Question

Identify Z in the sequence of reactions :
$C{H_3}C{H_2}CH = C{H_2}\xrightarrow[{{H_2}{O_2}}]{{HBr}}Y\xrightarrow{{{C_2}{H_5}ONa}}Z$
a.) $C{H_3} - {(C{H_2})_3} - O - C{H_2}C{H_3}$
b.) ${(C{H_3})_2} - CH - O - C{H_2}C{H_3}$
c.) $C{H_3} - {(C{H_2})_4} - O - C{H_3}$
d.) $C{H_3} - C{H_2} - CH(C{H_3}) - O - C{H_2}C{H_3}$

Answer 1

Hint: The hydrogen bromide in the presence of hydrogen peroxide follows anti Markovnikov’s rule to give additional product. This additional product then reacts with sodium salt to give ether. The alkyl bromide attaches with oxygen of the ${C_2}{H_5}ONa$ to give the product.

Complete answer:
We have a but-1-ene substrate. The reagents in the first step are HBr in presence of hydrogen peroxide. This leads to formation of Y which then reacts with sodium salt to give Z product.
The reagents in the first step i.e. the HBr in presence of hydrogen peroxide show additional reaction. This additional reaction follows anti Markovnikov’s rule and thus leads to formation of Y. The reaction can be written as -
$C{H_3}C{H_2}CH = C{H_2}\xrightarrow[{{H_2}{O_2}}]{{HBr}}C{H_3}C{H_2}C{H_2}C{H_2}Br$
Thus, the Y is 1- bromobutane. This then reacts with sodium salt to give product Z. The reaction for this step is-
$C{H_3}C{H_2}C{H_2}C{H_2}Br\xrightarrow{{{C_2}{H_5}ONa}}C{H_3}C{H_2}C{H_2}C{H_2} - O{C_2}{H_5}$

So, the option (a) is the correct answer.

Additional information:
> This reaction is regioselective in nature. The hydrogen peroxide used here is unstable and with a slight of sunlight, it breaks into OH free radicals.
> These OH free radicals form the Bromine free radicals which propagate the reaction.

Note: It must be noted that Anti Markovnikov rule says that the incoming group should bind to less substituted carbon atom rather than more substituted one. Markovnikov's rule is opposite to this. According to Markovnikov’s rule, the incoming group should bind to the more substituted carbon atom. If this rule would have followed, then we would have got 2- bromobutane as the product.