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JEE • CHEMISTRY • ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 255 questions

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181 Easy
181 of 255
Remembering +4 | -1
In the α−\alpha - halogenation of aliphatic acids (HVZ reaction) the catalyst used is:
View answer and explanation
Correct option: B
P
Option B (P) is correct because phosphorus is used as a catalyst in the Hell-Volhard-Zelinsky (HVZ) reaction.
182 Medium
182 of 255
Understanding +4 | -1
The strongest acid is:
View answer and explanation
Correct option: D
CHF2COOH\mathrm{CH}{\mathrm{F}_2}\mathrm{COOH}
Option D (CHF2COOH\mathrm{CH}{\mathrm{F}_2}\mathrm{COOH}) is correct because the acidity increases with the number of fluorine atoms due to the strong -I effect of fluorine.
183 Medium
183 of 255
Understanding +4 | -1
The correct order of acidic strengths of the carboxylic acids is
View answer and explanation
Correct option: D
Acetic acid < benzoic acid < formic acid
Option D (Acetic acid < benzoic acid < formic acid) is correct because electron-donating groups decrease acidity, and formic acid is more acidic than benzoic acid due to resonance stabilization.
184 Medium
184 of 255
Understanding +4 | -1
Read the following statements and choose the correct option

(i) The carbonyl carbon atom is sp2\mathrm{sp}^2-hybridised

(ii) The carbonyl carbon is an electrophilic (Lewis acid) centre

(iii) The carbonyl oxygen is a nucleophilic (Lewis base) centre

(iv) Carbonyl compounds are non- polar in nature.
View answer and explanation
Correct option: B
(i), (ii) and (iii) are correct
Option B ((i), (ii) and (iii) are correct) is correct because carbonyl carbon is sp2sp^2 hybridized, electrophilic, and carbonyl oxygen is nucleophilic, while carbonyl compounds are polar.
185 Easy
185 of 255
Remembering +4 | -1
Acetaldehyde is used:
View answer and explanation
Correct option: D
All are correct
Option D (All are correct) is correct because acetaldehyde is used in the preparation of dyes, chloral, and paraldehyde.
186 Medium
186 of 255
Understanding +4 | -1
On warming formic acid with ammoniacal silver nitrate, the product formed is:
View answer and explanation
Correct option: B
Metallic silver
Option B (Metallic silver) is correct because formic acid reduces ammoniacal silver nitrate to metallic silver.
187 Medium
187 of 255
Analyzing +4 | -1
Arrange the following in decreasing order of their acidic strength

A) CH3CH2OHC{H_3}C{H_2}OH B) CH3COOHC{H_3}COOH C) ClCH2COOHClC{H_2}COOH
D) FCH2COOHFC{H_2}COOH E) C6H5CH2COOH{C_6}{H_5}C{H_2}COOH
View answer and explanation
Correct option: A
D > C > E > B > A
The correct answer is B (D > C > B > E > A) because the acidity increases with the electron-withdrawing effect of the substituents, with fluorine being the strongest electron-withdrawing group among the given options.
188 Medium
188 of 255
Application +4 | -1
3-pentanol on reaction with aluminium tertiary butoxide in the presence of acetone gives
View answer and explanation
Correct option: C
3-pentanone
Option C (3-pentanone) is correct because 3-pentanol is oxidized to 3-pentanone using aluminium tertiary butoxide in the presence of acetone (Oppenauer oxidation).
189 Medium
189 of 255
Application +4 | -1
Reaction between (C2H5)2Cd  and  CH3COCl{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_2}\mathrm{Cd }\;\mathrm{and }\;\mathrm{C}{\mathrm{H}_3}\mathrm{COCl} l leads to the formation of
View answer and explanation
Correct option: B
Ethyl methyl ketone
The correct answer is Ethyl methyl ketone because the reaction between a dialkylcadmium reagent and an acyl chloride leads to the formation of a ketone.
190 Medium
190 of 255
Analyzing +4 | -1
Which of the following reactions is not an
example for nucleophilic addition - elimination
reaction?
View answer and explanation
Correct option: B
CH3CHO+NaHSO3⇌CH3−OH∣C∣H−OSO2Na{CH}_3 {CHO}+{NaHSO}_3 \rightleftharpoons C{H_3} - \begin{array}{cccccccccccccccccccc} {OH} \\ | \\ C \\ | \\ H \end{array} - OS{O_2}Na
Option B is correct because the reaction with NaHSO3NaHSO_3 is a simple nucleophilic addition, without the elimination of any molecule like water.
191 Hard
191 of 255
Analyzing +4 | -1
The correct order of decreasing boiling points of CH3CONH2  (A),CH3COCl  (B),CH3COOH  (C),  and (CH3CO)2O  (D),\mathrm{C}{\mathrm{H}_3}\mathrm{CON}{\mathrm{H}_2}\;(A),\mathrm{C}{\mathrm{H}_3}\mathrm{COCl}\;(B),\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\;\mathrm{(C),}\;\mathrm{and}\,{\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{CO}} \right)_2}\mathrm{O}\;(D), is:
View answer and explanation
Correct option: A
A>D>C>BA > D > C > B
Option A (A>D>C>BA > D > C > B) is correct because the boiling points are determined by intermolecular forces: amides have strong hydrogen bonding, followed by anhydrides, carboxylic acids (dimer formation), and then acyl chlorides.
192 Medium
192 of 255
Remembering +4 | -1
Semicarbazide is:
View answer and explanation
Correct option: C
NH2CONHNH2\mathrm{N}{\mathrm{H}_2}\mathrm{CONHN}{\mathrm{H}_2}
Option C (NH2CONHNH2\mathrm{N}{\mathrm{H}_2}\mathrm{CONHN}{\mathrm{H}_2}) is correct because it is the correct chemical formula for semicarbazide.
193 Medium
193 of 255
Analyzing +4 | -1
CH3−CO∣∣−OH→NH3(A)→Al2O3(B)→H3O+(D),    D   isC{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - OH\xrightarrow{{N{H_3}}}\left( A \right)\xrightarrow{{A{l_2}{O_3}}}\left( B \right)\xrightarrow{{{H_3}{O^ + }}}\left( D \right),\,\,\,\,D \,\,\,{\text{is}}
View answer and explanation
Correct option: B
CH3COOHC{H_3}COOH
The correct answer is CH3COOHC{H_3}COOH (Option B) because the reaction sequence involves ammonolysis, dehydration, and hydrolysis, ultimately converting the starting material to acetic acid.
194 Easy
194 of 255
Understanding +4 | -1
An acyl halide is formed when PCl5\mathrm{PC}{\mathrm{l}_5} reacts with an:
View answer and explanation
Correct option: A
Acid
Option A is correct because acyl halides are formed when PCl5 reacts with carboxylic acids.
195 Medium
195 of 255
Remembering +4 | -1
Which of the following will not give Iodoform test:
View answer and explanation
Correct option: D
Benzyl alcohol
Option D (Benzyl alcohol) is correct because it lacks the required CH3CO−CH_3CO- or CH3CH(OH)−CH_3CH(OH)- group to give a positive iodoform test.
196 Hard
196 of 255
Analyzing +4 | -1
Which of the following products is not possible in the reaction given below?

(C2H5COO)2Ca+(C6H5COO)2Ca→ distillation  Dry \left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{COO}\right)_2 \mathrm{Ca}+\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}\right)_2 \mathrm{Ca} \xrightarrow[\text { distillation }]{\text { Dry }}
View answer and explanation
Correct option: D
C6H5CH2COCH3\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COCH}_3
The correct answer is D (C6H5CH2COCH3\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COCH}_3) because dry distillation of calcium salts of carboxylic acids produces ketones, and this option is not a ketone.
197 Medium
197 of 255
Understanding +4 | -1
Ethanal reacts with alkali to give 3-hydroxybutanal. This reaction is:
View answer and explanation
Correct option: D
Aldol condensation
Option D (Aldol condensation) is correct because ethanal reacting with alkali to form 3-hydroxybutanal is a classic example of aldol condensation.
198 Easy
198 of 255
Remembering +4 | -1
Vinegar is a solution of acetic acid which is
View answer and explanation
Correct option: C
6−8%6 - 8\mathrm{ }\%
Option C (6−8%6 - 8\mathrm{ }\%) is correct because vinegar is defined as a 6-8% solution of acetic acid in water.
199 Hard
199 of 255
Analyzing +4 | -1
Statement I: The boiling points of aldehydes and ketones are lower than hydrocarbons and ethers of comparable masses.

Statement II: It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions.
View answer and explanation
Correct option: D
Statement II is correct, Statement I is incorrect.
Statement II is correct, Statement I is incorrect. Aldehydes and ketones have higher boiling points than hydrocarbons and ethers due to dipole-dipole interactions.
200 Medium
200 of 255
Remembering +4 | -1
Cannizzaro reaction occurs with
View answer and explanation
Correct option: B
C6H5CHO\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}
Option B (C6H5CHO\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}) is correct because Cannizzaro reaction is given by aldehydes which do not have alpha-hydrogen atoms, and benzaldehyde fits this criterion.
201 Medium
201 of 255
Understanding +4 | -1
The rate of esterification of acetic acid with methyl alcohol (I), ethyl alcohol (II), isopropyl alcohol (III) and tertiary butyl alcohol (IV) follows in the order
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Correct option: A
I >> II >> III >> IV
Option A (I >> II >> III >> IV) is correct because the rate of esterification decreases with increasing steric hindrance around the alcohol's hydroxyl group.
202 Medium
202 of 255
Understanding +4 | -1
Which of the following reactant give Tollen’s reagent and Fehling’s solution test?
View answer and explanation
Correct option: A
CH3CHO\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}
Option A (CH3CHO\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}) is correct because only aldehydes react with both Tollen’s reagent and Fehling’s solution.
203 Medium
203 of 255
Remembering +4 | -1
Which one is a polyprotic acid?
View answer and explanation
Correct option: B
Oxalic acid
Option B (Oxalic acid) is correct because oxalic acid (COOHCOOH) is a dibasic acid, meaning it can donate more than one proton, thus it is a polyprotic acid.
204 Medium
204 of 255
Understanding +4 | -1
The reaction of a carboxylic acid gives effervescences of CO2\mathrm{C}{\mathrm{O}_2} with NaHCO3\mathrm{NaHC}{\mathrm{O}_3}. The CO2\mathrm{C}{\mathrm{O}_2} comes from:
View answer and explanation
Correct option: B
NaHCO3\mathrm{NaHC}{\mathrm{O}_3}
Option B (NaHCO3\mathrm{NaHC}{\mathrm{O}_3}) is correct because the carbon dioxide (CO2\mathrm{C}{\mathrm{O}_2}) released in the reaction comes from the bicarbonate ion (HCO3−\mathrm{HC}{\mathrm{O}_3}^-) in NaHCO3\mathrm{NaHC}{\mathrm{O}_3}.
205 Medium
205 of 255
Application +4 | -1
Which reagent is useful in separating benzoic acid from phenol?
View answer and explanation
Correct option: D
5  %  NaHCO35\;\mathrm{\% }\;\mathrm{NaHC}{\mathrm{O}_3}
Option D (5  %  NaHCO35 \;\mathrm{\% }\;\mathrm{NaHC}{\mathrm{O}_3}) is correct because benzoic acid reacts with sodium bicarbonate while phenol does not, allowing for separation.
206 Medium
206 of 255
Understanding +4 | -1
R−COOH→2) H3O+1) LiAlH4/eitherproduct isR - COOH\xrightarrow[{2)\,{H_3}{O^ + }}]{{1)\,LiAl{H_4}/either}}{\text{product}}\,{\text{is}}
View answer and explanation
Correct option: A
R−CH2−OHR - C{H_2} - OH
Option A (R−CH2−OHR - C{H_2} - OH) is correct because LiAlH4LiAl{H_4} reduces carboxylic acids to primary alcohols.
207 Medium
207 of 255
Application +4 | -1
HCHO+C6H5CHO→con. NaOHproducts  areHCHO + {C_6}{H_5}CHO\xrightarrow[{}]{{con.\,NaOH}}{\text{products}}\,\,{\text{are}}
View answer and explanation
Correct option: D
HCOONa+C6H5CH2OHHCOONa + {C_6}{H_5}C{H_2}OH
Option D (HCOONa+C6H5CH2OHHCOONa + {C_6}{H_5}C{H_2}OH) is correct because formaldehyde undergoes oxidation to form formate, while benzaldehyde undergoes reduction to form benzyl alcohol in the Cannizzaro reaction.
208 Medium
208 of 255
Analyzing +4 | -1
Incorrect method of preparation for alcohols from
the following is:
View answer and explanation
Correct option: D
Ozonolysis of alkene
Option D is correct because ozonolysis of alkenes typically produces aldehydes, ketones, or carboxylic acids, but not alcohols directly. The other methods (Grignard reaction, alkyl halide hydrolysis, and hydroboration-oxidation) are valid alcohol synthesis methods.
209 Medium
209 of 255
Remembering +4 | -1
Cannizaro reaction is performed by
View answer and explanation
Correct option: D
Formaldehyde and benzaldehyde
The correct answer is D (Formaldehyde and benzaldehyde) because Cannizzaro reaction requires aldehydes without alpha-hydrogens, and both formaldehyde and benzaldehyde meet this criterion.
210 Easy
210 of 255
Remembering +4 | -1
Wolff-Kishner’s reaction is:
View answer and explanation
Correct option: A
Reduction of carbonyl compound into hydrocarbons
Option A (Reduction of carbonyl compound into hydrocarbons) is correct because the Wolff-Kishner reaction reduces carbonyl groups (C=O) to methylene groups (CH2).
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