The number of optically active isomer(s) of the compound ${C H}_{3} C H B r C H B r C O O H$ is(are):

zero

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one

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three

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four

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Solution

The correct option is C four
Total number of optical active isomers $= 2^{n}$
$n =$ number of chiral centres
Here, $n = 2$ , so, $2^{2} = 4$
So, the number of optical isomers $= 4$

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C H_{3} C H (O H) C O O H

The total number of optically active isomers for $C H_{2} O H (C H O H)_{3} C H O$ are:

{C H}_{2} O H {(C H O H)}_{3} C H O

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