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 $$2 \mathrm{NaBr}+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{NaHSO}_4+\mathrm{SO}_2+\mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}$$
Which of the products will be major product in the reaction given below? Explain
$\mathrm{CH}_3 \mathrm{CHICH}_3$ is the major product of the reaction. The mechanism of the reaction is as follows
$$\mathrm{HI} \rightleftharpoons \mathrm{H}^{+}+\mathrm{I}^{-}$$
Why is the solubility of haloalkanes in water very low?
Haloalkanes are slightly soluble in water. For the solubility to haloalkane in water, energy is required to overcome the attractions between its own molecules and to break the bonds between water molecules.
Less energy is released when new attractions are set up between the haloalkanes and the water molecules as these are not as strong as the original hydrogen bonds in water.