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Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very <u>strong base</u> (<em>sodium ethoxide</em>). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an <u>E2 mechanism</u>, therefore, the hydrogen that is removed must have an <u>angle of 180º</u> with respect to the leaving group (the "OH"). This is known as the <u>anti-periplanar configuration</u>.
The hydrogen that has this configuration is the one that placed with the <u>dashed bond</u> (<em>red hydrogen</em>). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
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