[hemchaystud101]

Here is the reaction:

1-butyne was treated with 1 mole sodium amide and the product of this reaction was then added to a solution (3R)-1-iodo-3-methylhexane in tetrahydrofuran. the product of that reaction was then treated with hydrogen gas over Pd on calcium carbonate that had been pretreated with quinoline and lead acetate.

My final answer is 3-methylnonane. (with the methyl as a solid line and the hydrogren dashed) ... is this correct? if not can you explain how to go through this?

[discodermolide]

Here is the reaction:

1-butyne was treated with 1 mole sodium amide and the product of this reaction was then added to a solution (3R)-1-iodo-3-methylhexane in tetrahydrofuran. the product of that reaction was then treated with hydrogen gas over Pd on calcium carbonate that had been pretreated with quinoline and lead acetate.

My final answer is 3-methylnonane. (with the methyl as a solid line and the hydrogren dashed) ... is this correct? if not can you explain how to go through this?

OK, you first form the anion of the alkyne which adds to the iodo-methylhexane; then the hydrogenation partially reduces the triple bond to the double bond. Perhaps you can work out the correct structure from this info?

[hemchaystud101]

That makes more sense. So then I would get 7-methyl-non-4-ene?