Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 2
ChemistrySingle correct+3 / -1Question 32 of 57

JEE Advanced 2021 · Paper 2, Chemistry question 13

Passage

Paragraph

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:

  • HX3C−H(g)→HX3CX ∙ (g)+HX ∙ (g)\ce{H3C-H(g) -> H3C^.(g) + H^.(g)}, ΔH∘=105\Delta H^\circ = 105 kcal mol−1^{-1}
  • Cl−Cl(g)→ClX ∙ (g)+ClX ∙ (g)\ce{Cl-Cl(g) -> Cl^.(g) + Cl^.(g)}, ΔH∘=58\Delta H^\circ = 58 kcal mol−1^{-1}
  • HX3C−Cl(g)→HX3CX ∙ (g)+ClX ∙ (g)\ce{H3C-Cl(g) -> H3C^.(g) + Cl^.(g)}, ΔH∘=85\Delta H^\circ = 85 kcal mol−1^{-1}
  • H−Cl(g)→HX ∙ (g)+ClX ∙ (g)\ce{H-Cl(g) -> H^.(g) + Cl^.(g)}, ΔH∘=103\Delta H^\circ = 103 kcal mol−1^{-1}

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^{-1})
(P) H−CH(CH3)2\mathbf{H{-}C}\mathrm{H(CH_3)_2} (i) 132
(Q) H−CH2Ph\mathbf{H{-}C}\mathrm{H_2Ph} (ii) 110
(R) H−CH=CH2\mathbf{H{-}C}\mathrm{H{=}CH_2} (iii) 95
(S) H−C≡CH\mathbf{H{-}C}{\equiv}\mathrm{CH} (iv) 88
Choose one option
Answer and worked solution

Official answer: A

Ranking the C-H bonds. The more stable the radical formed, the weaker the bond. More s-character makes the bond stronger.

  • Benzylic C-H is weakest (resonance-stabilised radical): 88, so Q - iv.
  • Secondary alkyl C-H: 95, so P - iii.
  • Vinylic C-H (sp2sp^2): 110, so R - ii.
  • Acetylenic C-H (spsp): 132, so S - i.

This matches option (A).

Topics:Reactive intermediatesBond enthalpy
← Previous questionNext question →