Question

Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide what lithium diorganocopper (Gilman) reagent is needed to convert 1-bromopropane into 1-hexene.

Answer 1

`(CH=CH_2)_2Cu Li`

Step-by-step explanation:

Gilman reagent is also known as the Lithium organocuprates. They are used as nucleophiles in the organic synthesis.

Gilman reagent is used for C - C bond formation.

H H H

| | |

`$$CH_3CH_2CH_2Br` -----→ H - C -- C -- C -- C =
`$CH_2$`

(1 bromo propane) | | | |

H H H H

(1 pentene)

Normally in C--C formation by Gilman reagent, the following reactions takes place :

R -- Br +
`$(Br')_2Cu Li \rightarrow $` R--R' + R' Cu + Li Br

In 1 pentene, i.e.

`$CH_3CH_2CH_2 $`
`$CH=CH_2$`

(from 1 bromo (from Gilman reagent)

propane)

So Gilman reagent is
`(CH=CH_2)_2Cu Li`

Organo lithium compound is
`$CH_2 = CHLi$`