Question

An alkyl bromide produces a single alkene when it reacts with sodium ethoxide and ethanol. This alkene undergoes hydrogenation and produces 2-methyl butane. What is the identity of the alkyl bromide?
A. 1-bromo-2,2-dimethylpropane
B. 1-bromobutane
C. 2-bromo-2-methylbutane
D. 1-bromo-2-methylbutane

Answer 1

Hint: The conversion of an alkyl halide into an alkene happens in presence of potassium hydroxide. This reaction is termed the elimination reaction. This reaction is used in the laboratory to produce alcohol.

Complete Step by Step Solution:
Here, an alkyl halide undergoes an elimination reaction in the presence of ethanol. In an elimination reaction, there is the removal of several atoms from a molecule. The formed alkene when undergoing a hydrogenation reaction gives 2-methyl butane. So, the haloalkane is 1-bromo-2-methylbutane. So, the elimination reaction,

Fig: Elimination reaction of 1-bromo-2-methylbutane

In the above elimination reaction, one hydrogen atom and one bromine atom are removed from the 1-bromo-2-methylbutane. And the products formed are 2-methylbut-1-ene, sodium bromide, and ethanol.

In the second step, 2-methylbut-1-ene undergoes a hydrogenation reaction. In this reaction, hydrogen undergoes a reaction with a compound in presence of nickel (catalyst) to form an alkane. 2-methylbut-1-ene undergoes a hydrogenation reaction to form 2-methylbutane. The reaction is,

Fig: Hydrogenation reaction
Therefore, the alkyl halide used in the given reaction is 1-bromo-2-methylbutane.
Hence, option C is right.

Note: The elimination reaction is of three types, namely E1, E2, E1cb. The E1 reaction is a two-step removal process known by the name of unimolecular elimination. The E2 reaction is a one step removal process and known by the name of bimolecular elimination.