2D NMR Analysis (HMBC) - Assigning Peaks Using HMBC (Part 2 of 2)

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  • Опубліковано 2 січ 2025

КОМЕНТАРІ • 48

  • @jojojojo9969
    @jojojojo9969 4 роки тому +3

    Very very useful , I wish if all could explain in that clear and comprehensive way ,, bravo and thanks

  • @nurmukhtar1416
    @nurmukhtar1416 4 роки тому +4

    hello, thank you so much for this video! you have no idea how helpful this is to a graduate student like me :)

  • @gutterball10
    @gutterball10 6 років тому +16

    figuring out these correlations reminds me of solving a Sudoku puzzle.

    • @vovaktovam4451
      @vovaktovam4451 2 роки тому

      I have had the same thought when working on my HMBC spectrum of quino-4H-one, although my brain does not overheat after sudokus

  • @doanhngo3472
    @doanhngo3472 4 роки тому +2

    Thank you for your explanation. I was so confused with the HMBC. Your video is really helpful!

  • @shadisadeghi1999
    @shadisadeghi1999 Рік тому +1

    Thank you so much for all of your videos they are so fascinating and helpful

  • @Errys
    @Errys 2 роки тому +1

    thanks dude, way better than how my professor explained this

  • @adamguzik6737
    @adamguzik6737 3 роки тому +2

    For all the people saying carbons (2) and (3) should be switched, you're right. I used ChemDraw to predict the 13C NMR spectrum and indeed, carbon 2 turned out to be 92 ppm and carbon 3: 183 ppm. Plus resonance structures confirm that carbon 2 is less shielded than carbon 3.
    docs.google.com/document/d/1IYKw0NF86H4-W5GRBOvMK5B9pkntqAv9Rj5jejxSbpw/edit?usp=sharing

  • @Kris_Lighthawk
    @Kris_Lighthawk 7 років тому +4

    Nice, but one question:
    There should have been a strong correlation between A and 4 in the HMBC and yet there is not even a weak one... Any explanation why that might be?
    Also, I am pretty sure you got 2 and 3 mixed up as as 3 is bound to both N and O which should unshield it a lot, putting it much more downfield than 2 who is only bound to other C

    • @TonyStJohn
      @TonyStJohn  7 років тому

      Hi Kris, you have a good point! I'm actually not 100% sure about this at this point in time, I'll have to get back to you, or maybe if you can figure out a good explanation you can comment here! As for your other question on the other video, I think the link you had was an old link to a bad video. This is the link you want: ua-cam.com/video/SMcNg9iugrY/v-deo.html

    • @armikatrina
      @armikatrina 5 років тому

      I agree with your assignment, Tony. Carbon 2 is beside two carbonyls so it's going to be much more deshielded than 3.

    • @s0t0e0f0a0n
      @s0t0e0f0a0n 5 років тому +3

      ​@@TonyStJohnI agree with Kris: C2 and C3 should be reversed.
      Why? Try not see this as just "in between two carbonyls" or attached to N and O. Inductive effects are always there, but resonance(conjugation) effects depend on the structure. Just because the protons are acidic if they are in between two carbonyls and it can stabilize a negative charge (by conjugation!) doesn't mean in a system like this it works the same.
      The ring on the right is aromatic. Carbon 3 is strongly electronwithdrawn bij conjugation from both carbonyls and by induction from the directly! attached N and O. The chemical shift is certainly is high but aromatic carbons attached to a single O are often near 150-160 ppm (compare simple anisole), this one had an extra N too so 170 is fine. Now, C2 is actually supplied with electrondensity by conjugation from the O and N !, inductive effect by O and N is much much weaker an extra bond away, the carbonyls do withdraw some electron density inductively, but there is no resonance structure which puts a positive charge on C2. The normal chemical shift (for a CH2) in between two carbonyls is around 60 ppm, and 100 ppm when in the enol form. This structure has more resemblance to the enol form with its sp2 carbon (or you can draw another resonance structure, to show the ring is truly aromatic). The signal is expected more downfield due to it being in an aromatic system (not because its between two carbonyls), but that is countered by the effective conjugation with N and O. 100 ppm seems perfectly fine.

  • @evacarloslopez4652
    @evacarloslopez4652 4 роки тому +1

    Hi Tony, I have a doubt. Why don´t the protons of NH2 correlate?

  • @anindyaganguly8385
    @anindyaganguly8385 5 років тому +2

    Hello Tony, Can you please do a NOESY analysis?

  • @dangriff12
    @dangriff12 4 роки тому

    Do one bond correlation splits mean that it is on the non aromatic part of the molecule? Basically I lost you at 10.03 there were no mentions about 1 bond correlations before that point.

  • @Sancho_top
    @Sancho_top 6 місяців тому

    Thanks you. Your video is helpful to me.

  • @elkOtooY
    @elkOtooY 4 роки тому

    What abot the case when I do not see protons from -NH2 in H NMR (proton exchange). I would assume there is no correlation in HSQC. The reason I am asking is that I want to confirm whether I have primary secondary or terciary amine in my molecule.

  • @gemma_skye
    @gemma_skye 7 років тому +3

    With the HMBC would we not also expect a strong correlation between the hydrogen A and carbon 4 as this is 3 bonds away? Why does this peak not show up?

    • @sandervandeputte2553
      @sandervandeputte2553 5 років тому

      Was thinking the same, but looks like it's still a mystery after 4 years :p

  • @suzetemariaameida2065
    @suzetemariaameida2065 2 роки тому

    hy i steel a little confuse about the carbon 2 and 3...i think the carbon 3 is not near 160 something???'🙄....Thank for your explanation.

  • @irenomegaputera3312
    @irenomegaputera3312 4 роки тому

    Very clear explanation. Thank you

  • @chramakanth9879
    @chramakanth9879 7 років тому +1

    Amazing ...well explained

  • @NCF19XX
    @NCF19XX 5 років тому

    The problem is my prof wants me to use hmbc as part of a structure elucidation and it sucks. Also she wants us to really just look at it for the quat carbons

  • @dettawavy
    @dettawavy 7 років тому +3

    woooooonnnnnndddddeerrffffffuuuuullllllll.... Wow, amazing... thanks so much from me and my friends

    • @funbaby5038
      @funbaby5038 4 роки тому

      Hey, can you please help me understand this. I have my exams in next week and im stuck in Hmqc and Hmbc

  • @scienceandtravel9625
    @scienceandtravel9625 3 роки тому

    Very good video. well done.

  • @andialexx
    @andialexx Рік тому

    This is still helpful 👏🏽

  • @scaters2010
    @scaters2010 3 роки тому

    really helpfull men, thanksss

  • @bellalilara2093
    @bellalilara2093 6 років тому

    WOW THAT HELPED ME A LOOOOT THANK UUUUU!!!

  • @pvijaykumar2360
    @pvijaykumar2360 4 роки тому

    Informative

  • @kushalavareddyyempalla4752
    @kushalavareddyyempalla4752 6 років тому

    Very helpfully..thank you very much

  • @dawsansanhtay3935
    @dawsansanhtay3935 2 роки тому

    good explanations

  • @antares1233062
    @antares1233062 7 років тому

    Well taught.
    Thanks!

  • @lisaonthemargins
    @lisaonthemargins 7 років тому +1

    Thanks buddy

  • @armandbernard4139
    @armandbernard4139 4 роки тому

    Thx a lot !

  • @johnhaggon5388
    @johnhaggon5388 6 років тому

    Thanks, Good info

  • @dawit7956
    @dawit7956 4 роки тому

    Can you please help me on my assignment

  • @matchem9178
    @matchem9178 4 роки тому

    THANK
    YOU.

  • @聂冬冬-x5d
    @聂冬冬-x5d 4 роки тому

    thank so much

  • @knutibktopplindenhoff4388
    @knutibktopplindenhoff4388 6 років тому

    God video

  • @mohamedchanfioumkouboi6443
    @mohamedchanfioumkouboi6443 6 років тому

    Thanks so much

  • @yogi5752
    @yogi5752 6 років тому

    Do more videos

  • @yogeshdubey786
    @yogeshdubey786 5 років тому

    Can you please provide this question?

  • @kusaluzumaki7730
    @kusaluzumaki7730 5 років тому

    Wow

  • @Muhammadislahmedia
    @Muhammadislahmedia 2 роки тому

    ...

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    @amparoclay 3 місяці тому

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  • @CottonBarton-x9r
    @CottonBarton-x9r 3 місяці тому

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  • @theuniverse5030
    @theuniverse5030 6 років тому

    Sir pls CI and FAB plss my exam on 15 fabuary 2019 pls