Question

An alkyl halide C5H11Br(A) reacts with ethanolic KOH to give an alkene (B) which reacts with Br2 to give a
compoundC)which on dehydrobromination gives an alkyne 'D' On treatment with sodium metal in liquid
ammonia one mole of 'D' gives one mole of sodium salt of 'D' and half a mole of hydrogen gas. Complete
hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the reactions involved.​

Answer 1

The compound A is 1 bromopentane,

B is 1 pentene, C is 1,2 dibromopentane and D is 1 pentyne.

Explanation:

• The reaction between the alkyl bromide and the alcoholic KOH gives the corresponding alkene.

• This alkene reacts with bromine in carbon disulphide to form the dibromo product, the dibromoalkane.

• This dibromoalkane reacts with the alcoholic KOH again to give an alkyne.

• This alkyne then reacts with sodium to form half mole of hydrogen and a sodium alkylide.

• As the end product reacted with sodium, then it's a terminal alkyne.

• So the products are correspondingly 1 bromopentane, 1 pentene, 1,2 dibromopentane and 1 pentyne respectively.

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