Select the single best answer. determine the following type of reaction: ch3―ch2―ch(br)―ch3 ch3―ch═ch―ch3 + nabr + h2o

Deshydrohalogenation

Explanation:

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Now, in the reaction we can see that we have an atom of Bromine in carbon 2, and in the product appears as NaBr. This means that the Br was substracted by elimination. It's an elimination because the final product do not have a substituent where the bromine was, (Like another nucleophyle such OH or another halide). If you look closely the final product, we can see that one hydrogen in carbon 3, is no longer there. So this electrophyle was also substracted, in this case, by a base (Such NaOH), so in this case, it's ocurring an elimination reaction via E2 (One step, bimolecular). So, as the final product has been substracted the nucleophyle and electrophyle, this treaction is a deshydrohalogenation (an atom of hydrogen and a halide were substracted). The mechanism of this, you can see it in the picture.

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