1,4-addition with cuprates (Gilman): practice and mechanism
Cuprates are soft nucleophiles and attack α,β-unsaturated ketones at the β-carbon (1,4-addition). The C=O group is kept and the methyl group sits on the β-carbon.
Example
Cuprates are "soft" nucleophiles. With α,β-unsaturated carbonyl compounds they do not attack the carbonyl carbon (1,2) but the β carbon (1,4 or conjugate addition). After protonation the C=O group is kept and the C=C bond is gone.
Mechanism step by step
- 1
The cuprate (CH₃)₂CuLi is a soft nucleophile and coordinates to the C=C bond of the enone.
- 2
A methyl group is transferred to the β-carbon; a lithium enolate forms.
- 3
Acidic workup protonates the enolate at the α-carbon: the saturated ketone with the methyl group in the β-position forms.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That is 1,2-addition to the C=O group, typical of Grignard and organolithium reagents. Cuprates add 1,4 to the β-carbon.
- The double bond is not simply hydrogenated: the cuprate brings a methyl group that is added to the β-carbon.
- The methyl group goes to the β-carbon, not the α-carbon. The α-carbon is only protonated on workup.
Often confused with: Grignard reaction · Michael addition
Frequently asked questions
Why does Grignard attack 1,2 but the cuprate 1,4?
Grignard reagents are hard nucleophiles and go for the strongly positive carbonyl carbon. Cuprates are soft and prefer the softer, conjugated β-carbon.
Do cuprates also react with haloalkanes?
Yes, the Corey–House synthesis: R₂CuLi + R′–X → R–R′. It forms C–C bonds between two alkyl groups.