By passing water gas $\left( {{\rm{CO}} + {{\rm{H}}_2}} \right)$ through an electric discharge at low pressure, we get:

${\rm{C}}{{\rm{H}}_3}{\rm{CHO}}$

${\rm{C}}{{\rm{O}}_2}\;{\rm{and}}\;{{\rm{H}}_2}{\rm{O}}$

When ${\rm{HCHO}}$ is treated with ${{\rm{C}}_6}{{\rm{H}}_5}{\rm{CHO}}$ in presence of NaOH, the products are:

${\rm{HCOONa }}\;{\rm{and}}\;{{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}\;and{\rm{ }}HCOONa$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_2}{\rm{OH }}\;{\rm{and}}\;{{\rm{C}}_6}{{\rm{H}}_5}{\rm{COONa}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH }}\;{\rm{and}}\;{{\rm{C}}_6}{{\rm{H}}_5}{\rm{COONa}}$

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JEE • CHEMISTRY • ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

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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 1-30 of 255 questions

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1 Medium

1 of 255

Remembering +4 | -1

By passing water gas

\left( {\mathrm{CO} + {\mathrm{H}_2}} \right)

through an electric discharge at low pressure, we get:

View answer and explanation

Correct option: A

HCHO

Option A (HCHO) is correct because water gas (CO + H2) passed through an electric discharge yields formaldehyde (HCHO).

2 Medium

2 of 255

Understanding +4 | -1

Addition of alcohols to aldehydes and ketones takes place in presence of dry

\mathrm{HCl}

gas because it

(i) Protonates the oxygen of the carbonyl compounds

(ii) Increases the electrophilicity of the carbonyl carbon

(iii) Removes the excess moisture from the reaction

(iv) Helps the reaction to move in the forward direction

View answer and explanation

Correct option: B

(i), (ii), (iii) and (iv)

Option B ((i), (ii), (iii) and (iv)) is correct because dry HCl protonates the carbonyl oxygen, increases the electrophilicity of the carbonyl carbon, removes excess moisture, and helps the reaction move forward.

3 Medium

3 of 255

Remembering +4 | -1

Benzyl alcohol and sodium benzoate is obtained by the action of sodium hydroxide on benzaldehyde. This reaction is known as

View answer and explanation

Correct option: B

Cannizaro’s reaction

Option B (Cannizaro’s reaction) is correct because the reaction of benzaldehyde with NaOH to form benzyl alcohol and sodium benzoate is a Cannizaro's reaction.

4 Medium

4 of 255

Understanding +4 | -1

Which one of the following reactions is a condensation reaction?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

\to

mesityl oxide

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

\to

mesityl oxide) is correct because it involves the elimination of a water molecule, which is characteristic of a condensation reaction.

5 Easy

5 of 255

Understanding +4 | -1

Carboxylic acids dimerise due to

View answer and explanation

Correct option: C

Intermolecular hydrogen bonding

The correct answer is C (Intermolecular hydrogen bonding) because carboxylic acids form dimers through intermolecular hydrogen bonds between the carbonyl oxygen and the hydroxyl hydrogen of two molecules.

6 Medium

6 of 255

Understanding +4 | -1

The number of aldehyde isomers for

{C_4}{H_8}O

will be

View answer and explanation

Correct option: C

Option C (2) is correct because there are two aldehyde isomers for

{C_4}{H_8}O

: butanal and 2-methylpropanal.

7 Medium

7 of 255

Understanding +4 | -1

Which of the following compounds does not have a carboxyl group?

View answer and explanation

Correct option: C

Picric acid

The correct answer is C (Picric acid) because picric acid is 2,4,6-trinitrophenol and contains a hydroxyl group rather than a carboxyl group.

8 Easy

8 of 255

Analyzing +4 | -1

The most acidic among the following is:

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

) is correct because carboxylic acids are generally more acidic than alcohols and phenols due to resonance stabilization of the carboxylate anion.

9 Medium

9 of 255

Remembering +4 | -1

When the aldehyde group is attached to a ring, which of the following suffix is added after the full name of the cycloalkane?

View answer and explanation

Correct option: B

Carbaldehyde

Option B (Carbaldehyde) is correct because when an aldehyde group is directly attached to a ring, the suffix "carbaldehyde" is used.

10 Medium

10 of 255

Application +4 | -1

When

\mathrm{HCHO}

is treated with

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CHO}

in presence of NaOH, the products are:

View answer and explanation

Correct option: D

\mathrm{HCOONa }\;\mathrm{and}\;{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

Option D (

\mathrm{HCOONa }\;\mathrm{and}\;{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

) is correct because it represents the products of a crossed Cannizzaro reaction between formaldehyde and benzaldehyde.

11 Medium

11 of 255

Understanding +4 | -1

Stephen’s reduction will covert

\mathrm{CH_3CH_2CN}

View answer and explanation

Correct option: B

EtCHO

Option B (EtCHO) is correct because Stephen's reduction converts nitriles to aldehydes, so

\mathrm{CH_3CH_2CN}

becomes

\mathrm{CH_3CH_2CHO}

12 Medium

12 of 255

Analyzing +4 | -1

Which of the following statement(s) is/are true regarding preparation of aldehydes and ketones?

(1) Both can be prepared by the oxidation of the concerned alcohol with copper at about

250^{\circ} \mathrm{C}

.

(ii) Both can be prepared by the oxidation of the concerned alcohol by Oppenauer oxidation.

(iii) Both can be prepared by the oxidation of respective alcohol with acidic dichromate.

View answer and explanation

Correct option: A

(i) only

Option A ((X) < (Y) < (Z)) is correct because metallic character decreases across a period, so the ionic oxide is formed by an element on the left, the amphoteric oxide by an element in the middle, and the giant molecule oxide by an element on the right.

13 Easy

13 of 255

Understanding +4 | -1

Acetic acid will be obtained on oxidation of

View answer and explanation

Correct option: A

Ethanol

Option A (Ethanol) is correct because the oxidation of ethanol yields acetic acid.

14 Easy

14 of 255

Analyzing +4 | -1

Which of the following is a better reducing agent for the following reduction?

R\mathrm{COOH} \to R\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

View answer and explanation

Correct option: D

{\mathrm{B}_2}{\mathrm{H}_6}/{\mathrm{H}_3}{\mathrm{O}^ + }

Option D (B2H6/H3O+) is correct because diborane is a selective reducing agent that reduces carboxylic acids to primary alcohols without affecting other functional groups.

15 Hard

15 of 255

Analyzing +4 | -1

NaAl{H_4}

reduces an ester into

HCHO

and

\left({C{H_3}} \right){ _2}CHOH.

This ester changes to ______ on reaction with

C{H_3}MgBr/{H_3}{O^ + }.

View answer and explanation

Correct option: B

C{H_3}CHO

Option C (

C{H_3}CHO

) is correct because the ester is

H - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - O - CH{\left( {C{H_3}} \right)_2}

, and reaction with

CH_3MgBr/H_3O^+

yields acetaldehyde.

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16 Medium

16 of 255

Analyzing +4 | -1

Arrange the following carboxylic acids in the decreasing order of their acidities.

1. PhCOOH

2.

\mathrm{o}-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}

\mathrm{p}-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}

\mathrm{m}-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}

View answer and explanation

Correct option: C

2>3>4>1

The correct answer is C (

2>3>4>1

) because electron-withdrawing groups increase acidity, and the ortho effect further enhances the acidity of o-nitrobenzoic acid.

17 Easy

17 of 255

Remembering +4 | -1

General formula of carbonyl compound is:

View answer and explanation

Correct option: A

{\mathrm{C}_n}{\mathrm{H}_{2n}}\mathrm{O}

Option A (

{\mathrm{C}_n}{\mathrm{H}_{2n}}\mathrm{O}

) is correct because carbonyl compounds (aldehydes and ketones) have the general formula

{\mathrm{C}_n}{\mathrm{H}_{2n}}\mathrm{O}

18 Medium

18 of 255

Remembering +4 | -1

The product formed by the reaction of an aldehyde with a primary amine is:

View answer and explanation

Correct option: A

Schiff base

Option A (Schiff base) is correct because the reaction of an aldehyde with a primary amine forms a Schiff base, which contains a C=N bond.

19 Medium

19 of 255

Analyzing +4 | -1

Assertion: The acetate ion is resonance stabilized.

Reason: Acetate ion is more basic than the methoxide ion.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

Option C is correct because the acetate ion is resonance stabilized, but it is a weaker base than the methoxide ion.

20 Medium

20 of 255

Understanding +4 | -1

The number of aldehyde isomers for

{C_4}{H_8}O

will be

View answer and explanation

Correct option: C

Option C (2) is correct because there are two aldehyde isomers for

{C_4}{H_8}O

: butanal and 2-methylpropanal.

21 Medium

21 of 255

Understanding +4 | -1

Riemer-Tiemann reaction and Kolbe’s reaction are respectively

View answer and explanation

Correct option: B

Both are electrophilic aromatic substitution reactions

Option B (Both are electrophilic aromatic substitution reactions) is correct because both Riemer-Tiemann and Kolbe's reactions involve electrophilic attack on an aromatic ring.

22 Hard

22 of 255

Application +4 | -1

5.26 {~g}

mixture of ethanol and acetaldehyde when heated with Fehling's solution gave

1.2 {~g}

of a red ppt. The percentage of acetaldehyde in the mixture is....

View answer and explanation

Correct option: A

Option A (7) is correct because the percentage of acetaldehyde in the mixture is calculated to be approximately 7% based on the amount of red precipitate formed.

23 Medium

23 of 255

Application +4 | -1

The product formed in the following ozonalysis is

C{H_2}BrCH = CHC{H_3}\xrightarrow{{{O_3}}}

View answer and explanation

Correct option: B

C{H_3}CHO

The correct answer is B (

C{H_3}CHO

) because the ozonolysis of

C{H_2}BrCH = CHC{H_3}

yields acetaldehyde.

24 Easy

24 of 255

Remembering +4 | -1

Action of acetylene on dilute

{H}_2 {SO}_4

gives

View answer and explanation

Correct option: B

Acetaldehyde

Option B (Acetaldehyde) is correct because acetylene reacts with dilute sulfuric acid in the presence of mercuric sulfate to form acetaldehyde.

25 Medium

25 of 255

Remembering +4 | -1

Aldehydes on reaction with hydroxylamine gives :

View answer and explanation

Correct option: A

Aldoxime

Option A (Aldoxime) is correct because aldehydes react with hydroxylamine to form aldoximes.

26 Medium

26 of 255

Analyzing +4 | -1

Which order is correct about acidity

View answer and explanation

Correct option: B

{C}_6 {H}_5 {COOH}>{CH}_3 {COOH}>{C}_5 {H}_5 {OH}

Option B (

{C}_6 {H}_5 {COOH}>{CH}_3 {COOH}>{C}_5 {H}_5 {OH}

) is correct because carboxylic acids are more acidic than phenols, and the acidity of carboxylic acids is influenced by substituents.

27 Hard

27 of 255

Analyzing +4 | -1

2-pentanone and 3-pentanone can be distinguished by one of the following:

View answer and explanation

Correct option: D

Iodoform test

Option D (Iodoform test) is correct. 2-pentanone (CH₃COCH₂CH₂CH₃) gives a positive iodoform test because it contains a methyl ketone group, while 3-pentanone (CH₃CH₂COCH₂CH₃) does not.

28 Medium

28 of 255

Analyzing +4 | -1

Aqueous

10 \% \, {NaHCO}_3

solution is used as a reagent for identifying '

{A}

'. Which of the following compounds yield '

{A}

' on hydrolysis?

View answer and explanation

Correct option: A

{CH}_3{COOC}_2 {H}_5

Option A (

{CH}_3{COOC}_2 {H}_5

) is correct because esters produce carboxylic acids upon hydrolysis, which can be identified by reaction with

NaHCO_3

29 Medium

29 of 255

Analyzing +4 | -1

Acetone and acetaldehyde can be identified by treatment with:

View answer and explanation

Correct option: D

\mathrm{Ag}\left( {\mathrm{N}{\mathrm{H}_3}} \right)_2^ +

Option D is correct because Tollen's reagent (Ag(NH₃)₂⁺) can distinguish between acetone and acetaldehyde, as only aldehydes reduce Tollen's reagent to form a silver mirror.

30 Medium

30 of 255

Analyzing +4 | -1

The correct order of acidic strength of given carboxylic acids is

\begin{array}{cccccccccccccccccccc} {{\text{CC}}{{\text{l}}_{\text{3}}}{\text{ - COOH}}}&{{\text{N}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - COOH}}}&{{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ - COOH}}} \\ {{\text{(a)}}}&{{\text{(b)}}}&{{\text{(c)}}} \end{array}

View answer and explanation

Correct option: D

a > b > c

The correct answer is D (

a > b > c

) because the electron-withdrawing effect of

CCl_3

is stronger than

NO_2CH_2

, which is stronger than

C_6H_5

Revision notes and key takeaways

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Key Takeaways

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