WBJEE · Chemistry · General Organic Chemistry
The stability of Me \(\mathcal{C}=\mathrm{CH}_{2}\) is more than that of \(\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) due to
- A inductive effect of the Me groups
- B resonance effect of the Me groups
- C hyperconjugative effect of the Me groups
- D resonance as well as inductive effect of the Me groups
Answer & Solution
Correct Answer
(C) hyperconjugative effect of the Me groups
Step-by-step Solution
Detailed explanation
2-hyperconltjugate forms. Therefore, \(\mathrm{Me} \underline{\mathcal{C}}=\mathrm{CH}_{2}\) is more stable than \(\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\)
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