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The number of optically active isomer(s) of the compound $C{{H}_{3}}CHBrCHBrCOOH$ is (are):
A. Zero
B. One
C. Three
D. Four


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Last updated date: 20th Jun 2024
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Answer
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Hint: To know the number of optical isomers is present in a compound first we should know the number of chiral centers is present in the given molecule. Chiral center means a carbon atom should be bonded to four different atoms. Each chiral center will produce two different optical isomers for a compound

Complete step by step answer:
- In the question it is given that we have to find the number optically active isomers possible for $C{{H}_{3}}CHBrCHBrCOOH$ .
- First of all we should draw the structure of the given compound.
- The structure of the given compound is as follows.

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- Now we have to find the number of chiral centers in the above structure.
- The chiral centers present in the given molecule can be represented with star marks as follows.

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- Means in the given compound there are two chiral centers.
- The formula to calculate the number of optically active isomers from the number of chiral centers is as follows.
- Total number of optically active isomers = ${{2}^{n}}$
Here n = number of chiral centers present for a particular molecule.
- We know that in the given molecule there are two chiral centers then substitute it in the above equation to get a number of optically active isomers.
- Total number of optically active isomers for the given compound $={{2}^{n}}={{2}^{2}}=4$ .
- Therefore the given molecule contains 4 optically active isomers.

- So, the correct option is D.

Note: We have to check which carbon is chiral and which carbon is achiral during the calculation of number of optically active isomers. Optically active means they will rotate the plane polarized light from one direction to another.