Which of the following is the major organic product formed in the E2 dehydrohalogenation of $C_2{H}_5\mathrm{Br}$ (bromoethane) with a strong base such as $\mathrm{NaOH}$?

• A. ${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{Na}$ (sodium ethyl)
• B. ${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{Br}$ (bromoethane, starting material)
• C. ${\mathrm{CH}}_3\mathrm{CH}{2}_2\mathrm{OH}$ (ethanol)
• D. $\mathrm{CH}{2}_2=\mathrm{CH}{2}_2$ (ethene)

Answer: (D)

Verified step by step guidance

1. Identify the type of reaction: E2 dehydrohalogenation involves the elimination of a hydrogen atom and a halogen atom (in this case, bromine) from adjacent carbon atoms, resulting in the formation of a double bond (alkene).
2. Recognize the substrate: The starting material is bromoethane (\(\mathrm{CH_3CH_2Br}\)), which is a primary alkyl halide with bromine attached to the second carbon.
3. Consider the base: A strong base such as sodium hydroxide (\(\mathrm{NaOH}\)) will abstract a proton (H) from the carbon adjacent to the carbon bearing the bromine, facilitating elimination.
4. Determine the elimination site: Since bromoethane has only two carbons, the proton removed by the base must come from the carbon adjacent to the one bonded to bromine, leading to the formation of a double bond between the two carbons.
5. Write the product: The elimination results in the formation of ethene (\(\mathrm{CH_2=CH_2}\)), where the bromine and a hydrogen are removed, creating a carbon-carbon double bond.