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Chapter 18: Problem 423

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

Expert verified
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

01

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. \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCHO} + \mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}^{-}\mathrm{O} + \mathrm{Na}^+ + \mathrm{H_2O} \]
02

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: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}^{-}\mathrm{O} + \mathrm{H_2O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCO} + \mathrm{OH}^{-} \]
03

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: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCO} + \mathrm{OH}^{-} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHOHCOO}^{-} + \mathrm{H_2O} \]
04

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.

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