The major product of the following reaction, if it occurs by S$$_N$$2 mechanism is:

The substrate given in the question is $$(S)\;CH_3-CH(Br)-CH_2-CH_3$$ ( $$(S)$$-2-bromobutane ). It is treated with aqueous hydroxide ion, and the statement in the question specifically says that the substitution proceeds by an $$S_N2$$ mechanism.

Step 1 : Nature of the mechanism
An $$S_N2$$ (bimolecular nucleophilic substitution) reaction is characterised by
  • a single, concerted transition state and
  • backside attack of the nucleophile on the carbon that bears the leaving group.
Because of the backside attack, the configuration at the stereogenic carbon is inverted. This stereochemical change is called Walden inversion.

Step 2 : Applying inversion to the present molecule
The nucleophile here is $$HO^-$$ and the leaving group is $$Br^-$$. The stereogenic carbon is $$C_2$$ of the butane chain. Since the starting configuration about $$C_2$$ is $$(S)$$, after complete inversion the configuration must become $$(R)$$.

Step 3 : Writing the product
Replacing $$Br$$ by $$OH$$ with inversion therefore gives $$(R)\;CH_3-CH(OH)-CH_2-CH_3,$$ i.e. $$(R)$$-butan-2-ol.

Thus the major product is $$(R)$$-2-butanol.

Option A which is: $$(\mathbf{R})$$-butan-2-ol