Question

Phenylglyoxal, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCHO}\), is converted by aqueous sodium hydroxide into sodium mandelate, \(\mathrm{C}_{6} \mathrm{H}_{5}\) CHOHCOONa. Suggest a likely mechanism for this conversion.

Short answer

The conversion of phenylglyoxal to sodium mandelate involves the following steps: 1) Deprotonation of glyoxal moiety by sodium hydroxide, forming an enolate anion and a sodium cation; 2) Protonation of the enolate anion with water, forming an alcohol group; 3) Formation of a carboxylate group from the carbonyl group in the presence of hydroxide ion; 4) Interaction of the negatively charged carboxylate group with the sodium cation, forming the final product, sodium mandelate.

Step-by-step solution

Deprotonate the glyoxal moiety with sodium hydroxide

First, the glyoxal moiety of phenylglyoxal needs to be deprotonated. Sodium hydroxide, being a strong base, readily abstracts the acidic hydrogen of the glyoxal moiety, resulting in a carbonyl with an enolate anion and a sodium cation. \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCHO} + \mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}^{-}\mathrm{O} + \mathrm{Na}^+ + \mathrm{H_2O} \]

Protonate the enolate anion

The enolate anion generated in step 1 can further react with water molecules to regain a proton, resulting in the formation of an alcohol group: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}^{-}\mathrm{O} + \mathrm{H_2O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCO} + \mathrm{OH}^{-} \]

Formation of carboxylate group from carbonyl

Finally, the carbonyl group present in the molecule reacts with the hydroxide ion formed in step 2 to form a carboxylate group: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCO} + \mathrm{OH}^{-} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCOO}^{-} + \mathrm{H_2O} \]

Add sodium cation to form the sodium mandelate

The negatively charged carboxylate group formed in step 3 interacts with the sodium cation that was released in step 1, forming the final product, sodium mandelate: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCOO}^{-} + \mathrm{Na}^+ \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCOONa} \] The proposed mechanism involves the abstraction of a proton by sodium hydroxide, followed by the generation of an enolate anion, and finally, the formation of a carboxylate group to produce sodium mandelate.

Key concepts

Enolate Formation

Enolate formation is a fundamental step in many organic reactions. When phenylglyoxal interacts with sodium hydroxide, an enolate ion is formed.
The presence of a carbonyl group in phenylglyoxal makes the hydrogen atom next to it (alpha hydrogen) more acidic.
This happens because the carbonyl group can stabilize the negative charge that forms when the hydrogen is removed.

• The acidic hydrogen is abstracted by the hydroxide ion from sodium hydroxide.
• This results in the formation of an enolate ion, which can be represented as \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}^{-}\mathrm{O} \).

In the enolate ion, the negative charge is delocalized between the oxygen and the alpha carbon. This stabilization is crucial for the intermediate's reactivity in the subsequent steps.

Carboxylate Group Formation

The formation of a carboxylate group is the final step in converting phenylglyoxal to sodium mandelate. After the formation of an alcohol from the enolate ion, further reaction with sodium hydroxide produces the carboxylate group.
This step involves attacking the newly formed hydroxyl group (from the alcohol) with an hydroxide ion.

• In this process, the hydroxide ion deprotonates the alcohol, and a negatively charged carboxylate ion, \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCOO}^{-} \), is formed.
• This process ensures the conversion of the carbonyl group to a more stable carboxylate structure.

The negatively charged carboxylate anion can then easily pair with sodium ions to form the desired sodium mandelate product.

Sodium Hydroxide Reactivity

Sodium hydroxide, a common strong base, plays a crucial role in the mechanism of forming sodium mandelate. Its reactivity stems from the presence of hydroxide ions, which are highly nucleophilic.

• The hydroxide ion is capable of abstracting moderately acidic protons due to its strong base characteristics.
• Its nucleophilic nature allows it to attack electrophilic centers, facilitating bond formations and rearrangements.

In the mechanism at hand, sodium hydroxide not only helps in forming the enolate ion but also assists in converting the carbonyl group into a carboxylate group. This multifaceted reactivity of sodium hydroxide is central to the transformation of phenylglyoxal in this reaction.

Proton Abstraction

Proton abstraction is a significant step in the process of enolate formation. In organic chemistry, removing a proton often leads to the formation of reactive intermediates that drive the reaction forward.
In this mechanism, proton abstraction by sodium hydroxide occurs twice:

• The first abstraction removes an alpha hydrogen, allowing the formation of an enolate ion from phenylglyoxal.
• The second abstraction further modifies the resulting alcohol into a carboxylate group.

Each of these proton abstractions is aided by the strong base ability of the hydroxide ion. This ability to pull away protons is especially vital in reactions where shifts in electron density are needed to stabilize intermediates and enable bond formation.