Compound $X$ is bubbled through bromine dissolved in $C C l_{4}$ & the product formed is $C H_{2} B r - C H_{2} B r$ .The compound formed when steam reacts with $X$ in the presence of phosphoric acid is :

C H_{3} O H

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C H_{2} = C H O H

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C H_{3} - C H_{2} O H

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C H_{3} - C H O

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Solution

The correct option is C $C H_{3} - C H_{2} O H$
Since product formed in first step is $C H_{2} B r - C H_{2} B r$ in presence of chloroform, therefore the compound $X$ is $C H_{2} = C H_{2}$
The next step is addition reaction as follows:
$C H_{2} = C H_{2} H_{2} O / H^{+} - ---- \to C H_{3} - C H_{2} O H$

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