Nitro group being electron withdrawing, deactivates the benzene nucleus to such an extent. that an electrophile cannot attack on benzene ring easily due to deactivation of benzene ring. Hence becomes incapable to give Friedel-Craft’s reaction. NOTE
Nitrobenzene because of its unreactivity towards Friedel-Craft's reaction is used as a solvent for this reaction.
262.
In the presence of peroxide, $$HCl$$ and $$HI$$ do not give anti-Markownikoff’s addition of alkenes because :
A
One of the steps is endothermic in $$HCl$$ and $$HI$$
B
Both $$HCl$$ and $$HI$$ are strong acids
C
$$HCl$$ is oxidizing and the $$HI$$ is reducing
D
All the steps are exothermic is $$HCl$$ and $$HI$$
Answer :
One of the steps is endothermic in $$HCl$$ and $$HI$$
Anti-Markownikoff addition is possible only in case of $$HBr$$ and not in $$HCl$$ and $$HI.$$ In $$HBr$$ both the chain initiation and propagation steps are exothermic, while in $$HCl,$$ first step is exothermic, and second step is endothermic and in $$HI,$$ no step is exothermic. Hence $$HCl$$ and $$HI$$ do not undergo anti-Markownikoff’s addition.
263.
The reaction of propene with $$HOCl$$ proceeds via the addition
of
Nitrosyl chloride adds on olefins according to
Markovnikof’s rule, where $$N{O^ + }$$ constitutes the positive
part of the addendum.
266.
In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the specie
No explanation is given for this question. Let's discuss the answer together.
267.
Consider the following sequence of reactions
\[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[700K]{C{{l}_{2}}}A\xrightarrow[420K,12\,atm]{N{{a}_{2}}C{{O}_{3}}}B\]
Compound $$'B'$$ is
268. \[+\,Cl-C{{H}_{2}}C{{H}_{2}}-C{{H}_{3}}\xrightarrow{AlC{{l}_{3}}}P\] \[\xrightarrow[\left( \text{ii} \right){{H}_{3}}{{O}^{+}}]{\left( \text{i} \right)\frac{{{O}_{2}}}{\Delta }}Q+\text{phenol}\]
The major products $$P$$ and $$Q$$ are