Optically pure (R)-2-halo compound reacts to give a product that is 70% inverted and 30% retained in spatial arrangement (an $S_{N} 1$ reaction with partial ion-pair shielding). The enantiomeric excess of the product, in favour of the inverted enantiomer, is

Answer: A

Enantiomeric excess

=

(major

-

minor)

= 70% - 30% = 40%

. Equivalently, 60% of the product is racemic (30% of each enantiomer) and the remaining 40% is the inverted enantiomer in excess.