JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 28

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Multiple Choice Questions

Question 1: The aldehyde which reacts with NaOH to produce an alcohol and a sodium salt is
(a) \( \ce{HCHO} \)
(b) \( \ce{CH3CHO} \)
(c) \( \ce{CH3CH2CHO} \)
(d) \( \ce{CH3CH2CH2CHO} \)
Show Answer & Explanation

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

Explanation:
1. Only methanal lacks alpha-hydrogens and gives Cannizzaro products.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 2: The reaction \( \ce{OHC-CHO + OH^-} \) giving \( \ce{HOCH2-COO^-} \) is
(a) Aldol condensation
(b) Knoevenagel reaction
(c) Intramolecular Cannizzaro reaction
(d) None of these
Show Answer & Explanation

Answer: (c) Intramolecular Cannizzaro reaction

Explanation:
1. Glyoxal has no alpha-hydrogen; one CHO is oxidised and the other reduced within the same molecule, an internal Cannizzaro reaction.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 3: In Wolff-Kishner reduction, the carbonyl group of aldehydes and ketones is converted into
(a) \( \ce{>CH2} \) group
(b) \( \ce{-CH3} \) group
(c) \( \ce{-CH2OH} \) group
(d) \( \ce{>CHOH} \) group
Show Answer & Explanation

Answer: (a) \( \ce{>CH2} \) group

Explanation:
1. Hydrazone formation followed by loss of N2 replaces C=O by CH2.

Teacher's Note:
1. Clemmensen (Zn-Hg/HCl, acid) and Wolff-Kishner (N2H4/KOH, base) both convert C=O to CH2; choose by the acid or base sensitivity of the rest of the molecule.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 4: Which of the following react with \( \ce{NaHSO3} \)?
(a) \( \ce{CH3COCH3} \)
(b) \( \ce{CH3CHO} \)
(c) \( \ce{HCHO} \)
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Aldehydes and methyl ketones such as acetone form bisulphite addition compounds.

Teacher's Note:
1. Aldehydes and methyl ketones give crystalline bisulphite adducts; hindered ketones such as acetophenone do not.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 5: Fehling solution is
(a) \( \ce{CuSO4} \) + lime
(b) \( \ce{CuSO4} \)(aq)
(c) \( \ce{CuSO4 + Na2CO3} \)
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. Fehling's solution is copper sulphate with alkaline Rochelle salt; copper sulphate with sodium carbonate and citrate is Benedict's solution.

Teacher's Note:
1. Fehling's solution is reduced by aliphatic aldehydes (red Cu2O) but not by ketones or aromatic aldehydes.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 6: Wolff-Kishner reduction reduces
(a) -COOH group
(b) \( \ce{-C#C-} \) group
(c) -CHO group
(d) -O- group
Show Answer & Explanation

Answer: (c) -CHO group

Explanation:
1. It converts the carbonyl of aldehydes and ketones to CH2.

Teacher's Note:
1. Clemmensen (Zn-Hg/HCl, acid) and Wolff-Kishner (N2H4/KOH, base) both convert C=O to CH2; choose by the acid or base sensitivity of the rest of the molecule.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 7: A compound has a vapour density of 29. In the presence of iodine, on warming with an aqueous solution of alkali, it gives a yellow precipitate. The compound is
(a) \( \ce{CH3CH2CHO} \)
(b) \( \ce{CH3CHOHCH3} \)
(c) \( \ce{CH3COCH3} \)
(d) \( \ce{CH3CH2COOH} \)
Show Answer & Explanation

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

Explanation:
1. Molar mass is 2 x 29 = 58.
2. Acetone has M = 58 and gives iodoform.
3. Propanal is also 58 but gives no iodoform, and propan-2-ol and propanoic acid have M = 60 and 74.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 8: Which of the following responds to the positive iodoform test?
(a) Butanol
(b) Butan-1-al
(c) Butanol-2
(d) 3-Pentanone
Show Answer & Explanation

Answer: (c) Butanol-2

Explanation:
1. Butan-2-ol has the CH3CH(OH)- group, which is oxidised to the methyl ketone and gives iodoform.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 9: The pair of compounds in which both give a positive test with Tollens' reagent is
(a) Glucose and sucrose
(b) Fructose and sucrose
(c) Acetophenone and hexanal
(d) Glucose and fructose
Show Answer & Explanation

Answer: (d) Glucose and fructose

Explanation:
1. Glucose has an aldehyde group, and fructose isomerises in alkali to an aldose, so both reduce Tollens' reagent.
2. Sucrose has no free carbonyl and acetophenone is a ketone.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 10: The silver mirror test can be used to distinguish between
(a) Ketone and acid
(b) Phenol and acid
(c) Aldehyde and acid
(d) Alcohol and phenol
Show Answer & Explanation

Answer: (c) Aldehyde and acid

Explanation:
1. An aldehyde (other than methanoic acid) reduces Tollens' reagent, whereas a carboxylic acid does not.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 11: Paraldehyde is
(a) A trimer of formaldehyde
(b) A trimer of acetaldehyde
(c) A hexamer of formaldehyde
(d) A hexamer of acetaldehyde
Show Answer & Explanation

Answer: (b) A trimer of acetaldehyde

Explanation:
1. Acetaldehyde with a trace of acid trimerises to the cyclic compound paraldehyde.

Teacher's Note:
1. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 12: Paraldehyde is used as a
(a) Medicine (hypnotic)
(b) Poison
(c) Polymer
(d) Dye
Show Answer & Explanation

Answer: (a) Medicine (hypnotic)

Explanation:
1. Paraldehyde, the trimer of acetaldehyde, is used as a sedative and hypnotic.

Teacher's Note:
1. Aldol needs alpha-hydrogen; Cannizzaro needs none.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 13: Formalin is an aqueous solution of
(a) Formic acid
(b) Formaldehyde
(c) Furfuraldehyde
(d) Fluorescein
Show Answer & Explanation

Answer: (b) Formaldehyde

Explanation:
1. Formalin is about 40 percent aqueous methanal.

Teacher's Note:
1. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 14: The methyl ketone group is identified by
(a) Iodoform test
(b) Fehling solution
(c) Tollens' reagent
(d) Schiff's reagent
Show Answer & Explanation

Answer: (a) Iodoform test

Explanation:
1. The CH3CO- group gives yellow iodoform with iodine and alkali.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 15: Which of the following does not give the Fehling solution test?
(a) Acetone
(b) Propanal
(c) Ethanal
(d) Butanal
Show Answer & Explanation

Answer: (a) Acetone

Explanation:
1. Ketones do not reduce Fehling's solution, whereas aliphatic aldehydes do.

Teacher's Note:
1. Fehling's solution is reduced by aliphatic aldehydes (red Cu2O) but not by ketones or aromatic aldehydes.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 16: A compound A (\( \ce{C5H10Cl2} \)) gives \( \ce{C5H10O} \) on hydrolysis. The latter reacts with \( \ce{NH2OH} \), forms iodoform, but does not give Fehling's test. A is
(a) \( \ce{CH3-CCl2-CH2CH2CH3} \)
(b) \( \ce{CH3CH2-CCl2-CH2CH3} \)
(c) \( \ce{CH3CH2CH2CH2CHCl2} \)
(d) \( \ce{CH3-CHCl-CHCl-CH2CH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3-CCl2-CH2CH2CH3} \)

Explanation:
1. A gem-dichloride hydrolyses to a carbonyl compound.
2. A methyl ketone that gives iodoform and no Fehling is pentan-2-one, which arises from 2,2-dichloropentane.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 17: CH3CHO reacts with HCN to give A. Compound A on hydrolysis gives
(a) Propanoic acid
(b) Propylamine
(c) Pyruvic acid
(d) 2-Hydroxypropanoic acid (lactic acid)
Show Answer & Explanation

Answer: (d) 2-Hydroxypropanoic acid (lactic acid)

Explanation:
1. Acetaldehyde cyanohydrin, CH3CH(OH)CN, hydrolyses to CH3CH(OH)COOH.

Teacher's Note:
1. Aldol needs alpha-hydrogen; Cannizzaro needs none.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 18: Which one does not give the Cannizzaro reaction?
(a) Benzaldehyde
(b) 2-Methylpropanal
(c) p-Methoxybenzaldehyde
(d) 2,2-Dimethylpropanal
Show Answer & Explanation

Answer: (b) 2-Methylpropanal

Explanation:
1. 2-Methylpropanal has an alpha-hydrogen, so it undergoes aldol reactions rather than Cannizzaro reaction.

Teacher's Note:
1. Cannizzaro reaction: an aldehyde with no alpha-hydrogen, in strong alkali, gives equal amounts of alcohol and carboxylate through hydride transfer.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 19: Which of the following will fail to react with potassium dichromate and dilute sulphuric acid?
(a) Ethanol
(b) Ethanal
(c) Propan-2-ol
(d) Propanone
Show Answer & Explanation

Answer: (d) Propanone

Explanation:
1. Alcohols and aldehydes are oxidised.
2. A ketone like propanone resists mild oxidation.

Teacher's Note:
1. For multi-step conversions write the intermediate after each reagent.
2. Check whether a reagent attacks the carbonyl, an alcohol or a C=C bond first.

Question 20: Acetone and acetaldehyde are differentiated by
(a) \( \ce{NaOH + I2} \)
(b) \( \ce{Ag(NH3)2^+} \)
(c) \( \ce{HNO2} \)
(d) \( \ce{I2} \)
Show Answer & Explanation

Answer: (b) \( \ce{Ag(NH3)2^+} \)

Explanation:
1. Tollens' reagent is reduced by acetaldehyde, not by acetone.
2. Both give iodoform with NaOH and I2.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. Check whether a reagent attacks the carbonyl, an alcohol or a C=C bond first.

Question 21: Which of the following will react with water?
(a) \( \ce{CHCl3} \)
(b) \( \ce{Cl3CCHO} \)
(c) \( \ce{CCl4} \)
(d) \( \ce{ClCH2CH2Cl} \)
Show Answer & Explanation

Answer: (b) \( \ce{Cl3CCHO} \)

Explanation:
1. The electron-withdrawing CCl3 group makes the carbonyl so electrophilic that chloral adds water to form the stable hydrate, chloral hydrate.

Teacher's Note:
1. Draw each structure and mark alpha-hydrogens before choosing a named reaction.
2. For multi-step conversions write the intermediate after each reagent.

Question 22: An organic compound A has molecular formula \( \ce{C3H6O} \) and gives the iodoform test. On saturation with dilute HCl it gives B of molecular formula \( \ce{C9H14O} \). A and B respectively are
(a) Propanal and mesitylene
(b) Propanone and mesityl oxide
(c) Propanone and 2,6-dimethylhepta-2,5-dien-4-one
(d) Propanone and mesitylene oxide
Show Answer & Explanation

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

Explanation:
1. Acetone gives the iodoform test.
2. Three molecules of acetone lose two molecules of water to give phorone (\( \ce{C9H14O} \)), 2,6-dimethylhepta-2,5-dien-4-one.

Teacher's Note:
1. The iodoform test is positive for CH3CO- groups and CH3CH(OH)- groups (ethanol, propan-2-ol, ethanal, methyl ketones); methanal, benzophenone and 3-pentanone do not respond.
2. For multi-step conversions write the intermediate after each reagent.

Question 23: Compound A (\( \ce{C3H8O} \)) is treated with acidified potassium dichromate to form B (\( \ce{C3H6O} \)). B forms a shining silver mirror with ammoniacal silver nitrate. B with an aqueous solution of semicarbazide hydrochloride and sodium acetate gives C. The structure of C is
(a) \( \ce{CH3CH2CH=NNHCONH2} \)
(b) \( \ce{CH3C(CH3)=NNHCONH2} \)
(c) \( \ce{CH3CH=NCONHNH2} \)
(d) \( \ce{CH3CH2CH=NCONHNH2} \)
Show Answer & Explanation

Answer: (a) \( \ce{CH3CH2CH=NNHCONH2} \)

Explanation:
1. A is propan-1-ol and B is propanal, which gives a silver mirror.
2. Its semicarbazone is \( \ce{CH3CH2CH=N-NH-CO-NH2} \).

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. Draw each structure and mark alpha-hydrogens before choosing a named reaction.

Question 24: Which is not true about acetophenone?
(a) Reacts to form 2,4-dinitrophenylhydrazone
(b) Reacts with Tollens' reagent to form silver mirror
(c) Reacts with \( \ce{I2/NaOH} \) to form iodoform
(d) On oxidation with alkaline \( \ce{KMnO4} \) followed by hydrolysis gives benzoic acid
Show Answer & Explanation

Answer: (b) Reacts with Tollens' reagent to form silver mirror

Explanation:
1. Acetophenone is a ketone and does not reduce Tollens' reagent; the other statements are true.

Teacher's Note:
1. Tollens' reagent, [Ag(NH3)2]+, is reduced to silver by aldehydes and not by ketones.
2. For multi-step conversions write the intermediate after each reagent.

Question 25: The enol form of acetone, after treatment with \( \ce{D2O} \), gives
(a) \( \ce{CH3-C(OD)=CH2} \)
(b) \( \ce{CD3-CO-CD3} \)
(c) \( \ce{CH2=C(OH)-CH2D} \)
(d) \( \ce{CD2=C(OD)-CD3} \)
Show Answer & Explanation

Answer: (b) \( \ce{CD3-CO-CD3} \)

Explanation:
1. Repeated enolisation with D2O exchanges all six alpha-hydrogens, and the final keto form is acetone-d6, \( \ce{CD3COCD3} \).

Teacher's Note:
1. Check whether a reagent attacks the carbonyl, an alcohol or a C=C bond first.
2. Draw each structure and mark alpha-hydrogens before choosing a named reaction.

Question 26: The appropriate reagent for converting (3-hydroxycyclopentyl) methyl ketone to 3-ethylcyclopentanol is
(a) Zn(Hg), HCl
(b) \( \ce{NH2NH2/OH^-} \)
(c) \( \ce{H2/Ni} \)
(d) \( \ce{NaBH4} \)
Show Answer & Explanation

Answer: (b) \( \ce{NH2NH2/OH^-} \)

Explanation:
1. The acyl group must become ethyl without affecting OH. Wolff-Kishner is basic and leaves the alcohol, whereas acid in Clemmensen could dehydrate it, and H2/Ni or NaBH4 only give an alcohol.

Teacher's Note:
1. Clemmensen (Zn-Hg/HCl, acid) and Wolff-Kishner (N2H4/KOH, base) both convert C=O to CH2; choose by the acid or base sensitivity of the rest of the molecule.
2. Draw each structure and mark alpha-hydrogens before choosing a named reaction.

Question 27: Which of the following has the most acidic hydrogen?
(a) 3-Hexanone
(b) 2,4-Hexanedione
(c) 2,5-Hexanedione
(d) 2,3-Hexanedione
Show Answer & Explanation

Answer: (b) 2,4-Hexanedione

Explanation:
1. In a 1,3-diketone the CH2 between the two carbonyls gives a doubly stabilised enolate, so 2,4-hexanedione is most acidic.

Teacher's Note:
1. For multi-step conversions write the intermediate after each reagent.
2. Check whether a reagent attacks the carbonyl, an alcohol or a C=C bond first.

Question 28: Which of the following will be most readily dehydrated in acidic conditions?
(a) 4-Hydroxypentan-2-one
(b) Pentan-2-ol
(c) 3-Hydroxypentan-2-one
(d) 5-Hydroxyhexan-2-one
Show Answer & Explanation

Answer: (a) 4-Hydroxypentan-2-one

Explanation:
1. A beta-hydroxy ketone loses water easily because the alkene formed is conjugated with the carbonyl (enone).

Teacher's Note:
1. Check whether a reagent attacks the carbonyl, an alcohol or a C=C bond first.
2. Draw each structure and mark alpha-hydrogens before choosing a named reaction.

Question 29: Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is
(a) MeCOCl
(b) MeCHO
(c) MeCOOMe
(d) MeCOOCOMe
Show Answer & Explanation

Answer: (a) MeCOCl

Explanation:
1. The chlorine in acetyl chloride is strongly electron-withdrawing and there is little resonance donation from it, so the carbonyl carbon is the most electrophilic.

Teacher's Note:
1. Draw each structure and mark alpha-hydrogens before choosing a named reaction.
2. For multi-step conversions write the intermediate after each reagent.

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