Question

Methyl alcohol is treated with conc $HI$ and red $P$. The main product obtained is?
(A) $C{H_3}OH$
(B) $C{H_4}$
(C) $C{H_2}{I_2}$
(D) $P{H_3}$

Answer 1

In order to solve this question we must know the properties of $HI$ and red phosphorus. $HI$ and red phosphorus is a reducing agent. As we know a reducing agent which is also called a reductant or reducer is an element or compound that loses or donates an electron to an electron recipient oxidizing agent in a redox chemical reaction. Hence, we can say that a reducing agent is thus oxidized when it loses electrons in the redox reaction.

Complete step-by-step answer: Hydroiodic acid and red Phosphorus is one of the strongest reducing agents. It can reduce all functional groups like carboxylic acid, acid anhydride, amide, carbonyle, acid halide, nitrile and ether to produce alkanes. Hence, methanol when made to react with $HI$ and red phosphorus gives methane.
$C{H_3}OH\xrightarrow[{conc{\text{ HI}}}]{{red{\text{ P}}}}C{H_4}$

Additional information:
The $HI/P$ reduction is called Nagai method. This reduction process is used to reduce various chemical compounds which contain at least one OH group or halocarbons to their respective alkanes.

Note: This reduction process involves two steps:
a.$SN2$ nucleophilic substitution of the $OH$ group by Iodine, which is done by using concentrated hydroiodic acid.
b.Reduction of the alkyl iodide by hydriodic acid.
In the first step, hydroiodic acid converts the hydroxy compound to iodide.
This reduction process is known to be powerful enough to remove oxygen from any organic compound and if reaction conditions are harsh enough it can even remove nitrogen as well. The reduction of terminal alkyl iodide however is longer and has a lower yield.