Question

Identify the reagents used for the following conversion

• A. A = LiAlH₄, B = NaOH_(aq), C = NH₂–NH₂/KOH, ethylene glycol
• B. A = LiAlH₄, B = NaOH_(alc), C =Zn/HCl
• C. A = DIBAL-H, B= NaOH_(aq), C = NH₂–NH₂/KOH, ethylene glycol
• D. A = DIBAL-H, B = NaOH_(alc), C = Zn/HCl

Answer 1

The Correct Option is D

To identify the reagents for the given chemical conversion, let's analyze each step: 

1. Step A: The first conversion involves the reduction of an ester to an aldehyde.
   • DIBAL-H (Diisobutylaluminium hydride) is a reagent known for its ability to reduce esters to aldehydes specifically under controlled conditions (low temperatures).
   • LiAlH₄ (Lithium Aluminium Hydride) is a strong reducing agent that typically reduces esters all the way to alcohols, which doesn't match the reaction type observed.
2. Step B: The conversion from an alcohol to an alkene involves a dehydration step.
   • NaOH (alc) (Alcoholic sodium hydroxide) is often used in elimination reactions to form alkenes through dehydration of alcohols.
   • NaOH (aq) is typically involved in substitution rather than elimination.
3. Step C: The conversion from an intermediate to an alkene involves reduction and removal of an oxygen group.
   • Zn/HCl is a known reducing system that can be used for converting ketones or aldehydes by reducing them, often used in the Clemmensen reduction which removes carbonyl groups.
   • NH₂–NH₂/KOH, ethylene glycol (Wolff-Kishner reduction) is another reduction method that converts ketones to alkanes, but typically under different conditions and solvents.

Conclusion: The correct reagents are:

• A = DIBAL-H
• B = NaOH (alc)
• C = Zn/HCl

Thus, the correct answer is A = DIBAL-H, B = NaOH_(alc), C = Zn/HCl.

Answer 2

In this reaction, the following steps occur:

Step A: Reduction of ester to aldehyde using DIBAL-H (Diisobutylaluminum hydride), a selective reducing agent that reduces esters to aldehydes without affecting other functional groups.

Step B: Aldol condensation, which involves the formation of an intramolecular aldol product. This reaction is catalyzed by NaOH (alc) (sodium hydroxide in alcoholic solution).

Step C: Reduction of the product to alcohol using Zn/HCl in the Clemmensen reduction method, which reduces aldehydes to alcohols.

Thus, the correct reagents are 4, which corresponds to:  
- \( A = \text{DIBAL-H} \)  
- \( B = \text{NaOH (alc)} \)  
- \( C = \text{Zn/HCl} \).

The Correct Answer is: A = DIBAL-H, B = NaOH_(alc), C = Zn/HCl