Question

Which compounds can be prepared in high yield by halogenation of an alkane? (a) 2-Chbropentane (b) Chlorocyclopentane (c) 2 -Bromo-2-methytheplane (d) \((R)-2\) - Bromo-3-methyylbutane (e) 2 -Bromo-2,4-trimethyipentane (f) lodoethane

Short answer

Answer: Chlorocyclopentane

Step-by-step solution

Option (a): 2-Chloropentane

To obtain 2-Chloropentane, a chlorine atom would replace a hydrogen atom at the second carbon atom of pentane. The reaction would be a free radical halogenation, which involves a radical chain mechanism. This process is not regioselective, meaning that it could also produce other isomers such as 3-Chloropentane. Therefore, the production of 2-Chloropentane might not be in high yield.

Option (b): Chlorocyclopentane

Chlorocyclopentane can be prepared by halogenation of cyclopentane, where a chlorine atom replaces a hydrogen atom on the cyclopentane ring. As the ring structure allows only one possible isomer to form, this reaction would produce Chlorocyclopentane in high yield.

Option (c): 2-Bromo-2-methylpentane

To obtain 2-Bromo-2-methylpentane, a bromine atom would replace a hydrogen atom on the second carbon atom of 2-methylpentane. However, bromine is relatively selective in halogenation reactions and mainly reacts with the allylic or benzylic carbon atoms. In this case, the reaction might not result in high yield for this particular compound.

Option (d): \((R)-2\)-Bromo-3-methylbutane

To obtain \((R)-2\)-Bromo-3-methylbutane, a bromine atom would replace a hydrogen atom on the second carbon atom of 3-methylbutane. However, the reaction would produce a mixture of enantiomers as it is not stereoselective. Consequently, the desired \((R)\)-enantiomer would not be produced in high yield.

Option (e): 2-Bromo-2,4-trimethylpentane

A bromine atom would replace a hydrogen atom on the second carbon atom of 2,4-trimethylpentane to obtain the compound. However, similar to option (c), bromine is selective, and it would most likely react with the allylic or benzylic carbon atoms. The production of this compound might not be in high yield.

Option (f): Iodoethane

Iodoethane can be prepared by halogenation of ethane, where an iodine atom replaces a hydrogen atom. However, iodination is energetically unfavorable and usually occurs very slowly under standard conditions. Therefore, the reaction might not produce Iodoethane in high yield. In conclusion, out of the given options, only Chlorocyclopentane (Option b) can be prepared in high yield by halogenation of an alkane.