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.
Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.
Resonance in halobenzene
From the above resonating structure it is very clear that electron density is rich at ortho and para position. Therefore, it is ortho and para directing not meta directing.
Classify the following compounds as primary, secondary and tertiary halides.
(a) 1-bromobut-2-ene
(b) 4-bromopent-2-ene
(c) 2-bromo-2-methylpropane
The structural formula of the given compounds are
In compound (i), carbon atom to which halogen is bonded, further bonded to two hydrogens and one carbon of hydrocarbon chain. So, it is primary halide. In compound (ii), $\alpha$-carbon is bonded with one hydrogen and two carbons of two hydrocarbons chain. So, it is secondary halide. In compound (iii) $\alpha$-carbon is bonded to three alkyl group, so it is tertiary halide.