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.
Identify the products A and B formed in the following reaction
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}-\mathrm{CH}-\mathrm{CH}_3+\mathrm{HCl} \longrightarrow \mathrm{A}+\mathrm{B}$
In the given reaction, addition occur and the following two products (A and B) are possible
Further the carbocation formed from compound $(A)$ is slightly less stable than carbocation leading to the formation of compound $(B)$ therefore the amount of 2-chloropentane $(B)$ will be slightly more than that of 3- chloropentane (A).
Which of the following compounds will have the highest melting point and why?
In compound (II), both the $\mathrm{CH}_3$ groups and Cl atoms at para-position to each other. Therefore, compound (II) is more symmetrical and it fits in the crystal lattice better than the other two isomers and hence it has the highest melting point than the others.
Write down the structure and IUPAC name for neo-pentylbromide.
The structure of neo-pentylbromide is
IUPAC name $\Rightarrow$ I-bromo -2,2-dimethylpropane
Common name $\Rightarrow$ neo-pentylbromide