Q6-37E - Organic Chemistry (Mcmurry)

Question

Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination of neopentane, ${\left({\mathbf{CH}}_{\mathbf{3}}\right)}_{\mathbf{4}}\mathbf{C}$, is a good way to prepare neopentyl chloride, ${\left({\mathbf{CH}}_{\mathbf{3}}\right)}_{\mathbf{3}}{\mathbf{CCH}}_{\mathbf{2}}\mathbf{Cl}$. Explain.

Step-by-Step Solution

Answer

Pentane      1-chloropentane     2-chloropentane   chloropentane

Neopentane                                       Neopentyl chloride/

                                                        1-Chloro-2,2-dimethyl propane

Radical chlorination of alkane

The radical chlorination of alkane generally depends upon the type of C-H bond present in the molecule. This chlorination of alkane is not useful because the mixture of the products often results when more than one kind of C-H  bond is present in the molecule.

Radical chlorination in pentane

In the case of pentane, three kinds of C-H bonds are present, which gives three different types of products which are represented as follows:

Pentane     1-chloropentane  2-chloropentane    3-chloropentane

Radical chlorination in neopentane

Whereas in the case of neopentane, only one kind of C-H bond is present, which gives only one type of product that is represented as follows:

Neopentane                                                   Neopentyl chloride/

                                                        1-Chloro-2,2-dimethyl propane

By observing the given two radical chlorination reactions, neopentane is an excellent way to prepare neopentyl chloride. In neopentane, a single monosubstituted product is formed because of one type of C-H bond.