CNMR Spectrometry in Organic Chemistry

[@nikhilsrivastav1537]

Hello sir. Since a and e both contains CH3 group but u have just shown the e carbon peak only DEPTmode . Could you tell the reason.

[@AttachMatthewAkunna]

Thank you for this video,it has helped to understand much about NMR especially in my country where learning difficulty

[@shakalpb1164]

Hey, thank you for this video, It helped me understanding the principles of C-NMR.

[@hirakhan5378]

For the DEPT mode, why is there no spike for the "a" carbon when looking for CH3? Why does it only detect the "e" carbon?

[@chabarek1]

Any answer for this question Please?

[@RepublikSivizien]

because it detects the "b", which actuall dept would not detect. Sorry, but this is crap, better look here: en.wikipedia.org/wiki/Carbon-13_nuclear_magnetic_resonance#Distortionless_enhancement_by_polarization_transfer_spectra

[@Kris_Lighthawk]

There should indeed have been a spike for a, they just made a error in the video and forgot it. This is also clear if you look at the next video where there is indeed a spike for a...

[@TalhaKhan-nr8oj]

@@Kris_Lighthawk thank you very much :)

[@sunilsahoo3829]

in DEPT mode what anout a?

[@humairak743]

I like how you refer to the carbon as a “he” 😂😂

[@amacgregor92]

Can't assume a carbon's gender these days

[@mohamadhanan5706]

LOL i was looking for this comment

[@Canaanabolaanan]

Your DEPT CH3 carbon graph omitted Carbon A. (11:22)

[@Dorito8111]

I noticed the same thing

[@jcmick8430]

can DEPT show splitting at the same time or should I just pull up that readout at the same time?

[@francislao6770]

Can I ask sir why is the carbon in the carbonyl group without attached hydrogen shows less intense signal compared to carbons in a carbonyl group with hydrogen attached? Thank you!

[@dlvivlviv]

Unprotonated carbons receive very little NOE, and their signals are always weak, only about 10% as strong as signals from protonated carbons.

[@sadafkabir4320]

As the shifting occurs like 3>2>1 then how come d is more shifted than a?

[@aroojshahid5658]

forkan saroar because d is further away from electronegative group as compared to that of a which is near

[@Canaanabolaanan]

Proximity to Oxygen dominates. Then if a tertiary vs a primary hydrogen were equidistant from oxygen, the tertiary H would be further shifted.

[@rupalisarde5720]

sir ,how one can predict 13 c nmr decoupled spectra in complex example.plz post video on same topic by using complex example thanku

[@kennedywilliamsdrums]

This was an awesome video.

[@summiakalsoom6861]

As "e" Carbon has more hydrogen,then why it's intensity is low

[@adamvoigt14]

Hey man! Thanks for the lecture! Also, you kind of sound like John Malkovich!

[@amnayousf8710]

Need furthur videos by you on C-13 NMR

[@christophervillota5188]

thanks for the vid clear and simple

[@ran9628]

you saved my Life!

[@tahakhan-we7ix]

u r best teacher thankuu so much

[@scitubechannel5302]

I am learning here..

[@Danielwceccon]

Im a chemistry student. My teacher said the carbon with 2 hidrogen is closer to zero and the CH3 and she is doctor in espectroscopy.

[@Knowbee]

Hi Valentim. Thanks for the comment. However, your teacher is incorrect. The CH3 carbon IS closer to zero. More specifically the CH3 carbon peaks at 30.32ppm and the CH2 carbon peaks at 52.80. Google: chemicalbook cnmr 4-methyl-2-pentanone to see the actual spectrum. But feel free to let me know if you see any mistakes in my lectures!

[@alimohammedzaki1830]

Thank you 😊

[@ladylay6232]

Thank you so much .

[@emilyh3644]

Thank you!

[@Lovejassar]

what about a hydrogen in DEPT mode

[@Knowbee]

DEPT is strictly an CNMR thing. At least as taught in a typical Organic Chemistry Class.

[@ahmeddsaad8357]

wonderful

[@nawafalhosine2957]

Thank u

[@Chem-of9mi]

thanks

[@Knowbee]

you're very welcome!

[@abdulqadoosqaisrani1083]

level

[@syuhadazulkiflee9248]

Thank you

[@nourahd6994]

Thanks