Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?
Iodination reactions are reversible in nature.
$$\mathrm{C}_6 \mathrm{H}_6+\mathrm{I}_2 \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{I}+\mathrm{HI}$$
In this above reaction, hydrogen iodide is formed apart from the required product. It has to be removed from the reaction mixture in order to prevent the backward reaction. To carry out the reaction in the forward direction, HI formed during the reaction is removed by oxidation using oxidising agent. Such as $\mathrm{HIO}_3$ or $\mathrm{HNO}_3$. The reaction is as follow
$$\begin{gathered} 5 \mathrm{HI}+\mathrm{HIO}_3 \longrightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O} \\ 2 \mathrm{HI}+2 \mathrm{HNO}_3 \longrightarrow \mathrm{I}_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O} \end{gathered}$$
By using a suitable oxidising agent HI is oxidised to give $\mathrm{I}_2$.
Out of o- and p-dibromobenzene which one has higher melting point and why?
p-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to symmetry, $p$ - isomer fits in the crystal lattice better than the o-isomer. Hence, p-dibromobenzene has higher melting point.
Which of the compounds will react faster in $\mathbf{S}_{\mathbf{N}} \mathbf{1}$ reaction with the ${ }^{-} \mathbf{O H}$ ion?
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl}$ or $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{Cl}$
$S_N 1$ mechanism depends upon the stability of carbocation that is formed as intermediate in the mechanism.
$\mathrm{CH}_6-\mathrm{CH}_2-\mathrm{Cl}$ will form $\mathrm{C}_6 \mathrm{H}_5^{+} \mathrm{CH}_2$ carbocation as intermediate.
This carbocation is resonance stabilised and will react faster in $\mathrm{S}_{\mathrm{N}} 1$ reaction.
While carbocation formed in $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$ is $\mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{H}_2$. This carbocation is highly unstable and not give $\mathrm{S}_{\mathrm{N}} 1$ reaction with ${ }^{-} \mathrm{OH}$ ion.
Why iodoform has appreciable antiseptic property?
lodoform liberate $\mathrm{I}_2$ when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of $\mathrm{I}_2$ not because of iodoform itself.
$$\mathrm{\mathop {CH{I_3}}\limits_{Iodine} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{skin}^{Contact\,with}} \mathop {{I_2}}\limits_{Iodine}}$$ (responsible for antiseptic property)
Haloarenes are less reactive than haloalkanes and haloalkenes. Explain.
Due to resonance, $\mathrm{C}-\mathrm{X}$ bond in haloarenes and haloalkenes have some double bond character. This partial double bond character of $\mathrm{C}-\mathrm{X}$ bond strengthen the bond. So, haloarenes and haloalkenes are less reactive than haloalkanes. Lets see the resonating structure of the haloarenes and haloalkenes
Now, more the number of resonating structure higher will be the stability of the compound and lesser will be the reactivity. In haloarenes, more resonating structures are observed than the haloalkenes. So, haloarenes are less reactive than haloarenes. In haloalkanes, this $\mathrm{C}-\mathrm{X}$ bond is purely single bond.