How to calculate coupling constants

[@JerryXGamer]

youtube videos like this are saving me right now. 10x better and more clear than our professor that presented this

[@lindsayharding8930]

Thanks for the feedback, I’m really glad you found it helpful!

[@abdelhamedmohamed2969]

Thank you so much, looks so simple but it is not mentioned often.

[@AlphaNumeric123]

Really clear video, very practical, thank you

[@bahmanramazani1099]

Thank you so much for your time and valuable video, I really appreciate it.

[@eddielookingbeautifulbravo8]

Best vid ever. Thanks for help.

[@katherineworks3233]

If only my lab instructors could explain these concepts even half as clearly as you.

[@AnalyticalScienceTutor]

Thanks for the feedback, glad I could help!

[@leonardomonteiro2187]

Thank you very much. With this video, I understand...

[@lydiac.pascual8767]

Thank you so much! Very clear explanation.

[@whatsupdude4813]

Great explanation,plz keep it up!

[@annamayinzenhofer9331]

great video! two questions: 1: how do you get 7.2 Hz from plugging in 500 MHz at 3:41 .. wouldn't it be 7.1 * 10^6? Hz ? 2: I thought the coupling constant was supposed to be independent from the external magnetic field? why do we multiply it by 500 then?

[@AnalyticalScienceTutor]

Hi, the ppm cancel out the MHz, as you have 10^-6 x 10^6. So we just omit both from the calculation for shorthand. The coupling constant in Hz IS independent of the applied field, but because we are starting our calculation with ppm, which are dependent on field, then we need to take the field into account. 1 ppm on a 500 MHz spectrometer is a different number of Hz (500) from what it would be on a 400 MHz instrument (400 Hz) so we need to take this into account in the conversion from ppm to Hz.

[@stwjeongguk3171]

i was confused at that too. thank youuuuuu

[@yohonepaligyan2014]

To calculate the distance you have said at 1:16 that whether we calculate from left hand peak or right hand peak we should get the same answer but we dont. Even in my work that I have to submit I am not getting same. Is this normal in HNMR?

[@jaymysupj4384]

Good. Thank you. *-*

[@ShumetAdugna]

it is good

[@sultanaafroza7387]

Thank u so much Mam

[@altafshaikh3280]

very usefull

[@devadasuchapala741]

Thanks a lot

[@SavedSis]

Can you do this with carbon nmr?

[@AnalyticalScienceTutor]

Hi, carbon-13 NMR spectra are almost always run without coupling (it is removed in the experiment) as it tends to make the signals overlap (and therefore it gets really difficult to see what's going on). We would normally use 2-dimensional NMR methods to determine carbon coupling constants.

[@renubalot3170]

Txs ma'am

[@elaissasanjose5668]

So if I get a lower Hz value, thats a cis?

[@elaissasanjose5668]

Thank you 😊

[@shadowknight743]

Np

[@ganeshgaur2002]

thanks a lot