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Hydrogenation of Alkynes

Statement: Ac...

Question

$S t a t e m e n t$ : Action of sodium in liquid $N H_{3}$ followed by $B r_{2} / C C l_{4}$ on 2-butyne produces meso-2,3-dibromobutane.
If the given statement is true then enter 1, if false enter 0.

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Solution

Reduction of 2-butyne with sodium in liquid ammonia gives trans-2-butene. Addition of one molecule of bromine to the carbon-carbon double bond of trans-2-butene gives meso-2,3-dibromobutane.
Hence the given statement is $t r u e$ and correct entry is $1$ .

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Similar questions

S t a t e m e n t

H_{2} / L i n d l a r^{'} s

catalyst followed by

B r_{2} / C C l_{4}

on 2-butyne produces a pair of diastereomers of 2,3-dibromobutane.
If the given statement is true the enter 1 if false enter 0.

S t a t e m e n t

: 2-butyne reaccts with one equivalent of

B r_{2}

C C l_{4}

to produce cis-2,3-dibromo-2-butene.
If true enter 1, if false enter 0.

S t a t e m e n t

: When 2-butyne reacts with triethylamine followed by

B r_{2}

then meso-2,3-dibromobutane is produced.

If the statement is true enter 1, else enter 0.

S t a t e m e n t

: 1 mole of propyne reacts with 1 mole of

B r_{2}

C C l_{4}

to produce trans

C H_{3} C B r = C H B r

If the given statement is true then enter 1, if it is false enter 0.

S t a t e m e n t

: One mole of propyne reacts with sodium in presence of liquid ammonia to produce propane.

If the given statement is true then enter 1, if false enter 0.

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