Question

Identify the product [C] formed at the end of the reaction below: $ 1,1,2,2-\text{Tetrabromopropane} + 2 \text{Zn (s)}/ \text{Ethanol} \rightarrow [B] $ $ [B] + 2 \text{moles of HBr} \rightarrow [C] $

• A. 2,2-Dibromopropane
• B. 1,1-Dibromopropane
• C. 1,3-Dibromopropane
• D. 1,2-Dibromopropane

Hint:

In reactions involving zinc and ethanol, look for the reduction of halides followed by electrophilic addition of HBr to the alkene to form the final dibromide product.

Answer 1

The Correct Option is A

This question involves the reaction of 1,1,2,2-Tetrabromopropane with zinc in ethanol, followed by the addition of HBr. The reactions can be broken down as follows: 1. First 
Step (Reduction with Zinc in Ethanol): 
When 1,1,2,2-Tetrabromopropane reacts with zinc in ethanol, a reduction occurs, and two bromine atoms are removed, creating a bromoalkene. This intermediate compound formed in step 1 will have a double bond and two bromine atoms on adjacent carbons.
2. Second Step (Reaction with HBr): 
The product of the first reaction then reacts with HBr. In this step, the alkene undergoes an electrophilic addition, where the hydrogen (H) from HBr adds to one carbon, and the bromine (Br) adds to the other carbon of the double bond. Since the two bromine atoms were initially on adjacent carbons in the original tetrabromopropane, the final product is 2,2-Dibromopropane. Thus, the final product is 2,2-Dibromopropane.