Tiffeneau–Demjanov rearrangement productsFavorskii rearrangementDetermination of products in reactions involving carbocation rearrangement?Regioselectivity of the Beckmann rearrangementHoffmann's rearrangementNumber of products on dehydrobrominationBenzilic acid rearrangementRing contraction in benzilic rearrangementRetention of configuration in Hofmann rearrangementMechanism of Fritsch–Buttenberg–Wiechell rearrangement8a-methyl-1,2,3,4,4a,8a-hexahydronaphthalen-4a-ylium carbocation rearrangement
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Tiffeneau–Demjanov rearrangement products
Favorskii rearrangementDetermination of products in reactions involving carbocation rearrangement?Regioselectivity of the Beckmann rearrangementHoffmann's rearrangementNumber of products on dehydrobrominationBenzilic acid rearrangementRing contraction in benzilic rearrangementRetention of configuration in Hofmann rearrangementMechanism of Fritsch–Buttenberg–Wiechell rearrangement8a-methyl-1,2,3,4,4a,8a-hexahydronaphthalen-4a-ylium carbocation rearrangement
$begingroup$
With reference to the paper J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 (ResearchGate link). It states that
cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.
However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?
organic-chemistry rearrangements
edited 32 secs ago
Melanie Shebel♦
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
$endgroup$
add a comment |
$begingroup$
With reference to the paper J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 (ResearchGate link). It states that
cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.
However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?
organic-chemistry rearrangements
edited 32 secs ago
Melanie Shebel♦
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
$endgroup$
$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago
add a comment |
$begingroup$
With reference to the paper J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 (ResearchGate link). It states that
cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.
However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?
organic-chemistry rearrangements
edited 32 secs ago
Melanie Shebel♦
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
$endgroup$
With reference to the paper J. Am. Chem. Soc. 1951, 73 (5), 2293–2295 (ResearchGate link). It states that
cis-2-Aminocyclohexanol on treatment with nitrous acid undergoes a pinacol rearrangement, yielding both cyclohexanone and cyclopentylmethanal as products.
However, it goes on to state that the trans epimer would specifically yield the latter as the major product. Can someone explain this (preferably via mechanism)?
organic-chemistry rearrangements
organic-chemistry rearrangements
edited 32 secs ago
Melanie Shebel♦
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
edited 32 secs ago
Melanie Shebel♦
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
edited 32 secs ago
Melanie Shebel♦
3,48173273
edited 32 secs ago
Melanie Shebel♦
3,48173273
edited 32 secs ago
Melanie Shebel♦
3,48173273
3,48173273
asked 14 hours ago
Sameer ThakurSameer Thakur
696
asked 14 hours ago
Sameer ThakurSameer Thakur
696
asked 14 hours ago
Sameer ThakurSameer Thakur
696
696
$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago
add a comment |
$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago
$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago
$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for migrating group to be anti to the leaving group (N2+), for orbital overlap reasons.
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the OH is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when OH is equatorial, the C-C bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the OH and N2 are axial there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the N2, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.
Apologies for hand-drawing - doing this on my phone...
answered 7 hours ago
PCKPCK
3987
$endgroup$
add a comment |
$begingroup$
I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as there 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.
1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293
answered 6 hours ago
user55119user55119
4,76211242
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for migrating group to be anti to the leaving group (N2+), for orbital overlap reasons.
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the OH is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when OH is equatorial, the C-C bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the OH and N2 are axial there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the N2, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.
Apologies for hand-drawing - doing this on my phone...
answered 7 hours ago
PCKPCK
3987
$endgroup$
add a comment |
$begingroup$
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for migrating group to be anti to the leaving group (N2+), for orbital overlap reasons.
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the OH is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when OH is equatorial, the C-C bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the OH and N2 are axial there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the N2, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.
Apologies for hand-drawing - doing this on my phone...
answered 7 hours ago
PCKPCK
3987
$endgroup$
add a comment |
$begingroup$
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for migrating group to be anti to the leaving group (N2+), for orbital overlap reasons.
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the OH is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when OH is equatorial, the C-C bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the OH and N2 are axial there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the N2, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.
Apologies for hand-drawing - doing this on my phone...
answered 7 hours ago
PCKPCK
3987
$endgroup$
Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for migrating group to be anti to the leaving group (N2+), for orbital overlap reasons.
You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the OH is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when OH is equatorial, the C-C bond is anti and the cyclopentane product is formed.
Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the OH and N2 are axial there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the N2, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.
Apologies for hand-drawing - doing this on my phone...
answered 7 hours ago
PCKPCK
3987
answered 7 hours ago
PCKPCK
3987
answered 7 hours ago
PCKPCK
3987
answered 7 hours ago
PCKPCK
3987
3987
add a comment |
add a comment |
$begingroup$
I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as there 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.
1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293
answered 6 hours ago
user55119user55119
4,76211242
$endgroup$
add a comment |
$begingroup$
I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as there 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.
1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293
answered 6 hours ago
user55119user55119
4,76211242
$endgroup$
add a comment |
$begingroup$
I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as there 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.
1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293
answered 6 hours ago
user55119user55119
4,76211242
$endgroup$
I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as there 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3.
As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. Clearly, the diaxial conformation of 1 is not a factor.
Assuming that the diazonium group has an A-value similar to that of cyanide (0.2 kcal/mol; OH, 1.0 kcal/mol), the equilibrium constant for the two cis conformations at ambient temperature is 3.8, which favors conformation 4b. If indeed the formation of aldehyde 3 predominates over cyclohexanone 5, then hydride migration must be a slower process than ring contraction. In the event, orbital overlap of the red bonds in 4 is essential.
1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293
answered 6 hours ago
user55119user55119
4,76211242
answered 6 hours ago
user55119user55119
4,76211242
answered 6 hours ago
user55119user55119
4,76211242
answered 6 hours ago
user55119user55119
4,76211242
4,76211242
add a comment |
add a comment |
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$begingroup$
I would suggest that you draw out three-dimensional structures for both starting materials.
$endgroup$
– Zhe
8 hours ago