Q3P - Organic Chemistry (Mcmurry)

Question

Draw and name all monochloro products you would expect to obtain from radical chlorination of 2-methylpentane. Which, if any, are chiral?

Step-by-Step Solution

Answer

Chlorination at 1,3,4 carbon produces chiral products as an enantiomeric racemic mixture.

Step 1: Radical chlorination

It is a very less selective but highly reactive process. 

Radical chlorination means, “The removal of one hydrogen from alkane and addition of one chlorine on that place to form a chloroalkane”. A radical is formed as an intermediate.

Step 2: Chirality and achirality

Chiral carbon is an asymmetric carbon atom, to which four different groups are attached, whereas in achiral carbon, two or more groups may be same.

Chiral carbon have non-superimposable mirror image, whereas the mirror image of achiral carbon is superimposable.

Step 3: Enantiomeric racemic mixture

Racemic mixture contains equal proportions (i.e. 50-50) of dextro and leavo rotatory enantiomers. It is optically inactive, because enantiomers are mirror images and they rotates the plane polarized light in equal but opposite direction.

Step 4: Radical chlorination of 2-methylpentane

2-methylpentane has total five sites or carbons for chlorination, so five products are formed. When chlorination occurs at 1,3 and 4 carbon, then the product formed is chiral. When chlorination occurs at 2 and 5 carbon, then achiral products are formed. All products form an enantiomeric racemic mixture.

The products are as follows: