
Tischenko reaction yields ester in the presence of catalyst used which is:
(a) \[LiAl{H_4}\]
(b) N-Bromosuccinimide
(c) \[Al{(O{C_2}{H_5})_3}\]
(d) \[Zn - Hg/HCl\]
Answer
164.1k+ views
Hint: The Tischenko reaction involves the dimerization of an aldehyde and produces ester in the presence of a metal catalyst. It is the modified form of the Cannizzaro reaction.
Complete step-by-step answer:
The Tischenko reaction is named after Russian chemist Vyacheslav Tischenko, who discovered that aluminium alkoxides are effective catalysts in reaction.
The Tischenko reaction is an organic chemical reaction that involves the disproportionation of an aldehyde. with no alpha hydrogen atom (non enolizable aldehydes) in the presence of metal alkoxide such as \[Al{(O{C_2}{H_5})_3}\].
In the Tishchenko reaction, two moles of aldehydes undergo disproportionation to produce an ester.
\[2C{H_3}CHO + Al{(O{C_2}{H_5})_3} \to C{H_3}COO{C_2}{H_5} + {H_2}O\]
It can be carried out in the aldehyde reactant as the solvent and driven practically until completion with facile further purification of the product by extraction.
The \[LiAl{H_4}\]i.e. Lithium Aluminium Hydride or LAH is a very strong and unselective reducing agent.
It can reduce:
Aldehydes, ketones, esters, carboxylic acid chlorides and carboxylic acid to alcohols.
It can reduce the amides (\[RCON{H_2}\]), and nitriles (\[RCN\])to amines (\[RC{H_2}N{H_2}\]).
N-Bromosuccinimide or NBS is a reagent, and it can be used for the bromination of various organic compounds. The bromination by NBS follows a radical mechanism.
\[Zn - Hg/HCl\] used in Clemmensen reduction. It can reduce the\[ - CO - \]group to \[ - C{H_2}\]group.
The above points indicate that \[Al{(O{C_2}{H_5})_3}\]is the reagent for the conversion of aldehyde to ester. Therefore, option (c) will be the correct answer.
Note:\[Al{(O{C_2}{H_5})_3}\]contains bonding through O and thus it does not have metal-carbon bond.
Complete step-by-step answer:
The Tischenko reaction is named after Russian chemist Vyacheslav Tischenko, who discovered that aluminium alkoxides are effective catalysts in reaction.
The Tischenko reaction is an organic chemical reaction that involves the disproportionation of an aldehyde. with no alpha hydrogen atom (non enolizable aldehydes) in the presence of metal alkoxide such as \[Al{(O{C_2}{H_5})_3}\].
In the Tishchenko reaction, two moles of aldehydes undergo disproportionation to produce an ester.
\[2C{H_3}CHO + Al{(O{C_2}{H_5})_3} \to C{H_3}COO{C_2}{H_5} + {H_2}O\]
It can be carried out in the aldehyde reactant as the solvent and driven practically until completion with facile further purification of the product by extraction.
The \[LiAl{H_4}\]i.e. Lithium Aluminium Hydride or LAH is a very strong and unselective reducing agent.
It can reduce:
Aldehydes, ketones, esters, carboxylic acid chlorides and carboxylic acid to alcohols.
It can reduce the amides (\[RCON{H_2}\]), and nitriles (\[RCN\])to amines (\[RC{H_2}N{H_2}\]).
N-Bromosuccinimide or NBS is a reagent, and it can be used for the bromination of various organic compounds. The bromination by NBS follows a radical mechanism.
\[Zn - Hg/HCl\] used in Clemmensen reduction. It can reduce the\[ - CO - \]group to \[ - C{H_2}\]group.
The above points indicate that \[Al{(O{C_2}{H_5})_3}\]is the reagent for the conversion of aldehyde to ester. Therefore, option (c) will be the correct answer.
Note:\[Al{(O{C_2}{H_5})_3}\]contains bonding through O and thus it does not have metal-carbon bond.
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