Question

The major product formed in the given reaction is: \[ \text{TiCl}_3(\text{DME})_2 \quad \text{Zn-Cu} \quad \text{DME, r.t.} \quad \longrightarrow \quad 1,2-\text{Dimethoxyethane}. \]

• A. \(\text{C}_6\text{H}_{12}\text{O}_2\) (1,2-Butanediol)
• B. \(\text{C}_5\text{H}_8\text{O}\) (Tetrahydrofuran)
• C. \(\text{C}_6\text{H}_{12}\text{O}_2\) (1,2-Butanediol)
• D. \(\text{C}_4\text{H}_8\text{O}_2\) (1,2-Dimethoxyethane)

Hint:

TiCl\(_3\) and Zn-Cu are commonly used to couple aldehydes or ketones to form diols.
The reaction type here involves coupling aldehydes to form a 1,2-diol product.

Answer 1

The Correct Option is A

In this reaction, TiCl\(_3\)(DME)\(_2\) with Zn-Cu is used to reduce the aldehyde groups. This type of reaction typically results in the formation of 1,2-butanediol via a coupling reaction between two molecules of an aldehyde. The reaction conditions involve the use of DME as a solvent and mild room temperature conditions. Thus, the major product formed in this case is 1,2-butanediol, corresponding to option (A).