Practice Start Test

Aldehydes, Ketones and Carboxylic Acids

JEE Chemistry · 33 questions · Page 1 of 4 · Click an option or "Show Solution" to reveal answer

Q1
Which one of the following reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon?
A Ethyl acetate
B Butan -2-one
C Acetamide
D Acetic acid
Correct Answer
Option B
Solution

It is Clemmensen's reduction

Q2
Which one of the following reactions will not yield propionic acid?
A CH3CH2COCH3 + OI−-/H3O+
B CH3CH2CH3 + KMnO4(Heat), OH−-/H3O+
C CH3CH2CCl3 + OH−-/H3O+
D CH3CH2CH2Br + Mg, CO2 dry ether/H3O+
Correct Answer
Option D
Solution

All gives propanoic acid as product but option (d) gives butanoic as product

Q3
The correct statement about the synthesis of erythritol (C(CH2OH)4) used in the preparation of PETN is :
A The synthesis requires four aldol condensations between methanol and ethanol.
B The synthesis requires two aldol condensations and two Cannizzaro reactions.
C The synthesis requires three aldol condensations and one Cannizzaro reaction.
D Alpha hydrogens of ethanol and methanol are involved in this reaction.
Correct Answer
Option C
Solution

The synthesis of erythritol from aldehyde requires three aldol condensation and one Cannizzaro reaction.

Q4
2, 4-DNP test can be used to identify :
A ether
B amine
C aldehyde
D halogens
Correct Answer
Option C
Solution

2,4 DNP test is used to identify – (CO) – group.

It gives addition reaction with carbonyl compounds.

So, it can be used to identify aldehyde in the given option.

It gives yellow/ orange ppt with carbonyl containing compounds.

Q5
Number of molecules from below which cannot give iodoform reaction is : Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol.
A 4
B 3
C 5
D 2
Correct Answer
Option A
Solution

The molecules listed below do not undergo the iodoform reaction: Butanal 2-Pentanone Pentanal 3-Pentanol

Q6
When CH2 = CH - COOH is reduced with LiAlH4, the compound obtained will be
A CH2 = CH - CH2OH
B CH3 - CH2 - CH2OH
C CH3 - CH2 - CHO
D CH3 - CH2 - COOH
Correct Answer
Option A
Solution
LiAlH4LiAl{H_4}

can reduce

COOHCOOH

group and not the double bond.

Q7
On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is
A CH3COOC2H5 + NaCl
B CH3Cl + C2H5COONa
C CH3COCl + C2H5OH + NaOH
D CH3COONa + C2H5OH
Correct Answer
Option A
Solution

There is no reaction hence the resultant mixture contains

CH3COOC2H5+NaCl.C{H_3}COO{C_2}{H_5} + NaCl.
Q8
Given below are two statements : Statement I : Ethyl pent-4-yn-oate on reaction with CH3MgBr gives a 3∘^\circ-alcohol. Statement II : In this reaction one mole of ethyl pent-4-yn-oate utilizes two moles of CH3MgBr. In the light of the above statements, choose the most appropriate answer from the options given below :
A Both Statement I and Statement II are false.
B Statement I is false but Statement II is true.
C Statement I is true but Statement II is false.
D Both Statement I and Statement II are true.
Correct Answer
Option C
Solution

Statement I is true But it consume 3 moles of CH3MgBr. So Statement II is false.

Q9
Experimentally reducing a functional group cannot be done by which one of the following reagents?
A Pt-C/H2
B Na/H2
C Pd-C/H2
D Zn/H2O
Correct Answer
Option B
Solution

Na in presence of H2, will not release electron which are required for reduction.

H2 gas also not get adsorbed on Na.

Hence Na/H2 cannot be used as a reducing agent.

Q10
Assertion A : Enol form of acetone [CH3COCH3] exists in < 0.1% quantity. However, the enol form of acetyl acetone [CH3COCH2OCCH3] exists in approximately 15% quantity. Reason R : Enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone. Choose the correct statement :
A Both A and R are true but R is not the correct explanation of A
B A is true but R is false
C A is false but R is true
D Both A and R are true and R is the correct explanation of A
Correct Answer
Option D
Solution

Acetyl acetone in enol form have intramolecular H-bonding, which is absent in acetone.

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