Since n is even in both the case,
Number of O.I.A = 2n
Number of meso forms=2(n–2)/2
Number of (O.I.A) racemic form=n/2
Total optical isomers=2n–1 + 2(n–2)/2
i. Number of asymmetric C atoms = 2
Different terminal groups : (–COOH) and (Me)
Number of O.I.A = 2n =22 =4
Number of meso forms =0
Number of (O.I.A) racemic form = 4/2 = 2
(racemate mixture)
∴ Total optical isomers = 4
Optically inactive forms = 2.
ii. Number of asymmetric C atoms = 2. Same terminal groups: (–COOH)
Number of O.I.A = 2n-1 = 22-t =2
Number of meso forms = 2(n-2)/2 = 2° = 1
Number of (O.I.A) racemic form = 2/2 =1
Total optical isomers = 2+1=3