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    Hey guys. Nitroacetanilide which has a NO2 group at one position on a benzene ring and an amide on another. I'm trying to figure out what isomer would be most prominent. The nitrogen on the Nitro group has a positive charge. If the amide group was 1 would the Nitro group prefer carbon 2, 3 or 4?
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    (Original post by 'bob')
    Hey guys. Nitroacetanilide which has a NO2 group at one position on a benzene ring and an amide on another. I'm trying to figure out what isomer would be most prominent. The nitrogen on the Nitro group has a positive charge. If the amide group was 1 would the Nitro group prefer carbon 2, 3 or 4?
    There is no such thing as 'preference' as regards isomers. I suspect that you are thinking about directing groups.

    If you start off with aniline then any nitration will take place at the 2,4,6 positions.

    BUT, if you start off with nitrobenzene further substitution will occur at the 3,5 positions
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    (Original post by charco)
    There is no such thing as 'preference' as regards isomers. I suspect that you are thinking about directing groups.

    If you start off with aniline then any nitration will take place at the 2,4,6 positions.

    BUT, if you start off with nitrobenzene further substitution will occur at the 3,5 positions
    Hey yh. Sorry thats what i meant. 2,4 and 6. Is there a reason why it might prefer the 4 (para) position.
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    (Original post by 'bob')
    Hey yh. Sorry thats what i meant. 2,4 and 6. Is there a reason why it might prefer the 4 (para) position.
    Statistically there is more chance of a 2,6 substitution (two identical sites), but the 4 position is less sterically hindered.

    I would guess that you get about 60% ortho and 40% para substiutution...

    But the NO2 group is large, so the steric effects could easily drive a greater % to the para position.
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    It's less easy to predict which substitution will dominate between the ortho and para positions of an ortho / para director. There are 2 ortho positions so statistically substitution there is more likely, but due to the steric repulsion of the electrophile approaching the ortho position (even more so when the side-group is large), we often get more para substitution, which is the case with the nitration of acetanilide.
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    (Original post by charco)
    Statistically there is more chance of a 2,6 substitution (two identical sites), but the 4 position is less sterically hindered.

    I would guess that you get about 60% ortho and 40% para substiutution...

    But the NO2 group is large, so the steric effects could easily drive a greater % to the para position.
    Yeah it seems like steric hindrance definitely pushes the ortho positions out of favour when the primary group is an amide with an additional methyl group. But why is there no reactivity at the meta position then?
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    (Original post by 'bob')
    Yeah it seems like steric hindrance definitely pushes the ortho positions out of favour when the primary group is an amide with an additional methyl group. But why is there no reactivity at the meta position then?
    The amine group exerts a +M effect on the benzene ring, i.e. it is able to donate electron density into the ring both activating it and directing substitution towards the o,p positions. That is not to say that there won't be any m substitution only that it is very much a minor product.

    The reason lies in the resonance forms of the intermediate positive ion. In o and p sbstitution one of the intermediate resonance forms has the positive charge on the carbon that holds the amine group. This is a stabilised form. Hence it is easier to go via the o or p substitution than the m.

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    Thanks! That was really helpful
 
 
 
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