JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 26

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Chapter-wise Objective Questions: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: The reaction \( \ce{C6H5CHO + CH3CHO} \) giving \( \ce{C6H5CH=CH-CHO} \) is known as
(a) Perkin's reaction
(b) Claisen-Schmidt condensation
(c) Benzoin condensation
(d) Cannizzaro's reaction
Show Answer & Explanation

Answer: (b) Claisen-Schmidt condensation

Explanation:
1. A crossed aldol condensation between an aromatic aldehyde (no alpha-hydrogen) and an aliphatic aldehyde or ketone is the Claisen-Schmidt condensation, giving cinnamaldehyde here.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 2: When two molecules of acetaldehyde condense in the presence of dilute alkali, it forms
(a) Acetal
(b) Sodium formate
(c) Aldol
(d) Mesitylene
Show Answer & Explanation

Answer: (c) Aldol

Explanation:
1. The product is 3-hydroxybutanal, called aldol.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 3: Acetaldehyde on treatment with dilute NaOH followed by heating gives
(a) \( \ce{CH3CH2CH2CH2OH} \)
(b) \( \ce{CH3CH2CH2CHO} \)
(c) \( \ce{CH3-CH=CHCHO} \)
(d) \( \ce{CH3-CH=CHCH2OH} \)
Show Answer & Explanation

Answer: (c) \( \ce{CH3-CH=CHCHO} \)

Explanation:
1. Aldol on heating loses water to give crotonaldehyde, but-2-enal.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 4: Benzaldehyde on reaction with acetophenone in the presence of sodium hydroxide solution gives
(a) \( \ce{C6H5CH=CHCOC6H5} \)
(b) \( \ce{C6H5COCH2C6H5} \)
(c) \( \ce{C6H5CH=CHC6H5} \)
(d) \( \ce{C6H5CH(OH)COC6H5} \)
Show Answer & Explanation

Answer: (a) \( \ce{C6H5CH=CHCOC6H5} \)

Explanation:
1. Claisen-Schmidt condensation gives chalcone, 1,3-diphenylprop-2-en-1-one.

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 5: Aldehydes and ketones can be converted to hydrocarbons by
(a) \( \ce{LiAlH4} \)
(b) \( \ce{H2/Pd-BaSO4} \)
(c) \( \ce{Na-Hg/HCl} \)
(d) \( \ce{NH2NH2/C2H5ONa} \)
Show Answer & Explanation

Answer: (d) \( \ce{NH2NH2/C2H5ONa} \)

Explanation:
1. Wolff-Kishner reduction (hydrazine with ethoxide) removes the carbonyl oxygen.
2. LiAlH4 gives alcohols, Pd-BaSO4 acts on acid chlorides, and Na-Hg is not a Clemmensen reagent.

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. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 6: An important reaction of acetone is autocondensation in presence of concentrated sulphuric acid to give the aromatic compound
(a) Mesitylene
(b) Trioxan
(c) Mesityl oxide
(d) Phorone
Show Answer & Explanation

Answer: (a) Mesitylene

Explanation:
1. Three acetone molecules condense with loss of three water molecules to give 1,3,5-trimethylbenzene (mesitylene).

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 7: Identify the organic compound which on heating with strong solution of NaOH is partly converted into an acid salt and partly into alcohol
(a) Benzyl alcohol
(b) Acetaldehyde
(c) Acetone
(d) Benzaldehyde
Show Answer & Explanation

Answer: (d) Benzaldehyde

Explanation:
1. Benzaldehyde undergoes Cannizzaro reaction, giving sodium benzoate and benzyl alcohol.

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. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 8: Which of the following does not give a brick red precipitate with Fehling solution?
(a) Acetone
(b) Acetaldehyde
(c) Formalin
(d) D-Glucose
Show Answer & Explanation

Answer: (a) Acetone

Explanation:
1. Ketones such as acetone do not reduce Fehling'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 9: Acetaldehyde and acetone can be distinguished by
(a) Molisch test
(b) Bromoform test
(c) Solubility in water
(d) Tollens' test
Show Answer & Explanation

Answer: (d) Tollens' test

Explanation:
1. Acetaldehyde reduces Tollens' reagent to silver; acetone does not.
2. Both give the haloform test and both dissolve in water.

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 10: Which compound is soluble in \( \ce{H2O} \)?
(a) \( \ce{HCHO} \)
(b) \( \ce{CH3CHO} \)
(c) \( \ce{CH3COCH3} \)
(d) All
Show Answer & Explanation

Answer: (d) All

Explanation:
1. Low members of aldehydes and ketones hydrogen-bond to water and mix with it.

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 11: Which gives lactic acid after hydrolysis with HCN?
(a) \( \ce{HCHO} \)
(b) \( \ce{CH3CHO} \)
(c) \( \ce{C6H5CHO} \)
(d) \( \ce{CH3COCH3} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3CHO} \)

Explanation:
1. Acetaldehyde cyanohydrin hydrolyses to 2-hydroxypropanoic acid (lactic acid).

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: \( \ce{CH3CHO} \) reacts with aqueous NaOH solution to form
(a) 3-Hydroxybutanal
(b) 2-Hydroxybutanal
(c) 4-Hydroxybutanal
(d) 3-Hydroxybutanol
Show Answer & Explanation

Answer: (a) 3-Hydroxybutanal

Explanation:
1. Aldol addition gives \( \ce{CH3CH(OH)CH2CHO} \).

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 13: Fehling's solution reacts with \( \ce{HCHO} \) to form a precipitate of
(a) White colour
(b) Yellow colour
(c) Red colour
(d) Blue colour
Show Answer & Explanation

Answer: (c) Red colour

Explanation:
1. Methanal reduces Cu(II) to red \( \ce{Cu2O} \).

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 14: The product of \( \ce{CH3MgI + HCHO} \) after hydrolysis is
(a) \( \ce{CH3CHO} \)
(b) \( \ce{CH3OH} \)
(c) \( \ce{C2H5OH} \)
(d) \( \ce{CH3-O-CH3} \)
Show Answer & Explanation

Answer: (c) \( \ce{C2H5OH} \)

Explanation:
1. Methanal gives a primary alcohol with one more carbon than the Grignard alkyl group: ethanol.

Teacher's Note:
1. Grignard reagents give a secondary alcohol with aldehydes (primary with HCHO) and a tertiary alcohol with ketones.
2. Use the tests: Tollens' and Fehling's for aldehydes, iodoform for CH3CO- and CH3CH(OH)-, 2,4-DNP for any carbonyl.

Question 15: A on heating at 800°C gives \( \ce{CH2=C=O} \). The reactant A is
(a) \( \ce{CH3CH2CHO} \)
(b) \( \ce{CH3CHO} \)
(c) \( \ce{CH3COCH3} \)
(d) \( \ce{C2H5OH} \)
Show Answer & Explanation

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

Explanation:
1. Pyrolysis of acetone gives ketene and methane.

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 16: Only an aldehyde or ketone having which of the following can undergo aldol condensation?
(a) At least one beta hydrogen
(b) At least one alpha hydrogen
(c) An aromatic ring
(d) No alpha hydrogen
Show Answer & Explanation

Answer: (b) At least one alpha hydrogen

Explanation:
1. The enolate needed for aldol condensation is formed by removing an alpha-hydrogen.

Teacher's Note:
1. Aldol addition needs at least one alpha-hydrogen; dilute alkali gives the beta-hydroxy carbonyl compound, and heating dehydrates it to an enone.
2. Compare reactivity with steric hindrance and electron donation: HCHO > RCHO > ArCHO > R2CO > Ar2CO.

Question 17: Clemmensen's reduction of ketones is carried out with
(a) \( \ce{H2} \) with Pd catalyst
(b) Glycol with KOH
(c) \( \ce{LiAlH4} \) in water
(d) Zn-Hg with HCl
Show Answer & Explanation

Answer: (d) Zn-Hg with HCl

Explanation:
1. Zinc amalgam and concentrated HCl reduce the carbonyl group to a methylene group.

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 18: Identify the product Y in the sequence \( \ce{CH3CHO + CH3MgI} \) (ether) giving X, then \( \ce{H2O/H^+} \) giving Y
(a) \( \ce{CH3OH} \)
(b) \( \ce{CH3CH2OH} \)
(c) \( \ce{(CH3)2CHOH} \)
(d) \( \ce{(CH3)3COH} \)
Show Answer & Explanation

Answer: (c) \( \ce{(CH3)2CHOH} \)

Explanation:
1. Acetaldehyde with CH3MgI gives propan-2-ol after hydrolysis.

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 19: What is the name of the reaction when benzaldehyde changes into benzyl alcohol?
(a) Friedel-Crafts reaction
(b) Kolbe's reaction
(c) Wurtz reaction
(d) Cannizzaro's reaction
Show Answer & Explanation

Answer: (d) Cannizzaro's reaction

Explanation:
1. With strong alkali benzaldehyde disproportionates to benzyl alcohol and benzoate.

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 20: The reagent that gives an orange coloured precipitate with acetaldehyde is
(a) \( \ce{NH2OH} \)
(b) \( \ce{NaHSO3} \)
(c) Iodine
(d) 2,4-DNP
Show Answer & Explanation

Answer: (d) 2,4-DNP

Explanation:
1. 2,4-Dinitrophenylhydrazine gives orange-yellow hydrazone crystals with carbonyl compounds.

Teacher's Note:
1. Ammonia derivatives (NH2OH, NH2NH2, PhNHNH2, semicarbazide, 2,4-DNP) condense with C=O to give C=N products by addition then loss of water.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 21: Which one is used in the manufacture of mirrors?
(a) Red lead (litharge)
(b) Ammoniacal \( \ce{AgNO3} \)
(c) Ammoniacal \( \ce{AgNO3} \) + red lead
(d) Ammoniacal \( \ce{AgNO3} \) + \( \ce{HCHO} \)
Show Answer & Explanation

Answer: (d) Ammoniacal \( \ce{AgNO3} \) + \( \ce{HCHO} \)

Explanation:
1. Silver mirrors are made by reducing Tollens' reagent with formaldehyde.

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 22: When \( \ce{CH3COCH3} \) reacts with \( \ce{Cl2} \) and NaOH, which of the following is formed?
(a) \( \ce{CHCl3} \)
(b) \( \ce{CCl2H2} \)
(c) \( \ce{CCl4} \)
(d) \( \ce{CH3Cl} \)
Show Answer & Explanation

Answer: (a) \( \ce{CHCl3} \)

Explanation:
1. Chlorine and NaOH form hypochlorite, which brings about haloform reaction of methyl ketones to give chloroform.

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 23: Which gives a difference between aldehyde and ketone?
(a) Fehling's solution
(b) Tollens' reagent
(c) Schiff's reagent
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Aldehydes reduce Fehling's and Tollens' reagents and restore Schiff's colour, while ketones do not.

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 24: An aldehyde turns pink with
(a) Benedict reagent
(b) Schiff reagent
(c) Fehling reagent
(d) Tollens' reagent
Show Answer & Explanation

Answer: (b) Schiff reagent

Explanation:
1. Schiff's reagent (decolourised magenta) turns pink with aldehydes.

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 25: The reaction of acetaldehyde with concentrated \( \ce{KMnO4} \) gives
(a) \( \ce{CH3COOH} \)
(b) \( \ce{CH3CH2OH} \)
(c) \( \ce{CH3COCH3} \)
(d) \( \ce{CH3OH} \)
Show Answer & Explanation

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

Explanation:
1. Acetaldehyde is oxidised to acetic acid.

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 26: When acetaldehyde is heated with Tollens' reagent, the following is obtained
(a) Methyl alcohol
(b) Silver acetate
(c) Silver mirror
(d) Formaldehyde
Show Answer & Explanation

Answer: (c) Silver mirror

Explanation:
1. Acetaldehyde reduces the silver complex to metallic silver.

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 27: The boiling point of acetone is
(a) 56°C
(b) 60°C
(c) 100°C
(d) 90°C
Show Answer & Explanation

Answer: (a) 56°C

Explanation:
1. Acetone boils at about 56°C.

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 28: Urotropine is
(a) Hexamethylene tetramine
(b) Hexaethylene tetramine
(c) Hexamethylene diamine
(d) None of these
Show Answer & Explanation

Answer: (a) Hexamethylene tetramine

Explanation:
1. Urotropine is hexamethylenetetramine, \( \ce{(CH2)6N4} \).

Teacher's Note:
1. Urotropine, (CH2)6N4, comes from methanal and ammonia; it is a urinary antiseptic.
2. Aldol needs alpha-hydrogen; Cannizzaro needs none.

Question 29: Magenta is
(a) Alkaline phenolphthalein
(b) Methyl red
(c) p-Rosaniline hydrochloride
(d) Red litmus
Show Answer & Explanation

Answer: (c) p-Rosaniline hydrochloride

Explanation:
1. Magenta (fuchsin) is p-rosaniline hydrochloride; sulphurous acid treatment gives Schiff's reagent.

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.

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