The ratio of elimination to substitution is exactly the same (26% elimination) for 2-bromo-2-methylbutane and 2-iodo-2-methylbutane in 80% ethanol/20% water at 25°C. By what mechanism does elimination most likely occur in these compounds under these conditions?

Question

Chemistry

Gilmore

21 November, 20:39

The ratio of elimination to substitution is exactly the same (26% elimination) for 2-bromo-2-methylbutane and 2-iodo-2-methylbutane in 80% ethanol/20% water at 25°C. By what mechanism does elimination most likely occur in these compounds under these conditions?

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Answers (1)

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Alessandra · Answer 1

The reaction proceeds through the SN-1 mechanism.

Explanation:

Based on the question given, the ratio of elimination to substitution is the same (26% elimination) for 2-bromo-2-methylbutane and 2-iodo-2-methylbutane in 80% ethanol/20% water at 25°C.

The formation of the same ratio of products from different starting materials is only possible when both reactions are proceeding with thesame intermediates.

In SN-1 and E-1 mechanism, reaction proceeds via a carbocation mechanism, whereas in the case of SN-2 mechanism, the nucleophile attacks the back side of the leaving group.

In the case of E-2 mechanism, both the beta and halide leaves at the same time to form an alkene.

Hence, the substitution reaction given in the question proceeds through the carbocation intermediate. This implies that the reaction proceeds through the SN-1 mechanism.