Which can reduce $R{\rm{COOH}}\, \to \,\,R{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$ ?

${\rm{B}}{{\rm{H}}_3}/{\rm{THF/}}{{\rm{H}}_3}{{\rm{O}}^ + }$

${\rm{Na}}/{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

${{\rm{H}}_2}/{\rm{ }}catalyst$

Chlorobenzene $\xrightarrow[X]{{{\rm{Reaction\;\;}}}}{\rm{\;Phenol\;}}\xrightarrow[Y]{{{\rm{Reaction\;\;}}}}$ Salicyladehyde $X\;and\;Y$ reactions are respectively……….

Fires rearrangement and Kolbe-Schmidt

Back to ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Your Progress

0 / 34 Answered

Correct

Incorrect

Accuracy

JEE • CHEMISTRY • ALDEHYDES & KETONES & CARBOXYLIC ACIDS

Aldehydes & ketones & carboxylic acids practice · Hard

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →

Showing 1-30 of 34 questions

Preparing practice and saved progress…

1 Hard

1 of 34

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.

2 Hard

2 of 34

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.

3 Hard

3 of 34

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.

4 Hard

4 of 34

Understanding +4 | -1

The decreasing order of solubility of methanal (A), propanaldehyde (B), benzaldehyde (C) and acetophenone (D):

View answer and explanation

Correct option: A

A > B > C > D

Option A is correct. Solubility decreases with increasing molecular weight due to the increasing hydrophobic character of alkyl or aryl groups. Methanal (A) is the most soluble, followed by propanaldehyde (B), benzaldehyde (C), and acetophenone (D).

5 Hard

5 of 34

Analyzing +4 | -1

The ease of hydrolysis with an alkali in the compounds

\begin{array}{l} \mathrm{C}{\mathrm{H}_3}\mathrm{COClC}{\mathrm{H}_3}\;\mathrm{CO} - \mathrm{O} - \mathrm{COC}{\mathrm{H}_3}\\ \;I\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;II\\ \mathrm{C}{\mathrm{H}_3}\mathrm{COO}{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_3}\mathrm{CON}{\mathrm{H}_2}\\ \;III\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;IV \end{array}

Is of the order

View answer and explanation

Correct option: A

I>II>III>IV

Option A (I>II>III>IV) is correct because acid chlorides are most reactive towards hydrolysis, followed by anhydrides, esters, and amides.

6 Hard

6 of 34

Understanding +4 | -1

Assertion (A) Acyl halide are more reactive than acid substance amide toward nucleophilic substitution.

Reason

(\mathrm{R}) \mathrm{X}^{-}

are better leaving group than

\mathrm{NH}_2

View answer and explanation

Correct option: B

Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Option B (Both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the better leaving group is not the ONLY reason for the higher reactivity of acyl halides.

7 Hard

7 of 34

Analyzing +4 | -1

Statement-I: Ketones can not be prepared by Rosenmund reduction.

Statement-II: Aldehydes are less reactive than ketones towards nucleophilic addition.

View answer and explanation

Correct option: A

Statement-I is true and statement-II is false

Option A (Statement-I is true and statement-II is false) is correct because ketones cannot be prepared by Rosenmund reduction, and aldehydes are more reactive than ketones towards nucleophilic addition.

8 Hard

8 of 34

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 distinguishes between aldehydes and ketones, with aldehydes giving a silver mirror test.

9 Hard

9 of 34

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.

10 Hard

10 of 34

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}}

Option C is correct. The presence of electron-withdrawing groups increases acidity, and the absence of electron-donating groups makes it the most acidic.

11 Hard

11 of 34

Analyzing +4 | -1

Order of Acidic strength of following carboxylic acids is

\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).

12 Hard

12 of 34

Understanding +4 | -1

Which can reduce

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

View answer and explanation

Correct option: C

\mathrm{B}{\mathrm{H}_3}/\mathrm{THF/}{\mathrm{H}_3}{\mathrm{O}^ + }

Option C (

\mathrm{B}{\mathrm{H}_3}/\mathrm{THF/}{\mathrm{H}_3}{\mathrm{O}^ + }

) is correct because

BH_3

in THF followed by hydrolysis is a reagent that can reduce carboxylic acids to primary alcohols.

13 Hard

13 of 34

Analyzing +4 | -1

{\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.

14 Hard

14 of 34

Remembering +4 | -1

Wacker method is used to convert alkene into corresponding………using

\mathrm{PbC}{\mathrm{l}_2}

View answer and explanation

Correct option: C

Aldehyde

Option C (Aldehyde) is correct because the Wacker method is used to convert alkenes into corresponding aldehydes.

15 Hard

15 of 34

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 is correct because the conversion of benzaldehyde to 3-phenylprop-2-en-1-oic acid (cinnamic acid) using acetic anhydride in the presence of a base is the Perkin condensation.

Explore Other Chapters in CHEMISTRY

View All →

P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs CHEMICAL BONDING & MOLECULAR STRUCTURE 514 MCQs SOLUTIONS 480 MCQs

16 Hard

16 of 34

Understanding +4 | -1

Statement I : Compounds containing -CHO group are easily oxidised to corresponding carboxylic acids
Statement II : Carboxylic acids can be reduced to alcohols by treatment with

LiAl{H_4}

View answer and explanation

Correct option: A

Both Statement I and II are correct

The correct answer is A (Both statement I and II are correct) because aldehydes can be oxidized to carboxylic acids, and carboxylic acids can be reduced to alcohols using

LiAlH_4

17 Hard

17 of 34

Understanding +4 | -1

Given below are some statements concerning formic acid, which of them is true?

View answer and explanation

Correct option: B

It is reducing agent

Option B (It is reducing agent) is correct because formic acid can reduce Tollen’s reagent, Fehling solution, mercuric chloride and

\mathrm{KMnO}_4

etc.

18 Hard

18 of 34

Understanding +4 | -1

Electrophile involved in Reimer – Tiemann reaction is

View answer and explanation

Correct option: C

:CC{l_2}

Option C is correct because the electrophile involved in the Reimer-Tiemann reaction is dichlorocarbene (:

{CC{l_2}}

19 Hard

19 of 34

Remembering +4 | -1

Magenta is:

View answer and explanation

Correct option: C

\mathrm{p - }

rosaniline hydrochloride

Option C (

\mathrm{p - }

rosaniline hydrochloride) is correct because Magenta is

\mathrm{p - }

rosaniline hydrochloride, a dye that is decolorized by

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}

to give Schiff’s reagent.

20 Hard

20 of 34

Analyzing +4 | -1

A compound A has a molecular formula

{\mathrm{C}_2}\mathrm{C}{\mathrm{l}_3}\mathrm{OH}\mathrm{.}

It reduces Fehling’s solution and on oxidation, gives a monocarboxylic acid B. A can be obtained by the action of chlorine on ethyl alcohol. A is

View answer and explanation

Correct option: B

Chloral

Option B (Chloral) is correct because chloral (CCl3CHO) reduces Fehling's solution and gives a monocarboxylic acid on oxidation, and it can be obtained by chlorination of ethanol.

21 Hard

21 of 34

Understanding +4 | -1

Which of the following is not true about the urea?

View answer and explanation

Correct option: C

It cannot be used for all types of crops and soils

Option C (It cannot be used for all types of crops and soils) is correct because urea can be used for all types of crops and soils.

22 Hard

22 of 34

Understanding +4 | -1

Benzaldehyde reacts with excess of anhydrous ethyl alcohol in the presence of

HCl

, gives:

View answer and explanation

Correct option: C

{C_6}{H_5}CH{\left( {O{C_2}{H_5}} \right)_2}

Option C (

{C_6}{H_5}CH{\left( {O{C_2}{H_5}} \right)_2}

) is correct because benzaldehyde reacts with excess ethanol in the presence of HCl to form an acetal.

23 Hard

23 of 34

Understanding +4 | -1

Chlorobenzene

\xrightarrow[X]{{\mathrm{Reaction\;\;}}}\mathrm{\;Phenol\;}\xrightarrow[Y]{{\mathrm{Reaction\;\;}}}

Salicyladehyde

X\;and\;Y

reactions are respectively……….

View answer and explanation

Correct option: C

Dow and Reimer-Tiemann

Option C is correct. Chlorobenzene is converted to phenol by Dow's process, and phenol is converted to salicylaldehyde by the Reimer-Tiemann reaction.

24 Hard

24 of 34

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 the boiling points are determined by intermolecular forces: amides have strong hydrogen bonding, followed by anhydrides, carboxylic acids (dimer formation), and then acyl chlorides.

25 Hard

25 of 34

Analyzing +4 | -1

Which of the following products is not possible in the reaction given below?

\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

\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COCH}_3

The correct answer is D (

\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.

26 Hard

26 of 34

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.

27 Hard

27 of 34

Remembering +4 | -1

Fehling’s solution is used in the detection of:

View answer and explanation

Correct option: C

Aldehydic group

Option C (Aldehydic group) is correct because Fehling's solution is a reagent used to detect the presence of aldehydes, which are oxidized to carboxylic acids.

28 Hard

28 of 34

Understanding +4 | -1

Which of the following have high melting points?

View answer and explanation

Correct option: A

Acids containing even number of carbon atoms

Option A (Acids containing even number of carbon atoms) is correct because even-numbered carbon acids pack more efficiently in the crystal lattice, leading to higher melting points.

29 Hard

29 of 34

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 is correct because the reactivity of acid derivatives towards nucleophilic acyl substitution depends on the leaving group ability, with acid chlorides being the most reactive and amides the least.

30 Hard

30 of 34

Remembering +4 | -1

Acetophenone on oxidation by perbenzoic acid gives phenyl acetate. The reaction is named as:

View answer and explanation

Correct option: A

Baeyer-Villiger oxidation

Option A (Baeyer-Villiger oxidation) is correct because the reaction of acetophenone with perbenzoic acid to form phenyl acetate is a Baeyer-Villiger oxidation, which involves the insertion of an oxygen atom into a carbonyl compound.

Revision notes and key takeaways

Review ALDEHYDES & KETONES & CARBOXYLIC ACIDS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 34 single-correct-answer MCQs on ALDEHYDES & KETONES & CARBOXYLIC ACIDS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

Showing 1-30 of 34 questions

Your Progress

Aldehydes & ketones & carboxylic acids practice · Hard

Explore Other Chapters in CHEMISTRY

Key Takeaways

Continue Your Practice

Your practice