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.
Compound ' A ' with molecular formula $\mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}$ is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound 'A' only. When another optically active isomer ' $B$ ' of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ' $A$ ' and ' $B$ '.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.
(i) As the rate of reaction depends upon the concentration of compound ' $A$ ' $\left(\mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}\right)$ only therefore, the reaction is proceeded by $\mathrm{S}_{\mathrm{N}} 1$ mechanism and the given compound will be tertiary alkyl halide i.e., 2-bromo-2-methylpropane and the structure is as follow
Optically active isomer of $(A)$ is 2-bromobutane $(B)$ and its structural formula is
(ii) As compound (B) is opically active therefore, compound (B) must be 2-bromobutane. Since, the rate of reaction of compound (B) depends both upon the concentration of compound $(B)$ and KOH , hence, the reaction follow $\mathrm{S}_{\mathrm{N}} 2$ mechanism. In $\mathrm{S}_{\mathrm{N}} 2$ reaction, nucleophile attack from, the back side, therefore, the product of hydrolysis will have opposite configuration
Write the structures and names of the compounds formed when compound ' $A$ ' with molecular formula $\mathrm{C}_7 \mathrm{H}_8$ is treated with $\mathrm{Cl}_2$ in the presence of $\mathrm{FeCl}_3$.
When compound ' $A$ ' with molecular formula, $\mathrm{C}_7 \mathrm{H}_8$ is treated with $\mathrm{Cl}_2$ in the presence of $\mathrm{FeCl}_3 \mathrm{o}$-chlorotoluene or $p$-chlorotoluene will be formed as the compound A with molecular formula $\mathrm{C}_7 \mathrm{H}_8$ is toluene.
