The compound A, B and C in the following reaction sequence are (in the order A, B, C):
$C H_{3} - C H_{2} - O H P B r_{3} - -- \to (A) a l c . K O H - ----- \to (B) B r_{2} / C C l_{4} - ----- \to (C)$

C_{2} H_{5} B r, C H_{3} - C H_{2} - O H, C H_{3} C H B r_{2}

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C_{2} H_{5} B r, C H \equiv C H, C H_{2} = C H B r

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C_{2} H_{5} B r, C H_{2} = C H_{2}, B r - C H_{2} - C H_{2} - B r

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C_{2} H_{5} B r, C H_{3} C H_{2} O H, B r C H_{2} - C H_{2} B r

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Solution

The correct option is C $C_{2} H_{5} B r, C H_{2} = C H_{2}, B r - C H_{2} - C H_{2} - B r$

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Similar questions

C H_{3} - C H_{2} - O H P B r_{3} - -- \to (A) a l c . K O H - ----- \to (B) B r_{2} / C C l_{4} - ----- \to (C)

C H_{3} - C H_{2} - O H P B r_{3} - -- \to (A) a l c . K O H - ----- \to (B) B r_{2} / C C l_{4} - ----- \to (C)

C H_{3} - C H_{2} - C H_{2} - C l (a) p r o p y l c h l o r i d e) A l c . K O H - ----- \to C H_{3} - C H = C H_{2} (P r o p e n e b)

C H_{3} C H_{2} - O H A \to C H_{3} - C H_{2} - C l B \to C H_{2} = C H_{2} .

{C H}_{2} = {C H}_{2} c o l d - ------ \to d i l . {K M n O}_{4} A e x c e s s - --- \to {P C l}_{5} B

(A)

(B)

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