Q16 / 180 · KCET 2013 · Chemistry
The number of optical isomers of the compound $\mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CHBr}-\mathrm{COOH}$ is
- A0
- B1
- C3
- Correct answerD4
Solution
Compounds having similar physical and chemical properties but differing only in the behaviour towards polarised light are called optical isomers.
The number of possible stereoisomers when molecule cannot be divided into two equal halves i.e., molecule has no symmetry.
(i) The number of $d$ - and $l$-forms $a=2^{n}$
(where $n=$ number of asymmetric carbon atoms) $\therefore a=2^{n}=2^{2}=4$
Total number of meso forms, $m=0$
Total number of optical isomers
$\begin{aligned}
&=a+m \\
&=4+0=4
\end{aligned}$

Here (i) and (iii), (i) and (iv), (ii) and (iii), (ii) and (iv) are diastereomeric pairs while (i) and (ii), (iii) and (iv) are enantiomers.

Here (i) and (iii), (i) and (iv), (ii) and (iii), (ii) and (iv) are diastereomeric pairs while (i) and (ii), (iii) and (iv) are enantiomers.
Hence, the correct answer is option: 'D'.