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.
Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
Lewis acids are electron deficient species. They are responsible for, inducing heterolytic fission in halogen molecule.
Role of Lewis acid is to produce an electrophile. The eletrophile produce will attack on electron rich benzene ring to produce aryl bromides and chlorides.
$$\begin{aligned} \mathrm{AlCl}_3+\mathrm{Cl}_2 & \longrightarrow\left[\mathrm{AlCl}_4\right]^{-}+\mathrm{Cl}^{+} \\ \mathrm{AlBr}_3+\mathrm{Br}_2 & \longrightarrow\left[\mathrm{AlBr}_4\right]^{-}+\mathrm{Br}^{+} \end{aligned}$$
This electrophile will further attack on benzene.
Which of the following compounds (i) and (ii) will not react with a mixture of NaBr and $\mathrm{H}_2 \mathrm{SO}_4$. Explain why?
(i) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
(ii) 
Partial double bond character of a bond increses the strength of the bond and hence, decreases the stability. Phenol will not react with a mixture of NaBr and $\mathrm{H}_2 \mathrm{SO}_4$ because it is resonance stabilised. Due to resonance, partial double bond character arises in $\mathrm{C}-\mathrm{O}$ bond of phenol and it becomes more stable than alcohol. $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\right)$.
Reaction $$2 \mathrm{NaBr}+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{NaHSO}_4+\mathrm{SO}_2+\mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}$$
