Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by Parallex)
    S and T both have 6 peaks. For S, there are 5 non-equivalent H peaks in the ring structure and 1 on the OH. For T, the same applies for the ring and the methyl group.
    Oh! Got you. Thanks so much, that's really helped. Quite sneaky - I forgot that there's obviously another H atom bonded to the carbon which has the CH3 attached to it
    Offline

    16
    ReputationRep:
    (Original post by Parallex)


    Equivalent carbon groups are the same colour. The compound is symmetrical about the O bonds so you only need to consider one half of the structure. Splitting the compound horizontally there is also a symmetry, so the groups below are equivalent to the groups above. There are 3 non-equivalent carbon groups in the molecule.
    ah brilliant thanks.

    Can you somehow explain 6bi to me, i've been trying to understand the ms for a while. I don't quite understand how the further substitutions work. Like what happened to the 3 hydrogens from the original ch3br
    Offline

    2
    ReputationRep:
    (Original post by Super199)
    ah brilliant thanks.

    Can you somehow explain 6bi to me, i've been trying to understand the ms for a while. I don't quite understand how the further substitutions work. Like what happened to the 3 hydrogens from the original ch3br
    Here's the mechanism so you can understand what's happening.


    Notice that the lone pair remains on the N atom even after the first and second substitution. This means that it can still act as a nucleophile, so the same reaction that occurred in step 1 can occur again. In the 3rd substitution, the quaternary ammonium salt is formed and it isn't a nucleophile any longer. It bonds ionically with Br-.
    Offline

    1
    ReputationRep:
    Hi,
    I don't understand how to qualitatively explain the effects of added OH-/ H+ to a buffer solution. Can somebody please explain this?

    Thank you
    Offline

    2
    ReputationRep:
    can anyone else just not get their head around any analytical??
    Offline

    2
    ReputationRep:
    Is it plane polarised or polarized light?

    In a few of the recent papers polarized, whereas in the book and a 2012 paper it says polarised...
    Offline

    1
    ReputationRep:
    Plane polarised

    (Original post by shiney101)
    Is it plane polarised or polarized light?

    In a few of the recent papers polarized, whereas in the book and a 2012 paper it says polarised...
    Offline

    7
    ReputationRep:
    can someone help please?!

    1e

    http://filestore.aqa.org.uk/subjects...4-QP-JUN14.PDF
    Offline

    3
    ReputationRep:
    (Original post by DiamondsForever)
    Hi,
    I don't understand how to qualitatively explain the effects of added OH-/ H+ to a buffer solution. Can somebody please explain this?

    Thank you
    HA disassociates Into H+ and A-

    So if you add H+ = There is a large amount of A- in the buffer due to the salt - SO an increase in H+ will cause it to react with A- to form HA( equillbrium shifts to the left)

    Adding OH- = OH- can react with H+ reducing the amount of H+ so equilibrium shifts to right
    Offline

    13
    ReputationRep:
    Can anyone please explain how to go about tackling this nmr type question i literally just dont no what to do Name:  ImageUploadedByStudent Room1465762096.034779.jpg
Views: 188
Size:  124.2 KB


    Posted from TSR Mobile
    Offline

    5
    (Original post by Parallex)
    Only one of the indicators has pH 6 within its indicating range and that's 4-nitrophenol. At lower pH it is colourless and as pH increases it becomes yellow. The result is a mix between this and hence light yellow.

    do you ignore the other indicators?
    Offline

    2
    ReputationRep:
    (Original post by Superbubbles)
    do you ignore the other indicators?
    It's not that you ignore them, it's that none of the other indicators have pH 6 within their indicating range.
    Offline

    7
    ReputationRep:
    HELP PLEASE!

    Give an example of the use of a buffer solution.
    Offline

    12
    ReputationRep:
    (Original post by Lilly1234567890)
    can someone help please?!

    1e

    http://filestore.aqa.org.uk/subjects...4-QP-JUN14.PDF
    If Kc has increased, that means the forward reaction has been favoured (if Kc<1, reverse reaction is favoured, if Kc>1, forward reaction is favoured).

    The forward reaction is exothermic (indicated by delta H<0).

    Therefore, for the forward reaction to have been favoured (due to Kc having increased), the temperature must have decreased, as the equilibrium would shift to oppose the decrease in temperature by favouring the forward reaction.

    Therefore, T1 is higher.
    Offline

    12
    ReputationRep:
    (Original post by Lilly1234567890)
    HELP PLEASE!

    Give an example of the use of a buffer solution.
    Lactic acid and sodium lactate as an acidity regulator (appeared on June 13).

    The ratio [HA]/[A-] remains fairly constant in the buffer. Any H+ added will react with A- to form HA and the equilibrium HA <------> A- + H+ will shift to oppose the change.
    Offline

    12
    ReputationRep:
    Hi all, I am doing this exam - I'm hoping it will be good, I find CHEM4 much easier than CHEM5. Occasionally, nmr catches me out and I really have to be careful not to make silly errors counting carbon or proton environments.

    I'd kill for that A*!!
    Offline

    12
    ReputationRep:
    (Original post by Sexybadman)
    Can anyone please explain how to go about tackling this nmr type question i literally just dont no what to do Name:  ImageUploadedByStudent Room1465762096.034779.jpg
Views: 188
Size:  124.2 KB


    Posted from TSR Mobile
    Integration value is the number of hydrogens for that particular peak, relative to the smallest non-zero number (which will usually be 1).

    Use your data sheet to see what the peaks consist of - then start drawing. Trailing bonds on your data sheet will take the place of either hydrogens or other groups - use the integration value to deduce this.
    Offline

    13
    ReputationRep:
    (Original post by Cadherin)
    Integration value is the number of hydrogens for that particular peak, relative to the smallest non-zero number (which will usually be 1).

    Use your data sheet to see what the peaks consist of - then start drawing. Trailing bonds on your data sheet will take the place of either hydrogens or other groups - use the integration value to deduce this.
    Still struggling to understand what to do can you show maybe a worked example ?


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Sexybadman)
    Still struggling to understand what to do can you show maybe a worked example ?


    Posted from TSR Mobile
    What paper and which question is it? I c ant see it properly on my laptop
    Offline

    3
    ReputationRep:
    A few quick questions:

    1. How do you work out the major peak of a compound?
    2. How does Sn/HCl or Ni/H+ or similar reduce benzene to hexane?
    3. When would you ever use PKa in a paper?
    4. Anhydride to amide?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.