1HNMR question

[greenseeking]

---

Members don't see this ad.

How many signals are observed in this 1HNMR?

I counted 12 but the correct answer is 14!!!

EDIT- Also, the answer key randomly put another Hydrogen on the "N" next to the first carbonyl group to the left. How do we know that we are supposed to randomly insert an H onto a molecule? I thought this question would be as simple as just counting the unique numbers of hydrogens that are already present in the figure.

 

[NoQuarter]

Possibly a mistake, I am counting 13.

 

[doy]

edit:

Hydrogens by the sulfur are not equal because they are not able to freely rotate. So the Hydrogen on the sulfur side will feel a different magnetic force (IE shift a tad bit downfield) than the H on the other side.

Sorry for the horrendous lack of photoshop skills. but it will do.

 

[greenseeking]

edit:

Hydrogens by the sulfur are not equal because they are not able to freely rotate. So the Hydrogen on the sulfur side will feel a different magnetic force (IE shift a tad bit downfield) than the H on the other side.

Sorry for the horrendous lack of photoshop skills. but it will do.

Thanks, this was really great! I have two questions though:
How did you know to put the extra H on the nitrogen?
And the two hydrogens on the ring near the sulfur: Isn't the ring planar? so the hydrogens would be feeling equal forces bc they are each above and below the ring.

thanks!

 

[BerkReviewTeach]

How did you know to put the extra H on the nitrogen?

The same way you knew to put two hydrogens on the methylene carbons. Convention is to not explicitly draw all of the Hs and have the viewer fill them in via the Octet rule. That N only has two bonds shown, so there must be an H making the third bond.

And the two hydrogens on the ring near the sulfur: Isn't the ring planar? so the hydrogens would be feeling equal forces bc they are each above and below the ring.

That particular carbon is sp³-hybridized and it is next to an sp³-hybridized S that is not involved in resonance, so the ring cannot be planar. And to emphasize doy's point, those two hydrogens are diastereomeric, meaning they are not equivalent. One cannot rotate to become the other and one is above the ring (pseudo-axial) and the other is below the ring (pseudo-equatorial), so they feel different local magetic fields from the molecule.

And the reality is that my explanation (using diastereomeric hydrogens) is going beyond the scope of the MCAT. What you should have experienced when solving this question was that at a cursory first glance there are 13 unique hydrogens. None of the unique ones could be equivalent to any of the other 12, so there are at least 13. This should lead you to pick choice D, as it is the only answer that fits "at least 13". If you wish, you can look closer (although you dont need to as you've already found the best answer). As doy pointed out, the two Hs on the C of the six-membered ring are not equivalent, which leads to 14.