1

JEE Main 2019 (Online) 9th January Morning Slot

The compounds A and B in the following reaction are, respectively :

A
A = Benzyl alcohol, B = Benzyl cyanide
B
A = Benzyl chloride, B = Benzyl cyanide
C
A = Benzyl alcohol, B = Benzyl isocyanide
D
A = Benzyl chloride, B = Benzyl isocyanide

Explanation

2

JEE Main 2019 (Online) 9th January Morning Slot

The major product of following reaction is :

A
RCOOH
B
RCONH2
C
RCHO
D
RCH2NH2

Explanation

3

JEE Main 2019 (Online) 9th January Evening Slot

The major product obtained in the following reaction is :

A
B
C
D

Explanation

Nucleophilicity of NH2 $>$ OH
4

JEE Main 2019 (Online) 10th January Morning Slot

The major product of the following reaction is

A
B
C
D

Explanation

Alcoholic KOH performs $\alpha$ - $\beta$ elimination with halides.

The carbon which is attached to Br atom is called $\alpha$ carbon.

Here, two $\alpha$ carbon presents. Let's call them $\alpha$1 and $\alpha$2.

In case of $\alpha$1 carbon, H elimination can take place either from $\beta$1 or $\beta$2 carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $\beta$1 carbon then a $\pi$ bond is created which will perticipate in resonance with benzene ring and product will be more stable. But if H is removed from $\beta$2 carbon then the created $\pi$ bond will not perticipate in resonance with benzene ring so product will be less stable.

In case of $\alpha$2 carbon, H elimination can take place either from $\beta$2 or $\beta$3 carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $\beta$2 carbon then a $\pi$ bond is created which will perticipate in resonance with the $\pi$ bond associated with the $\alpha$1 carbon so product will be more stable. But if H is removed from $\beta$3 carbon then the created $\pi$ bond will not perticipate in any resonance so product will be less stable.