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Elimination requires a strong base to remove a $$\beta$$-hydrogen and form an alkene.
$$NaOEt/EtOH$$ and $$NaOH$$ (with heat or alcohol) act as strong bases, so they favor elimination.
$$NaI$$ provides $$I^-$$, which is a good nucleophile but a weak base. Hence, it undergoes substitution instead of elimination.
Therefore, $$NaI$$ is not suitable for the reaction.
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