The molecule is completely symmetrical so both NH should give same peak; they are chemically identical.

is there another proton on the carbonyl carbon? Your carbonyl carbon only has 3 bonds.

It is doubtful whether the given structure is exact (carbonyl carbon is not fully saturated). Hope there is another substitution on one or both (different substituents) carbonyl carbon(s). Have you recorded any other spectral analysis like mass spectroscopy. Unless we have differential substitution, we may receive single peak in 1H NMR for such symmetrical molecules.

@Krishna and @Greg, your observations are highly insightful. Except if the NH are in different chemical environment, they should resonate as a single peak with a relative integration of 2.

@Lubna, you may check the integration of the peak to ascertain whether the other peak is from an impurity or the expected product. Also, are the peaks broad or sharp?

I really appreciate@greg @Krishna and @nathaneal sir for giving inputs.

Regarding the structure , yes carbonyl has got substitution. One of the carbonyl is attached to hydroxy benzene and other carbonyl group is attached with thiophene molecule. I have attached an file where the exact structure of molecule is displayed. I request you to kindly go through it and provide me with your helpful suggestion.

@KRISHNA sir, I have carried out mass characterization, C13 also

@Nathanael sir, regarding the peaks of nmr I felt one peak to be bit broader and another bit sharp .following is the nmr of molecule.i have attached an file.plz go through it.

Dear Afroz,

The peaks at chemical shift values, 11.27 ppm (A) and 10.59 ppm (B) or your compound are corresponds to the amide HN< near to the acylthiophene (A) and the amide HN< near to the salicylyl function (B). There will be two reasons for this assignment.

Reason 1: Acylthiophene group is seems to be more electron deficient on comparison with the salicylyl group which results high chemical shift value for amide HN< of acylthiophene side.

Reason 2: The chelation of 'HO-' function of salicylyl group with its 'C=O' group can obviously decrease the electron deficiency of C=O group leading to resonate the HN< group at high frequency (low chemical) value.

I guess, this information may be useful in solving your problem.

Regards,

Dr. Katta Venkateswarlu.

I'm uploading this chembio draw prediction which may also be helpful. The -OH should be downfield low intensity peak around 11.9 ppm, the NH close to phenol should be the peak at 11.27 ppm and the peak at 10.59 should be the peak close to the thiophene moiety. Hope this is helpful.

@venkatesh sir, I am very thankful for your valuable information. Sir my question is should I go for D2O exchange characterization to give more stronger justification for the obtained nmr values for two NH groups . Because I am at the verge of submitting my thesis and I am worried reviewer may raise the question regarding this.

@ NATHANAEL sir, plz do upload the chembio draw prediction. I will surely refer to it.

Please find attached the chembio prediction.

Dear Afroz,

D2O exchange experiments may not give you any information related to the positions of the HN< groups. Although, HMBC and NOESY experiments are sometimes helpful but these seems to be unnecessary in your case. The answer I was provided above may be sufficient if the adjudicator(s) of your Thesis rise any question.

Here is the NMR predictions at Chemdraw after changing the orientation of -OH group and hence, it may suggested to don't depend on these estimations.

Here is the good one

Thank you @venkateswarlu & Nathanael sir for your beneficial inputs. I m getting the clarity now.

And also speaking about same environment and plane of symmetry as mentioned by @Greg sir and @Krishna sir, I assume that just one peak for NH proton will be resonated. The structure is mentioned here. As we can see both the NH has got same chemical condition.that is pyridine and carbonyl group .Plz give your inputs.

Please do not rely on prediction in ChemDraw. It is, at best, an approximation and it is almost worthless for predicting exchangeable protons. Clemens, once again, has the most reliable proposal and that is to perform an HMBC.

I agree with Clemmen's recommendation to use 2D NMR. Based on an MM simulation, there are two lowest-energy conformers: One formed an intramolecular hydrogen bond (IHB) between the phenol-OH and the nearest carbonyl. The other conformers formed an IHB with one of the nearest NH. Thus, it is possible to discriminate between the two NH by acid-base titration. Considering that aryl hydrazides are more acidic than phenol, the addition of small aliquots of a base, such as t-BuOK (not exceeding one equivalent), should selectively deprotonate one of the N protons. The removal of the NH proton associated with phenol will drastically alter the phenolic OH signal.

@Clemens, @John, @bright sir, thank u for your valuable suggestions.

I agree with John Hollerton and Clemens Anklin

Dear Dr. Lubna Afroz,

Please specify exactly:

(i) which is the molecular formula of your compound, not just the functional group of interest. For example, does it also contain a phenol type OH group?

(ii) Which is the NMR solvent used? The chemical shifts of NH groups are strongly dependent upon the hydrogen bond ability of solvent.

(iii) Which are the NMR integrals of the NH groups?

With kindest regards,

I.P. Gerothanassis

I agree with John Hollerton, Clemens Anklin, and Grant Simpson that HMBC would likely be a great way to assign these N-H peaks definitively.

Dear Lubna,

1H-13C HMBC is definitely a good idea...just a warning: most likely you will see for O=C-NH-NH-C=O the strong peak from the "left" NH to the "right" C=O and vice versa!

Best wishes

Alfred

dear Lubna Afroz,

you are quite sure that the spectrum uploaded on Dec. 23 represents the formula uploaded the same day? I would expect a significant greater number of signals. I do not know whether it is worth trying assignment of individual NH and OH protons if the remainder of the spectrum/molecule is unclear. You might also have to consider that your molecule is an amide. This class of compounds frequently exhibits double sets of signals due to restricted rotation about the C-N (amide) bonds.

Two -NH proton in NMR will give rise to single broad signal due to quadruple moment.If there exists cis-trans isomerism considering benzene ring in the central position. then two NH may give two signal at a different regions around 10 delta ppm.

dear colleague: hello

two NH in (11.27 and 10.59 ppm)respectly , its false you have note true molecullar @structure because NH in general was situed in range of ( 5 to 4.5 ).

best regards

Dear colleague,

According to the structure molecule, there is resonance accouter state and transfer between N-H and C=O so the structure will be as following (HO-C=N-N=C-OH) in the end the two peaks for (O-H) groups not to (N-H) because OH group appearance in this values (10-11 ppm) and N-H appearance in 4-5 ppm.

Regards

dear colleage : according of structure (HO-C=N-N=C-OH) you have not NH ?????????????????? but you have OH in the end and are symetric??????? You have two peaks OH equal ????

Dear Lubna Afroz your interesting technical question has been asked more than two months agod. Has your problem been solved in the meantime? To the best of my knowledge the molecule / ligand shown in your screenshot (December 23) has first been reported in the following research article:

Self-Assembly of a Nanoscopic Prism via a New Organometallic Pt3 Acceptor and Its Fluorescent Detection of Nitroaromatics

Article Self-Assembly of a Nanoscopic Prism via a New Organometallic...

Unfortunately this article has not yet been posted as public full text on RG. However, the first author has an RG profile. Thus you can easily contact him directly via RG at https://www.researchgate.net/profile/Sushobhan-Ghosh-2 and request the full text of the paper from him. Of course you can also discuss your NMR data directly with him.

Good luck with your work and best wishes, Frank Edelmann

Respected Frank T. Edelmann Sir, Thanks in tons for suggesting me the research paper which has been a great help and definitely will discuss the related spectra with the main author.

Respected Abdullah Jawad Kadhim sir, Madhukar Baburao Deshmukh sir, Alfred Ross sir, thank you all for your valuable inputs.

Dear Wolfgang Baumann, Spectrum uploaded on 23rd December doesnt represents the formula mentioned on the very same day. If u go through the previous discussions there you will find the thiophene based structure related to the spectrum. Thank you.

Dear Abdelkader Rahmouni sir, there are no hydroxyl groups on either end of the molecule. I have uploaded the structure sir. If you go through the previous discussions, u will find the structure along with the NMR spectrum. Thank you sir.

dear lubna ,you have two hydroxyl group in the end (HO-C=N-N=C-OH)?????????????????? you have no (NH) ???????????????? So, (11.25 and 10.59 ppm) its for (OH) group ?????????? Its clear and simple .

best regards