\(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgI} \longrightarrow(\mathrm{A})+\mathrm{CH}_{4}\) \((\mathrm{A})+\mathrm{CH}_{3} \mathrm{I} \longrightarrow(\mathrm{B})+\mathrm{MgI}_{2}\) (A) and (B) are (a) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{4}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) and \(\mathrm{CH}_{4}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\) and \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{I}\) and

Short Answer

Expert verified
Answer: Product (A) is H-C≡C-MgI, and product (B) is CH3-C≡C-CH3.

Step by step solution

01

Reaction of alkyne with Grignard reagent

Grignard reagents react with alkynes to form alkyl magnesium halide. In our given reaction, HC≡CH (ethyne) is reacting with CH3MgI (methylmagnesium iodide) to form product (A) and CH4 (methane). To determine the structure of product (A), we must know where the bond will be formed. Grignard reagents are strong nucleophiles, and they usually attack the least substituted carbon atom in the triple bond.
02

Determine structure of product (A)

The least substituted carbon in HC≡CH is the terminal carbon bonded to hydrogen. Thus, the CH3 group of CH3MgI will attack the terminal carbon of HC≡CH, and the magnesium will attack the other carbon of the triple bond. This leads to the formation of H-C≡C-MgI as product (A).
03

Reaction of product (A) with alkyl halide

Product (A) is H-C≡C-MgI, and now it reacts with CH3I (methyl iodide) to form product (B) and MgI2 (magnesium iodide). As the Grignard reagent acts as a strong nucleophile, it will attack the carbon atom of CH3I, breaking the carbon-iodine bond.
04

Determine structure of product (B)

The attack of C≡C-MgI on CH3I leads to the formation of a new C-C bond, resulting in the structure CH3-C≡C-CH3 as product (B). Therefore, the correct answer is option (c) where product (A) is \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{MgI}\), and product (B) is \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\).

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