Practice Problem 8

Methyllithium (CH₃Li) can be used to form bonds between carbon and either main-group metals or transition metals:

HgCl₂(s) + 2 CH₃Li(l) Hg(CH₃)₂(l) + 2 LiCl(s)

WCl₆(s) +  6 CH₃Li(l) W(CH₃)₆(l) + 6 LiCl(s)

It can be used also to form bonds between carbon and other nonmetals:

PCl₃(s) +  3 CH₃Li(l) P(CH₃)₃(l) + 3 LiCl(s)

or between carbon atoms:

CH₃Li(l) + H₂CO(g) [CH₃CH₂OLi] CH₃CH₂OH(l)

Use Lewis structures to explain the stoichiometry of the following oxidation-reaction, which is used to synthesize methyllithium:

CH₃Br(l) + 2 Li(s) CH₃Li(l) + LiBr(s)

Answer

The active metals are reducing agents in all of their chemical reactions.  In this reaction, lithium metal reduces a carbon atom from an oxidation state of -2 to -4, as shown below:

Reduction: CH₃Br + 2 e^-  CH₃Li

The other half of the reaction therefore must involve the net loss of a total of two electrons:

Oxidation: 2 Li 2 Li⁺ + 2 e^-

Two moles of lithium are therefore consumed for each mole of CH₃Br in this reaction.  One of the Li⁺ ions formed in this reaction combines with the negatively charged carbon atom to form methyllithium and the other participates from solution with the Br ^- ion as LiBr:

CH₃Br(l) + 2 Li(s) CH₃Li(l) + LiBr(s)