For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e., the carbon atoms depicted with stereobonds). Show the product stereochemistry clearly. If there is more than one organic product, both products may be drawn in the same box. Ignore elimination involving C3 and any carbon atom other than C4.

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Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an anti-elimination. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a specific stereoisomer, in other words, it is not possible to assign a Z or E configuration for this alkene.

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