Question

The number of isomers possible for a dibromo derivative (Molecular weight = 186 u) of an alkene is (Br = 80 u):

• A. 2
• B. 3
• C. 4
• D. 6

Hint:

When working with organic compounds, the molecular weight and positioning of functional groups like bromine can help identify the number of possible isomers. For alkenes, the position of the halogen atoms (in this case, bromine) on the carbon chain is key to identifying isomeric forms.

Answer 1

The Correct Option is B

Step 1: Understand the molecular weight and alkene structure The molecular weight of the alkene is given as 186 u, and since the bromine atom weighs 80 u, two bromine atoms will contribute \(2 \times 80 = 160\) u to the molecular weight. Hence, the alkene portion of the molecule must have a molecular weight of \(186 - 160 = 26\) u. The alkene with a molecular weight of 26 u will have the formula \( C_2H_4 \), corresponding to the simplest alkene, ethene (\(C_2H_4\)).
Step 2: Consider possible isomers of the dibromo derivative In the dibromo derivative of the alkene, we have two bromine atoms to place across the ethene molecule. Bromine atoms can be placed at different positions, leading to different isomers. These isomers can be:
1. 1,2-Dibromoethene
2. 1,3-Dibromoethene
3. 1,4-Dibromoethene
These are the three possible isomers of the dibromo derivative of ethene.
Step 3: Conclusion Therefore, the number of possible isomers of the dibromo derivative is 3.