Match List I with List II
LIST I

A.Gatterman-Koch reaction
B.Stephens reaction
C.Hoffmann Bromamide degradation D.Fittig reaction LIST II I.Diphenyl

II.Ethyl amine

III.Benzaldehyde

IV.Acetaldehyde

Choose the correct answer from the options given below:

This question was previously asked in

Shift 27/05/2023 3:30 PM - 6:30 PM

A-IV, B-III, C-II, D-I

A-III, B-IV, C-II, D-I

A-III, B-II, C-IV, D-I

A-IV, B-III, C-I, D-II

Correct Answer

A-III, B-IV, C-II, D-I

Let's match the reactions in List I with their respective products in List II:

A. Gatterman-Koch reaction:

This reaction involves the formylation of aromatic compounds. It converts benzene to benzaldehyde (C6H5CHO).
So, A matches with III (Benzaldehyde).

B. Stephens reaction:

The Stephens reaction is used to convert a nitrile into an amine. In this case, it converts ethyl nitrile (CH3CH2CN) to ethylamine (C2H5NH2).
B matches with Iv (Ethylamine).

C. Hoffmann Bromamide degradation:

This reaction converts amides to primary amines. It converts acetamide (CH3CONH2) to methylamine (CH3NH2).
C matches with II (Acetaldehyde).

D. Fittig reaction:

The Fittig reaction is a coupling reaction between two aryl halides to form a biphenyl compound. For example, it can convert two phenyl halides to diphenyl.
D matches with I (Diphenyl).

So, the correct matching is as follows:

Practice on the go with our mobile app

A group 14 element is converted into n-type semiconductor by doping it with :

CUET ki Practice Means DuBuddy Pe Practice