1

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

The decreasing order of ease of alkaline hydrolysis for the following esters in

A
III > II > IV > I
B
IV > II > III > I
C
III > II > I > IV
D
II > III > I > IV

## Explanation

More is the electrophilic character of carbonyl group of ester faster is the alkaline hydrolysis.

Since –I effect groups increase the electrophilic nature of carboxyl carbon whereas +I effect groups decrease the electrophilic nature of carboxyl carbon. Therefore, the correct decreasing order is III > II > I > IV.
2

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

What is the IUPAC name of the following compound ?

A
4–Bromo-3-methylpent-2-ene
B
3–Bromo 1, 2-dimethylbut-1-ene
C
3–Bromo-3-methyl-1, 2-dimethylprop-1-ene
D
2-Bromo-3-methylpent-3-ene

## Explanation

Note : While numbering, double bond should get least possible number on the longest path.
3

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

A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol–1]
A
33.6
B
0.84
C
8.4
D
16.8

## Explanation

2NaHCO3 + (COOH)2 $\to$ (COONa)2 + 2H2O + 2CO2

$\therefore$ 1 mole of CO2 is produced by 1 mole of NaHCO3.

Given, volume of CO2 produced = 0.25 ml

25 L of CO2 contains 1 mol

$\therefore$ 0.25 ml of CO2 contains = ${1 \over {25 \times {{10}^3}}} \times 0.25$ moles

= 10-5 moles

$\therefore$ Moles of NaHCO3 = 10-5 moles

Mass of NaHCO3 = 10-5 $\times$ 84 g

$\therefore$ % of NaHCO3 in a tablet

= ${{84 \times {{10}^{ - 5}}} \over {10 \times {{10}^{ - 3}}}} \times 100$ = 8.4 %
4

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

In the following compound,

the favourable site/s for protonation is/are :
A
(b), (c) and (d)
B
(a) and (e)
C
(a) and (d)
D
(a)

## Explanation

At (a) position lone pair of N atom is in resonance with $\pi$ bond so lone pair can't donate easily to H+ ion.

At (b) position N atom is directly connected with double bond so lone pair of N atom don't perticipate in resonance so lone pair can be donated easily to H+ ion.

At (c) position N atom is directly connected with double bond so lone pair of N atom don't perticipate in resonance so lone pair can be donated easily to H+ ion.

At (d) position N atom is directly connected with double bond so lone pair of N atom don't perticipate in resonance so lone pair can be donated easily to H+ ion.

At (e) position lone pair of N atom is in resonance with $\pi$ bond so lone pair can't donate easily to H+ ion.

$\therefore$ (b), (c) and (d) are the favourable sites for protonation.