Relevant Reaction:
Sodium methoxide reacts with bromoethane to form methoxyethane as the product.

Explanation:
Sodium methoxide (NaOCH3) is a strong base, commonly used in organic reactions for its ability to deprotonate acidic hydrogen atoms. On the other hand, bromoethane (C2H5Br) is an alkyl halide, which can undergo nucleophilic substitution reactions with strong bases.

In this reaction, sodium methoxide acts as a nucleophile, attacking the carbon atom of bromoethane. This results in the displacement of the bromine atom by the methoxy group (CH3O-). The overall reaction can be represented as follows:

CH3O^- + C2H5Br → CH3OC2H5 + Br^-

The product formed is methoxyethane (CH3OC2H5), also known as ethyl methyl ether. It is an ether compound consisting of an ethyl group (C2H5) attached to an oxygen atom, which in turn is attached to a methyl group (CH3). The bromide ion (Br^-) is the leaving group in this reaction.

This reaction is an example of a nucleophilic substitution reaction, specifically an SN2 (substitution nucleophilic bimolecular) reaction. In SN2 reactions, the nucleophile attacks the carbon atom at the same time as the leaving group departs. This leads to the inversion of stereochemistry at the carbon center, resulting in the formation of the product.

It is important to note that the reaction requires heating to proceed efficiently. The heat provides the necessary activation energy for the reaction to occur at a reasonable rate. Without heating, the reaction may be slow or might not occur at all.

In conclusion, the reaction between sodium methoxide and bromoethane leads to the formation of methoxyethane (ethyl methyl ether) as the product.