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
A hydrocarbon of molecular mass $72 \mathrm{~g} \mathrm{~mol}^{-1}$ gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.
Since, the molar mass of hydrocarbon is $72 \mathrm{~g} \mathrm{~mol}^{-1}$ thats why the hydrocarbon is $\mathrm{C}_5 \mathrm{H}_{12}$ i.e., pentane.
On photo chlorination, it gives monochloro derivatives so, all the hydrogen atoms must be equivalent and the structure of the compound will be
Name of the alkene which will yield/chloro 1-methylcyclohexane by its reaction with HCl . Write the reactions involved.
Two alkenes are possible
Addition takes place in accordance with Markownikoff's rule i.e., negative part of the adding molecule will get attached to that carbon which has lesser number of hydrogen atom.