Question

The reactions which cannot be applied to prepare an alkene by elimination, are

Choose the correct answer from the options given below:

• A. B & E Only
• B. B, C & D Only
• C. A, C & D Only
• D. B & D Only

Hint:

In elimination reactions, strong bases such as NaOEt or NaOMe can induce E2 eliminations, while aqueous KOH typically favors SN2 reactions. Additionally, oxidation reactions like in Reaction D do not lead to alkenes via elimination.

Answer 1

The Correct Option is D

The question asks which reactions cannot be used to prepare an alkene by elimination. Elimination reactions typically involve the removal of a small molecule from adjacent carbon atoms and formation of a double bond.

1. Option A shows a dehydrohalogenation reaction using NaOEt, a typical strong base, where elimination of HBr from the substrate can occur, forming an alkene. This is a valid elimination reaction.
2. Option B shows a reaction using KOH(aq), which typically leads to nucleophilic substitution rather than elimination. Thus, it is generally not used to prepare alkenes effectively in this context.
3. Option C involves NaOMe, another strong base, typical for promoting elimination (dehydrohalogenation) and forming an alkene, making it a valid elimination process.
4. Option D features potassium hydroxide in the presence of phenol, leading to a different reaction mechanism, such as nucleophilic aromatic substitution or other condensation reactions, not alkene formation typically.
5. Option E shows a dehydration of alcohol at 573 K, which is a standard elimination reaction, forming an alkene by removing water (H₂O).

Based on the above analysis, Options B and D do not typically facilitate the formation of alkenes via elimination reactions. Therefore, the correct answer is B & D Only.

Answer 2

Step 1: Reviewing Reaction A:
Reaction A involves the NaOEt base, which induces an E2 elimination reaction. This reaction forms an alkene by eliminating the bromine from the cyclohexane ring.
Thus, Reaction A is valid for elimination.
Step 2: Reviewing Reaction B:
Reaction B involves the aqueous KOH, which can induce E2 elimination on the alkyl halide to form an alkene. However, Reaction B is invalid because KOH (aqueous) is more likely to induce nucleophilic substitution (SN2) rather than elimination under these conditions.
Reaction B is not valid for elimination because aqueous KOH typically favors nucleophilic substitution over elimination.
Step 3: Reviewing Reaction C:
Reaction C involves sodium methoxide (\( \text{NaOMe} \)), which can induce an E2 elimination reaction to form an alkene by abstracting a proton from the \(\beta\)-carbon. This reaction is valid for elimination. Reaction C is valid for elimination.
Step 4: Reviewing Reaction D:
Reaction D involves the oxidation of phenol (\( \text{C}_6\text{H}_5\text{OH} \)) with \( \text{Na}_2\text{Cr}_2\text{O}_7 \) and \( \text{H}_2\text{SO}_4 \), which leads to the formation of a quinone, not an alkene. This is an oxidation reaction, not an elimination reaction.
Reaction D is not valid for elimination.
Step 5: Reviewing Reaction E:
Reaction E involves the use of Cu at 573K, which can dehydrogenate the alcohol to form an alkene via E1 elimination. This reaction works and forms an alkene. Reaction E is valid for elimination.
Conclusion:
Reactions B and D are the reactions that cannot be applied to prepare an alkene by elimination. Reaction B favors nucleophilic substitution, and Reaction D involves oxidation rather than elimination.