Question

Methanol and ethanol are distinguished by
A) Treating with Schiff’s reagent
B) Treating with Lucas reagent
C) Heating with iodine and alkali
D) Treating with $Cr{O_2}$in dil. ${H_2}S{O_4}$\[\]

Answer 1

Hint: We know that methanol consists of one carbon and ethanol consists of two carbon, hence the structure of methanol is $C{H_4}$ and that of ethanol is ${C_2}{H_6}$ . Ethanol is a poor acid as compared to methanol. Both these alcohols cannot be distinguished physically as they appear the same that is why we use chemical tests to distinguish them.

Complete step-by-step answer: So, here in the above Question we will come to know an idea about what is distinguished between the methanol and ethanol.
Methanol and ethanol are distinguished by a type of test known as Iodoform When ethanol is warmed with iodine in the presence of $NaOH$, it forms a yellow-colored precipitate but methanol does not react positively to iodoform test.
The iodoform test is positive for acetaldehyde, methyl ketones, secondary alcohols that contain alcohol groups in alpha position along with ethanol.
 If we attach the aldehyde group then, if the $C{H_3}$ group is present then only it will give the iodoform test. Suppose if we put ${C_2}{H_6}$ or $C{H_3}C{H_2}$ group then it will not give the iodoform test which also means iodoform can easily distinguish between the methanol and ethanol.
The reaction proceeds as:
$C{H_3}C{H_2}OH + {I_2} + NaOH \to CH{I_3} + HCO{O^ - }N{a^ + }$
The yellow precipitate is triiodomethane (iodoform) which can be recognized by an antiseptic smell.

Hence, option C) is the correct answer.

Note:Both methanol and ethanol react with sodium, liberating ${H_2}$ gas and forming sodium alkoxide. Both of them react with Lucas reagent very slowly only on boiling. None of them reacts with tollens reagent. Ethanol is the only primary alcohol which gives the iodoform test.