Q.$(R)$-2-bromobutane is treated with aqueous $\text{NaOH}$ under conditions that favour the $S_N2$ pathway. Predict the configuration of the butan-2-ol formed and explain why the descriptor changes even though the spatial arrangement simply inverts.

The $S_N2$ mechanism is a single, concerted step in which a nucleophile attacks the halogen-bearing carbon from the side directly opposite the leaving group (backside attack), passing through one transition state where the carbon is partially bonded to both incoming and outgoing groups. Because both the substrate and nucleophile take part in this one rate-determining step, the rate law is second order overall, rate $= k[\text{RX}][\text{Nu}^-]$. The backside approach forces the other three substituents to flip to the opposite face as the reaction proceeds -- Walden inversion -- so the product's spatial arrangement is the mirror image of the substrate's; if the leaving group and incoming nucleophile share the same CIP priority rank (as $\text{Br}$ and $\text{OH}$ typically do), this spatial inversion also flips the $R/S$ descriptor. $S_N2$ is fastest for methyl and primary substrates (unhindered backside), is favoured by strong nucleophiles and polar aprotic solvents (which leave the nucleophile unsolva …