Correct match of the {C}-{H} bonds (shown in bold) in Column J with their BDE in Column K is(S) {H}-{C} {CH} Column J Molecule Column K BDE (kcal mol ^{-1} ) (P) {H}-{CH}({CH}₃)₂ (i) 132 (Q) {H}-{CH}₂{Ph} (ii) 110 (R) {H}-{CH}={CH}₂ (iii) 95 (S) {H}-{C} {CH} (iv) 88 [JEE Adv 2021 P2]

Correct Answer is : P−iii,Q−iv,R−ii,S−iP-{iii}, Q-{iv}, R-{ii}, S-{i}P−iii,Q−iv,R−ii,S−i

Correct match of the {C}-{H} bonds (shown in bold) in Column J with their BDE in Column K is(S) {H}-{C} {CH} Column J Molecule Column K BDE (kcal mol ^{-1} ) (P) {H}-{CH}({CH}₃)₂ (i) 132 (Q) {H}-{CH}₂{Ph} (ii) 110 (R) {H}-{CH}={CH}₂ (iii) 95 (S) {H}-{C} {CH} (iv) 88 [JEE Adv 2021 P2]

Correct Answer is : P−iii,Q−iv,R−ii,S−iP-{iii}, Q-{iv}, R-{ii}, S-{i}P−iii,Q−iv,R−ii,S−i

Test Index

JEE Advanced 2021 Paper 2

Paragraph for next 2 questions

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 :

\mathrm{Cl}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{\bullet}(\mathrm{g})\;\Delta H^{\circ}=58\;\mathrm{kcal}\;\mathrm{mol}^{-1}

\mathrm{H}_{3}\mathrm{C}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{H}_{3}\mathrm{C}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{\bullet}(\mathrm{g})\;\Delta H^{\circ}=85\;\mathrm{kcal}\;\mathrm{mol}^{-1}

\mathrm{H}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{H}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{*}(\mathrm{g})\;\Delta H^{\circ}=103\;\mathrm{kcal}\;\mathrm{mol}^{-1}

Section: Chemistry

Question : 32 of 57

Marks: +1, -0

Correct match of the

\mathrm{C}-\mathrm{H}

bonds (shown in bold) in Column

J

with their BDE in Column

K

is(S)

\mathrm{H}-\mathrm{C}\equiv \mathrm{CH}

Column J Molecule	Column K BDE (kcal mol $^{-1}$ )
(P) $\mathrm{H}-\mathrm{CH}(\mathrm{CH}_3)_2$	(i) 132
(Q) $\mathrm{H}-\mathrm{CH}_2\mathrm{Ph}$	(ii) 110
(R) $\mathrm{H}-\mathrm{CH}=\mathrm{CH}_2$	(iii) 95
(S) $\mathrm{H}-\mathrm{C}\equiv \mathrm{CH}$	(iv) 88

[JEE Adv 2021 P2]

$P-\text{iii}, Q-\text{iv}, R-\text{ii}, S-\text{i}$
$P-\text{i}, Q-\text{i}, R-\text{iii}, S-\text{iv}$
$P-\text{iii}, Q-\text{ii}, R-\text{i}, S-\text{iv}$
$P-\text{ii}, Q-\text{i}, R-\text{iv}, S-\text{iii}$

Validate

Solution:

Stability of free radical

\propto \frac{1}{\text{Bond energy}}

(Q)

\mathrm{Ph}-\mathrm{CH}_2-\mathrm{H} \rightarrow \underset{\text{Most stable dueto resonance}}{\mathrm{Ph}-\overset{\bullet}{\mathrm{CH}}_2 + \mathrm{H}^{\bullet}}

(R)

\mathrm{CH}_2=\mathrm{CH}-\mathrm{H} \rightarrow \underset{\text{Less stable}}{\mathrm{CH}_2=\overset{\bullet}{\mathrm{C}}\mathrm{H} + \mathrm{H}^{\bullet}}

(S)

\mathrm{CH}\equiv \mathrm{CH}-\mathrm{H} \rightarrow \underset{\text{More \% S-Character decreases stability of free radical}}{\mathrm{CH}=\overset{\bullet}{\mathrm{C}} + \mathrm{H}^{\bullet}}

Qrequire least BDE and Srequired maximum BDE So, order of BDE is

Q< P< R< S

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