What is the enthalpy change for the isomerization reaction: CH 2 =CH-CH 2 -CH=CH-CH=CH 2 (A) → CH 2 =CH-CH=CH-CH=CH-CH 3 (B) (reagent: NaNH 2 , Δ) Magnitude of resonance energies of A and B are 50 and

Question

What is the enthalpy change for the isomerization reaction:

CH₂=CH-CH₂-CH=CH-CH=CH₂ (A) → CH₂=CH-CH=CH-CH=CH-CH₃ (B) (reagent: NaNH₂, Δ)

Magnitude of resonance energies of A and B are 50 and 70 kJ/mol, respectively. A and B contain identical bonds (same Kekulé bond inventory), so their bond-enthalpy contributions are equal.

TestHub · Accepted Answer

Isomerization alters neither the number nor the type of bonds (three C=C, four C-C, and ten C-H bonds on both sides). Therefore, the Kekulé (bond-enthalpy) components of the two enthalpies cancel out. Only the resonance stabilization differs. The change in enthalpy, Δ H, is given by:

Δ H = RE_A - RE_B

Given RE_A = 50 kJ and RE_B = 70 kJ, we calculate:

Δ H = 50 - 70 = -20 kJ

This corresponds to option (d).

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