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Compound A and compound B are constitutional isomers with molecular formula C4H9Cl. Treatment of compound A with sodium methoxide gives trans-2-butene as the major product, while treatment of compound B with sodium methoxide gives a different disubstituted alkene as the major product.Draw the structure of compound A.

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cristisp93

Compound A will be 2-chloro-butane which in the presence of sodium methoxide will undergo the elimination of the halogen with a hydrogen from the neighboring atom carbon (not the one with which the halogen in bound),  which have less hydrogen atoms, forming the trans-2-butene.

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sebassandin

Answer:

2-chlorobutane.

Explanation:

Hello,

In this case, the treatment of the substance A with sodium methoxide to yield trans-2-butene stands for a dehydrohalogenation in the presence of a strong base (the sodium methoxide), such chemical reaction is undergone by alkyl halides, in this case with chlorine, based on its molecular formula. In such a way, on the attached picture, you will find both the structure of the 2-chlorobutane and its chemical reaction to yield the trans-2-butene throughout the usage of sodium methoxide.

Best regards.

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