
Bouveault-Blanc reduction reaction involves
A) Reduction of a carbonyl compound with $Na/Hg$ and $HCl$
B) Reduction of an acyl halide with ${H_2}/Pd$
C) Reduction of an anhydride with $LiAl{H_4}$
D) Reduction of an ester with $Na/{C_2}{H_5}OH$
Answer
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Hint: Bouveault-Blanc reduction was used widely as before the availability of the reducing agent lithium aluminium hydroxide and other related reagents. This reaction particularly involves the use of alkali metal as a reagent.
Complete step by step answer:
1) Bouveault-Blanc reaction was used particularly for the reduction of esters before the introduction of the reducing agent lithium aluminium hydroxide, which is now widely used for the reduction of esters.
2) It reduces esters into alcohol in presence of alkali metal sodium and ethyl alcohol which gives the reaction product as terminal alcohol of that ester compound.
3) The general schematic representation of the Bouveault-Blanc reduction reaction will go as shown below:
$R - CO - O{R^1}\xrightarrow[{EtOH}]{{N{a^0}}}R - C - OH + {R^1}OH$
4) In the above reaction $N{a^0}$ acts as a single electron reducing agent and ethyl alcohol that is $EtOH$ acts as a proton donor.
Hence, by analyzing the above steps we have came a conclusion that Bouveault-Blanc reduction reaction involves the reduction of an ester with the reagent $Na/{C_2}{H_5}OH$ which shows option D as a correct answer.
Additional information: This reaction method is widely used as an inexpensive replacement for lithium aluminium hydroxide as reductions of esters in industrial chemical reactions. It was the only alternative method for reduction of esters which was in use before the development of the metal hydride reducing reagents.
Note:
In Bouveault-Blanc reduction reaction, while the reaction is proceeding if there is a proper source of the proton is not provided like ethyl alcohol is used in the above reaction then the dimerization of intermediate will take place instead of the alcohol product.
Complete step by step answer:
1) Bouveault-Blanc reaction was used particularly for the reduction of esters before the introduction of the reducing agent lithium aluminium hydroxide, which is now widely used for the reduction of esters.
2) It reduces esters into alcohol in presence of alkali metal sodium and ethyl alcohol which gives the reaction product as terminal alcohol of that ester compound.
3) The general schematic representation of the Bouveault-Blanc reduction reaction will go as shown below:
$R - CO - O{R^1}\xrightarrow[{EtOH}]{{N{a^0}}}R - C - OH + {R^1}OH$
4) In the above reaction $N{a^0}$ acts as a single electron reducing agent and ethyl alcohol that is $EtOH$ acts as a proton donor.
Hence, by analyzing the above steps we have came a conclusion that Bouveault-Blanc reduction reaction involves the reduction of an ester with the reagent $Na/{C_2}{H_5}OH$ which shows option D as a correct answer.
Additional information: This reaction method is widely used as an inexpensive replacement for lithium aluminium hydroxide as reductions of esters in industrial chemical reactions. It was the only alternative method for reduction of esters which was in use before the development of the metal hydride reducing reagents.
Note:
In Bouveault-Blanc reduction reaction, while the reaction is proceeding if there is a proper source of the proton is not provided like ethyl alcohol is used in the above reaction then the dimerization of intermediate will take place instead of the alcohol product.
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