Do SN2 reactions always invert absolute configuration?

Question

I know that an $\mathrm{S_N2}$ reaction proceeds through complete inversion of configuration, but is it necessary that the absolute configuration (R/S) of the chiral centre change after an $\mathrm{S_N2}$ attack ?

Answer 1

For the ease of thinking about this, let's imagine $\mathrm{S_N2}$ attack of $\ce{CN-}$ on bromochlorofluoroiodomethane, where iodide is being displaced.

Absolute stereochemistry is retained in this case because the highest priority substituent is becoming the lowest priority substituent. If you were to find the absolute stereochemistry of $\ce{2 ->3->4}$ in (R)-bromochlorofluoroiodomethane, which becomes $\ce{1->2->3}$ after attack, you would see that it is S and after attack becomes R. You can see that here:

Answer 2

The assigned configuration is based on the substituents at the stereocenter. If what is entering has a different prority than what is leaving relative to the other groups, there may be no change in R/S.

For example:

The SN2 reaction of cyanide with MeCClHCH(SEt)2; in this case S alkyl chloride-->S nitrile (or R-->R) because the -CH(SEt)2 substituent is higher priority than -CN. The -Cl was 1st priority in the starting material.

Answer 3

To determine whether the $S_N2$ substitution inverts $R/S$ configuration, just do some simple arithmetic.

Let $Nu$ be the nucleophile, $L$ be the leaving group, and $R_1,R_2, R_3$ be the remaining groups. $L$ will be higher in priority than 0, 1, 2, or 3 of the groups, and similarly for $Nu$. You work out the difference between the two correct numbers for your specific reaction. If it's even, you invert absolute configuration, if it's odd, you don't. For example, in @Ringo's example above the leaving group, iodide, is higher than 3 of the retained groups but cyanide is higher than 0. These numbers have an odd difference --> absolute configuration is retained.