1.

The number of optical isomers for the compound, CH_(3)-CH(Br)-CH(Br)C_(2)H_(5) is

Answer»

1
2
4
3

Solution :The compound `CH_(3)-CH(Br)CH(Br)C_(2)H_(5)` cannot be divided into two identical HALVES. It has 2 chiral CENTRES. The formula for number of optically active form is `2^(n)`. Where n is the number of chiral centres.
Hence `2^(2) = 4`
Hence the number of isomers are 4 so choice (a), (b) and (d) are INCORRECT while (C) is correct.


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