Aryl Halides and Phenols - Result Question 13
####16. Phenol reacts with methyl chloroformate in the presence of $NaOH$ to form product $A$. $A$ reacts with $Br _2$ to form product $B$. $A$ and $B$ are respectively
(2018 Main)
(a)
and
(b)
(c)
(d)
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Answer:
Correct Answer: 16. (d)
Solution:
- Given,
In the above road map, first reaction appears as acid base reaction followed by $S _N AE$ (Nucleophilic substitution through Addition and Elimination). Both the steps are shown below
(i) Acid base reaction
$$ \widehat{O}^{OH} \xrightarrow{-OH} \widehat{O}^{O^{-}}+H _2 O $$
(ii) $S _N A E$
In the product of $S _N AE$ the attached group is ortho and para-directing due to following cross conjugation
Cross conjugation due to which lone pair of oxygen 1 will be easily available to ring resulting to higher electron density at 2 , 4, 6 position with respect to group. However from the stability point of view ortho positions are not preferred by substituents as group $-O-\underset{O}{C}-O-CH _3$ is bulky.
Hence, on further bromination of $S _N AE$ product para bromo derivative will be the preferred product i.e.
$$ \overbrace{}^{O-O}+Br _2 \rightarrow O _{Br}^{C-O-CH _3}+HBr $$
