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

Answer:

$(CH=CH_2)_2Cu Li$

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 $CH_2 = CHLi$