Question

The best method for the preparation of 2,2-dimethylbutane is via the reaction of:
(A) $M{{e}_{3}}CBr$ and $MeC{{H}_{2}}Br$ in $Na/ether$
(B) ${{(M{{e}_{3}}C)}_{2}}CuLi$ and $MeC{{H}_{2}}Br$
(C) ${{(MeC{{H}_{2}})}_{2}}CuLi$ and $M{{e}_{3}}CBr$
(D) $M{{e}_{3}}CMgI$ and $MeC{{H}_{2}}I$

Answer 1

Hint: As we know that 2,2-dimethylbutane (also known as neohexane) is indeed the most compact and branched of the hexane isomers and the only one with a quaternary carbon and a butane as backbone or parent chain. It is best prepared by Corey–House synthesis. So here will explain this reaction and find the reactants used for the preparation of 2,2-dimethylbutane.

Complete answer:
Let us discuss about Corey–House synthesis followed by the preparation method of 2,2-dimethylbutane as follows:-
-Corey–House synthesis: It is the coupling reaction or can say good synthetic way where we can join two alkyl groups together to produce higher alkanes. It is the reaction between the organocopper reagent (which is generally a lithium dialkylcuprate) and an alkyl halide or an aryl halide.
-It follows the ${{S}_{N}}2$ reaction mechanism that involves the nucleophilic substitution reaction of the leaving group (which usually consists of halide substituents or other electron withdrawing substituents) with a nucleophile in a given compound.
-For the preparation of 2,2-dimethylbutane ($M{{e}_{3}}C-C{{H}_{2}}-C{{H}_{3}}$), we would require an organocopper reagent which has 3 methyl group along with the alkyl halide that have ethyl as its alkyl group. The reaction takes places as follows:-
${{(M{{e}_{3}}C)}_{2}}CuLi+MeC{{H}_{2}}Br\xrightarrow{{{S}_{N}}2}M{{e}_{3}}C-C{{H}_{2}}-C{{H}_{3}}+BrM{{e}_{3}}C-CuLi$

So from the above data we can conclude that the best method for the preparation of 2,2-dimethylbutane is via the reaction of: (B) ${{(M{{e}_{3}}C)}_{2}}CuLi$ and $MeC{{H}_{2}}Br$.

Note:
-Remember that this organocopper reagent is also known as Gilman’s reagent which can be ${{1}^{\circ }}$, ${{2}^{\circ }}$ or ${{3}^{\circ }}$ but alkyl halide is limited to be only of ${{1}^{\circ }}$.
-Also this method of preparation is always better than Wurtz reaction.