JEE Challenger
More from Organic Chemistry - Basic Principles
ChemMedAdv 2021 • Paper 2

Matching Carbon Hydrogen Bond Dissociation Energies for Organic Molecules

Select the correct match of the C−H\mathbf{C-H} bonds (highlighted in bold) in Column J\textbf{J} with their corresponding bond dissociation energies (BDE) in Column K\textbf{K}.

Column J (Molecule)Column K (BDE in kcal mol−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\begin{array}{|l|l|} \hline \text{Column J (Molecule)} & \text{Column K (BDE in kcal mol}^{-1}\text{)} \\ \hline \text{(P) } \mathbf{H-CH(CH_3)_2} & \text{(i) } 132 \\ \hline \text{(Q) } \mathbf{H-CH_2Ph} & \text{(ii) } 110 \\ \hline \text{(R) } \mathbf{H-CH=CH_2} & \text{(iii) } 95 \\ \hline \text{(S) } \mathbf{H-C\equiv CH} & \text{(iv) } 88 \\ \hline \end{array}

Options

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

Topics & Concepts

Organic Chemistry - Basic PrinciplesAlkanes, Alkenes, AlkynesResonance

Step-by-Step Solution

To determine the correct match between the given C-H\text{C-H} bonds and their corresponding Bond Dissociation Energies (BDE), we utilize the principle that the C-H\text{C-H} bond dissociation energy is inversely proportional to the stability of the free radical formed after homolytic cleavage, and directly related to the percentage of ss-character of the carbon atom involved.

BDE∝1Stability of Free Radical\text{BDE} \propto \frac{1}{\text{Stability of Free Radical}}

  1. (Q) H−CH2Ph\mathbf{H-CH_2Ph} (Benzylic C-H\text{C-H} bond):

    • Homolytic cleavage forms the benzyl radical (PhC˙H2\text{Ph}\dot{\text{C}}\text{H}_2), which is strongly stabilized by resonance over the benzene ring.
    • Because the resulting radical is exceptionally stable, this bond has the lowest bond dissociation energy.
    • Therefore, Q→(iv) 88 kcal mol−1\mathbf{Q \rightarrow (iv)\ 88\text{ kcal mol}^{-1}}.

  2. (P) H−CH(CH3)2\mathbf{H-CH(CH_3)_2} (Isopropyl C-H\text{C-H} bond):

    • Homolytic cleavage forms a secondary (2∘2^\circ) alkyl radical, (CH3)2C˙H(\text{CH}_3)_2\dot{\text{C}}\text{H}, which is stabilized by hyperconjugation involving 66 α\alpha-hydrogens.
    • It is less stable than the benzyl radical but more stable than sp2sp^2 and spsp radicals.
    • Therefore, P→(iii) 95 kcal mol−1\mathbf{P \rightarrow (iii)\ 95\text{ kcal mol}^{-1}}.

  3. (R) H−CH=CH2\mathbf{H-CH=CH_2} (Vinylic C-H\text{C-H} bond):

    • The carbon atom is sp2sp^2 hybridized (33.3%33.3\% ss-character). Higher ss-character increases the electronegativity of carbon, pulling the electron density closer and resulting in a shorter, stronger C-H\text{C-H} bond.
    • The vinyl radical (CH2=C˙H\text{CH}_2=\dot{\text{C}}\text{H}) is localized and unstable.
    • Therefore, R→(ii) 110 kcal mol−1\mathbf{R \rightarrow (ii)\ 110\text{ kcal mol}^{-1}}.

  4. (S) H−C≡CH\mathbf{H-C\equiv CH} (Alkynyl C-H\text{C-H} bond):

    • The carbon atom is spsp hybridized (50%50\% ss-character). This high ss-character imparts maximum electronegativity to carbon, making the C-H\text{C-H} bond extremely strong and difficult to cleave.
    • Therefore, it possesses the highest bond dissociation energy among the given compounds.
    • Thus, S→(i) 132 kcal mol−1\mathbf{S \rightarrow (i)\ 132\text{ kcal mol}^{-1}}.

Comparing our results with the given choices:

  • P→iii\mathbf{P \rightarrow iii}
  • Q→iv\mathbf{Q \rightarrow iv}
  • R→ii\mathbf{R \rightarrow ii}
  • S→i\mathbf{S \rightarrow i}

This corresponds to Option (A).

PreviousNext