I know that benzaldehyde decomposes into benzoic acid and benzyl alcohol via Cannizzaro Reaction..
But, can benzyl alcohol go through a Cannizzaro Reaction to form benzaldehyde and benzoic acid?
Yes, benzyl alcohol can undergo Cannizzaro reaction to form benzaldehyde and benzoic acid. The Cannizzaro reaction is a disproportionation reaction in which an aldehyde or ketone simultaneously oxidizes and reduces itself.
To understand how benzyl alcohol can undergo Cannizzaro reaction, let's look at the structure of benzyl alcohol:
Benzyl alcohol: C6H5CH2OH
In the presence of a strong base, such as sodium hydroxide (NaOH), one molecule of benzyl alcohol can act as both a reducing agent and an oxidizing agent. The hydroxyl group (OH) of benzyl alcohol is a good nucleophile, capable of attacking the carbonyl group of another molecule of benzyl alcohol.
Here's how the Cannizzaro reaction occurs:
1. A strong base, such as NaOH, abstracts a proton from the hydroxyl group (OH) of one molecule of benzyl alcohol, resulting in the formation of the alkoxide ion (C6H5CH2O-).
2. Another molecule of benzyl alcohol is attacked by the alkoxide ion, resulting in the transfer of a hydride ion (H-) from the carbon atom to the oxygen atom. This process oxidizes the alcohol group (OH) to an aldehyde group (CHO) and simultaneously reduces the carbonyl group (C=O) to a hydroxyl group (-OH).
Overall reaction:
2C6H5CH2OH → C6H5CHO + C6H5CH2O- + H2O
C6H5CH2O- + H2O → C6H5COOH + C6H5CH2OH
As a result, one molecule of benzyl alcohol is oxidized to form benzaldehyde (C6H5CHO), while another molecule is reduced to form benzoic acid (C6H5COOH). In this way, benzyl alcohol undergoes the Cannizzaro reaction to simultaneously form benzaldehyde and benzoic acid.
It is important to note that the Cannizzaro reaction is more favorable for aldehydes with no alpha hydrogens (such as benzaldehyde) than for aldehydes with alpha hydrogens (such as formaldehyde). So, while benzyl alcohol can undergo the Cannizzaro reaction, benzaldehyde is the preferred substrate for this reaction.