Question

. In addition to the deshielding attributes of electron withdrawing groups, π-conjugation or aromatic moieties have a substantial deshielding effect on protons in NMR. The physical reasoning for this effect has to do with the fact that the externally applied Bo magnetic field induces delocalized electron currents in the π orbitals, which in turn generate their own magnetic field that adds constructively to the strength of Bo at the location of protons in the plane of conjugation. This effect is illustrated in the figure below for a proton on a benzene ring. π Electron current field lines from Brins Bo Assign all of the peaks in the spectrum below to the specific protons on its corresponding molecule, taking careful note of which parts of the molecule participate in π-conjugation. Again, provide some justification for each of your assignments. 2.29 2.22 7.78 7.29 5. 4.1 6.18 9.36 10 8 4 2 0 PPM NH OH он

Answer 1

Hey

The peak at 9.36 and 6.18 are smaller in heights so, giving indication of hydrogen of OH and NH

the NH peak should be more downfield because of the ring current of benzene ring in vicinity of NH proton.

so, the Nh peak appears at 9.36 and The 2 OH peak appears at 6.18.also, the height of peak at 6.18 is bigger because it is due to 2 protons of OH
One point to add is they never couple with any other proton because thay are exchangable protons, they are in motion, so the neighbouring protons never feel their presence and so they are never splitted.and didnt participate in causing splitting of other protons.
The region of 7-8 is due to aromatic protons.
The benzenoid proton which is beta to C=O group is more downfield(7.8) because resonance creates a positive charge on beta carbon(making it electron deficient) so, making this proton downfield than the other.(7.29)(they split each other thus a doublet for both)

The methyl groups attached to benzene shows peak at 2.3-2.5

so, we get singlet for 6 proton at 2.2 and 2.3

The carbons to which OH are attached also bears 2 protons
since electronegative atom is attached, it withdraws electron density from it making them downfield to about4-5.

Now, how to identify which H corresponds to 4.1 and which to 5.1

The electronegativity of O>N so, C=O withdraws more than C=N so,
proton bearing C=O is more downfield 5.1 than other 4.1

Hope it helped
thanks!!!