1

JEE Advanced 2021 Paper 2 Online

MCQ (Single Correct Answer)
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :



Cl $$-$$ Cl (g) $$\to$$ Cl$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 58 kcal mol$$-$$1

H3C $$-$$ Cl (g) $$\to$$ H3C$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 85 kcal mol$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 103 kcal mol$$-$$1
For the following reaction

CH4(g) + Cl2(g) $$\buildrel {light} \over \longrightarrow $$ CH3Cl(g) + HCl (g)

the correct statement is
A
Initiation step is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$58 kcal mol$$-$$1.
B
Propagation step involving $${}^ \bullet C{H_3}$$ formation is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$2 kcal mol$$-$$1.
C
Propagation step involving CH3Cl formation is endothermic with $$\Delta$$H$$^\circ$$ = +27 kcal mol$$-$$1.
D
The reaction is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$25 kcal mol$$-$$1.

Explanation

The reaction is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$25 kcal mol$$-$$1

CH4(g) + Cl2(g) $$\buildrel {hv} \over \longrightarrow $$ CH3Cl(g) + HCl(g)

(a) Initiation step is breaking of Cl$$-$$Cl bond which requires energy and hence, this step is endothermic.

(b) Propagation step involving formation of CH3 is also endothermic as bond between C$$-$$H is breaking.

(c) Propagation step involving formation of CH3Cl is exothermic as bond is formed between C$$-$$Cl

(d) $$\mathop H\limits^ \bullet $$ + $$\mathop {Cl}\limits^ \bullet $$ $$\to$$ HCl, $$\Delta$$H4$$^\circ$$ = $$-$$103 kcal/mol

Enthalpy of reaction, $$\Delta$$H$$^\circ$$ = $$\Delta$$H$$_1^o$$ + $$\Delta$$H$$_2^o$$ + $$\Delta$$H$$_3^o$$ + $$\Delta$$H$$_4^o$$

= 58 + 105 $$-$$ 85 $$-$$ 103

= $$-$$25 kcal mol$$-$$1

Hence, the reaction is exothermic with $$\Delta$$H$$^o$$ = $$-$$25 kcal/mol

2

JEE Advanced 2021 Paper 2 Online

MCQ (Single Correct Answer)
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :
Correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is

Column J
Molecule
Column K
BDE (kcal $$mo{l^{ - 1}}$$)
(P) H-CH($$C{H_3}$$)$$_2$$ (i) 132
(Q) H-CH$$_2$$Ph (ii) 110
(R) H-CH=CH$$_2$$ (iii) 95
(S) H-C $$ \equiv $$ CH (iv) 88
A
P - iii, Q - iv, R - ii, S - i
B
P - i, Q - ii, R - iii, S - iv
C
P - iii, Q - ii, R - i, S - iv
D
P - ii, Q - i, R - iv, S - ii

Explanation

A more stable radical means lesser the bond dissociation energy.

Radicals formed are

Radical formed from Q (PhCH2$$-$$H) is most stable due to resonance.

Stability of free radical decreases with increase in % s-character.

P radical $$\to$$ sp3 $$\to$$ % s-character 25%

R radical $$\to$$ sp2 $$\to$$ % s-character 33%

S radical $$\to$$ sp $$\to$$ % s-character 50%

Thus, order of stability of free radical is

Q > P > R > S

Order of bond dissociation energy is

S > R > P > Q

3

JEE Advanced 2021 Paper 1 Online

MCQ (Single Correct Answer)
Among the following, the conformation that corresponds to the most stable conformation of meso-butane-2,3-diol is
A
B
C
D
4

JEE Advanced 2020 Paper 1 Offline

MCQ (Single Correct Answer)
The Fischer projection of D-erythrose is shown below :



D-erythrose and its isomers are listed as P, Q, R, and S in Column - I. Choose the correct relationship of P, Q, R, and S with D-erythrose from Column - II.
A
P Q R S
2 3 2 2
B
P Q R S
3 1 1 2
C
P Q R S
2 1 1 3
D
P Q R S
2 3 3 1

Explanation

First of all, we need to convert wedge-dash formula into Fischer projection formula.



Now we can clearly see, that compound P is same as given compound, thus compound P is identical to D-erythrose.

Compound Q and D-erythrose are not mirror images of one another and are non-superimposable on one another. Thus, they are diastereomers.

Compound R and D-erythrose are also not mirror images of one another and are non-superimposable on one another. Thus, they are diastereomers.

Compound S and D-erythrose are chiral molecules. They are mirror images of one another. Furthermore, the molecules are non-superimposable on one another. This means that the molecules cannot be placed on top of one another and give the same molecule. Chiral molecules with one or more stereocenters can be enantiomers.

Hence, the correct option is (c).

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