Question

In Victor Meyer’s test, the colour given by $1^0$ , $2^0$ and $3^0$ alcohols are respectively
A) red, colourless, blue
B) red, blue, colourless
C) colourless, red, blue
D) red, blue, violet

Answer 1

Hint: Victor Maeyer’s test is used to differentiate primary, secondary and tertiary alcohols. In Victor Maeyer’s test, alcohols are treated with the reagents $P+I_2$ , $AgN{O}_2$, $HONO$ and $NaOH$ in the steps one by one. The colours obtained with the alkali helps us to distinguish the alcohols.

Complete step by step solution:
$1^0$ , $2^0$ and $3^0$ alcohols can be distinguished using the Victor Maeyer’s test. Victor Maeyer’s test is done in four steps with the reagents $P+I_2$ , $AgN{O}_2$, $HONO$ and $NaOH$ respectively.
In Victor Maeyer’s test, the alcohols are treated in the following steps.
Step 1: Firstly, the sample alcohol is treated with the reagent $P+I_2$ to get the iodoalkane as a product.
Step 2: The iodoalkane obtained is then treated with $AgN{O}_2$ solution to get the nitroalkane.
Step 3: The nitroalkane obtained is then treated with $HONO$ (nitrous acid).
Step 4: The resulting solution in step 3 is then treated with the alkali like $NaOH$ and the colour is obtained.
Let us now do Victor Maeyer’s test of primary, secondary and tertiary alcohols.
- Victor Maeyer’s test of $1^o$ alcohols:

- Victor Maeyer’s test for $2^o$ alcohols:

- Victor Maeyer’s test for $3^o$ alcohols:

Thus, we can conclude from the above reactions of Victor Maeyer’s test of $1^0$, $2^0$ and $3^0$ alcohols that, primary alcohols ($1^o$) gives the red colour solution with alkali, secondary alcohols ($2^o$) gives the blue colour solution with alkali, and tertiary alcohols ($3^o$) gives the colourless solution.

Thus, option (B) is the correct answer.

Note: Nitrous acid (HONO) is the combination of two reagents $NaN{O}_3$ and $H_{2}S{O}_4$. A key point to note is that Victor Maeyer’s test is not given by phenols because the procedure involves the breaking of OH bonds with carbon but in case of phenol, a carbon-oxygen bond is much stronger, having a partial bond character.