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Draw the alkyne formed when 3,4-dichloroheptane is treated with an excess of strong base such as sodium amide?

November 27, 2022 0

I have already tried the answer CH3-CH=CH-CH=CH-CH2-CH3

Why is the above answer wrong, and what is the correct answer?

You were right in saying that the strong base will cause an elimination reaction at the chlorinated carbons and the compound you listed may be one possible product of the reaction. However, the question asks for the alkyne formed. Alkynes have a triple bond so the product is CH3-CH2C-(triple bond)-C-CH2-CH2-CH3 called 3-heptyne. This is the major product of the reaction.

Source(s): Biochem major