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JEE • CHEMISTRY • ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

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61 Easy
61 of 255
Remembering +4 | -1
Which of the following contain an aldehyde?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because vanilla beans, meadowsweet, and cinnamon all contain aldehydes, specifically vanillin, salicylaldehyde, and cinnamaldehyde, respectively.
62 Medium
62 of 255
Understanding +4 | -1
The yield of ester in esterification can be increased by

CH3CH2OH+CH3COOH⇌CH3COOCH2CH3+H2O\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOCH}_2 \mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}
View answer and explanation
Correct option: D
all the above factors
Option D (all the above factors) is correct because removing water or increasing the concentration of either reactant will shift the equilibrium towards the product side, increasing the yield of the ester.
63 Medium
63 of 255
Remembering +4 | -1
Ketones can be prepared by:
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Correct option: D
None of the above
Option D (None of the above) is correct because Rosenmund's reduction and Stephen's reduction are both methods used to prepare aldehydes, not ketones.
64 Medium
64 of 255
Remembering +4 | -1
Cannizzaro reaction is given by
View answer and explanation
Correct option: D
PhCHOPhCHO
Option D (PhCHOPhCHO) is correct because Cannizzaro reaction requires the absence of alpha-hydrogens, which is the case for benzaldehyde (PhCHOPhCHO).
65 Hard
65 of 255
Analyzing +4 | -1
CH3COCH3  and  CH3CH2CHO\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}\;and\;\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CHO} can be distinguished by
View answer and explanation
Correct option: B
Tollen’s reagent
Option B (Tollen’s reagent) is correct because it distinguishes between aldehydes and ketones, with aldehydes giving a silver mirror test.
66 Hard
66 of 255
Analyzing +4 | -1
A compound, containing only carbon, hydrogen and oxygen, has a molecular weight of 44. On complete oxidation it is converted into a compound of molecular weight 60. The original compound is
View answer and explanation
Correct option: A
An aldehyde
Option A (An aldehyde) is correct because the molecular weight increases by 16 upon oxidation, indicating the addition of one oxygen atom.
67 Medium
67 of 255
Remembering +4 | -1
Acetone is prepared by:
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Correct option: C
Pyrolysis of calcium acetate
Option C (Pyrolysis of calcium acetate) is correct because acetone is prepared by the dry distillation (pyrolysis) of calcium acetate.
68 Medium
68 of 255
Understanding +4 | -1
What is the product in the reaction

CH3CONH2→NaOH2/HClX?\mathrm{C}{\mathrm{H}_3}\mathrm{CON}{\mathrm{H}_2}\xrightarrow{{\mathrm{NaO}{\mathrm{H}_2}/\mathrm{HCl}}}X?
View answer and explanation
Correct option: A
CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}
Option A (CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}) is correct because the reaction of an amide with nitrous acid (generated in situ from NaNO2/HCl) results in the formation of a carboxylic acid.
69 Hard
69 of 255
Analyzing +4 | -1
Most acidic compound among the following is
View answer and explanation
Correct option: C
{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - }}\mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {\text{O}} \\ {{\text{||}}} \end{array}} {\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - }}\mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {\text{O}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ - }}\mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {\text{O}} \\ {{\text{||}}} \end{array}} {\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - }}\mathop {\text{C}}\limits^{\begin{array}{cccccccccccccccccccc} {\text{O}}
Option C is correct. The presence of electron-withdrawing groups increases acidity, and the absence of electron-donating groups makes it the most acidic.
70 Easy
70 of 255
Remembering +4 | -1
Chloral belongs to the class of:
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Correct option: B
Aldehydes
Option B (Aldehydes) is correct because Chloral has the formula CCl3CHO\mathrm{CC}{\mathrm{l}_3}\mathrm{CHO}, which contains the -CHO functional group characteristic of aldehydes.
71 Medium
71 of 255
Understanding +4 | -1
Acetic acid on warming with hydrazoic acid in presence of conc.H2SO4conc.{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} gives:
View answer and explanation
Correct option: B
CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}
Option B (CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}) is correct because acetic acid reacts with hydrazoic acid in the presence of conc.H2SO4conc.{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} to give methylamine.
72 Medium
72 of 255
Analyzing +4 | -1
Various products formed on oxidation of 2,5-dimetylhexan-3-one are

i) CH3−C∣CH3H−COOH{\text{C}}{{\text{H}}_{\text{3}}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{C}}{{\text{H}}_{\text{3}}}} \end{array}} {\text{H}} - {\text{COOH}}

ii) CH3−C∣CH3H−CH2−COOH{\text{C}}{{\text{H}}_{\text{3}}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{C}}{{\text{H}}_3}} \end{array}} {\text{H}} - {\text{C}}{{\text{H}}_2} - {\text{COOH}}

iii) CH3COOH\mathrm{CH}_3 \mathrm{COOH}

iv) HCOOH\mathrm{HCOOH}
View answer and explanation
Correct option: C
(i), (ii), (iii) and (iv)
Option C (i, ii, iii, and iv) is correct because the oxidation of 2,5-dimethylhexan-3-one leads to the formation of various carboxylic acids due to cleavage at different bonds near the carbonyl group.
73 Medium
73 of 255
Understanding +4 | -1
Carboxylic acids are more acidic than phenol and alcohol because of
View answer and explanation
Correct option: D
resonance stabilization of their conjugate base
Option D (resonance stabilization of their conjugate base) is correct because the carboxylate ion, the conjugate base of carboxylic acids, is resonance-stabilized, making carboxylic acids more acidic than phenols and alcohols.
74 Easy
74 of 255
Remembering +4 | -1
Formalin is:
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Correct option: C
Formaldehyde + methanol + water
Option C (Formaldehyde + methanol + water) is correct because formalin is a solution of formaldehyde in water, typically stabilized with methanol to prevent polymerization.
75 Medium
75 of 255
Remembering +4 | -1
In Clemmensen reduction, carboxyl compound is treated with
View answer and explanation
Correct option: A
Zinc amalgam +HCl+ HCl
Option A (Zinc amalgam +HCl+ HCl) is correct because Clemmensen reduction uses zinc amalgam and hydrochloric acid to reduce carbonyl compounds to alkanes.
76 Hard
76 of 255
Analyzing +4 | -1
Order of Acidic strength of following carboxylic acids is
A)  C6H5COOHB) CH3CH2COOHC) CH3COOHD) C6H5CH2COOH\begin{array}{cccccccccccccccccccc} {{\text{A)}}\,\,{C_6}{H_5}COOH}&{{\text{B)}}\,C{H_3}C{H_2}COOH}&{{\text{C)}}\,C{H_3}COOH}&{{\text{D)}}\,{C_6}{H_5}C{H_2}COOH} \end{array}
View answer and explanation
Correct option: A
A > D > C > B
Option A (A > D > C > B) is correct because electron-withdrawing groups (EWG) increase acidity, and the phenyl group in benzoic acid (A) is more effective than in phenylacetic acid (D).
77 Medium
77 of 255
Analyzing +4 | -1
An optically active compound 'X{X}' has molecular formula C4H8O3{C}_4 {H}_8{O}_3. It evolves CO2{CO}_2 with NaHCO3{NaHCO}_3.
'X{X}' reacts with LiAlH4{LiAlH}_4 to give an achiral compound 'X{X}' is
View answer and explanation
Correct option: C
CH3−CH∣CH2OH−COOHC{H_3} - \mathop {CH}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_2}OH} \end{array}} - COOH
Option C is correct because the compound reacts with NaHCO3 to release CO2, indicating the presence of a carboxylic acid group, and reduction with LiAlH4 yields an achiral compound.
78 Medium
78 of 255
Remembering +4 | -1
Commercially carboxylic acids are reduced to
alcohols by converting them to the _____
View answer and explanation
Correct option: A
esters
Option A (esters) is correct because carboxylic acids are commercially reduced to alcohols by first converting them into esters, which are then reduced via catalytic hydrogenation.
79 Medium
79 of 255
Analyzing +4 | -1
Statement-1: pp-Methoxybenzaldehyde undergoes nucleophilic addition more readily than pp-nitrobenzaldehyde.

Statement-2: The carbon atom of the aldehyde group is more electrophilic in pp-nitrobenzaldehyde than in pp-Methoxybenzaldehyde.
View answer and explanation
Correct option: D
Statement-1 is False; Statement-2 is True
Option D (Statement-1 is False; Statement-2 is True) is correct because electron-withdrawing groups increase the electrophilicity of the carbonyl carbon, making nucleophilic addition more favorable.
80 Medium
80 of 255
Analyzing +4 | -1
C5H10O\mathrm{C}_5 \mathrm{H}_{10} \mathrm{O} is a carbonyl compound. The number of carbonyl structural isomers possible for this molecular formula are
View answer and explanation
Correct option: D
7
Option D (7) is correct because there are 7 structural isomers of carbonyl compounds with the formula C5H10O\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}.
81 Medium
81 of 255
Analyzing +4 | -1
Statement-I: CH2F−COOH{CH}_2 {F}-{COOH} is more acidic than CH2Cl−COOH{CH}_2 {Cl}-{COOH}

Statement-II: O2N−CH2−COOH{O}_2 {N}-{CH}_2-{COOH} is less acidic than NC−CH2−COOH{NC}-{CH}_2-{COOH}
View answer and explanation
Correct option: C
I is true and II is false
Option C (I is true and II is false) is correct because fluorine is more electronegative than chlorine, making fluoroacetic acid more acidic, and the nitro group has a stronger -I effect than the cyano group.
82 Medium
82 of 255
Remembering +4 | -1
The reaction

RCOOH+N3H→Conc.H2SO4RNH2+CO2+N2R\mathrm{COOH} + {\mathrm{N}_3}\mathrm{H}\xrightarrow{{\mathrm{Conc}\mathrm{.}{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}}}R\mathrm{N}{\mathrm{H}_2} + \mathrm{C}{\mathrm{O}_2} + {\mathrm{N}_2}

is called
View answer and explanation
Correct option: B
Schmidt reaction
Option B (Schmidt reaction) is correct because the reaction of a carboxylic acid with hydrazoic acid in the presence of concentrated sulfuric acid to form an amine is the Schmidt reaction.
83 Medium
83 of 255
Analyzing +4 | -1
Assertion (A): Ethyl acetoacetate forms butane-1,4-diol with NaBH4{NaBH}_4

Reason (R): NaBH4{NaBH}_4 can reduces ketone and esters group.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option 5 moles is correct because A is the limiting reagent and the stoichiometry of the reaction A + B + C -> D indicates that 1 mole of A produces 1 mole of D.
84 Medium
84 of 255
Understanding +4 | -1
The reaction, CH3CHO+H2N−NH2→CH3CH=N.  NH2C{H_3}CHO + {H_2}N - N{H_2} \to C{H_3}CH = N.\,\,N{H_2} is
View answer and explanation
Correct option: C
Addition-elimination
Option C (Addition-elimination) is correct because the reaction involves both the addition of the amine to the carbonyl group, followed by the elimination of water.
85 Hard
85 of 255
Understanding +4 | -1
Which can reduce RCOOH →  RCH2OHR\mathrm{COOH}\, \to \,\,R\mathrm{C}{\mathrm{H}_2}\mathrm{OH} ?
View answer and explanation
Correct option: C
BH3/THF/H3O+\mathrm{B}{\mathrm{H}_3}/\mathrm{THF/}{\mathrm{H}_3}{\mathrm{O}^ + }
Option C (BH3/THF/H3O+\mathrm{B}{\mathrm{H}_3}/\mathrm{THF/}{\mathrm{H}_3}{\mathrm{O}^ + }) is correct because BH3BH_3 in THF followed by hydrolysis is a reagent that can reduce carboxylic acids to primary alcohols.
86 Medium
86 of 255
Understanding +4 | -1
In which of the following reaction, carbon-carbon
bond formation does not occur:
View answer and explanation
Correct option: C
Haloform reaction
Option C (Haloform reaction) is correct because the haloform reaction involves the conversion of a methyl ketone to a carboxylic acid and a haloform, without forming a new carbon-carbon bond.
87 Medium
87 of 255
Analyzing +4 | -1
Among the following which has lowest pKa\mathrm{p}{K_a} values:
View answer and explanation
Correct option: B
HCOOHHCOOH
Option B (HCOOH) has the lowest pKa value because it is the strongest acid among the given options.
88 Medium
88 of 255
Remembering +4 | -1
Benzoin is
View answer and explanation
Correct option: D
α−\alpha - hydroxy ketone
Option D (α−\alpha - hydroxy ketone) is correct because benzoin is an α\alpha-hydroxy ketone formed from benzaldehyde.
89 Hard
89 of 255
Analyzing +4 | -1
C6H5CHO+HCN→C6H5CH(CN)OH;{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CHO} + \mathrm{HCN} \to {\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CH}\left( {\mathrm{CN}} \right)\mathrm{OH}; the product is:
View answer and explanation
Correct option: C
Racemate
The correct answer is C (Racemate) because the reaction forms a chiral center without stereochemical control, resulting in equal amounts of both enantiomers.
90 Medium
90 of 255
Analyzing +4 | -1
Among the following acids which has the lowest pKa\mathrm{p}{K_a} value?
View answer and explanation
Correct option: C
HCOOH\mathrm{HCOOH}
Option C (HCOOH\mathrm{HCOOH}) is correct because formic acid (HCOOH) is the strongest acid among the given options, and stronger acids have lower pKa\mathrm{p}{K_a} values.
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