Question

Both compounds A and B have molecular formula C6H14. Monochlorination of compound A results in the formation of two constitutional isomers. Monochlorination of compound B results in the formation of four constitutional isomers. Identify compounds A and B, and show the products of each monochlorination.

1) Draw compound A and the products of its monochlorination.

2)Draw compound B and the products of its monochlorination.

Answer 1

Compound A could be 3-methylpentane, which upon monochlorination forms two constitutional isomers. Compound B could be 2,3-dimethylbutane, and it forms four constitutional isomers when monochlorinated. This reflects the different positions at which hydrogen can be replaced by chlorine.

Step-by-step explanation:

The student's question involves identifying two isomeric hydrocarbons with the molecular formula C6H14 that generate different numbers of constitutional isomers when each is subjected to monochlorination. Compounds A and B are expected to be different structural isomers of hexane, which will form different chlorinated products based on their structure.

Compound A, which forms two constitutional isomers upon monochlorination, could be 3-methylpentane. This structure has two types of hydrogen atoms - those on the methyl group and those on the rest of the carbon chain. When chlorination occurs, either set of hydrogen atoms can be replaced with a chlorine atom, resulting in two different constitutional isomers.

Compound B, which forms four constitutional isomers upon monochlorination, could be 2,3-dimethylbutane. This structure has four different types of hydrogen atoms, thus chlorination at each type results in a different constitutional isomer.

The respective monochlorination reactions show the formation of two constitutional isomers from chlorinating compound A and four constitutional isomers from chlorinating compound B, based on the positions where the hydrogen is replaced by chlorine.