Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). 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
I hope it helps!
