Remembering that SN2 reactions go with 100 % inversion of configuration, while SN1 reactions lead to racemization, explain why the reaction of (Z

The question is incomplete; the complete question is;

Remembering that Sn2 reactions go with 100% inversion of configuration, while Sn1 reactions lead to racemization, explain why the reaction of (R)-2-butanol as in this experiment gives a mixture of about 75% (S)- 2 - bromobutane and about 25% (R)-2-bromobutane

The reaction occurs mostly by SN2 mechanism

Explanation:

The two major routes for the synthesis of an alkyl halide involves the use of an alkene or an alkanol reacting with a hydrogen halide in both cases as the starting materials. The both routes involves the formation of a planar carbocation intermediate which is flat and planar.

Note that (R)-2-butanol is a secondary alcohol. Recall that nucleophillic substitution at a secondary carbon atom occurs mostly by SN2 mechanism. This implies that reaction of this alcohol is expected to proceed mostly by SN2 mechanism. SN2 mechanism involves the inversion of configuration, hence, we expect to have an excess of the product (S)- 2 - bromobutane in which there is an inversion of configuration compared with 25% (R)-2-bromobutane in which there is retention of configuration. This goes a long way to show us that the reaction occurs mostly by SN2 mechanism.

Secondly , steric factors also favour the formation of 75% (S)- 2 - bromobutane over 25% (R)-2-bromobutane. There is less steric hindrance in the transition state leading to the formation of (S)- 2 - bromobutane than that leading to the formation of (R)-2-bromobutane.

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