Final answer:
The most plausible mechanism involves the reduction of a carbonyl group by NaBH4 where the hydride ion performs a nucleophilic attack on the carbonyl carbon, leading to the formation of an alcohol, assuming that the MeO(CH2)2OMe includes an implicit carbonyl group.
Step-by-step explanation:
The student's question concerns the mechanism that accounts for the formation of products when MeO(CH2)2OMe is reduced by NaBH4 in methanol. It appears that the question suggests the compound MeO(CH2)2OMe to have a carbonyl group, even though it is not explicitly stated in the formula.
Assuming that this is the case and considering the context, option B) seems to be the most likely scenario where the carbonyl group is reduced by the hydride ion (H-), which is provided by NaBH4. This hydride performs a nucleophilic attack on the carbon of the carbonyl group, thereby reducing it to an alcohol - the immediate product of the addition of hydride to a carbonyl.
This mechanism is consistent with a general nucleophilic addition reaction where a hydride donor like NaBH4 adds hydride to the electrophilic carbonyl carbon, forming a tetrahedral intermediate. The subsequent steps typically involve protonation processes to yield the final alcohol product.
NaBH4 is specifically known for its ability to reduce aldehydes and ketones to alcohols, but it is less reactive toward other carboxylic acid derivatives such as acid chlorides that LiAlH4 can reduce. Given that ether groups are generally not susceptible to reaction with NaBH4, option C) can be dismissed.