Reaction of alkyl halides with strong unhindered bases

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The discussion centers on the reaction of primary alkyl halides with the unhindered base EtONa, questioning why the solution manual states that the only product is an alkene. Participants express confusion over the omission of the SN2 reaction, which they believe should occur in significant amounts. The conversation highlights the role of steric hindrance, suggesting that it can affect reactions beyond just the beta carbon. Comparisons are made to neopentylic structures, which cannot undergo SN2 due to steric factors. A clear explanation of the reaction mechanisms involved is requested for better understanding.
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Please look at the image. For (b) and (c), what the solution manual says makes no sense to me.
You're reacting a primary alkyl halide with an unhindered base EtONa. They say that the only product is the alkene. Why do they not consider the SN2 reaction that WILL take place in larger quantity?
I'm uploading the question too.
 

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There is no other SN2 reaction. There aren't any other leaving groups.
 
I am actually not too sure, but I think it is due to the steric hinderance. The hinderance does not always have to be on the beta carbon, if you compare the structure to a neopentylic structure, which can never participate in an Sn2, there are similarities, except that there are only two metyl substituents, instead of three.

In all honesty, if I would not have known the answer i would have gone for Sn2 as well so I would also love to see a good explanation.
 

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