First slide

Alkenes

Question

$C H_{3} - C H_{2} - C H = C H_{2} \to_{P e r o x i d e}^{H B r} A (m a j o r) \to_{S N^{2}}^{a q . K O H} B \overset{C o n c . H_{2} S O_{4} / Δ}{\to} C (M a j o r) .' C' i s$

Moderate

A

$C H_{3} - C H = C H - C H_{3}$
B

$C H_{3} - C H_{2} - C H = C H_{2}$
C

${(C H_{3})}_{2} C = C H_{2}$
D

$C H_{3} - C H = C H_{2}$

Solution

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