When ${\rm{C}}{{\rm{H}}_2} = CH-COOH$ is reduced by ${\rm{LiAl}}{{\rm{H}}_4}$, the compound obtained is:

${\rm{C}}{{\rm{H}}_2} = CH{\rm{C}}{{\rm{H}}_2}OH$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{COOH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CHO}}$

The compound ${\rm{X}}$, in the reaction is $X\xrightarrow{{{\rm{C}}{{\rm{H}}_3}{\rm{CHO}}}}Y\xrightarrow{{{\rm{Hydrolysis}}}}{\rm{Mg}}\left( {{\rm{OH}}} \right){\rm{I}} + {\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$

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

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CO}}$

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NEET • CHEMISTRY • ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

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Showing 1-30 of 137 questions

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1 Medium

1 of 137

Understanding +4 | -1

Acetone reacts with Grignard reagent to form

View answer and explanation

Correct option: B

\mathrm{2}^\circ

alcohol

Option B (

\mathrm{2}^\circ

alcohol) is correct because the reaction of acetone (a ketone) with a Grignard reagent followed by hydrolysis yields a secondary alcohol.

2 Easy

2 of 137

Remembering +4 | -1

Liquid obtained by distillation of red ant is

View answer and explanation

Correct option: C

Formic acid

Option C (Formic acid) is correct because formic acid is obtained by distillation of red ants.

3 Easy

3 of 137

Understanding +4 | -1

The compound which will not reduce Tollen's reagent is

View answer and explanation

Correct option: C

Acetone

Option C (Acetone) is correct because ketones generally do not reduce Tollen's reagent, while aldehydes and formic acid do.

4 Easy

4 of 137

Application +4 | -1

Consider the following reactions,

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\xrightarrow{{\mathrm{CaC}{\mathrm{O}_3}}}A\xrightarrow{{\mathrm{Heat}}}B

Compound B is:

View answer and explanation

Correct option: D

A ketone

Option D (A ketone) is correct because the reaction sequence leads to the formation of a ketone.

5 Hard

5 of 137

Understanding +4 | -1

Conversion of benzaldehyde to 3-phenylprop-2-en-1-oic acid is

View answer and explanation

Correct option: A

Perkin condensation

Option A (Perkin condensation) is correct because it involves the condensation of an aromatic aldehyde with an anhydride in the presence of a base to form an alpha, beta-unsaturated acid.

6 Medium

6 of 137

Analyzing +4 | -1

A mixture of calcium acetate and calcium formate on heating gives:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because heating a mixture of calcium acetate and calcium formate results in the formation of acetone, acetaldehyde, and formaldehyde through different reaction pathways.

7 Hard

7 of 137

Understanding +4 | -1

When acetaldehyde is heated with Fehling’s solution, it gives a red precipitate of:

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{u}_2}\mathrm{O}

Option D (

\mathrm{C}{\mathrm{u}_2}\mathrm{O}

) is correct because acetaldehyde reduces Fehling's solution to form cuprous oxide.

8 Hard

8 of 137

Analyzing +4 | -1

\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 can oxidize aldehydes (CH3CH2CHO) but not ketones (CH3COCH3).

9 Easy

9 of 137

Application +4 | -1

An organic compound contains hydrogen, oxygen, a single carbon atom and responds positively to Tollen’s reagent. The compound is:

View answer and explanation

Correct option: A

HCHO

Option A (HCHO) is correct because formaldehyde contains hydrogen, oxygen, a single carbon atom, and responds positively to Tollen's reagent.

10 Easy

10 of 137

Understanding +4 | -1

The Clemmensen reduction of benzaldehyde gives:

View answer and explanation

Correct option: C

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_3}

Option C (

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_3}

) is correct because Clemmensen reduction reduces the carbonyl group of benzaldehyde to a methylene group, resulting in toluene (

\mathrm{C_6H_5CH_3}

11 Easy

11 of 137

Understanding +4 | -1

Salicylic acid as compared to benzoic acid:

View answer and explanation

Correct option: A

Is more acidic

Option A (Is more acidic) is correct because the presence of the hydroxyl group at the ortho position in salicylic acid provides an intramolecular hydrogen bond, stabilizing the conjugate base and increasing acidity compared to benzoic acid.

12 Hard

12 of 137

Remembering +4 | -1

Stephen’s reduction is used to prepare aldehyde from

View answer and explanation

Correct option: B

Alkyl cyanides

Option B (Alkyl cyanides) is correct because Stephen's reduction involves the reduction of alkyl cyanides (nitriles) to aldehydes using

\mathrm{SnCl}_2

and HCl followed by hydrolysis.

13 Medium

13 of 137

Understanding +4 | -1

Benzene sulphonic acid on treating with

{\mathrm{P}_2}{\mathrm{O}_5}

gives:

View answer and explanation

Correct option: C

Acid anhydride

Option C (Acid anhydride) is correct because benzene sulfonic acid dehydrates to form its anhydride when treated with

P_2O_5

14 Easy

14 of 137

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.

15 Hard

15 of 137

Application +4 | -1

Percentage of Se(at. mass = 78.4 ) in peroxidase anhydrase enzyme is 0.5% by weight, then minimum molecular mass of peroxidase anhydrase enzyme is:

View answer and explanation

Correct option: A

1.576 \times {10^4}

Option A (

1.576 \times {10^4}

) is correct because it is calculated by setting up the equation: (atomic mass of Se / molecular mass of enzyme) * 100 = percentage of Se.

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16 Easy

16 of 137

Application +4 | -1

When

\mathrm{C}{\mathrm{H}_2} = CH-COOH

is reduced by

\mathrm{LiAl}{\mathrm{H}_4}

, the compound obtained is:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_2} = CH\mathrm{C}{\mathrm{H}_2}OH

Option B (

CH_2=CHCH_2OH

) is correct because

LiAlH_4

reduces the carboxylic acid group to a primary alcohol while leaving the alkene unaffected.

17 Easy

17 of 137

Remembering +4 | -1

Methyl ketones are characterised through:

View answer and explanation

Correct option: B

The iodoform test

Option B (The iodoform test) is correct because methyl ketones give a positive iodoform test, forming a yellow precipitate of iodoform (

CHI_3

18 Medium

18 of 137

Analyzing +4 | -1

The compound

\mathrm{X}

, in the reaction is

X\xrightarrow{{\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}}}Y\xrightarrow{{\mathrm{Hydrolysis}}}\mathrm{Mg}\left( {\mathrm{OH}} \right)\mathrm{I} + \mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{O}_2}

Option B (

\mathrm{C}{\mathrm{O}_2}

) is correct because the Grignard reagent formed from CO2 reacts with acetaldehyde to form a product that hydrolyzes to acetic acid.

19 Easy

19 of 137

Application +4 | -1

The conversion

\mathrm{PhCN} \rightarrow \mathrm{PhCOCH}_3

, can be achieved most conveniently by reaction with

View answer and explanation

Correct option: A

\mathrm{CH}_3 \mathrm{MgBr}

followed by hydrolysis

Option A (

\mathrm{CH}_3 \mathrm{MgBr}

followed by hydrolysis) is correct because Grignard reagents react with nitriles to form ketones after hydrolysis.

20 Easy

20 of 137

Application +4 | -1

Identify the organic compound which, on heating with strong solution of NaOH, partly converted into an, acid salt and partly into alcohol.

View answer and explanation

Correct option: D

Benzaldehyde

Option D (Benzaldehyde) is correct because it undergoes Cannizzaro reaction with strong NaOH, forming benzoate salt and benzyl alcohol.

21 Hard

21 of 137

Analyzing +4 | -1

The correct order of decreasing boiling points of

\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 > B

Option A (

A > D > C > B

) is correct because amides have the highest boiling points due to strong hydrogen bonding, followed by anhydrides, then carboxylic acids, and finally acid chlorides.

22 Easy

22 of 137

Analyzing +4 | -1

Under suitable conditions

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}\left( A \right),\;\,{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH}\left( B \right)

and

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COOH}\left( C \right)

can act as acids. The increasing order of their acidic strengths is:

View answer and explanation

Correct option: A

A < B < C

Option A (

A < B < C

) is correct because the acidity increases in the order: alcohol < phenol < carboxylic acid due to resonance stabilization of the conjugate base.

23 Easy

23 of 137

Understanding +4 | -1

Silver chloride dissolves in excess of

N{H_3}

solution. The cation present in solution is

View answer and explanation

Correct option: C

{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }

Option C (

{\left[ {Ag{{(N{H_3})}_2}} \right]^ + }

) is correct because silver chloride forms a complex ion with ammonia, specifically the diamminesilver(I) ion.

24 Easy

24 of 137

Remembering +4 | -1

Which reaction is used for detecting the presence of carbonyl group?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because hydrazine, phenyl hydrazine, and hydroxylamine all react with carbonyl groups to form derivatives that can be used for detection.

25 Medium

25 of 137

Understanding +4 | -1

Which of the following statement is correct?

View answer and explanation

Correct option: C

Boiling points of acids are higher than corresponding alcohols.

Option C (Boiling points of acids are higher than corresponding alcohols) is correct because carboxylic acids form stronger hydrogen bonds than alcohols due to the carbonyl group.

26 Hard

26 of 137

Understanding +4 | -1

Correct order of reactivity of acid derivatives towards a nucleophile is

View answer and explanation

Correct option: A

R\mathrm{COCl} > {\left( {R\mathrm{CO}} \right)_2}\mathrm{O} > R\mathrm{COO}R > R\mathrm{CON}{\mathrm{H}_2}

Option A (

RCOCl > (RCO)_2O > RCOOR > RCONH_2

) is correct because acid chlorides are the most reactive, followed by anhydrides, esters, and then amides, due to the leaving group ability.

27 Medium

27 of 137

Application +4 | -1

When benzyl chloride is boiled with aqueous solution of lead nitrate in current of carbon dioxide, the main product is:

View answer and explanation

Correct option: C

Benzaldehyde

Option C (Benzaldehyde) is correct because benzyl chloride is hydrolyzed to benzyl alcohol, which is then oxidized to benzaldehyde by lead nitrate.

28 Hard

28 of 137

Application +4 | -1

Butan-2-one can be converted to propanoic acid by:

View answer and explanation

Correct option: C

\mathrm{NaOH}/{\mathrm{I}_2}/{\mathrm{H}^ + }

Option C (

\mathrm{NaOH}/{\mathrm{I}_2}/{\mathrm{H}^ + }

) is correct because this represents the haloform reaction, which converts methyl ketones to carboxylic acids with one less carbon.

29 Easy

29 of 137

Remembering +4 | -1

Fehling’s solution is:

View answer and explanation

Correct option: C

Copper sulphate, Rochelle salt + NaOH

Option C (Copper sulphate, Rochelle salt + NaOH) is correct because Fehling's solution is an alkaline solution of copper(II) ions complexed with tartrate ions (from Rochelle salt).

30 Hard

30 of 137

Remembering +4 | -1

Mild oxidation of carboxylic acids occurs at…….position.

View answer and explanation

Correct option: C

\beta

Option C (

\beta

) is correct because mild oxidation of carboxylic acids typically occurs at the beta position.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 137 single-correct-answer MCQs on ALDEHYDES & KETONES & CARBOXYLIC ACIDS.
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