Alkene Reactions and Markovnikov's Rule

Problem # 1336

Determine a synthesis to prepare 2-chloro-4-methylpentane from 1-iodo-4-methylpentane.

from

1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction). Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.

Problem Details

MendelSet practice problem # 1336 submitted by reifa on October 26, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): Markovnikov's rule

Problem # 530

Let's work through anti and syn additions to alkenes.

Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).

Solution

The reaciton of an alkene with Br₂ is an anti-addition, and hydrogenation (H₂ or D₂) is a syn addition. From this we can figure our the relative stereochemistry of each product.

Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.

There are three ways for products not to be optically active:

products can be achiral
products can be a racemic mixture
products can be meso

In both a) and b) each product has a sterocenter, so the products can't be achiral.

In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.

In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.