Question

Which of the following test can be used to distinguished methanol from ethanol ?

• A. $\text{Heating with}{I}_2\text{and alkali}$
• B. $\text{Treating with schiff's reagent}$
• C. $\text{Treating with}Cr{O}_3\text{solution in dil.}{H}_{2}S{O}_4$
• D. $\text{Treating with Lucas reagent}$

Answer 1

The correct option is A $\text{Heating with}{I}_2\text{and alkali}$

Iodoform test: When Iodine and potassium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed.
Iodoform test is used to check the presence of carbonyl compounds with the structure $R-CO-C{H}_3$ or alcohols with the structure $R-CH\left(OH\right)-C{H}_3$ in a given unknown substance.
If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a $C{H}_{3}C=O$ group.
Methanol can be distinguished from ethanol by heating with $I_2$ and $NaOH$. Ethanol undergoes the reaction and forms a yellow precipitate of $CH{I}_3$. This is known as iodoform test.
$\underset{\text{ethanol}}{C{H}_{3}C{H}_{2}OH}\underset{+NaOH}{\overset{+I_2}{\to }}C{H}_{3}CHO\stackrel{+I_2}{\to }C{I}_{3}CHO$
$C{I}_{3}CHO\underset{HCOONa}{\overset{+NaOH}{\to }}CH{I}_3$ (Iodoform)
Methanol does not undergo above reaction.