8.2 Carboxylic acids and their derivative — Chemistry STD 12 Science — Question

Statement $I$ : Bromination of phenol in solvent with low polarity such as $\mathrm{CHCl}_3$ or $\mathrm{CS}_2$ requires Lewis acid catalyst.

Statement $II$ : The lewis acid catalyst polarises the bromine to generate $\mathrm{Br}^{+}$.

In the light of the above statements, choose the correct answer from the options given below :

$\begin{array}{*{20}{c}}
  {Ph - CH = CH - CH - C = O\xrightarrow[{PH = 4.5}]{{{H_2}N - OH}}} \\ 
  {{\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } \\ 
  {{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,H{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } 
\end{array}$ ?

1. Silicon doped with electron rich impurity is a $p-$type semiconductor.
2. Silicon doped with an electron rich impurity is an $n-$type semiconductor.
3. Delocalised electrons increase the conductivity of doped silicon.
4. An electron vacancy increases the conductivity of $n-$type semiconductor.

(en-ethylenediamine; $H _{2} N - CH _{2}- CH _{2}- NH _{2}$ )

Major product of above reaction is