Which of the following leads to formation of $1^{0}$ alcohol?

{C H}_{3} - C H = {C H}_{2} H_{2} O - - \to H^{+}

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{C H}_{3} - C H = {C H}_{2} N a {B H}_{4} / {O H}^{⊖} - --------- \to H g {(O A c)}_{2} / T H F

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{C H}_{3} - C H = {C H}_{2} H_{2} O_{2} / {O H}^{⊖} - ------ \to B_{2} H_{6} / T H F

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{C H}_{3} - C \equiv C - {C H}_{3} d i l . H_{2} {S O}_{4} - ----- \to d i l . H g {S O}_{4}

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Solution

The correct option is C ${C H}_{3} - C H = {C H}_{2} H_{2} O_{2} / {O H}^{⊖} - ------ \to B_{2} H_{6} / T H F$
Morbenzene rule $\to H^{+}$ goes to carbon having more hydrogen among the double bond ones.

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{CH}_{3} - {CH}_{2} - CH = {CH}_{3}

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

C H_{3} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

A g N O_{3}

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