Question

Of the isomeric hexanes, the isomers that give the minimum and the maximum number of monochloro derivatives are respectively:
(A) $2,3\text{-dimethylbutane and n-hexane}$
(B) $\text{3-methylpentane and 2,3-dimethylbutane}$
(C) $2,2\text{-dimethylbutane and 2-methylpentane}$
(D) $\text{2,3-dimethylbutane and 2-methylpentane}$
(E) $\text{2-methylpentane and 2,2-dimethylbutane}$

Answer 1

For every hydrogen atom present, a chlorine atom can be replaced. But not all of them will require substitution. Look out for the symmetry in the molecules.

Complete step by step solution:

First let us understand what we mean by monochloro derivatives. When a hydrocarbon is given, we can substitute any of its hydrogen atoms with a chlorine atom. A monochloro derivative is formed when only a particular carbon atom in the whole molecule has one chlorine atom. Say for example we have been given hydrocarbon ethane. We know that it has two carbon atoms. So to make a monochloro derivative out of it, we substitute any one hydrogen atom attached to any one carbon atom. It is as shown below:

As it is clearly visible, both the monochloro derivatives are the same compound because of the line of symmetry present in the ethane molecule. So in the compounds given above, we have to take care of this factor and only sort out those monochloro derivatives of a particular molecule which are different from each other structurally.
- Let us list out the structures of all the compounds given in a serial wise manner:
- $2,3\text{-dimethylbutane }$

- $2,2\text{-dimethylbutane}$

- $\text{2-methylpentane}$

There are only three monochloro derivatives of $2,3\text{-dimethylbutane}$and $\text{n-hexane}$ each because of the line of symmetry that passes in the middle, as shown below:

There are four monochloro derivatives for $2,2\text{-dimethylbutane}$. It has a horizontal line of symmetry as shown below:

There are four monochloro derivatives for $\text{3-methylpentane}$. It has a vertical line of symmetry as shown below:

There are five monochloro derivatives for $\text{2-methylpentane}$. The longest carbon chain remains the same both the ways as shown below:

So the isomers that give the minimum and the maximum number of monochloro derivatives respectively are $\text{2,3-dimethylbutane and 2-methylpentane}$, option (D).

Note: Finding the line of symmetry is important to solve this question. This was a simple case as the molecules only contained one such line but things could get complicated if we have more of them.
Only one chlorine atom can be substituted per carbon atom, any more and it would be wrong.