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aminophenol with R-COCl

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    If aminophenol was reacted with an acyl chloride in the presence of an AlCl3 catalst would all the following happen
    (a) nuclephillic addition-elimination on R-OH group
    (b) nuclephillic addition-elimination on R-NH2 group
    (c) electrophilic substitution of acyl group on arene ring?

    If not why?
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    (Original post by ilovemath)
    If aminophenol was reacted with an acyl chloride in the presence of an AlCl3 catalst would all the following happen
    (a) nuclephillic addition-elimination on R-OH group
    (b) nuclephillic addition-elimination on R-NH2 group
    (c) electrophilic substitution of acyl group on arene ring?

    If not why?
    C.

    With the formation of the AlCl4-, you get an electrophile forming, R-CO+. This electrophile is attacked by the delocalised ring of electrons in the benzene ring, and you get electrophilic substitution.

    The amine group is more of a red herring, leading you to think down the route of a nucleophilic attack
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    (Original post by thegodofgod)
    C.

    With the formation of the AlCl4-, you get an electrophile forming, R-CO+. This electrophile is attacked by the delocalised ring of electrons in the benzene ring, and you get electrophilic substitution.

    The amine group is more of a red herring, leading you to think down the route of a nucleophilic attack
    ok so If there was no AlCl3 would you get both (a) and (b)
    I am not entirely sure as if you react it with an anhydride the R-OH group is not affected. Why is this?
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    (Original post by ilovemath)
    ok so If there was no AlCl3 would you get both (a) and (b)
    I am not entirely sure as if you react it with an anhydride the R-OH group is not affected. Why is this?
    Not too sure about this, but I think that the delocalised electrons in the benzene ring cause a negative inductive effect on the -NH2 and -OH groups. Therefore, their lone pairs of electrons are less available to attack the delta + carbon atom in the acyl chloride. Therefore, the reaction would be slow, as aminophenol (may be ) a weak nucleophile.
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    (Original post by thegodofgod)
    Not too sure about this, but I think that the delocalised electrons in the benzene ring cause a negative inductive effect on the -NH2 and -OH groups. Therefore, their lone pairs of electrons are less available to attack the delta + carbon atom in the acyl chloride. Therefore, the reaction would be slow, as aminophenol (may be ) a weak nucleophile.
    well nitrogen is less electronegative than oxygen so there should be a greater electron density around the oxygen and hence oxygen SHOULD react with acyl chlorides / acid anhydrides.

    maybe someone else can suggest why this is not what happens
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    Lone pairs are conjugated into the benzene ring. This explains the reactivity.....

    1. As the lone pairs are conjugated they are not as available for nucleophillic attack of an electrophile.

    2. Because the lone pairs are conjugated into the ring, they can donate into the ring. This increases the electron density in the benzene ring making it more nucleophillic.

    3. The acid catalyst may form a lewis acid adduct with the NH2 group, making it non nucleophillic. This effect IS present with a protic acid, forming a positively charged R-NH3 + group which is no longer nucleophillic.

    Also, to correct a statement above.

    "well nitrogen is less electronegative than oxygen so there should be a greater electron density around the oxygen and hence oxygen SHOULD react with acyl chlorides / acid anhydrides."

    As the oxygen is more electronegative, the energy of the lone pairs are lower than for N. They are therefore les available for nucleophillic attack. This is why Nitrogen nucleophiles are more reactive than Oxygen nucleophiles, and why ammonia is more basic than water
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    (Original post by JMaydom)
    Lone pairs are conjugated into the benzene ring. This explains the reactivity.....

    1. As the lone pairs are conjugated they are not as available for nucleophillic attack of an electrophile.

    2. Because the lone pairs are conjugated into the ring, they can donate into the ring. This increases the electron density in the benzene ring making it more nucleophillic.

    3. The acid catalyst may form a lewis acid adduct with the NH2 group, making it non nucleophillic. This effect IS present with a protic acid, forming a positively charged R-NH3 + group which is no longer nucleophillic.

    Also, to correct a statement above.

    "well nitrogen is less electronegative than oxygen so there should be a greater electron density around the oxygen and hence oxygen SHOULD react with acyl chlorides / acid anhydrides."

    As the oxygen is more electronegative, the energy of the lone pairs are lower than for N. They are therefore les available for nucleophillic attack. This is why Nitrogen nucleophiles are more reactive than Oxygen nucleophiles, and why ammonia is more basic than water
    so you're saying that the reason
    OH-R'-NH2 (R' = phenyl ring and OH and NH2 are in 1,4 positions) reacts with R-COCl to produce OH-R-NH-CO-R is because the NH2 group is a better nucleophile so competes better for the electrophile (does this mean that the -OH group will NEVER react with an -NH2 group present?)
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    (Original post by ilovemath)
    so you're saying that the reason
    OH-R'-NH2 (R' = phenyl ring and OH and NH2 are in 1,4 positions) reacts with R-COCl to produce OH-R-NH-CO-R is because the NH2 group is a better nucleophile so competes better for the electrophile (does this mean that the -OH group will NEVER react with an -NH2 group present?)
    No, because the pKa of phenol is around 10 whereas aniline is around 30. Hence deprotonation of phenol is a lot, lot easier, and the phenoxide anion can act as a nucleophile through O or C, depending on the electrophile and reaction conditions. Presence of a mild base can encourage nucleophilic attack of the O rather than N or C.
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    (Original post by illusionz)
    No, because the pKa of phenol is around 10 whereas aniline is around 30. Hence deprotonation of phenol is a lot, lot easier, and the phenoxide anion can act as a nucleophile through O or C, depending on the electrophile and reaction conditions. Presence of a mild base can encourage nucleophilic attack of the O rather than N or C.
    yes but the attack is of the N not the O???
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    (Original post by ilovemath)
    yes but the attack is of the N not the O???
    I don't recall covering ambident nucleophiles at A level so you might not even need to know this but anyway. A deprotonated R-OH (ie RO-) is more nucleophilic than neutral R-NH2. Because it is easier for a base to deprotonate the O, this anion is formed rather than N-.

    However, different electrophiles would prefer to react with the oxygen or the carbon. I can't really explain why without going into molecular orbital theory which you won't have come across. But put very, very simply, if you have a molecule with two nucleophilic sites, one charged and one uncharged (such as phenoxide), then a charged/highly delta positive electrophile such as a carbonyl will generally prefer to react with the charged O, and an uncharged, unpolar electrophile such as methyl iodide would prefer to react with the carbon site.

    So, to answer the original question of phenol + ROCl.

    Phenol + ROCl ---> normal friedel crafts acylation
    phenol + ROCl + base ---> initial formation of a phenyl ester (although this will rearrange by something known as the Fries rearrangement if AlCl3 is present)
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    (Original post by illusionz)

    So, to answer the original question of phenol + ROCl.

    Phenol + ROCl ---> normal friedel crafts acylation
    phenol + ROCl + base ---> initial formation of a phenyl ester (although this will rearrange by something known as the Fries rearrangement if AlCl3 is present)
    No the original question was:
    NH2-R-OH (R = phenyl group) reacts with R'-OCl to produce R'-O-NH-R-OH
    I wanted to know why the NH group DID react but the OH group did not react
    (this reaction is how paracetamol is prepared)
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    (Original post by ilovemath)
    No the original question was:
    NH2-R-OH (R = phenyl group) reacts with R'-OCl to produce R'-O-NH-R-OH
    I wanted to know why the NH group DID react but the OH group did not react
    (this reaction is how paracetamol is prepared)
    Ah sorry.

    The N lone pair is higher in energy than the O lone pair, and so it is more nucleophilic. Because you have both the O and the N donating into the ring they are not as deactivated as they would be on their own. Use of an acid anhydride is used because it has a very delta positive carbon and so for the reasons I explained above, the N is a better nucleophile for this electrophile than the C. I can't really give a proper explanation and I know this isn't really good enough but you simply don't understand the theory behind it.

    The reaction would also give a mixture of different products which would have to be separated. You wouldn't get exclusive reaction through N.
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    (Original post by illusionz)
    Ah sorry.

    The N lone pair is higher in energy than the O lone pair, and so it is more nucleophilic. Because you have both the O and the N donating into the ring they are not as deactivated as they would be on their own. Use of an acid anhydride is used because it has a very delta positive carbon and so for the reasons I explained above, the N is a better nucleophile for this electrophile than the C. I can't really give a proper explanation and I know this isn't really good enough but you simply don't understand the theory behind it.
    so for an A-level examination it would be sufficient to say
    "The lone pair on the nitrogen is more nucleophillic so is attacked in preference by the R-COCl / andhyride carbon"

    another question: if excess R-OCl was used, would any of the OH groups be attacked?
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    (Original post by ilovemath)
    so for an A-level examination it would be sufficient to say
    "The lone pair on the nitrogen is more nucleophillic so is attacked in preference by the R-COCl / andhyride carbon"

    another question: if excess R-OCl was used, would any of the OH groups be attacked?
    In all honest I have no idea what they'd want you to say. In terms of N vs O that explanation is correct, but if you consider N vs C then it gets a bit more involved.
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    thanks
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    Messy :p:

    A-level questions are very good at omitting side reactions/unwanted products for the purpose of making the questions easier (justifiably so), so if a question gives you 1,4-aminophenol and tells you that it reacts with AlCl3 and an acyl chloride, then it simply wants the reaction with the aromatic ring (probably suggested by a given molecular formula or a previous part of the question). Truth is, is that you would likely end up with several different combinations of products, with acylations at various/multiple points. If it reacts without AlCl3 then you will get reaction at the -NH2 and -OH functional groups and the reaction would be faster at the amine component - if you had an excess then it's perfectly possible to get a significant amount of acylation at the -OH too (acyl chlorides are powerful electrophiles afterall, and this is how phenolic esters can be made).

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Updated: May 18, 2012
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