Question

Neopentyl bromide undergoes dehydrohalogenation to give alkenes even though it has no hydrogen. This is due to which of the following mechanisms?
A. ${E_2}$ mechanism
B. ${E_1}$ mechanism
C. Rearrangement of carbocations by ${E_1}$ mechanism
D. ${E_1}cB$ mechanism.

Answer 1

Write down the possible alkenes that can be formed by dehydrohalogenation of Neopentyl bromide. Check for the mechanism where the products are most stable and easier to form, also check whether the mechanism contributes to stability of carbocations in the reaction process.

Complete answer:
We can write the equation for dehydrohalogenation of Neopentyl bromide as
${\left( {C{H_3}} \right)_3}C - C{H_2}Br \to {\left( {C{H_3}} \right)_2}C = CH - C{H_3}$ this reaction is done in the presence of $KOH$
Here, we can see that in order to remove $HBr$ from the molecule and form a double bond, the second does not contain any Hydrogen atoms. So, when Bromine is removed from the first carbon, a carbocation is formed- ${\left( {C{H_3}} \right)_3}C - C{H_2}^ +$
The carbocation is unstable and the carbon attached to it is highly stable due to the three methyl groups attached to it. Hence one methyl group is transferred to the primary carbocation and the charge is transferred to the tertiary carbocation in order to stabilize the molecule forming ${\left( {C{H_3}} \right)_2}{C^ + } - C{H_2} - C{H_3}$
Now, a Hydrogen atom is removed from the adjacent carbon to form our desired product, alkene.
Hence in order to form an alkene, neopentyl bromide is undergoing rearrangement of carbocations in dehydrohalogenation through ${E_1}$ or elimination mechanism.

**Therefore the answer is Option C.

Note:**
In order to eliminate $HBr$ from the given neopentyl bromide compound, we use the ${E_1}$ mechanism. To obtain carbocation stability in this dehydrohalogenation process, the rearrangement of molecules between two adjacent carbon atoms occurs and this makes the elimination process possible.