Carboxylic Acids Test

Result

Carboxylic Acids Test - 9

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

Only One Option Correct Type
Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
The major organic product in the following reaction is

A

B

C

D

Solution
Question 2
1 / -0.25

Consider the following reaction,

A

B

C

D

Solution
Question 3
1 / -0.25

Which is the best hydride (H^- ) donor in the key step of Cannizaro reaction?

A

B

C

D

Solution

Dianion is better hydride donor. Also the electron donating group (CH₃ O) increases hydride (H^- ) donating a bility .
Question 4
1 / -0.25

Consider the following reaction,
Q.
The major organic product is

A

B
CH₃ CH₂ OH + CH₃ COOH

C
CH₃ CH₂ OH

D
CH₃ COOC₂ H₆

Solution

It is a simple Tischenko reaction
Question 5
1 / -0.25

What is the major organic product in the following reaction?

A
CH₃ —C(CH₂ OH)₃

B
CH₃ OH

C
HCOOH

D
Both (a) and (c)

Solution
Question 6
1 / -0.25

Predict the major organic product in the following reaction.

A

B

C

D

Solution

Intramolecular Cannizaro reaction occur.
Question 7
1 / -0.25

Whta is the major organic product formed in the follwing reaction ?

A

B

C

D

Solution
Question 8
1 / -0.25

For a Cannizaro reaction,

Rate law is derived as : Rate = k [RCHO]² [HO^- ]²
From the above rate law, it can be concluded that

A
reaction is pseudo 2nd orde

B
reaction involves hydride ion transfer

C
reaction does not show kinetic isotopic effect

D
hydride donor is a dianion

Solution

For formation of dianion hydride donor, an additional mole of NaOH is consumed

Hence, rate becomes 2nd order with respect to HO^- and fourth order overall.
Question 9
1 / -0.25

Consider the following reaction,

Q.
How the product Z can be prepared selectively using X and Y and other reagents?

A
Treating Y first with LDA followed by subsequent addition of X

B
Treating X first with LDA followed by subsequent addition of Y

C
Treating mixture of X and Y with LDA

D
Treating mixture of X and Y with NaOH and continuous removal of Z [LDA = Lithium diisopropyl amide [(CH₃ )₂ CH₂ NLi]

Solution
Question 10
1 / -0.25

What is the major product formed in the following intramolecular Cannizaro reaction?

A

B

C

D

Solution

—CHO para to electron withdrawing —NO₂ group is better hydride (H^- ) acceptor.