Question

Identify the product 'A' formed in the following reaction (Addition of $\text{Br}_{2}$ to an alkene).

• A. 2,3-dibromopentane
• B. 2-Bromo-3-methylbutane
• C. 3-Bromo-2-methylbutane
• D. 2,3-dibromo-2-methylbutane

Hint:

Electrophilic addition of $X_2$ to an alkene always results in a vicinal dihalide (halogens on adjacent, neighboring carbons). The double bond literally breaks and hands one halogen to each participating carbon.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The addition of bromine ($Br_2$) to an alkene is an electrophilic addition reaction where one bromine atom adds to each carbon of the double bond.

Step 2: Formula Application:
$R-CH=CH-R' + Br_2 \to R-CH(Br)-CH(Br)-R'$.

Step 3: Explanation:
Based on the typical context of this question involving 2-methylbut-2-ene: the $Br_2$ adds across the double bond at positions 2 and 3. This results in two bromine atoms on adjacent carbons, specifically 2,3-dibromo-2-methylbutane.

Step 4: Final Answer:
The product 'A' is 2,3-dibromo-2-methylbutane.