the bond through the NH2 of the hydrazine part.

Teacher's Note:
1. The amide NH2 of semicarbazide is not nucleophilic.
2. Semicarbazones are crystalline derivatives.

Answer: (c) propanone and 2,6-dimethylhepta-2,5-dien-4-one (phorone)

Explanation:
1. A gives the iodoform test and has formula \( \ce{C3H6O} \), so it is acetone.
2. Dry HCl condenses three molecules of acetone with loss of two water molecules, giving phorone, \( \ce{C9H14O} \).
3. Mesityl oxide is \( \ce{C6H10O} \) and mesitylene is \( \ce{C9H12} \).

Teacher's Note:
1. Two molecules give mesityl oxide, three give phorone.
2. Check the molecular formulae.

Answer: (a) cyclopentanecarbaldehyde

Explanation:
1. NaBH4 gives cyclopentanol, HBr gives cyclopentyl bromide, and the Grignard reagent adds to formaldehyde to give cyclopentylmethanol (C).
2. PCC oxidises the primary alcohol to cyclopentanecarbaldehyde (D).

Teacher's Note:
1. Grignard plus HCHO gives a primary alcohol with one more carbon.
2. PCC stops at the aldehyde.

Answer: (a) diethyl carbonate

Explanation:
1. The enolate of the ketone attacks diethyl carbonate and loses ethoxide.
2. This adds an ester group directly on the alpha carbon.

Teacher's Note:
1. This is a crossed Claisen condensation.
2. Ethyl formate would add CHO.

Answer: (a) (4-tert-butylphenyl)acetic acid

Explanation:
1. NBS brominates the benzylic methyl group.
2. The Grignard reagent of the benzyl bromide adds to \( \ce{CO2} \) to give a carboxylic acid with one more carbon.

Teacher's Note:
1. Benzylic bromination keeps the ring intact.
2. The acid is a phenylacetic acid.

Answer: (a) heptan-4-one

Explanation:
1. Claisen condensation gives the beta-keto ester \( \ce{CH3CH2CH2COCH(C2H5)COOEt} \).
2. Hydrolysis and heating decarboxylate the beta-keto acid, giving \( \ce{CH3CH2CH2COCH2CH2CH3} \), heptan-4-one.

Teacher's Note:
1. Beta-keto acids lose \( \ce{CO2} \) on heating.
2. The ketone carbon comes from the first ester carbonyl.

Answer: (a) 2,2-dimethylpropane-1,3-diol

Explanation:
1. Two successive alkylations at the active methylene give diethyl dimethylmalonate.
2. \( \ce{LiAlH4} \) reduces both ester groups to primary alcohols, giving \( \ce{HOCH2C(CH3)2CH2OH} \).

Teacher's Note:
1. Malonic ester has two acidic hydrogens.
2. LiAlH4 reduces esters to alcohols.