haloalkane2

The answer is INCORRECTION. Steric hindrance is NOT the answer of the question.

In the two reactions, the sodium alkoxides ((CH3)2CHONa and CH3ONa) can act as both nucleophiles to cause nucleophilic substitution (SN reaction) and bases to cause elimination (E reaction).

In (1), the reactant is CH3Br, which has only one C atom in each molecule, can only undergo SN­ reaction but NOT E reaction. There is no elimination product, and only substitution product is formed.
SN: (CH3)2CHONa + CH3Br → (CH3)2CH-O-CH3 + NaBr

In (2), the reactant is (CH3)2CHBr, which is a secondary bromoalkane, can undergo both SN reaction and E reaction. There is a significant amount of elimination product, which may even be the major product.
SN: (CH3)2CHBr + CH3ONa → (CH3)2CH-O-CH3 + NaBr
E: (CH3)2CHBr + CH3ONa → CH3-CH=CH3 + CH3OH + NaBr

Due to the presence of elimination reaction in (2), (1) is a better choice for obtaining the substitution product, (CH3)2CH-O-CH3.
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