Question

Which of the following isomers of the hexane can give two mono chlorination compounds:
A.$n -$hexane
B.$2,3 -$dimethyl butane
C.$2,2 -$dimethylbutane
D.$2 -$methyl pentane

Answer 1

Isomers: Those molecules which have the same molecular formulas but different in their atomic arrangement. The chemical and physical properties of isomers may or may not be the same to each other. Mono chlorination compounds are those compounds in which one hydrogen atom from the alkane is replaced by one chlorine atom.

Complete step by step answer:
Chlorination compounds: Those compounds in which hydrogen atoms from the alkane are replaced by chlorine atoms. If one hydrogen atom is replaced by one chlorine atom then it will be called mono chlorination. If two hydrogen atoms are replaced by two chlorine atoms then it is known as di chlorination and so on. Different chlorination products are considered when there are different names according to IUPAC rules. So, now we will check option by option.
For a) $n -$hexane: There will be three mono chlorination as there are three different types of hydrogen present in the $n -$hexane. Three monochlorination products will be as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl, C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(Cl)C}}{{\text{H}}_3}$ and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(Cl)C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}$

For b) $2,3 -$dimethyl butane: There will be two mono chlorination as there are two different types of hydrogen present in the $2,3 -$dimethyl butane. These two monochlorination products will be as:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{\text{)CH(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_{\text{2}}}{\text{Cl}}$ and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{\text{)C(Cl)(C}}{{\text{H}}_3}{\text{)C}}{{\text{H}}_3}$

For c) $2,2 -$dimethylbutane: There will be three mono chlorination as there are three different types of hydrogen present in the $2,2 -$dimethylbutane. Three monochlorination products will be as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C(C}}{{\text{H}}_{\text{3}}}{\text{)(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_{\text{2}}}{\text{Cl, C}}{{\text{H}}_{\text{3}}}{\text{CH(Cl)C(C}}{{\text{H}}_{\text{3}}}{\text{)(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3}$ and ${\text{C}}{{\text{H}}_2}{\text{(Cl)C}}{{\text{H}}_{\text{2}}}{\text{C(C}}{{\text{H}}_{\text{3}}}{\text{)(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3}$

And for d) $2 -$methyl pentane: There will be five mono chlorination as there are five different types of hydrogen present in the $2 -$methyl pentane. Five mono chlorination products will be as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_{\text{2}}}{\text{Cl, C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C(Cl)(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3},{\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(Cl)CH(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3},$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(Cl)C}}{{\text{H}}_{\text{2}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3}$ and ${\text{C}}{{\text{H}}_2}{\text{(Cl)C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{\text{)C}}{{\text{H}}_3}$
Hence the answer will be option B which has two monochlorination products.

Note:
The question is asking for two mono chlorination compounds which means the compounds formed after removing the hydrogen and adding the chlorine in place of hydrogen should give us different compounds. We can check whether the compounds formed after the reaction are the same or different by naming the compounds using the IUPAC nomenclature.