OC NMR Question

[Contach]

---

Members don't see this ad.

From those jerks at Kaplan:

9. Which of the following compounds could produce the spectrum presented below?

a) Acetaldehyde
b) Propanal
c) 1,2-butadiene
d) Ethanol

How do you do this?

 

[Lazerous]

Is it B?

 

[AndyK]

Propanal's the only one it can be, but the far left peak should be a doublet.

The three carbons are inequivalent, so there will be three peaks. C1 has one H -> doublet peak downfield due to proximity to Oxygen. C2 has two H's -> triplet peak. C3 has three H's -> quadruplet peak farthest upfield.

 

[Contach]

Is it B?

Indeed it is! can you explain the peaks and coupling patterns to me for b? thanks

 

[mistero]

i do not understand how it could possibly be B.

In b, the furthest one should be a triplet. the second closest should be a quartet and teh closes should be a triplet.

 

[artanis]

Aside from just structure elucidation, you have to know characteristics chemical shifts. Based on the data, one should gather this. A triplet beside a quartet is usually something like this:

-CH2CH3

The 2H experience (3+1) = 4 quartet
The 3H experience (2 +1) = 3 triplet

now this narrows it down to either Propanal, or Ethanol.
As we can see the last chemical shift is a Singlet. This is both possible with
-O-H, as well as COOH, with the H bonded to the C.
However, for alcohols, the chemical shift of the singlet H is around 3-5delta, but for an aldehyde, the singlet shift is at 9-10 delta. This is VERY characteristic of an aldehyde. That's how I did this question.

 

[Contach]

i agree with mistero, which is why I don't see how it matches up to Propanol.

I understand splitting.

Artanis I challenge you this:the CH2 experiences a quartet splitting because of the 3H on the methyl..Why doesn't the H on the C1 of the aldehyde also split the CH2? Therefore shouldnt it be a double of a quartet?

Also, ARtanis, the diagram (like mistero and AndyK stated) should have the aldehyde C1 H split into a triplet, not a singlet. This is due to the H's of the CH2.

I understand that O-H can dissociate and reassociate liberally and is therefore not be involved in splitting, but the H of an aldehyde is not acidic and would not do this....


The 3-5 ppm for O-H and 9-10 ppm for Aldehyde helps. Thanks - I will remember this.

 

[AndyK]

"Propanal has three types of H : the CH2 unit, the CH3 and the aldehyde H (H-C=O).
This means we see three sets of peaks.
The most distinctive peak is the aldehyde H. It is very deshielded (higher d / ppm, 9.79 ppm) due to its proximity to the p bond and the electronegative O atom. The coupling constant for aliphatic aldehydes is usually quite small (<3Hz). This often means they appear not to couple to their neighbours unless a higher resolution spectra is studied. So we will ignore that coupling here and assume it is a singlet.
The CH2 unit attached to the C=O is more deshielded (2.46 ppm) than the CH3 unit attached to the CH2 unit (1.11 ppm).
The CH2 unit has 3 H neighbours, so the n+1 rule means that we see the 4 lines of a quartet.
The CH3 unit has 2 H neighbours, so the n+1 rule means we see the 3 lines of a triplet.
Integrals have a 1:2:3 ratio"

http://www.chem.ucalgary.ca/courses/351/nmrsketch/13/nmr.html (click on "give up" to see the solution)

Great resource for NMR problems too...

 

[Contach]

The coupling constant for aliphatic aldehydes is usually quite small (<3Hz). This often means they appear not to couple to their neighbours unless a higher resolution spectra is studied. So we will ignore that coupling here and assume it is a singlet.

That's crazy. News to me.
Man. NMR is another achilles heel for me.

thx

 

[Tina324]

would someone be so kind to explain to me this problem from scratch? I have a prob with NMR because we only spent a day on it in class.

PLEASEEEEE😳 exam in less than a week!

 

[alanan84]

would someone be so kind to explain to me this problem from scratch? I have a prob with NMR because we only spent a day on it in class.

PLEASEEEEE😳 exam in less than a week!

Do you have your O chem book? It might be best for you to go over that chapter to nail out the basics. Real quickly, the number of peaks gives you the number of nonequivalent protons. Here's a good thread to help you be able to determine between equivalent and nonequivalent:
http://forums.studentdoctor.net/showthread.php?t=545830

The splitting of the peaks tells you how many protons are on adjacent carbons. (n+1)

Where the peaks show up tells you how deshielded they are.

 

[tranv117]

Just try to keep it simple and the question is pretty easy.

Propanal is the best choice here. As explained earlier, -CH2CH3 typically results in a quartet and triplet. Use the n+1 rule for H's within 3 bonds of each other. The methyl group Hs have the same spin, so consider those 3 as n=1. They'll couple to the other 2 Hs giving you a triplet. Same reasoning for the reverse, the other 2 Hs have the same spin so n=1. 1+3=4 so that gives you a quartet.

Now in propanal, you have 1 H left (O=C-H). Hs attached to carbonyls will be most deshielded and furthest downfield therefore you see that single peak. And YES it should only be a single peak. Count the bonds between the carbonyl H and the CH2 Hs. More than 3 bonds, so they won't couple.

Now for alcohol (O-H). This typically results in peak around 5ppm. Not as deshielded as O=C-H. Hope that helps. Really, try to keep it simple and HNMR, CNMR is pretty easy.

 

[mistero]

"Propanal has three types of H : the CH2 unit, the CH3 and the aldehyde H (H-C=O).
This means we see three sets of peaks.
The most distinctive peak is the aldehyde H. It is very deshielded (higher d / ppm, 9.79 ppm) due to its proximity to the p bond and the electronegative O atom. The coupling constant for aliphatic aldehydes is usually quite small (<3Hz). This often means they appear not to couple to their neighbours unless a higher resolution spectra is studied. So we will ignore that coupling here and assume it is a singlet.
The CH2 unit attached to the C=O is more deshielded (2.46 ppm) than the CH3 unit attached to the CH2 unit (1.11 ppm).
The CH2 unit has 3 H neighbours, so the n+1 rule means that we see the 4 lines of a quartet.
The CH3 unit has 2 H neighbours, so the n+1 rule means we see the 3 lines of a triplet.
Integrals have a 1:2:3 ratio"

http://www.chem.ucalgary.ca/courses/351/nmrsketch/13/nmr.html (click on "give up" to see the solution)

Great resource for NMR problems too...

Why isnt the quartet further downfield than the triplet, it is closer to the very electronegative O?

 

[AndyK]

Why isnt the quartet further downfield than the triplet, it is closer to the very electronegative O?

You're right, it should be... I guess they just goofed when making the spectrum.