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Synthesis of a weinreb amide from an acid



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Synthesis of mandelic acid from benzaldehydeCarbon decrease from 1-octyne to 1-hexeneAttack of amine on amideSynthesis of a bicyclo[2.2.2]octane from benzoic acidSynthesis of indigo from anthranilic acid - reaction not taking placePurification/Extraction of Sodium PermanganateEvaluation of a synthesis from aspirin to paracetamol from YouTubeIndigo synthesis from anthranilic acidICHO Synthesis Problem 2018Monitoring amino acid esterification










5












$begingroup$


The following shows the conversion of an acid to a weinreb amide:



enter image description here



It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.



In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $ce HCl$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?










share|improve this question









$endgroup$
















    5












    $begingroup$


    The following shows the conversion of an acid to a weinreb amide:



    enter image description here



    It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.



    In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $ce HCl$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?










    share|improve this question









    $endgroup$














      5












      5








      5





      $begingroup$


      The following shows the conversion of an acid to a weinreb amide:



      enter image description here



      It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.



      In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $ce HCl$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?










      share|improve this question









      $endgroup$




      The following shows the conversion of an acid to a weinreb amide:



      enter image description here



      It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.



      In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $ce HCl$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?







      synthesis amines






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Apr 13 at 9:31









      Tan Yong BoonTan Yong Boon

      4,35311148




      4,35311148




















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here






          share|improve this answer









          $endgroup$












          • $begingroup$
            Oh oops... Yes it is an anhydride... Very careless of me.
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:34










          • $begingroup$
            Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:37










          • $begingroup$
            I think the question has omitted the tertiary base used in both steps.
            $endgroup$
            – Waylander
            Apr 13 at 10:43











          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here






          share|improve this answer









          $endgroup$












          • $begingroup$
            Oh oops... Yes it is an anhydride... Very careless of me.
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:34










          • $begingroup$
            Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:37










          • $begingroup$
            I think the question has omitted the tertiary base used in both steps.
            $endgroup$
            – Waylander
            Apr 13 at 10:43















          6












          $begingroup$

          Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here






          share|improve this answer









          $endgroup$












          • $begingroup$
            Oh oops... Yes it is an anhydride... Very careless of me.
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:34










          • $begingroup$
            Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:37










          • $begingroup$
            I think the question has omitted the tertiary base used in both steps.
            $endgroup$
            – Waylander
            Apr 13 at 10:43













          6












          6








          6





          $begingroup$

          Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here






          share|improve this answer









          $endgroup$



          Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 13 at 10:12









          WaylanderWaylander

          6,88311424




          6,88311424











          • $begingroup$
            Oh oops... Yes it is an anhydride... Very careless of me.
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:34










          • $begingroup$
            Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:37










          • $begingroup$
            I think the question has omitted the tertiary base used in both steps.
            $endgroup$
            – Waylander
            Apr 13 at 10:43
















          • $begingroup$
            Oh oops... Yes it is an anhydride... Very careless of me.
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:34










          • $begingroup$
            Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
            $endgroup$
            – Tan Yong Boon
            Apr 13 at 10:37










          • $begingroup$
            I think the question has omitted the tertiary base used in both steps.
            $endgroup$
            – Waylander
            Apr 13 at 10:43















          $begingroup$
          Oh oops... Yes it is an anhydride... Very careless of me.
          $endgroup$
          – Tan Yong Boon
          Apr 13 at 10:34




          $begingroup$
          Oh oops... Yes it is an anhydride... Very careless of me.
          $endgroup$
          – Tan Yong Boon
          Apr 13 at 10:34












          $begingroup$
          Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
          $endgroup$
          – Tan Yong Boon
          Apr 13 at 10:37




          $begingroup$
          Is my reasoning regarding why an acid chloride is not generated instead in the first step correct? Reaction of the acid chloride with the amine in the 2nd step would generate HCl which would protonate the amine
          $endgroup$
          – Tan Yong Boon
          Apr 13 at 10:37












          $begingroup$
          I think the question has omitted the tertiary base used in both steps.
          $endgroup$
          – Waylander
          Apr 13 at 10:43




          $begingroup$
          I think the question has omitted the tertiary base used in both steps.
          $endgroup$
          – Waylander
          Apr 13 at 10:43

















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