$$t - BuOK$$ is a bulky strong base and causes elimination
(dehydrohlogenation) reaction
The given reaction proceeds via biomolecular elimination.
5.
Which of the following reaction$$(s)$$ can be used for the preparation of alkyl halides?
\[\begin{align}
& \text{(i)}\,C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{anh.\,ZnC{{l}_{2}}} \\
& \text{(ii)}\,C{{H}_{3}}C{{H}_{2}}OH+HCl\to \\
& \text{(iii)}\,{{\left( C{{H}_{3}} \right)}_{3}}COH+HCl\to \\
& \text{(iv)}{{\left( C{{H}_{3}} \right)}_{2}}CHOH+HCl\xrightarrow{anh.ZnC{{l}_{2}}} \\
\end{align}\]
In (i) and (iv) due to the presence of Lucas reagent $$\left( {HCl + anh.\,ZnC{l_2}} \right)$$ alcohols give alkyl halides while in (iii) alkyl halide is formed due to $${S_N}1$$ reaction.
6.
Which of the following is the correct order of decreasing $${S_N}2$$ reactivity?
In $${S_N}2$$ mechanism transition state is pentavelent. For
bulky alkyl group it will have sterical hinderance and smaller alkyl group will favour the $${S_N}2$$ mechanism. So the decreasing order of reactivity of alkyl halides is $$RC{H_2}X > {R_2}CX > {R_3}CX$$
7.
Tertiary alkyl halides are practically inert to substitution by \[{{S}_{N}}2\] mechanism because of
Due to steric hindrance tertiary alkyl halide do not react
by $${S_N}2$$ mechanism they react by $${S_N}1$$ mechanism.
$${S_N}2$$ mechanisam is followed in case of primary and secondary alkyl halides
The order is
$$C{H_3} - X > C{H_3} - C{H_2}X > {\left( {C{H_3}} \right)_2}CH.X > {\left( {C{H_3}} \right)_3}C - X$$
8.
The reaction of $${C_6}{H_5}CH = CHC{H_3}$$ with $$HBr$$ produces
When chlorine gas is reacted with propene at high temperature $$\left( {{{400}^ \circ }C} \right),$$ then substitution occurs in place of addition reaction. Hence, allyl chloride is formed
10.
Reactivity order of halides for dehydrohalogenation is
$$F, Cl,$$ $$Br$$ and $$I$$ are the elements of $$VII$$ $$A$$ group. In a group atomic radii increases from top to bottom and the bond dissociation energy decreases as
$$R - F > R - Cl > R - Br > R - I$$
So, during dehydrohalogenation $$R-I$$ bond breaks more easily than $$R-I$$ bond. Hence, order of reactivity will be
$$R - I > R - Br > R - Cl > R - F$$