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The compound C(2)FCI Brl has...

The compound `C_(2)FCI` Brl has

A

4 isomers

B

2 optical isomers

C

2 geometrical isomers

D

6 isomers

Text Solution

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The correct Answer is:
To solve the problem regarding the compound C2FClBrI and determine the number of isomers it has, we will analyze the types of isomers that can be formed, specifically focusing on geometrical and optical isomers. ### Step-by-Step Solution: 1. **Identify the Structure of the Compound**: The compound C2FClBrI consists of two carbon atoms (C) and four different halogen atoms (F, Cl, Br, I). The general formula suggests that we have a dihaloalkane with two different carbon atoms. 2. **Determine the Possibility of Geometrical Isomers**: Geometrical isomers occur due to the restricted rotation around a double bond or due to the presence of different groups attached to the same carbon atoms. In this case, since we have four different substituents (F, Cl, Br, I) attached to the two carbon atoms, we can have geometrical isomers. 3. **Check for Optical Isomers**: Optical isomers (enantiomers) are formed when a compound has a chiral center, which means it cannot be superimposed on its mirror image. For a compound to be optically active, it must not have a plane of symmetry. In the case of C2FClBrI, if we draw its structure, we can see that it has a plane of symmetry, making it optically inactive. 4. **Count the Geometrical Isomers**: To count the geometrical isomers, we can arrange the four different halogen atoms around the two carbon atoms in various configurations. The arrangements can be done as follows: - Place F, Cl, Br, and I in different positions around the carbon atoms. - Ensure that no two groups are identical in each arrangement. After systematically arranging the groups, we find that there are 6 unique configurations (isomers) that can be formed. 5. **Conclusion**: Based on the analysis, we conclude that the compound C2FClBrI has: - **6 geometrical isomers** (option 4). - **No optical isomers** due to the presence of a plane of symmetry. ### Final Answer: The compound C2FClBrI has a total of **6 isomers**. ---
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    plane of symmetry
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