Question

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 mol–1 )
P	\(\text{H--CH(CH}_3\text{)}_2\)	i	132
Q	\(\text{H--CH}_2\text{Ph}\)	ii	110
R	\(\text{H--CH}=CH_2\)	iii	95
S	\(\text{H--C≡CH}\)	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 – iii
• A. Initiation step is exothermic with \(\Delta\)H° = –58 kcal mol^–1
• B. Propagation step involving ^·CH₃ formation is exothermic with \(\Delta\)H° = –2 kcal mol^–1
• C. Propagation step involving CH₃Cl formation is endothermic with \(\Delta\)H° = +27 kcal mol^–1
• D. The reaction is exothermic with \(\Delta\)H° = –25 kcal mol^–1

Answer 1

The Correct Option is A

Order of stability of free radical
\(Q > P > R > S\)
\(\text{Stability of free radical } \alpha \frac{1}{\text{Bond energy}}\)

∴ Order of bond energy :
\(S > R > P > Q\)

Answer 2

Step (1) → Endothermic (bond breaking)

Step (2) → ∆H = 105 – 103 = 2 kcal/mol (Endothermic)

Step (3) → ∆H = 58 – 85 = –27 kcal/mol (Exothermic)

For complete reaction

\(\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \overset{\text{light}}{\longrightarrow} \text{CH}_3\text{Cl (g)} + \text{HCl(g)}\)

∆H = 58 + 105 – 85 – 103

= –25 kcal/mol