Concept: Elimination in haloalkanes

Elimination of HX from an alkyl halide to form an alkene is usually favored by:

• a strong base,
• often in alcoholic medium,
• reagents such as alcoholic $KOH$, $NaOEt/EtOH$, etc.

A reagent is not suitable for elimination if it does not act as a base to remove a $\beta$-hydrogen.

Check each option

Option A: $NaOH/H_2O$

• Aqueous $NaOH$ mainly favors nucleophilic substitution.
• However, under suitable conditions it can still give some elimination, though it is not the best reagent.
• So it is not completely unsuitable.

Option B: $NaOEt/EtOH$

• Sodium ethoxide in ethanol is a strong base in alcoholic medium.
• This is a classic elimination reagent.
• Hence, it is suitable.

Option C: $NaOH/H_2O$-$EtOH$

• Hydroalcoholic medium with base can promote elimination better than pure aqueous base.
• So this is also used for elimination.
• Hence, it is suitable.

Option D: $NaI$

• $NaI$ provides $I^-$, which is a good nucleophile but a weak base.
• It is used in substitution reactions such as Finkelstein reaction, not for elimination.
• Therefore, it is not suitable for elimination.

Conclusion

The reagent that is not suitable for elimination reaction is: $\boxed{D\;:\; NaI}$

Comparison with stored correct answer

Stored correct answer = $D$

My derived answer also = $D$

So they agree.