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    (Original post by reneetaylor)
    Yeah, didn't put concentrated just HCl, should be hopefully.



    I don't think the N2+ was replaced by the iodine. If you looked at the structure you'd see so. :/
    It had an OH or something I think too.
    I'm not sure, I'm probably wrong.
    (Original post by otrivine)
    I dont think it had

    because it was only N2+ going to benzene with X on top? anyone
    I'm nearly 100% sure there was another OH on the bottom of the benzene ring with the I on it
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    (Original post by Gulzar)
    What about NaNO2 and
    HCL
    yes that is acceptable

    i remember they accepted either HNO2 or NaNO2
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    (Original post by otrivine)
    thats third mark gone, was the structure of benzene

    C6H5COOCH(CH3)2
    For the last question? If so I got a that benzene structure with an ester attached to it and yeah 2 methyl groups

    3 marks is nothing otrivine so far, you can still get full ums.
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    Omg... I've done so bad in this. Put 1,2,3-tripropanol and didn't even attempt the last time (didn't have time). What do we think an A and B will be? How did everyone else find this people compared to past ones? I'm so gutted
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    (Original post by lucindaellaaa)
    I'm nearly 100% sure there was another OH on the bottom of the benzene ring with the I on it
    Yeah me too, the N2+ was still on it right?
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    (Original post by reneetaylor)
    For the last question? If so I got a that benzene structure with an ester attached to it and yeah 2 methyl groups

    3 marks is nothing otrivine so far, you can still get full ums.
    yay

    was the bezene the first bit, and then the COO and then the methyl groups,??????

    because I remember this gave 7
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    (Original post by Sinkim)
    Guys was it 7 carbon enviroments? that one mark question?
    yep
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    (Original post by Serpentine111)
    Omg... I've done so bad in this. Put 1,2,3-tripropanol and didn't even attempt the last time (didn't have time). What do we think an A and B will be? How did everyone else find this people compared to past ones? I'm so gutted
    I put diol instead of triol! Haha, I'm sure you haven't done as bad as you think though, just keep positive and TRY to enjoy your summer.

    Probably 47 for an A, 43 for a B 39 for a C.
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    (Original post by lucindaellaaa)
    I'm nearly 100% sure there was another OH on the bottom of the benzene ring with the I on it
    same here I agree too
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    (Original post by reneetaylor)
    Yeah me too, the N2+ was still on it right?

    I didn't put the N2+ because i know that when an Diazonium ion reacts with say, phenol, it doesn't have the N2+ on the phenol, just the Diazonium ion.
    So when it said reaction three, it was the diazonium ion plus the other benzene ring with the OH and the I, not the N2+

    But I could be wrong ofc :')
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    (Original post by otrivine)
    yes that is acceptable

    i remember they accepted either HNO2 or NaNO2
    i think they may only give one for hno2 as it isn't a reagent nano2 is
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    (Original post by otrivine)
    yay

    was the bezene the first bit, and then the COO and then the methyl groups,??????

    because I remember this gave 7
    Yep! By COO do you mean C=O and C-O(ester)?
    And that's the way I did it!

    I got 7 too, just counted the peaks in the Carbon NMR spectrum thing
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    (Original post by keepontrying)

    same here I agree too

    Wooo! I was panicking then thinking I'd lost a stupid mark for adding an OH
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    (Original post by lucindaellaaa)
    Wooo! I was panicking then thinking I'd lost a stupid mark for adding an OH
    I remember counting there from the OH and the iodine was like 4 carbons away from it , just to make sure I got the right structure so nope your right!
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    (Original post by Holz888)
    And how are you going to stop the -COOH group from being reduced instead of the ketone group?
    Carboxylic acids need a different reducing agent to be reduced, only the ketone part would be reduced by NaBH4
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    (Original post by Patel3000)
    i think they may only give one for hno2 as it isn't a reagent nano2 is
    well in the book and mark scheme their first option was HNO2
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    (Original post by lucindaellaaa)
    I didn't put the N2+ because i know that when an Diazonium ion reacts with say, phenol, it doesn't have the N2+ on the phenol, just the Diazonium ion.
    So when it said reaction three, it was the diazonium ion plus the other benzene ring with the OH and the I, not the N2+

    But I could be wrong ofc :')
    You're probably right, I'm pretty sure I misunderstood that whole concept!

    The only question which got me tbh, diazonium ion didn't click *_*

    Thanks for explaining
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    (Original post by reneetaylor)
    Yep! By COO do you mean C=O and C-O(ester)?
    And that's the way I did it!

    I got 7 too, just counted the peaks in the Carbon NMR spectrum thing
    yep thats what I mean


    then why TMS is added I said as a reference peak

    the integration i said the peak area or somethign like that is proportional to the number of protons in that environment ?
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    (Original post by JeevenSD)
    Carboxylic acids need a different reducing agent to be reduced, only the ketone part would be reduced by NaBH4
    So what was the reagent used?
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    (Original post by otrivine)
    yep thats what I mean


    then why TMS is added I said as a reference peak

    the integration i said the peak area or somethign like that is proportional to the number of protons in that environment ?
    Yep and yep
 
 
 
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