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    Name:  5555555.png
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Size:  40.9 KBFor this question, I don't get why B is the right answer. Doesn't the movement of curly arrows from the C-H bond into the pi system mean that a positive ion will form. I don't get why H will instead bond with O and not come off the ring ....


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    The curly arrow from the C-H bond to the pi system does mean that it would come off, however there is another curly arrow from the lone pair on the oxygen, showing the movement of electrons and formation of a bond there between O and H
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    (Original post by JN17)
    The curly arrow from the C-H bond to the pi system does mean that it would come off, however there is another curly arrow from the lone pair on the oxygen, showing the movement of electrons and formation of a bond there between O and H
    But something will come off right? Why doesn't it form the OH- ion since a pair of electrons is donated to the delocalised pi system...

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    (Original post by coconut64)
    But something will come off right? Why doesn't it form the OH- ion since a pair of electrons is donated to the delocalised pi system...

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    It's not going to form a negative ion as the reaction is carried out in acidic medium.
    For an OH- to be formed the S-O bond would have to break. There is no reason for it to do so.
    The hydrogen atom from the benzene ring is captured by a lone pair on the oxygen as it is right next to it, and this allows the delocalisation stabilization to be maintained.
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    I suppose it might be easier to show it with an additional step:
    Imagine after the second step, the pi system is reformed with the oxygen&lone pair of electrons, there would now be a H+ formed to the side.
    This H+ then reacts with the O- to form the OH you see in the final product.
    What the diagram has done is put these two steps together which I can see to be a bit confusing but otherwise the question would be too easy if they showed you the positive hydrogen&negative oxygen on the compound and asked what was formed
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    Thanks for the help
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    (Original post by JN17)
    I suppose it might be easier to show it with an additional step:
    Imagine after the second step, the pi system is reformed with the oxygen&lone pair of electrons, there would now be a H+ formed to the side.
    This H+ then reacts with the O- to form the OH you see in the final product.
    What the diagram has done is put these two steps together which I can see to be a bit confusing but otherwise the question would be too easy if they showed you the positive hydrogen&negative oxygen on the compound and asked what was formed
    You definitely shouldn't write another step as the reaction will pretty much totally proceed by the lone pair on oxygen attacking the H, as this intramolecular process will occur orders of magnitude more quickly than another base taking the proton.

    You can't just lose a proton, a Lewis base has to attack it.

    The point of a mechanism is to show how a reaction likely proceeds. Adding another step would be completely contrary to this.
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    (Original post by alow)
    You definitely shouldn't write another step as the reaction will pretty much totally proceed by the lone pair on oxygen attacking the H, as this intramolecular process will occur orders of magnitude more quickly than another base taking the proton.

    You can't just lose a proton, a Lewis base has to attack it.

    The point of a mechanism is to show how a reaction likely proceeds. Adding another step would be completely contrary to this.
    So what is the best way to think about this? So can I interpret as the H+ ion coming off and the H+ ion bonds to the o to form OH? Cheers
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    the other explanations even though being insightful dont give you a general rule to follow during your exams in case you encounter such a question again.
    listen to this, you wont have trouble after this. if you look closely whenever a bond is broken a curly arrrow is shown. you question why didnt the o come off and form an OH, you can see that in the first step a curly arrow is shown showing that the pi bond between s and o is no more. now in the final step do you see any curly arrow originating from the S-O bond? you dont, so it is safe to say an s-OH will be formed. this should make perfect sense to you and is something you can apply to several other situations.
 
 
 
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