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













1












$begingroup$


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)?










share|improve this question











$endgroup$











  • $begingroup$
    I would suggest that you draw out three-dimensional structures for both starting materials.
    $endgroup$
    – Zhe
    8 hours ago















1












$begingroup$


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)?










share|improve this question











$endgroup$











  • $begingroup$
    I would suggest that you draw out three-dimensional structures for both starting materials.
    $endgroup$
    – Zhe
    8 hours ago













1












1








1





$begingroup$


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)?










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 32 secs ago









Melanie Shebel

3,48173273




3,48173273










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
















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















5












$begingroup$

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.



enter image description here



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...






share|improve this answer









$endgroup$




















    4












    $begingroup$

    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.



    enter image description here



    1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293






    share|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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.



      enter image description here



      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...






      share|improve this answer









      $endgroup$

















        5












        $begingroup$

        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.



        enter image description here



        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...






        share|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          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.



          enter image description here



          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...






          share|improve this answer









          $endgroup$



          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.



          enter image description here



          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...







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 7 hours ago









          PCKPCK

          3987




          3987





















              4












              $begingroup$

              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.



              enter image description here



              1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293






              share|improve this answer









              $endgroup$

















                4












                $begingroup$

                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.



                enter image description here



                1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293






                share|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  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.



                  enter image description here



                  1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293






                  share|improve this answer









                  $endgroup$



                  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.



                  enter image description here



                  1) G. E. McCasland, J. Am. Chem. Soc., 1951, 73, 2293







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 6 hours ago









                  user55119user55119

                  4,76211242




                  4,76211242



























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