NS #5 Chem Phys: Question 20, 22 26

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Why is an amide a deshielding group in NMR? I thought that Electron Withdrawing Groups were deshielding and Electron Donating Groups were shielding. I see from various sources that amide would deshield, but I thought that amide was an electron donating group, similar to an alkyl group. Wouldn't only an amine be desheielding?

Thank you!

Not a chemist but I'll try to explain with some detail below.

NH2 protons of primary alkyl amines typically appear as a somewhat broadened signal at δ 1-2 in CDCl3. The broadening has several sources: partially averaged coupling to neighboring protons, intermolecular exchange with other NH or OH protons, and partially coalesced coupling to the quadrupolar 14N nucleus (I = 1), which usually has a short T1. In the example below, the CH2 group bonded to amino (δ 2.82) shows little indication of coupling to the NH2 protons, so NH exchange must be rapid on the NMR time scale. The amide proton at δ 7.1 is broadened by residual coupling to 14N, not by exchange, since the N-CH2 signals are a sharp quartet from accidental equivalence of the vicinal HN-CH2 and CH2-CH2 couplings.


Amide NH signals typically appear around δ 7, as in the example below. They are generally in slow exchange with other NH and OH signals. Thus, neighboring protons will show coupling to the NH proton, as in the examples, where the CH2 bonded to the amide nitrogen is a quartet and the N-Me group is a doublet. The amide N-H protons are typically broad from poorly resolved coupling to 14N, so the coupling to neighboring protons is usually not resolved in the NH signal.


THIS is a good summary of NMR basics. Most of it you will not need to worry about but it does have a good explanation, with images, of Amide/amine shifts. Esters and Amide can act as EDG and EWG due to the overall combination of inductive (movement of e density) and resonance (effect via the pi bond rearrangement). If the arene is connected via the heteroatom, the lone pairs can be donated via resonance to the arene and hence activate it. However, if the arene is connected via the C=O group then resonance withdrawal due to the carbonyl group will deactivate the arene.

Hope this helps, good luck!

I learned that only frequency of light mattered for electromagnetic absorption and not intensity? This is something that Einstein discovered isn't it? Wikipedia says, "Usually, the absorption of waves does not depend on their intensity (linear absorption), although in certain conditions (usually, in optics), the medium changes its transparency dependently on the intensity of waves going through, and saturable absorption (or nonlinear absorption) occurs." Why does the intensity change the absorption of light in this case?

Thank you!

As your wiki source says, much of the time, intensity will not influence abruption of waves. Howver, in this passage ,we are told this is not the case and that the very effect of the technique is dependent upon " ...energy is absorbed, thermal expansion followed by thermal contraction occurs, generating pressure waves within the absorbing material."

This absorption tells us that the intensity of the beam of electromagnetic radiation is attenuated in passing through a the tissue, converting the energy of the radiation to an equivalent amount of energy which appears within the medium; the radiant energy is converted into heat or some other form of molecular energy (pressure waves). Greater intensity = greeter E transfer = more expansion and contraction leads to greater pressure waves in the procedure.

We also have a dedicated NextStep forum HERE if you have more questions on this exam!

Hope this helps, good luck!


There was a formula that showed cos(theta), and I can see that I missed using it. But on a content review perspective, wouldn't a perpendicular movement still show changes in the doppler effect? I drew a diagram to illustrate my thinking. The distance between the observer and object are still decreasing and increasing as the object moves perpendicular to the observer, even if it's at an angle. Why wouldn't the dopper effect work for a perpendicular mover?

Thank you!

The question does not ask about the doppler shift, but about the PAD frequency shift. As for your perpendicular question, I assume you mean "at a right angle to the motion of the detector"

If so, there will be a Doppler effect as anytime 2 objects are moving relative to each-other you will have a Doppler shift.

However if we mean if the source is moving perpendicular to a line connecting the source and detector, then no, there will be no shift as this means the objects are moving parallel to each other.

If it the motion is initially at a right angle to the detector, and continues in a straight line, then yes there is a Doppler effect because the object is moving away from the observer.

But to be clear on a few things:

1) If it continues in a straight line, the angle is no longer 90 degrees but is instead constantly changing.

2) When moving at a right angle to the detector, it's impossible for it to be approaching the observer, it must be moving away, so the Doppler effect will always be a redshift.

3) If, instead of continuing in a straight line, it circles the observer, the angle will stay at a constant 90 degrees and there will be NO Doppler effect.

Hope this helps, good luck!