Question

How might conjugate addition reactions of lithium diorganocopper reagents be used to synthesize the following compounds?

Short answer

(a)

(b)

(c)

(d)

Step-by-step solution

Step-by-Step SolutionStep 1:Selection of reactant to form the conjugate addition product

To choose the reactant to form the conjugate addition product steps which follow are as follows:

• Aldehyde and ketone carbon numbered “1” and counted two carbons away from the carbonyl carbon. The double bond in the alpha, beta-unsaturated starting material is made by connecting the carbon numbers “2’’ and “3” by a double bond.
• The group bonded to the “3” carbon formed the alkyl lithium reagent.

Step 2:Synthesis of heptan-2-one (a)

2-heptanone is synthesized by reacting with the following reagent:

• but-3-en-2-one comes from forming a double bond between 2^nd and 3^rd carbon.
• Alkyl lithium reagent $\text{Li}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{Cu}$ comes from the group joining the 3^rd carbon.

Thus but-3-en-2-one reacting with $\text{Li}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{Cu}$followed by acidic workup results in the synthesis of 2-heptanone.

Synthesis of 2-heptanone

Synthesis of 3,3-dimethylcyclohexanone (b)

3,3-dimethylcyclohexanoneis synthesized by reacting with the following reagent:

• 3-methylcyclohex-2-enone comes from forming a double bond between 2^nd and 3^rd carbon.
• Alkyl lithium reagent $\text{Li}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{Cu}$ comes from the group joining the 3^rd carbon.

Thus reacting 3-methylcyclohex-2-enone with $\text{Li}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{Cu}$followed by acidic workup results in the synthesis of 3,3-dimethylcyclohexanone.

Synthesis of 3,3-dimethylcyclohexanone

Synthesis of 4-tert-Butyl-3-ethylcyclohexanone (c)

4-tert-Butyl-3-ethylcyclohexanone is synthesized by reacting with the following reagent:

• 4-tert-Butyl-cyclohex- 2-enone comes from forming a double bond between 2^nd and 3^rd carbon.
• Alkyl lithium reagent $\text{Li}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{Cu}$comes from the group joining the 3^rd carbon.

Thus reacting with 4-tert-Butyl-cyclohex- 2-enone$\text{Li}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{Cu}$ followed by acidic workup synthesizes 4-tert-Butyl-3-ethylcyclohexanone.

Synthesis of 4-tert-Butyl-3-ethylcyclohexanone

Synthesis of 8a-vinyloctahydronaphthalen (d)

8a-vinyloctahydronaphthalenis synthesized by reacting with the following reagent:

• 4,4a,5,6,7,8-hexahydronaphthalen-2(3H)-one comes from forming a double bond between 2^nd and 3^rd carbon.
• Alkyl lithium reagent$\text{Li}{\left({\text{CH}}_2\text{=CH}\right)}_{\text{2}}\text{Cu}$ comes from the group joining the 3^rd carbon.

Thus reacting,4a,5,6,7,8-hexahydronaphthalen-2(3H)-one with$\text{Li}{\left({\text{CH}}_2\text{=CH}\right)}_{\text{2}}\text{Cu}$ followed by acidic workup synthesizes 8a-vinyloctahydronaphthalen.

Synthes of 8a-vinyloctahydronaphthalen