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Byproducts of LiAlH4 reduction of amides
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Lillian Hafner
Byproducts of LiAlH4 reduction of amides
Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required: e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:
In the workup conditions, $\ce{H3O+}$ is followed by $\ce{^–OH}$. An aqueous acidic solution is often used to carry out the protonolysis step that follows the $\ce{LiAlH4}$ reduction (as shown in the following mechanism). The excess of acid that is typically used converts the amine, which is a base, into its conjugate-acid ammonium ion. Hydroxide is then required to neutralize this ammonium salt and thus give the neutral amine.
Although water itself rather than acid can be used in the protonolysis step, for practical reasons the acidic workup is more convenient. Thus, the extra neutralization step is required.
The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:
When less than stoichiometric quantities of $\ce{LiAlH4}$ are used for the reduction of primary amides, the corresponding nitrile may be formed:
Moreover, high selectivity toward the nitrile is also observed with amides containing electron-rich quinoline systems. If traces of water are present,
partial reduction of amides to aldehydes over $\ce{LiAlH4}$ can also comprise an undesirable competing reaction. Hydrolysis to the alcohol product has also been reported for secondary amide substrates.
GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References
Loudon, G. M.; Parise, J. Organic chemistry, 6th Ed.; Roberts and Company Publishers: Greenwood Village, Colorado, 2016. ISBN 978-1-936221-34-9.
Nystrom, R. F.; Brown, W. G. Journal of the American Chemical Society1948, 70 (11), 3738–3740. DOI 10.1021/ja01191a057.
Smith, A. M.; Whyman, R. Chemical Reviews2014, 114 (10), 5477–5510. DOI 10.1021/cr400609m.
Thiedemann, B.; Schmitz, C. M. L.; Staubitz, A. The Journal of Organic Chemistry2014, 79 (21), 10284–10295. DOI 10.1021/jo501907v.
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Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required: e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:
In the workup conditions, $\ce{H3O+}$ is followed by $\ce{^–OH}$. An aqueous acidic solution is often used to carry out the protonolysis step that follows the $\ce{LiAlH4}$ reduction (as shown in the following mechanism). The excess of acid that is typically used converts the amine, which is a base, into its conjugate-acid ammonium ion. Hydroxide is then required to neutralize this ammonium salt and thus give the neutral amine.
Although water itself rather than acid can be used in the protonolysis step, for practical reasons the acidic workup is more convenient. Thus, the extra neutralization step is required.
The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:
When less than stoichiometric quantities of $\ce{LiAlH4}$ are used for the reduction of primary amides, the corresponding nitrile may be formed:
Moreover, high selectivity toward the nitrile is also observed with amides containing electron-rich quinoline systems. If traces of water are present,partial reduction of amides to aldehydes over $\ce{LiAlH4}$ can also comprise an undesirable competing reaction. Hydrolysis to the alcohol product has also been reported for secondary amide substrates.
GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References
Loudon, G. M.; Parise, J. Organic chemistry, 6th Ed.; Roberts and Company Publishers: Greenwood Village, Colorado, 2016. ISBN 978-1-936221-34-9.
Nystrom, R. F.; Brown, W. G. Journal of the American Chemical Society1948, 70 (11), 3738–3740. DOI 10.1021/ja01191a057.
Smith, A. M.; Whyman, R. Chemical Reviews2014, 114 (10), 5477–5510. DOI 10.1021/cr400609m.
Thiedemann, B.; Schmitz, C. M. L.; Staubitz, A. The Journal of Organic Chemistry2014, 79 (21), 10284–10295. DOI 10.1021/jo501907v.
Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required: e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:
The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
$$\ce{2RCONR'R'' + LiAlH4 -> 2RCH2NR'R'' + LiAlO2}$$
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:
GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References
Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required: e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:
The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
$$\ce{2RCONR'R'' + LiAlH4 -> 2RCH2NR'R'' + LiAlO2}$$
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:
GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References
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