Draw the alkyne formed when 2,3-dichloropentane is treated with an excess of strong base such as sodium amide.

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When an alkyne such as 2,3-dichloropentane is treated with excess sodium amide, a two-step elimination reaction will happen. The reaction would then yield the final product in the form of 2-pentyne. Its structure is shown in the attached image.

We obtain the alkyne 2-pentyne through E2 reaction mechanism.

Explanation:

Alkyl halides are compounds that contain a carbon chain in which one or more hydrogen atoms have been replaced by a halogen atom (F, Cl, Br, I) and are characterized by having halogen atoms in adjacent carbons. In the dehydrohalogenation the alkyl halide is reacted with a strong base. As a result an alkyne is obtained.

In this case we use sodium amide to favor the type E2 reaction mechanism (type 2 elimination) and we obtain the following alkyne:

It is 4.92 e-4 sorry if i am wrong

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