Question

An alkyl bromide A on treatment with Na and ether gives a hydrocarbon B. B on treatment with HBr and peroxide gives $Br - {(C{H_2})_6} - Br$. Compound A is
(A) $C{H_3}C{H_2}C{H_2}Br$
(B) ${H_2}C = CH - C{H_2} - Br$
(C) $C{H_3} - CH = CHBr$
(D) $C{H_3} - C{H_2} - C{H_2} - C{H_2} - Br$

Answer 1

Hint: An alkyl halide on reaction with sodium in dry ether gives forms alkanes. Alkenes in reaction with hydrogen bromide give products according to the anti-Markovnikov rule. We will accordingly find the products and get the answer.

Complete step by step answer:
Here in the given question when an alkyl halide reacts with sodium in the ether it forms an alkane. when the alkene reacts with hydrogen bromide in presence of peroxide it forms alkyl halide according to anti mark rule.
We have been given that we will be doing reaction of hydrocarbon with hydrogen bromide and peroxide so first and fourth are itself alkyl halide so it will not undergo reaction so A and D are incorrect.
Coming to the second option that is allyl bromide we react it with sodium in ether taking two moles of it the reaction is as follows
$2C{H_2} = CH - C{H_2} - Br \to {(C{H_2} = CH - C{H_2})_2}$
Now we will react it with hydrogen in presence of peroxide it will form a product according to the anti-Markovnikov rule where the negative part of the reagent gets attached to that carbon having a higher number of hydrogen atoms.
So the reaction will be as follows
${(C{H_2} = CH - C{H_2})_2} \to {(Br - C{H_2} - C{H_2} - C{H_2})_2}$
Here we can see that the carbon chain is of six carbons and bromine is present at the first and second positions.
Therefore, option B is correct
Coming to the third option that is vinyl bromide it undergoes Wurtz reaction. Here, in the following product, anti-Markovnikov rule can’t be applied as the double-bonded carbon has hydrogen present on it. Therefore, option C is incorrect.
Hence we can conclude that compound A, i.e. alkyl bromide ${H_2}C = CH - C{H_2} - Br$

So, the correct answer is Option B.

Note: Wurtz reaction is used for the formation of hydrocarbons and increasing the number of carbon atoms. It follows a free radical mechanism and due to those side products such as alkenes are also possible.