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 ( kcalmol−1 )
(P) H−CH(CH3)2
(i) 132
(Q) H−CH2Ph
(ii) 110
(R) H−CH=CH2
(iii) 95
(S) H−C≡CH
(iv) 88
(A)P - iii, Q - iv, R - ii, S - i✓ Correct
(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
Answer: (A)
Solution
BDE decreases with increasing radical stability. (P) H−CH(CH3)2 forms a secondary radical (most stable among P, Q) → lowest BDE among P, Q → 95 (iii). (Q) H−CH2Ph forms a benzylic radical (resonance-stabilized) → lower BDE → 88 (iv). (R) H−CH=CH2 forms a vinylic radical (sp² carbon, less stable) → higher BDE → 110 (ii). (S) H−C≡CH forms an alkynic radical (sp carbon, least stable, highest s-character) → highest BDE → 132 (i). Order of BDE: Q < P < R < S, matching option (A).
JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Single correct option
