JEE Advanced 2021 Paper 2 Q13 Chemistry Thermodynamics & Thermochemistry Bond Enthalpy Medium

JEE Advanced 2021 Paper 2 · Q13 · Bond Enthalpy

Correct match of the C–H bonds (shown in bold) in Column J with their BDE in Column K is:

Column J (Molecule): (P) H–CH(CH$_3$)$_2$ (secondary C–H of isopropyl, sp$^3$, propan-2-yl radical formed) (Q) H–CH$_2$Ph (benzylic C–H of toluene, sp$^3$, benzyl radical formed) (R) H–CH=CH$_2$ (vinylic C–H of ethene, sp$^2$, vinyl radical formed) (S) H–C$\equiv$CH (acetylenic C–H of ethyne, sp, ethynyl radical formed)

Column K (BDE, kcal mol$^{-1}$): (i) 132; (ii) 110; (iii) 95; (iv) 88

  1. A. P – iii, Q – iv, R – ii, S – i
  2. B. P – i, Q – ii, R – iii, S – iv
  3. C. P – iii, Q – ii, R – i, S – iv
  4. D. P – ii, Q – i, R – iv, S – iii
Reveal answer + step-by-step solution

Correct answer:A

Solution

Two trends decide the BDE: 1. s-character of the C-H bond: higher s-character $\Rightarrow$ shorter, stronger bond. sp (50%) > sp$^2$ (33%) > sp$^3$ (25%). 2. Stability of the radical formed: more stable radical $\Rightarrow$ lower BDE. Benzyl > tertiary $\approx$ secondary alkyl (no resonance) for sp$^3$.

So: (S) H-C$\equiv$CH (sp, ethynyl radical, no resonance, sp s-character highest): BDE = 132 $\Rightarrow$ (i). (R) H-CH=CH$_2$ (sp$^2$, vinyl): BDE = 110 $\Rightarrow$ (ii). (P) H-CH(CH$_3$)$_2$ (sp$^3$, isopropyl/secondary radical, no resonance): BDE = 95 $\Rightarrow$ (iii). (Q) H-CH$_2$Ph (sp$^3$ but benzylic, resonance-stabilised benzyl radical, weakest of the four): BDE = 88 $\Rightarrow$ (iv).

Hence the matching is P-iii, Q-iv, R-ii, S-i, which is option (A).

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