Which of the above reaction conditions would be best to perform the above reaction?

H B r

(one equivalent) then

B r_{2}

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H_{2}, P d / C

then

B r_{2} / h v

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B r_{2}

(two equivalent)

H_{2}, P d / C

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O s O_{4} / H I O_{4}

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Solution

The correct option is C $H B r$ (one equivalent) then $B r_{2}$

The conditions that would be best to perform the given reaction are:
$H B r$ (1 equivalent) followed by bromine.

First, a molecule of HBr is added to carbon carbon triple bond according to Markownikoff's rule. Bromine atom is added to more substituted carbon atom and hydrogen atom is added to less substituted carbon atom. This is followed by the addition of a molecule of bromine to C=C double bond.

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H_{2} + B r_{2} \to 2 H B r

H_{2} + {B r}_{2} ⟶ 2 H B r

= K [H_{2}] {[{B r}_{2}]}^{\frac{1}{2}}

H_{2} + {B r}_{2} \to 2 H B r

\frac{d x}{d t} = k [H_{2}] {[{B r}_{2}]}^{1 / 2}

H_{2} + {B r}_{2} \to 2 H B r

\frac{d x}{d t} = k [H_{2}]^{1 / 2} {[{B r}_{2}]}^{1 / 2}

{B r}_{2}

H - C \equiv C - C H_{2} - C H_{2} - C H_{3} \to H_{3} C - C H = C H - C H_{2} - C H_{3}

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