Can a cyclic Amine form an Amide?












3












$begingroup$


This is a question from CAIE.



I need to understand why the secondary amine in serotonin molecule does not undergo condensation reaction with $ce{CH_3COCl}$ ?



Because, as far as I know, there should be an amide formation for this reaction.



Here's a picture of serotonin:



enter image description here



I must be lacking some information making that $ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $ce{CH_3COCl}$.






organic-chemistry amines amides







share|improve this question











$endgroup$












  • $begingroup$
    Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
    $endgroup$
    – YUSUF HASAN
    6 hours ago










  • $begingroup$
    Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
    $endgroup$
    – Amar30657
    5 hours ago










  • $begingroup$
    No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
    $endgroup$
    – YUSUF HASAN
    5 hours ago








  • 3




    $begingroup$
    Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
    $endgroup$
    – Waylander
    5 hours ago








  • 1




    $begingroup$
    @Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
    $endgroup$
    – Waylander
    5 hours ago


















3












$begingroup$


This is a question from CAIE.



I need to understand why the secondary amine in serotonin molecule does not undergo condensation reaction with $ce{CH_3COCl}$ ?



Because, as far as I know, there should be an amide formation for this reaction.



Here's a picture of serotonin:



enter image description here



I must be lacking some information making that $ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $ce{CH_3COCl}$.






organic-chemistry amines amides







share|improve this question











$endgroup$












  • $begingroup$
    Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
    $endgroup$
    – YUSUF HASAN
    6 hours ago










  • $begingroup$
    Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
    $endgroup$
    – Amar30657
    5 hours ago










  • $begingroup$
    No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
    $endgroup$
    – YUSUF HASAN
    5 hours ago








  • 3




    $begingroup$
    Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
    $endgroup$
    – Waylander
    5 hours ago








  • 1




    $begingroup$
    @Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
    $endgroup$
    – Waylander
    5 hours ago
















3












3








3





$begingroup$


This is a question from CAIE.



I need to understand why the secondary amine in serotonin molecule does not undergo condensation reaction with $ce{CH_3COCl}$ ?



Because, as far as I know, there should be an amide formation for this reaction.



Here's a picture of serotonin:



enter image description here



I must be lacking some information making that $ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $ce{CH_3COCl}$.






organic-chemistry amines amides







share|improve this question











$endgroup$




This is a question from CAIE.



I need to understand why the secondary amine in serotonin molecule does not undergo condensation reaction with $ce{CH_3COCl}$ ?



Because, as far as I know, there should be an amide formation for this reaction.



Here's a picture of serotonin:



enter image description here



I must be lacking some information making that $ce{NH}$ unsuitable for this reaction. And I know that the other amine and the phenol reacts with $ce{CH_3COCl}$.






organic-chemistry amines amides




organic-chemistry amines amides






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 53 mins ago









Night Writer

2,951323




2,951323










asked 6 hours ago









Amar30657Amar30657

696




696












  • $begingroup$
    Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
    $endgroup$
    – YUSUF HASAN
    6 hours ago










  • $begingroup$
    Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
    $endgroup$
    – Amar30657
    5 hours ago










  • $begingroup$
    No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
    $endgroup$
    – YUSUF HASAN
    5 hours ago








  • 3




    $begingroup$
    Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
    $endgroup$
    – Waylander
    5 hours ago








  • 1




    $begingroup$
    @Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
    $endgroup$
    – Waylander
    5 hours ago




















  • $begingroup$
    Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
    $endgroup$
    – YUSUF HASAN
    6 hours ago










  • $begingroup$
    Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
    $endgroup$
    – Amar30657
    5 hours ago










  • $begingroup$
    No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
    $endgroup$
    – YUSUF HASAN
    5 hours ago








  • 3




    $begingroup$
    Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
    $endgroup$
    – Waylander
    5 hours ago








  • 1




    $begingroup$
    @Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
    $endgroup$
    – Waylander
    5 hours ago


















$begingroup$
Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
$endgroup$
– YUSUF HASAN
6 hours ago




$begingroup$
Secondary amine group shown here is quite sterically hindered..Approach at Burgi-Dunitz trajectory would most probably be difficult
$endgroup$
– YUSUF HASAN
6 hours ago












$begingroup$
Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
$endgroup$
– Amar30657
5 hours ago




$begingroup$
Did you mean lone pair of electron in N is less available for carbocation to be attacked? @YUSUFHASAN
$endgroup$
– Amar30657
5 hours ago












$begingroup$
No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
$endgroup$
– YUSUF HASAN
5 hours ago






$begingroup$
No..I mean what you have in mind is a nucleophilic attack by NH on CH3COCl as the first step, right? So that would be retarded by steric hindrance..
$endgroup$
– YUSUF HASAN
5 hours ago






3




3




$begingroup$
Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
$endgroup$
– Waylander
5 hours ago






$begingroup$
Nothing to do with sterics. The lone pair on the nitrogen is delocalised into the aromatic system so is not available for nucleophilic attack. If you want to functionalise the indole-NH you generally have to formally deprotonate with strong base. Otherwise it reacts as an enamine through the 3-position.
$endgroup$
– Waylander
5 hours ago






1




1




$begingroup$
@Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
$endgroup$
– Waylander
5 hours ago






$begingroup$
@Amar30657 You need to read more about aromatic systems. en.wikipedia.org/wiki/Indole
$endgroup$
– Waylander
5 hours ago












2 Answers
2






active

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6












$begingroup$

The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise though there are some interesting techniques using carbonyl azoles catalysed by DBU reference and others using aldehyde and alcohol substrates such as here. Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position example






share|improve this answer









$endgroup$





















    2












    $begingroup$

    If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!






    share|improve this answer









    $endgroup$














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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      6












      $begingroup$

      The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise though there are some interesting techniques using carbonyl azoles catalysed by DBU reference and others using aldehyde and alcohol substrates such as here. Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position example






      share|improve this answer









      $endgroup$


















        6












        $begingroup$

        The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise though there are some interesting techniques using carbonyl azoles catalysed by DBU reference and others using aldehyde and alcohol substrates such as here. Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position example






        share|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise though there are some interesting techniques using carbonyl azoles catalysed by DBU reference and others using aldehyde and alcohol substrates such as here. Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position example






          share|improve this answer









          $endgroup$



          The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise though there are some interesting techniques using carbonyl azoles catalysed by DBU reference and others using aldehyde and alcohol substrates such as here. Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position example







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          WaylanderWaylander

          7,10311424




          7,10311424























              2












              $begingroup$

              If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!






              share|improve this answer









              $endgroup$


















                2












                $begingroup$

                If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!






                share|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!






                  share|improve this answer









                  $endgroup$



                  If you consider the lone pair on that N-atom and apply Hückel's (4n+2)π-e rule to check aromaticity , it satisfies all the conditions. So in order to achieve aromatic-stabilisation the nitrogen's valency is no more available for condensation!







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  ANBENZENEANBENZENE

                  1086




                  1086






























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