For the following reactions:
(A) ${CH}_{3} {CH}_{2} {CH}_{2} Br + KOH ⟶ {CH}_{3} CH = {CH}_{2} + KBr + H_{2} O$
(B)

(C)
Which of the following statement is correct?

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(A) is substitution, (B) and (C) are addition reactions.

(A) is elimination, (B)and (C) are substitution reactions.

(A) and (B) are elimination reactions and (C) is addition reaction.

(A) is elimination, (B) is substitution and (C) is addition reaction.

answer is D.

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Detailed Solution

$\begin{array}{r} {CH}_{3} {CH}_{2} {CH}_{2} Br + KOH ⟶ {CH}_{3} CH = {CH}_{2} + KBr + H_{2} O \end{array}$

Saturated compound is converted into unsaturated compound by removal of group of atoms hence, it is an elimination reaction.

- Br group is replaced by-OH group hence, it is a substitution reaction.

Addition of $B r_{2}$ converts an unsaturated compound into a saturated compound hence, it is an addition reaction.

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