Question

Name the reagent used for bromination of phenol to \[2,\,4,\,6\, - tribromophenol\] :
(A). $HBr$
(B). $NBS$
(C). $B{r_2},{H_2}{O_2}$
(D). $B{r_2},{H_2}O$

Answer 1

Hint: Try to solve the reaction with all the given reagents and you will see that $HBr$ is used to alkyl halide addition on double bond. On the other hand, $NBS$ is used for bromination but it will do bromination at allylic and benzylic position. Third option which is $B{r_2},{H_2}{O_2}$ bromine with hydrogen peroxide it also does bromination but hydrogen peroxide is an oxidising agent. Last we have, $B{r_2},{H_2}O$ it will do bromination at all the three positions.

Complete solution:
Let’s start with NBS, it is a brominating agent it means it will do bromination, introduce a bromide group but the position site is very important for this reagent. N-Bromosuccinimide NBS is used for bromination at allylic position or benzylic position.
$C{H_2} = CH - C{H_3}$ If we have propene then if we apply this reagent on propene it will do allylic and benzylic bromination. We can understand it with a reaction.
$C{H_2} = CH - C{H_3}\xrightarrow{{NBS}}\,C{H_2} = CH - C{H_2} - Br$
In case of phenol, there is no allylic or benzylic position available, so this reagent doesn't work on phenol.
Next, we have $HBr$ it will add on a double bond in such a way that the most negative atom which is bromine here will attack on carbon atoms having lesser number of hydrogens according to markovnikov rule. As the double bonds in phenol are making a stable aromatic structure thus, when we add $HBr$ in phenol, it will not give the required product.
Among $B{r_2},{H_2}{O_2}$ and $B{r_2},{H_2}O$ the difference is in the solvent. In the fourth option there is water present while in third one hydrogen peroxide is present. So we get the required product from the fourth option $B{r_2},{H_2}O$ .
${C_6}{H_5}OH\xrightarrow{{B{r_2},\,{H_2}O}}{C_6}{H_3}B{r_2}$
On reacting phenol with $B{r_2},{H_2}O$ , it introduces bromine at $2,\,4,\,6\,positions$ .
Hence, Option (D) is correct $B{r_2},{H_2}O$.

Note: For writing the product we have to write the whole mechanism so in case $B{r_2},{H_2}O$ solvent is water but in case of $B{r_2},{H_2}{O_2}$ we have hydrogen peroxide which is a very good reducing agent. So, for the product we want three bromines at $2,\,4,\,6\,positions$ which hydrogen peroxide can’t perform.