JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 13

Chemistry Objective Questions and Answers: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: The reaction by which benzaldehyde cannot be prepared is:
(a) benzoic acid + Zn/Hg and conc. HCl
(b) toluene + CrO2Cl2 in CS2, followed by H3O+
(c) benzoyl chloride + H2 over Pd/BaSO4
(d) benzene + CO + HCl in the presence of anhydrous AlCl3
Show Answer & Explanation

Answer: (a) benzoic acid + Zn/Hg and conc. HCl

Explanation:
1. Zn/Hg and HCl (Clemmensen conditions) reduce carbonyl groups and would not stop at the aldehyde; benzoic acid is not reduced this way. The other three are the Etard, Rosenmund and Gattermann–Koch reactions.

Teacher's Note:
1. Etard, Rosenmund and Gattermann–Koch all give benzaldehyde.

Question 2: Salicylic acid can be easily made by the reaction between:
(a) phenol and CO2
(b) benzoic acid and H2O2
(c) benzenediazonium chloride and CO2
(d) phenol and formic acid
Show Answer & Explanation

Answer: (a) phenol and CO2

Explanation:
1. Sodium phenoxide with CO2 under heat and pressure gives sodium salicylate (Kolbe–Schmidt reaction), which on acidification gives salicylic acid.

Teacher's Note:
1. Phenol plus CO2 under pressure.

Question 3: Toluene is converted into benzaldehyde by reagent P, and benzaldehyde is converted into Q by Zn–Hg/HCl. P and Q, respectively, are:
(a) CrO3/H3O+, benzyl alcohol
(b) CrO2Cl2/H3O+, toluene
(c) NaOCl/heat, toluene
(d) KMnO4/H+, benzyl alcohol
Show Answer & Explanation

Answer: (b) CrO2Cl2/H3O+, toluene

Explanation:
1. CrO2Cl2 (Etard reaction) oxidises the methyl group to –CHO and stops there. Clemmensen reduction converts –CHO into –CH3, giving back toluene.

Teacher's Note:
1. Etard reaction is controlled oxidation of toluene.

Question 4: CH2=CH–CH2–CHO is converted into CH3–CH2–CH2–CH2–OH. The appropriate reagent is:
(a) metal + H2
(b) LiAlH4
(c) NaBH4
(d) all of these
Show Answer & Explanation

Answer: (a) metal + H2

Explanation:
1. Both C=C and C=O must be reduced. Catalytic hydrogenation (metal and H2) does this; LiAlH4 and NaBH4 reduce only C=O and leave the isolated C=C.

Teacher's Note:
1. Metal hydrides do not reduce isolated C=C.

Question 5: Benzoyl chloride is converted into benzaldehyde by P. P will be:
(a) Pd/BaSO4 + H2
(b) LiAlH4
(c) NaBH4
(d) Pd + H2
Show Answer & Explanation

Answer: (a) Pd/BaSO4 + H2

Explanation:
1. A poisoned catalyst (Rosenmund) is needed to stop at the aldehyde. LiAlH4 and NaBH4 reduce further to the alcohol.

Teacher's Note:
1. Use partially poisoned Pd.

Question 6: Cyclohexanone undergoes aldol condensation followed by heating. The product is:
(a) 2-cyclohexylidenecyclohexan-1-one
(b) 2-cyclohexylcyclohexan-1-ol
(c) 2-(cyclohexylidene)cyclohexane-1,3-dione
(d) 2-(1-hydroxycyclohexyl)cyclohexan-1-one
Show Answer & Explanation

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

Explanation:
1. Aldol addition gives the β-hydroxy ketone, 2-(1-hydroxycyclohexyl)cyclohexanone. Heating dehydrates it to the conjugated enone, 2-cyclohexylidenecyclohexanone.

Teacher's Note:
1. Heating the aldol product gives the condensation product.

Question 7: Ph–CHO + (CH3CO)2O (CH3COONa, CH3COOH, heat) → A → (LiAlH4, H2O) B. B is:
(a) Ph–CH2–CH2–CH2–OH
(b) Ph–CH=CH–CH2–OH
(c) Ph–CH2–CH2–COOH
(d) Ph–CH=CH–CHO
Show Answer & Explanation

Answer: (b) Ph–CH=CH–CH2–OH

Explanation:
1. Perkin reaction gives cinnamic acid (A). LiAlH4 reduces the –COOH to –CH2OH without touching the conjugated C=C, giving cinnamyl alcohol, PhCH=CHCH2OH (B).

Teacher's Note:
1. LiAlH4 reduces acids to primary alcohols.

Question 8: 1-Ethyl-3-methylbenzene is heated with KMnO4. The product is:
(a) benzene-1,3-dicarboxylic acid (isophthalic acid)
(b) 3-(carboxymethyl)benzoic acid
(c) 3-methylbenzoic acid
(d) 3-ethylbenzoic acid
Show Answer & Explanation

Answer: (a) benzene-1,3-dicarboxylic acid (isophthalic acid)

Explanation:
1. KMnO4 oxidises every alkyl side chain that has a benzylic hydrogen to a –COOH group, so both the ethyl and methyl groups become –COOH.

Teacher's Note:
1. The whole side chain, whatever its length, is oxidised to –COOH.

Question 9: Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. This is due to their:
(a) formation of intramolecular H-bonding
(b) formation of carboxylate ion
(c) more extensive association through van der Waals forces
(d) formation of intermolecular H-bonding (dimers)
Show Answer & Explanation

Answer: (d) formation of intermolecular H-bonding (dimers)

Explanation:
1. Carboxylic acids form strong intermolecular hydrogen bonds and exist as dimers in the liquid and vapour state.

Teacher's Note:
1. Dimerisation roughly doubles the effective molar mass.

Question 10: Compound A, C8H10O, reacts with NaOI (from Y and NaOH) to give a yellow precipitate with a characteristic smell. A and Y are:
(a) 4-methylbenzyl alcohol and I2
(b) 2-phenylethanol and I2
(c) 1-phenylethanol and I2
(d) 2,4-dimethylphenol and I2
Show Answer & Explanation

Answer: (c) 1-phenylethanol and I2

Explanation:
1. A positive iodoform test needs a CH3CH(OH)– or CH3CO– group. Of the isomers, 1-phenylethanol, C6H5CH(OH)CH3, has it. The reagent NaOI comes from I2 and NaOH.

Teacher's Note:
1. Iodoform is the yellow precipitate with antiseptic smell.

Question 11: Benzoic acid is converted into propylbenzene. The correct sequence of reagents is:
(a) SOCl2, Cd(C2H5)2, NH2NH2/OH−
(b) SOCl2, C2H5MgBr, Zn–Hg/HCl
(c) SOCl2, CH3Li, NH2NH2/OH−
(d) Cd(CH3)2, SOCl2, NH2NH2/OH−
Show Answer & Explanation

Answer: (a) SOCl2, Cd(C2H5)2, NH2NH2/OH−

Explanation:
1. SOCl2 gives benzoyl chloride. Diethylcadmium converts it into propiophenone, C6H5COC2H5, and a Wolff–Kishner reduction gives propylbenzene, C6H5CH2CH2CH3.

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

Question 12: 3-Cyanobenzoic acid is treated with B2H6 to give A, which on treatment with DIBAL-H gives B. B is:
(a) 3-(hydroxymethyl)benzaldehyde
(b) 3-(hydroxymethyl)benzyl alcohol
(c) benzene-1,3-dicarbaldehyde
(d) 3-cyanobenzyl alcohol
Show Answer & Explanation

Answer: (a) 3-(hydroxymethyl)benzaldehyde

Explanation:
1. Diborane selectively reduces the –COOH group to –CH2OH without touching –CN, giving 3-cyanobenzyl alcohol (A). DIBAL-H then reduces –CN to –CHO after hydrolysis, giving 3-(hydroxymethyl)benzaldehyde.

Teacher's Note:
1. B2H6 reduces carboxylic acids selectively.

Question 13:
Assertion (A): Acetamide on reaction with Br2/NaOH gives CH3NH2.
Reason (R): The carbon of the amide functional group becomes part of Na2CO3.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both are true. In the Hofmann reaction the carbonyl carbon is lost as carbonate, which is why the amine has one carbon less than the amide.

Teacher's Note:
1. This is the Hofmann bromamide degradation.

Question 14: A carbonyl compound gives a pink colour with Schiff's reagent and a yellow precipitate when boiled with iodine and caustic alkali. It also gives a red precipitate with Fehling's solution. It is likely to be:
(a) formaldehyde
(b) propionaldehyde
(c) acetaldehyde
(d) crotonaldehyde
Show Answer & Explanation

Answer: (c) acetaldehyde

Explanation:
1. Schiff's test, Fehling's test: aldehyde. Iodoform test: CH3CO– group. Only acetaldehyde meets all three.

Teacher's Note:
1. Only acetaldehyde is an aldehyde with a methyl ketone group.

Question 15: Which of the following statements is wrong?
(a) All methyl ketones give a positive iodoform test
(b) Acetaldehyde is the only aldehyde that gives the iodoform test
(c) All secondary alcohols give a positive iodoform test
(d) Any alcohol that can be oxidised to an acetyl group gives a positive iodoform test
Show Answer & Explanation

Answer: (c) All secondary alcohols give a positive iodoform test

Explanation:
1. Only secondary alcohols of the type CH3CH(OH)R give iodoform, because only they are oxidised to a methyl ketone.

Teacher's Note:
1. Isopropyl alcohol gives iodoform but 3-pentanol does not.

Question 16: In the sequence acetone → (I2/alkali) A → (Ag, heat) B → (dil. H2SO4, Hg2+) C, the compounds A, B and C are:
(a) CHI3, H2C=CH2, CH3CH2OH
(b) CHI3, HC≡CH, CH3CHO
(c) CHI3, CH3–C≡CH, CH3COCH3
(d) CHI3, (CH3)2CHOH, CH3COCH3
Show Answer & Explanation

Answer: (b) CHI3, HC≡CH, CH3CHO

Explanation:
1. Haloform reaction gives iodoform (A). Heating iodoform with silver gives ethyne (B). Hydration of ethyne gives acetaldehyde (C).

Teacher's Note:
1. Iodoform plus silver powder gives acetylene.

Question 17: In the sequence RCOCl + H2 → (Pd–BaSO4) A → (HCN) B → (H3O+) C, the compounds A, B and C are:
(a) RCHO, RCH(OH)CN, RCH(OH)CH2NH2
(b) RCHO, RCH(OH)CN, RCH(OH)COOH
(c) RCHO, RCOCH2CN, RCOCH2COOH
(d) RCHO, RCH2CN, RCH2COOH
Show Answer & Explanation

Answer: (b) RCHO, RCH(OH)CN, RCH(OH)COOH

Explanation:
1. Rosenmund reduction gives the aldehyde (A), HCN adds to give the cyanohydrin (B), and hydrolysis gives the α-hydroxy acid (C).

Teacher's Note:
1. Cyanohydrin hydrolysis gives α-hydroxy acids.

Question 18: In the sequence propyne → (dil. H2SO4, Hg2+) A → (CHCl3) B, and A → (I2/NaOH) C, the compounds A, B and C are:
(a) CH3CH2CHO, CH3CH2CH2Cl, CHI3
(b) CH3COCH3, CCl3–CO–CH3, CHI3
(c) CH3COCH3, CCl3–C(CH3)2OH, CHI3
(d) CH3CH2CHO, CCl3–CH(OH)CH2CH3, CHI3
Show Answer & Explanation

Answer: (c) CH3COCH3, CCl3–C(CH3)2OH, CHI3

Explanation:
1. Hydration gives acetone (A). Acetone with chloroform (base) gives chloretone, CCl3C(CH3)2OH (B). The haloform reaction with I2/NaOH gives iodoform (C).

Teacher's Note:
1. Chloroform adds to the carbonyl group in base.

Question 19: In the sequence CH3COOC2H5 → (A) → CH3COOC(CH3)3 and → (B) → CH3CONHNH2, the reagents A and B are:
(a) isopropyl alcohol, H2N–NH2
(b) isopropyl alcohol, H2N–OH
(c) tert-butyl alcohol, H2N–NH2
(d) tert-butyl alcohol, H2N–OH
Show Answer & Explanation

Answer: (c) tert-butyl alcohol, H2N–NH2

Explanation:
1. Transesterification with tert-butyl alcohol gives the tert-butyl ester. Hydrazine replaces –OC2H5 to give the hydrazide CH3CONHNH2.

Teacher's Note:
1. Hydroxylamine would give a hydroxamic acid.

Question 20: Methylamine reacts with acetyl chloride to form:
(a) CH3NH2 (no reaction)
(b) CH3NHNa
(c) CH3NHCOCH3
(d) (CH3)2NCOCH3
Show Answer & Explanation

Answer: (c) CH3NHCOCH3

Explanation:
1. CH3NH2 + CH3COCl → CH3NHCOCH3 + HCl. The product is N-methylacetamide.

Teacher's Note:
1. Acylation of a primary amine gives an N-substituted amide.

Question 21: The intermediates formed during the reaction of RCONH2 with Br2 and KOH are:
(a) RCONHBr and RNCO
(b) RNHCOBr and RNCO
(c) RNH–Br and RCONHBr
(d) RCONBr2
Show Answer & Explanation

Answer: (a) RCONHBr and RNCO

Explanation:
1. N-bromination gives RCONHBr, which loses HBr to form the anion; migration of R gives the isocyanate RN=C=O. Hydrolysis gives RNH2.

Teacher's Note:
1. Hofmann bromamide rearrangement proceeds through an isocyanate.

Question 22: In the sequence CH3CH2Cl → (NaCN) X → (Ni/H2) Y → (acetic anhydride) Z, Z is:
(a) CH3CH2CH2NHCOCH3
(b) CH3CH2CH2NH2
(c) CH3CH2CH2CONHCH3
(d) CH3CH2CH2CONHCOCH3
Show Answer & Explanation

Answer: (a) CH3CH2CH2NHCOCH3

Explanation:
1. NaCN gives propanenitrile (X), hydrogenation gives propylamine (Y), and acetylation gives N-propylacetamide, CH3CH2CH2NHCOCH3.

Teacher's Note:
1. Amines are acetylated by acetic anhydride.

Question 23:
Assertion (A): Benzaldehyde with alkaline CuSO4 (Fehling's solution) gives a red-brown precipitate.
Reason (R): The red precipitate is due to Cu2O.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Aromatic aldehydes like benzaldehyde do not reduce Fehling's solution, so the assertion is false.
2. The red precipitate formed by aliphatic aldehydes is indeed Cu2O, so the reason is true.

Teacher's Note:
1. Aromatic aldehydes do not give Fehling's test.

Question 24: Butanenitrile, CH3CH2CH2C≡N, is hydrolysed with H3O+ to A, which is heated with NH3 to give B. Product B is:
(a) CH3CH2CH2COOH
(b) CH3CH2CH2CHO
(c) CH3CH2CH2NC
(d) CH3CH2CH2CONH2
Show Answer & Explanation

Answer: (d) CH3CH2CH2CONH2

Explanation:
1. Acid hydrolysis gives butanoic acid (A). The ammonium salt loses water on heating to give butanamide (B).

Teacher's Note:
1. Ammonium carboxylates dehydrate to amides.

Question 25: Sodium ethoxide reacts with ethanoyl chloride. The compound produced is:
(a) ethyl chloride
(b) ethyl ethanoate
(c) diethyl ether
(d) butan-2-one
Show Answer & Explanation

Answer: (b) ethyl ethanoate

Explanation:
1. C2H5ONa + CH3COCl → CH3COOC2H5 + NaCl. This is nucleophilic acyl substitution.

Teacher's Note:
1. The product is an ester.

Question 26: Trichloroacetaldehyde (chloral) undergoes a Cannizzaro reaction with NaOH. The mixture of products contains sodium trichloroacetate and another compound. The other compound is:
(a) 2,2,2-trichloropropanol
(b) chloroform
(c) 2,2,2-trichloroethanol
(d) trichloromethanol
Show Answer & Explanation

Answer: (c) 2,2,2-trichloroethanol

Explanation:
1. CCl3CHO has no α-hydrogen. One molecule is oxidised to CCl3COONa and the other is reduced to CCl3CH2OH, 2,2,2-trichloroethanol.

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

Question 27: Sodium phenoxide is heated with CO2 under pressure at 398 K to give B, and B is acidified and treated with acetic anhydride to give C. The major product C is:
(a) methyl salicylate
(b) 4-acetoxybenzoic acid
(c) 2-acetoxybenzoic acid (aspirin)
(d) 2,4-diacetylphenol
Show Answer & Explanation

Answer: (c) 2-acetoxybenzoic acid (aspirin)

Explanation:
1. Kolbe–Schmidt reaction gives salicylic acid, which is acetylated at the phenolic –OH by acetic anhydride, giving aspirin.

Teacher's Note:
1. Aspirin is acetylsalicylic acid.

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Multiple Choice Questions (MCQs) for JEE Chemistry Aldehydes Ketones and Carboxylic Acids

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