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    Name:  image.jpg
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Size:  421.6 KB Hi guys, can anyone help me please?
    I'm not sure which alcohol group on compound C reacts with carboxylic acid group?
    Also, I don't know if I've substituted the bromines onto the correct carbons in the ring?
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    The same here: where is the correct place to substitue the 2 Br on the organic product?

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    Thank you again
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    You probably should know that esterification of phenol -OH groups is very slow and is usually done with the likes of an acyl chloride rather than carboxylic acid. Therefore the esterification happens through the CH2-OH group.

    The Br substitution can happen at any point on the ring. The key to the reaction is that Br2 goes in and HBr comes out - don't think that both Br atoms of a Br2 molecule attach. Your product is fine, as long as you kick out 2HBr.
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    (Original post by Pigster)
    You probably should know that esterification of phenol -OH groups is very slow and is usually done with the likes of an acyl chloride rather than carboxylic acid. Therefore the esterification happens through the CH2-OH group.

    The Br substitution can happen at any point on the ring. The key to the reaction is that Br2 goes in and HBr comes out - don't think that both Br atoms of a Br2 molecule attach. Your product is fine, as long as you kick out 2HBr.
    Thank you Pigster

    I think these should be the answers then:

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    That'll do, pig. That'll do.
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    (Original post by Pigster)
    That'll do, pig. That'll do.
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    Thank you very much for your help ☺️
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    That bromination substitution pattern is actually incorrect. Probably a bit more advanced theory than you've been taught, but the bromine atoms would add para (opposite) to the substituents. Both the methoxy group and the hydroxyl are activating ortho/para directors, para position favoured over ortho due to sterics. After the first bromine is added, it will also act as an ortho para director. The position that satisfies the directing influences of the most substituent is ortho to the first bromine atom.
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    (Original post by VanEkerDr)
    That bromination substitution pattern is actually incorrect. Probably a bit more advanced theory than you've been taught, but the bromine atoms would add para (opposite) to the substituents. Both the methoxy group and the hydroxyl are activating ortho/para directors, para position favoured over ortho due to sterics. After the first bromine is added, it will also act as an ortho para director. The position that satisfies the directing influences of the most substituent is ortho to the first bromine atom.
    Hi

    Thanks for your help so far. I have of couple of questions:

    How do you know the para position is favoured over the ortho position for the first bromine atom?

    Secondly, I may be wrong, but I would have thought the first substituted bromine would be deactivating, therefore a meta director for the second bromine?

    Thanks again!
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    Hey

    No problem, I'll do my best to answer them. First off, the para position is favoured over the ortho position because of sterics. You've stated that you know about directing groups so you'll known that the ortho and para positions are stabilised through resonance. The para position is favoured only because bromine is a rather large atom, being in the 4th period, so sticking it in the para position means it's out of the way, ortho means it's next to another group.

    Secondly, you're correct in saying that halogens are deactivating. This is true, because they are electron withdrawing. However, because they have lone pairs (3 to be exact) they can still donate them and there are resonance forms where this is true, so they are actually ortho/para directors. They are the only example of deactivating ortho/para directors.
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    This is why I put 'can' in italics earlier in the thread. There is a preference for which positions and where those positions are is beyond A-level and was not credited/penalised - the M/S just wanted two Br atoms substituted with the release of 2HBr (rather than a botched addition or the loss of H2)

    Even though there are directing effects, the other products can form - they just are minor products.
 
 
 
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