Question

Identify (a) and (b)

• A. (a) = methane sulphonate $\left(b\right)=\frac{NaCN}{DMF}$
• B. (a) = TsCl $\left(b\right)=\frac{NaCN}{DMF}$
• C. (a) = Methanesulfonyl chloride $\left(b\right)=R-\stackrel{+}{N}\equiv \frac{\stackrel{\ominus }{C}}{DMF}$
• D. (a) = Methanesulfonyl chloride $\left(b\right)=\frac{NaCN}{DMF}$

Answer 1

The correct option is D

(a) = Methanesulfonyl chloride $\left(b\right)=\frac{NaCN}{DMF}$

With the reagent (a), there is no change in the stereochemical arrangement at the carbon of interest. In the next step, there is a Walden Inversion at the carbon where the leaving group is substituted with $-CN$. This step definitely has to be $S_{N}2$ and thus it would make sense to have a suitable solvent - preferably a polar aprotic one.

The first step basically converts the poor leaving group $-OH$ into a really good one so the next step - as in the $S_{N}2$ substitution can happen smoothly.