[Exproxsul]

K here we go, i redid the integration calculations. benzene ring has 5 protons,  the quintet at 2.5-3 has 1 and the doublet at 1 ppm has 6 protons.
The nmr spec is attached below.

The aromatic region has 5 protons, it does not mean that the phenyl ring has 5 attached to it, look at the symmetrical pattern, I suggested a 1,4 disubstituted phenyl ring.

i tried out all different 1,4 substituents , did not work out.

[discodermolide]

K here we go, i redid the integration calculations. benzene ring has 5 protons,  the quintet at 2.5-3 has 1 and the doublet at 1 ppm has 6 protons.
The nmr spec is attached below.

The aromatic region has 5 protons, it does not mean that the phenyl ring has 5 attached to it, look at the symmetrical pattern, I suggested a 1,4 disubstituted phenyl ring.

i tried out all different 1,4 substituents , did not work out.

I simulated the NMR of the compound I think it is, the shifts differ slightly but don't worry about that, look at the patterns.

[Exproxsul]

4-(propan-2-yl)phenol im assuming that small peak is the OH?

[discodermolide]

4-(propan-2-yl)phenol im assuming that small peak is the OH?

That's what I think it is, and the small peak is the OH, in your actual spectrum it is under the aromatic signals.
If you did an exchange with D₂O the OH would become OD and the integration would change from 5 to 4.
Or you could expand the aromatic portion of the spectrum and then you would see it a bit clearer.

[Exproxsul]

4-(propan-2-yl)phenol im assuming that small peak is the OH?

That's what I think it is, and the small peak is the OH, in your actual spectrum it is under the aromatic signals.
If you did an exchange with D₂O the OH would become OD and the integration would change from 5 to 4.
Or you could expand the aromatic portion of the spectrum and then you would see it a bit clearer.

Thank you very much for all the time and help you have given me. I really appreciate it.

[discodermolide]

4-(propan-2-yl)phenol im assuming that small peak is the OH?

That's what I think it is, and the small peak is the OH, in your actual spectrum it is under the aromatic signals.
If you did an exchange with D₂O the OH would become OD and the integration would change from 5 to 4.
Or you could expand the aromatic portion of the spectrum and then you would see it a bit clearer.

Thank you very much for all the time and help you have given me. I really appreciate it.

No problem