Question

$\oplus$ CH$_3$-CHO $\xrightarrow{CH_3OH/H^{\oplus} (excess)}$

Answer 1

CH$_3$-CH(OCH$_3$)$_2$ (1,1-dimethoxyethane)

Answer 2

Explanation of the solution:

This reaction is the acid-catalyzed formation of an acetal from an aldehyde and an alcohol.

1. Acetaldehyde (CH$_3$-CHO) is an aldehyde.
2. Methanol (CH$_3$OH) is an alcohol.
3. H$^{\oplus}$ (excess) indicates an acid catalyst and excess alcohol.

Aldehydes react with two equivalents of alcohol in the presence of an acid catalyst to form acetals. The reaction proceeds via a hemiacetal intermediate, which then reacts further with a second molecule of alcohol to form the stable acetal.

The carbonyl oxygen of acetaldehyde is protonated, making the carbonyl carbon more electrophilic. Methanol then attacks this carbon, forming a hemiacetal. The hydroxyl group of the hemiacetal is subsequently protonated and leaves as water, forming a resonance-stabilized carbocation (oxonium ion). This carbocation is then attacked by a second molecule of methanol, followed by deprotonation, to yield the acetal.

The overall reaction is:

$\text{CH}_3\text{-CHO} + 2\text{CH}_3\text{OH} \xrightarrow{\text{H}^{\oplus}} \text{CH}_3\text{-CH(OCH}_3)_2 + \text{H}_2\text{O}$

The product is 1,1-dimethoxyethane.