Question

Which of the following is the proper method to prepare n-hexane from n-propyl alcohol?
$C{H_3}C{H_2}C{H_2}OH\xrightarrow{X}C{H_3}C{H_2}C{H_2}Br\xrightarrow{Y}C{H_3}{(C{H_2})_4}C{H_3}$
A) $X \to HBr,Y \to HCN$
B) $X \to HBr,Y \to Na,ether$
C) $X \to B{r_2},Y \to C{H_3}CN$
D) $X \to B{r_2},Y \to KMn{O_4}$

Answer 1

In this question we can see that OH of the alcohol is replaced with Br (Halogen). This is known as the Nucleophilic substitution reaction. In this reaction a positively charged nucleophile attacks an electron rich centre and replaces the leaving group. We’ll discuss this further in the question.

Complete Step By Step Answer:
To make alkyl halides from alcohol, we need to add a strong acid like HBr, HCl, HI which will replace the -OH by alkyl halide. Alcohols are basically poor leaving groups, but the addition of strong acid leads to the formation of an excellent leaving group. If the counterion ${X^ - }$ is nucleophilic it substitutes the position of -OH. The mechanism can be shown as:
$R - OH + H - X \to R - O{H_2}^ + \to R - X + {H_2}O$
Therefore, among the given options X has to be HBr. The reaction can be given as:
$C{H_3}C{H_2}C{H_2}OH\xrightarrow{{HBr}}C{H_3}C{H_2}C{H_2}Br$
Next is formation of alkane from alkyl halide. When alkyl halide is treated with sodium metal in presence of dry ether, it forms alkane with double the no. of carbon atoms present in the alkyl halide. This reaction is known as the Wurtz Reaction. In this case three carbon alkyl halides are converted to 6 carbon alkane. Hence Wurtz reaction must have happened. The reaction can be shown as:
$C{H_3}C{H_2}C{H_2}Br\xrightarrow{{Na,Ether}}C{H_3}{(C{H_2})_4}C{H_3}$
The overall reaction hence is: $C{H_3}C{H_2}C{H_2}OH\xrightarrow{{HBr}}C{H_3}C{H_2}C{H_2}Br\xrightarrow{{Na,Ether}}C{H_3}{(C{H_2})_4}C{H_3}$
The correct option is B).

Note:
There are many other ways for the conversion of alkanes like the Hydrolysis of Grignard reagent, reduction of aldehydes and ketones by zinc amalgam and conc. HCl, By reduction of alkyl halide using $LiAl{H_4}/{H_2} - Pd$ or nascent hydrogen.