Organic Chemistry 1 · Alkyne transformations
Prop-1-ynylcyclohexane from ethynylcyclohexane and CH3I
Conditions: NaNH2, NH3(l); then CH3I
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The mechanism, step by step
Next plays each step's curved arrows and explains them. The written steps are below.
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Amide removes the terminal alkyne proton
The only acidic hydrogen of ethynylcyclohexane is the terminal alkyne C–H (pKa about 25); the ring C–H bonds (pKa about 50) are untouched. Amide ion removes it to give cyclohexylacetylide and ammonia.
The curved arrows
- The amide lone pair takes the terminal alkyne hydrogen.
- The sp C–H pair stays on carbon as the cyclohexylacetylide lone pair.
Cyclohexylacetylide attacks iodomethane
The acetylide carbon attacks the methyl carbon of iodomethane from the side opposite iodine, and iodide leaves in the same concerted step. The bulky cyclohexyl group is on the nucleophile, far from the reacting carbon, so it does not slow the substitution.
The curved arrows
- The acetylide lone pair forms the new bond to the methyl carbon by backside attack.
- The C–I pair leaves on iodine as iodide, an excellent leaving group.
Prop-1-ynylcyclohexane
Prop-1-ynylcyclohexane is an internal alkyne with cyclohexyl on one end and methyl on the other. Iodomethane has no β-hydrogens, so the basic acetylide has no elimination pathway and methylation is clean.
What you'll practice
Deprotonate ethynylcyclohexane with amide and draw the SN2 methylation by iodomethane that gives prop-1-ynylcyclohexane.
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