Carboxylic Acids Test

Result

Carboxylic Acids Test - 10

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Question 1
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One or More than One Options Correct Type

Direction (Q. Nos. 1-5) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Consider the cross Cannizaro reaction given below

Q.

The expected product(s) is/are

A

B

C
C₆ H₅ CH₂ OH

D
All are correct

Solution

(I) is formed by nucleophilic attack in the first step. (I) may undergo intramolecular hydride transfer giving option (a) or may undergo intermolecuiar hydride transfer to C₆ H₅ CHO giving option (c) and option (d).
Question 2
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In the Cannizaro reaction mentioned below

Q.

The possible product(s) is/ar

A
CH₃ OH

B
CH₃ OD

C
HCOOD

D
All are correct

Solution
Question 3
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What is/are the expected organic product(s) in the following reaction?

A

B

C
HCOOH

D
Option B,C are Correct

Solution

It is a case of cross Cannizaro reaction in which formaldehyde is always oxidised and the other aldehyde is reduced. So, correct option B and D.
Question 4
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Consider the following reaction,

Q.

The expected organic product(s) is/are

A
CH₃ COOCH₂ CH₂ CH

B
CH₃ CH₂ COOCH₂ CH

C
CH₃ COOCH₂ CH₃

D
All are correct

Solution

It is a case of cross Tischenko reaction.

Here, both can be oxidised or both can be reduced. In the next step ester formation occur by all possible combination of acid and alcohols.
Question 5
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Consider the following reaction sequence,

Q.

The expected organic product(s) is/are

A

B

C

D
Option A,C are Correct

Solution
Question 6
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Comprehension Type

Direction (Q. Nos. 6 - 8) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer mong the four given options (a), (b), (c) and (d).

Passage

When aldehydes lacking a-hydrogen is treated with concentrated solution of a strong base, Cannizaro reaction takes place. In this reaction, one molecule of aldehyde is oxidised while other is reduced. The widely accepted mechanism is

Step II

Q.

What is the rate law derived from the above mechanism?

A
R = k [A —CHO] [HO^- ]²

B
R = k [A —CHO]² [HO^- ]²

C
R = k [A —CHO]² [HO^- ]

D
R = k [A —CHO]²

Solution

The intermediate “I ”can be determined as
Substituting [I] in Eq (i) gives
R = k₁ K [ACHO]² [HO^- ] = k [ACHO]² [OH^- ']
Question 7
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When aldehydes lacking a-hydrogen is treated with concentrated solution of a strong base, Cannizaro reaction takes place. In this reaction, one molecule of aldehyde is oxidised while other is reduced. The widely accepted mechanism is

Step II

Q.

Which of the following observed fact establishes the correctness of the above mechanism?

A
If A —COD is used, reaction occur at same rate provided all other conditions are same

B
In mixed Cannizaro reaction, if one aldehyde is H₂ CO, it is always oxidise

C
If reaction is carried out in D₂ O in the presence of NaOD, some C —D bond formation occur

D
If reaction is carried out in D₂ Oin the presence of NaOD, no C —D bond formation occur

Solution

When reaction is carried out in D,O , no C —D bond formation occur. It establishes the correctness of above mechanism that reaction proceed by H^- (hydride) transfer in the slow rate determining step.
Question 8
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When aldehydes lacking a-hydrogen is treated with concentrated solution of a strong base, Cannizaro reaction takes place. In this reaction, one molecule of aldehyde is oxidised while other is reduced. The widely accepted mechanism is

Step II

Q.

In certain experim ental condition, rate is found to be proportional to square of concentration of base (OH^- ).This indicates that

A
with change in condition, Step I may become slow, rate determining step

B
dianion intermediate is acting as hydride (H^- ) don

C
Both Step I and Step II occur at comparable rate

D
mechanism changes with nature of aldehyde

Solution

Formation of dianion involves two moles of HO^- per mole of aldehyde
Question 9
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Matching List Type

Direction (Q. Nos. 9 and 10) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

Q.

Consider the reactions of Column I and match with the products of Column II. Mark the correct option from the codes given below.

A
a

B
b

C
c

D
d

Solution

(i) is better H^- acceptor, reduced to FCH₂ OH.
(ii) CH₂ O is oxidised and C₆ H₅ CHO is reduced in cross Cannizaro reaction.
(iii) Electron withdrawing —NO₂ makes —CHO a better hydride acceptor, hence p-nitrobenzaldehyde is reduced and benzaldehyde is oxidised.
(iv) p-nitrobenzaldehyde is reduced, p-methoxy benzaldehyde becomes better hydride donor after attack by HO^- , hence oxidised.
Question 10
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Match the reactions of Column I with the type of reactions from Column II. Mark the correct option from the codes given below.

A
a

B
b

C
c

D
d

Solution

(i) Initially, aldol reaction followed by Cannizaro reaction giving C(CH₂ OH)4 + HCOOH.
(ii) F —CHO undergo Cannizaro reaction due to absence of α-H.
(iii) It has difficulty in aldol condensation, hence undergo Cannizaro reaction predominantly.
(iv) All aldol, Cannizaro and Claisen reaction occur.
Question 11
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One Integer Value Correct Type

Direction (Q. Nos. 11 - 13) This section contains 3 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Consider the following Cannizaro reaction,

Q.

How many different products are formed?

A
6

B
5

C
4

D
1

Solution

Possible products are
Question 12
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Consider the following Cannizaro reaction,

Q.

How many different products are expected in this reaction?

A
0

B
5

C
9

D
8

Solution

Possible products are
Question 13
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Consider the following reaction,

Q.

If CH₃ OH is formed exclusively by hydride transfer from a dianion intermediat what is the overall order of reaction?

A
0

B
9

C
4

D
5

Solution

If dianion is hydride donor, the rate law is
R = k [H₂ CO]² [OH^- ]²
Hence, reaction is overall fourth order.