1

### 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
2

### 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.
3

### JEE Main 2019 (Online) 10th January Evening Slot

An aromatic compound 'A' having molecular formula C7H6O2 on treating with aqueous ammonia and heating forms compound 'B'. The compound 'B' on reaction with molecular bromine and potassium hydroxide provides compound 'C' having molecular formula C6H7N.. The structure of 'A' is
A
B
C
D

## Explanation

4

### JEE Main 2019 (Online) 11th January Evening Slot

A compound 'X' on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound 'X' is :
A
CH3CH2CH2CONH2
B
CH3CON(CH3)2
C
CH3CH2COCH2NH
D
CH3COCH2NHCH3

## Explanation

Br2/NaOH(Hoffman Hypobromide reagent) converts amide into primary amine having one carbon atom less, which gives carbylamine test. In the Amide there should be
-CONH2 group to produce primary amine.