Question

The starting substance for the preparation of $C{H_3}I$ is
A. $C{H_3}OH$
B. ${C_2}{H_5}OH$
C. $C{H_3}CHO$
D. ${(C{H_3})_2}CO$

Answer 1

Hint:The chemical substance with the formula $CH_3I$ is also known as methyl iodide and is frequently referred to as "MeI”. The preparation of alkyl halide is generally carried out by nucleophilic substitution. In which a leaving group gets replaced with an upcoming group without reducing carbon count.

Complete step-by-step answer:Let we know the starting substance by the following reaction which is stated below,
$C{H_3}OH + HI + ZnC{l_2} \to C{H_3}I + {H_2}O$

In the above-given reaction, methanol whose chemical formula is written as $C{H_3}OH$ reacts with hydrogen iodide $(HI)$ in the presence of $ZnC{l_2}$ to give methyl iodide $C{H_3}I.$

Therefore, the correct answer is $C{H_3}OH$

Option ‘A ’ is correct

Additional Information:Methanol is also called wood alcohol. It is an organic compound and the simplest alcohol. It is volatile, light, and flammable. It has physical and chemical properties similar to that of ethanol.
It must be noted that Methanol is a prospective energy source since it can be stored more easily as a liquid than hydrogen.

Note: The nucleophilic substitution takes place when the -OH group of methanol gets replaced by the -Cl group which acts as a nucleophile in the presence of hydrogen chloride and zinc chloride. However, this method is not applicable for the preparation of aryl halides because the C-O bond in phenols has a partial double bond character and is not easy to break double bond in comparison to single bond.