
An alkane ${C_7}{H_{16}}$ is produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide. The name of the product is:
A. 3-methylhexane
B. 2-ethylpentane
C. 3-ethylpentane
D. n-heptane
Answer
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Hint: The lithium di(3-pentyl)cuprate is commonly known as Gilman reagent which is used for coupling reaction. On reacting with ethyl bromide carbon-carbon bond is formed by replacing the halogen of ethyl bromide with alkyl group. In this reaction nucleophilic substitution reaction takes place where the Gilman reagent acts as a nucleophile.
Complete step by step answer:
Lithium di(3-pentyl)cuprate is known as Gilman reagent. The Lithium di(3-pentyl)cuprate is formed by reacting organolithium reagent with cuprous iodide. Gilman reagents act as a carbanion. The lithium di(3-pentyl)cuprate (Gilman reagent) gives coupling reaction with alkyl halide. In the reaction of lithium di(3-pentyl)cuprate (Gilman reagent) with alkyl halide, results in the replacement of halogen atom by a alkyl group to form carbon-carbon bond of hydrocarbon alkane, lithium halide and alkyl copper.
In this question, lithium di(3-pentyl)cuprate reacts with ethyl bromide to form an alkane as a main product because of carbon-carbon coupling.
The reaction of lithium di(3-pentyl)cuprate reacts with ethyl bromide is shown below.
${[{(C{H_3}C{H_2})_2}CH]_2}CuLi + BrC{H_2}C{H_3} \to {C_2}{H_5} - CH({C_2}{H_5}) - {C_2}{H_5} + {(C{H_3}C{H_2})_2}CHCu + LiBr$
In this reaction, lithium di(3-pentyl)cuprate reacts with ethyl bromide to form 3-ethyl pentane as a main product, pentan-3-yl copper and lithium bromide.
The Gilman reagent acts as a nucleophile for nucleophilic substitution reaction ${S_N}2$ reaction.
Thus, the name of alkane ${C_7}{H_{16}}$ produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide is 3-ethyl pentane.
Therefore, the correct option is C.
Note:
The Gilman reagent lithium di(3-pentyl)cuprate gives contrast property with the Grignard reagent. The alkyl halide on reacting with Grignard reagent gives very less quantity of the desired product as compared to Gilman reagent.
Complete step by step answer:
Lithium di(3-pentyl)cuprate is known as Gilman reagent. The Lithium di(3-pentyl)cuprate is formed by reacting organolithium reagent with cuprous iodide. Gilman reagents act as a carbanion. The lithium di(3-pentyl)cuprate (Gilman reagent) gives coupling reaction with alkyl halide. In the reaction of lithium di(3-pentyl)cuprate (Gilman reagent) with alkyl halide, results in the replacement of halogen atom by a alkyl group to form carbon-carbon bond of hydrocarbon alkane, lithium halide and alkyl copper.
In this question, lithium di(3-pentyl)cuprate reacts with ethyl bromide to form an alkane as a main product because of carbon-carbon coupling.
The reaction of lithium di(3-pentyl)cuprate reacts with ethyl bromide is shown below.
${[{(C{H_3}C{H_2})_2}CH]_2}CuLi + BrC{H_2}C{H_3} \to {C_2}{H_5} - CH({C_2}{H_5}) - {C_2}{H_5} + {(C{H_3}C{H_2})_2}CHCu + LiBr$
In this reaction, lithium di(3-pentyl)cuprate reacts with ethyl bromide to form 3-ethyl pentane as a main product, pentan-3-yl copper and lithium bromide.
The Gilman reagent acts as a nucleophile for nucleophilic substitution reaction ${S_N}2$ reaction.
Thus, the name of alkane ${C_7}{H_{16}}$ produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide is 3-ethyl pentane.
Therefore, the correct option is C.
Note:
The Gilman reagent lithium di(3-pentyl)cuprate gives contrast property with the Grignard reagent. The alkyl halide on reacting with Grignard reagent gives very less quantity of the desired product as compared to Gilman reagent.
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