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in Chemistry by (74.3k points)

Write down the number of asymmetric carbon atoms in each optically active compound and report the number of isomers. 

i. CH3 (CHOH)2 COOH

ii. COOH(CHOH)2 COOH

1 Answer

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Best answer

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

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