Identity of the Alkyl Bromide

To determine the identity of the alkyl bromide, we need to analyze the given information and follow the reaction steps.

Step 1: Reaction with Sodium Ethoxide and Ethanol
When the alkyl bromide reacts with sodium ethoxide (NaOC2H5) and ethanol (C2H5OH), it produces a single alkene. This suggests that the alkyl bromide is a primary alkyl bromide, as primary alkyl bromides typically undergo elimination reactions to form a single alkene.

Step 2: Hydrogenation of the Alkene
The alkene produced in the previous step undergoes hydrogenation, resulting in the formation of 2-methyl butane. Hydrogenation is the addition of hydrogen gas (H2) across the double bond of an alkene, resulting in the formation of an alkane.

Analysis
Based on the given information, we can conclude the following:

1. The alkyl bromide is a primary alkyl bromide, as it undergoes elimination to form a single alkene.
2. The alkene produced in the first step is an unsaturated hydrocarbon, as it undergoes hydrogenation to form an alkane (2-methyl butane).

Final Conclusion
Based on the given information and analysis, the identity of the alkyl bromide can be determined as 2-bromobutane.

Explanation
Here is a detailed explanation of the reaction steps and the reasoning behind the conclusion:

1. The alkyl bromide reacts with sodium ethoxide and ethanol. This reaction is an example of an elimination reaction, specifically the E2 mechanism. In this mechanism, the alkyl bromide acts as the electrophile, and the ethoxide ion acts as the nucleophile. The ethoxide ion attacks the beta-carbon of the alkyl bromide, resulting in the formation of a new carbon-carbon double bond and the expulsion of the bromide ion. Since a single alkene is formed, it suggests that the alkyl bromide is primary, as primary alkyl halides typically undergo E2 elimination to form a single alkene.

2. The alkene produced in the first step undergoes hydrogenation. Hydrogenation is a reaction that involves the addition of hydrogen gas across a double bond, resulting in the formation of an alkane. In this case, the alkene is hydrogenated to form 2-methyl butane. The presence of a methyl group suggests that the original alkene had a substituent attached to one of the carbons.

3. Considering the information from both steps, we can conclude that the alkyl bromide is 2-bromobutane. This compound is a primary alkyl bromide, and upon elimination, it forms a single alkene. The resulting alkene, when hydrogenated, produces 2-methyl butane, which has a methyl group attached to one of the carbons.

In summary, based on the given information and the reaction steps, the identity of the alkyl bromide is determined to be 2-bromobutane.