When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When the isomeric halide (R)-2-bromo-2,5- dimethylnonane is dissolved in CH3OH under the same conditions, nucleophilic substitution forms an optically active solution. Draw the products formed in each reaction, and explain why the difference in optical activity is observed.

See explanation

Explanation:

The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.

On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.

Answer taken away. (Sorry, I put the wrong one)