Regioselectivity in iodolactonisation of γ,δ-unsaturated carboxylic acid











up vote
3
down vote

favorite












Iodolactonisation of 3,3-dimethylpent-4-enoic acid



In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.





References




  1. Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.


  2. Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.











share|improve this question




























    up vote
    3
    down vote

    favorite












    Iodolactonisation of 3,3-dimethylpent-4-enoic acid



    In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.





    References




    1. Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.


    2. Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.











    share|improve this question


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Iodolactonisation of 3,3-dimethylpent-4-enoic acid



      In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.





      References




      1. Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.


      2. Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.











      share|improve this question















      Iodolactonisation of 3,3-dimethylpent-4-enoic acid



      In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.





      References




      1. Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.


      2. Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.








      organic-chemistry nucleophilic-substitution regioselectivity stereoelectronics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 12 hours ago









      orthocresol

      37.7k7110226




      37.7k7110226










      asked 13 hours ago









      Avnish Kabaj

      3,61931344




      3,61931344






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          6
          down vote



          accepted










          There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.



          Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.



          However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.



          For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]



          6-endo-tet and 6-exo-tet reactions



          It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.



          The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):



          σ* orbitals for three different cases



          The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).



          As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.



          5-membered vs 6-membered ring formation



          A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:



          Acid-catalysed epoxide opening





          References




          1. Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.

          2. Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.






          share|improve this answer




























            up vote
            3
            down vote













            Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.



            Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.



            enter image description here



            Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.






            share|improve this answer





















              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "431"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f106846%2fregioselectivity-in-iodolactonisation-of-%25ce%25b3-%25ce%25b4-unsaturated-carboxylic-acid%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              6
              down vote



              accepted










              There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.



              Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.



              However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.



              For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]



              6-endo-tet and 6-exo-tet reactions



              It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.



              The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):



              σ* orbitals for three different cases



              The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).



              As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.



              5-membered vs 6-membered ring formation



              A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:



              Acid-catalysed epoxide opening





              References




              1. Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.

              2. Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.






              share|improve this answer

























                up vote
                6
                down vote



                accepted










                There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.



                Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.



                However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.



                For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]



                6-endo-tet and 6-exo-tet reactions



                It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.



                The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):



                σ* orbitals for three different cases



                The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).



                As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.



                5-membered vs 6-membered ring formation



                A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:



                Acid-catalysed epoxide opening





                References




                1. Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.

                2. Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.






                share|improve this answer























                  up vote
                  6
                  down vote



                  accepted







                  up vote
                  6
                  down vote



                  accepted






                  There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.



                  Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.



                  However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.



                  For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]



                  6-endo-tet and 6-exo-tet reactions



                  It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.



                  The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):



                  σ* orbitals for three different cases



                  The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).



                  As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.



                  5-membered vs 6-membered ring formation



                  A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:



                  Acid-catalysed epoxide opening





                  References




                  1. Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.

                  2. Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.






                  share|improve this answer












                  There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.



                  Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.



                  However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.



                  For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]



                  6-endo-tet and 6-exo-tet reactions



                  It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.



                  The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):



                  σ* orbitals for three different cases



                  The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).



                  As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.



                  5-membered vs 6-membered ring formation



                  A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:



                  Acid-catalysed epoxide opening





                  References




                  1. Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.

                  2. Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 12 hours ago









                  orthocresol

                  37.7k7110226




                  37.7k7110226






















                      up vote
                      3
                      down vote













                      Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.



                      Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.



                      enter image description here



                      Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.






                      share|improve this answer

























                        up vote
                        3
                        down vote













                        Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.



                        Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.



                        enter image description here



                        Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.






                        share|improve this answer























                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.



                          Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.



                          enter image description here



                          Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.






                          share|improve this answer












                          Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.



                          Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.



                          enter image description here



                          Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 6 hours ago









                          Vice Chemist

                          2513




                          2513






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Chemistry Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f106846%2fregioselectivity-in-iodolactonisation-of-%25ce%25b3-%25ce%25b4-unsaturated-carboxylic-acid%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              サソリ

                              広島県道265号伴広島線

                              Setup Asymptote in Texstudio