H-nmr

[bustedgrill]

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Okay. I checked for other threads that really break H-NMR down well, but not much. Guys and gals if anyone has time could you break this down barney style? I mean from front to back, I've read my Ochem book, BB and it is still not making any sense. Thank you so much.

 

[nze82]

Okay. I checked for other threads that really break H-NMR down well, but not much. Guys and gals if anyone has time could you break this down barney style? I mean from front to back, I've read my Ochem book, BB and it is still not making any sense. Thank you so much.

It's really not that complicated.
The number of peaks always corresponds to the the number of hydrogens that are not equivalent. What do I mean by that? Look at the following example:

-The hydrogens on the two red methyl groups are magnetically equivalent (Start from the 1st methyl group and go counter clockwise. You'll see, H, CH3, H, CH3, Cl. Start from the 2nd methyl group and go counter-clockwise. You'll see the same exact substituents). So, they both represent one peak.

-The two green hydrogens are also magnetically equivalent. So, they both represent one peak.

-The hydrogens on the blue methyl group are magnetically different from those of the red methyl groups and the two green hydrogens (Look at the neighboring groups moving in clockwise and counter clockwise fashion, and you'll see that these neighboring groups are not like the neighboring groups of the blue and green hydrogens). So, these protons represent our 3rd peak.


So far, we have determined that we have 3 peaks. Now let's look at the splitting pattern:

Protons that are a maximum of 3 bonds away from one another can split each other in n+1 fashion (n = # of protons). What does this mean?

-Look at the red hydrogens. The green hydrogen is 2 bonds away, so it can split the peak representing the red hydrogens. How does it split it?
There is only 1 green hydrogen next to each of these methyl groups. So:
n + 1 = 1+1 = 2 (The peak representing the the red hydrogens will be a doublet).

-Look at the green hydrogen. The green hydrogen has 6 neighboring protons (3 belong to the red methyl group and 3 belong to the blue methyl group). So the splitting pattern is 6 + 1 = 7

-Look at the blue hydrogens. There are 2 neighboring hydrogens (The green ones). So the splitting pattern is 2 + 1 = 3

In conclusion, you'll have 3 peaks:
-One doublet
-One triplet
-And one with 7 splits

Here are couple of more examples:

Hope this helps!

 

[bustedgrill]

It's really not that complicated.
The number of peaks always corresponds to the the number of hydrogens that are not equivalent. What do I mean by that? Look at the following example:

-The hydrogens on the two red methyl groups are magnetically equivalent (Start from the 1st methyl group and go counter clockwise. You'll see, H, CH3, H, CH3, Cl. Start from the 2nd methyl group and go counter-clockwise. You'll see the same exact substituents). So, they both represent one peak.

-The two green hydrogens are also magnetically equivalent. So, they both represent one peak.

-The hydrogens on the blue methyl group are magnetically different from those of the green methyl groups and the two green hydrogens (Look at the neighboring groups moving in clockwise and counter clockwise fashion, and you'll see that these neighboring groups are not like the neighboring groups of the blue and green hydrogens). So, these protons represent our 3rd peak.


So far, we have determined that we have 3 peaks. Now let's look at the splitting pattern:

Protons that are a maximum of 3 bonds away from one another can split each other in n+1 fashion (n = # of protons). What does this mean?

-Look at the red hydrogens. The green hydrogen is 2 bonds away, so it can split the peak representing the red hydrogens. How does it split it?
There is only 1 green hydrogen next to each of these methyl groups. So:
n + 1 = 1+1 = 2 (The peak representing the the red hydrogens will be a doublet).

-Look at the green hydrogen. The green hydrogen has 6 neighboring protons (3 belong to the red methyl group and 3 belong to the blue methyl group). So the splitting pattern is 6 + 1 = 7

-Look at the blue hydrogens. There are 2 neighboring hydrogens (The green ones). So the splitting pattern is 2 + 1 = 3

In conclusion, you'll have 3 peaks:
-One doublet
-One triplet
-And one with 7 splits

Here are couple of more examples:

Hope this helps!

OMG! Thank you so much! This is the best break down on NMR that I have ever seen! After D-school you should start writing text books, because you broke this down so well. Thank you agian for taking time out of your day to help a stranger.

Best of luck
-Mike

 

[nze82]

You're very welcome!🙂

 

[whawha]

I have a question about the last molecule. How come the symmetry is not taken into account?

 

[suhshi]

for the cyclohexane molecule im pretty sure they would all display doublets right because each hydrogen has only one neighbor

if you count the number of neighbors for each hydrogen, its only neighbor is the hydrogen attached to the carbon making up the cyclohexane

correct me if im wrong

 

[Extirpator]

for the cyclohexane molecule im pretty sure they would all display doublets right because each hydrogen has only one neighbor

if you count the number of neighbors for each hydrogen, its only neighbor is the hydrogen attached to the carbon making up the cyclohexane

correct me if im wrong

I was just about to post the same thing. That first molecule is entirely off. I think you forgot to count the C-H bonds... as well as not realizing that it's not a benzene and thus there are hydrogens not pictured.

 

[276640]

Yeah, the person that posted above me is correct.

 

[whawha]

Hm, so,

For the 1st molecule,

We will have six peaks. (Doublet by the proton on the carbon that chlorine is attached to, doublet by the protons on ortho-methyl, quadruplet by the proton on ortho-carbon, triplet by protons on meta carbon, 6-splitting by the proton on para-carbon, doublet by the protons on para-methyl)

(I know it's not a benzene, but i just used o, m, p names 🙂)

I think that the second molecule is correct as it is.

For the 3rd molecule,

We will have two peaks. (one triplelet either by #1 or #3 carbon depending on which side of the symmetry you are looking at, doublet by #2 carbon)

Please correct me if im wrong!!!

Thanks!

 

[vlct0ria]

I think nze just meant to draw a benzene instead of cyclohexane...in that case all his statements would be correct

 

[whawha]

I think nze just meant to draw a benzene instead of cyclohexane...in that case all his statements would be correct

Yes, thats that I thought too. But I still wanna make sure if I am interpreting the spectrua correctly (first as drawn and the last one)

 

[vlct0ria]

Yes, thats that I thought too. But I still wanna make sure if I am interpreting the spectrua correctly (first as drawn and the last one)

For the cyclohexane...

I agree we will have 6 peaks...but wouldn't the proton attached to the ring where Cl is be a triplet (because of the 2 protons attached to the ring on either side of it)? I agree about the doublet for the protons in the red methyl groups...and would the protons attached to the ring at the red methyls be split into 7 (3 H on the methyl, 2 H attached to the ring below it, and 1 H attached to the ring at the Cl)? The green CH2 groups would be a triplet (H attached to the ring at the blue methyl plus H attached to the ring at red methyl)....and the then protons that are in the blue methyl group would just be a doublet, right?

Someone correct me if I severely messed that up!

 

[whawha]

For the cyclohexane...

I agree we will have 6 peaks...but wouldn't the proton attached to the ring where Cl is be a triplet (because of the 2 protons attached to the ring on either side of it)? I agree about the doublet for the protons in the red methyl groups...and would the protons attached to the ring at the red methyls be split into 7 (3 H on the methyl, 2 H attached to the ring below it, and 1 H attached to the ring at the Cl)? The green CH2 groups would be a triplet (H attached to the ring at the blue methyl plus H attached to the ring at red methyl)....and the then protons that are in the blue methyl group would just be a doublet, right?

Someone correct me if I severely messed that up!

Don't we need to look at only one side when we have symmetry? So, doublet instead of triplet? If you look #209 of Destroyer 2009, you will see what I mean. Please confirm this with me 🙂

 

[vlct0ria]

Don't we need to look at only one side when we have symmetry? So, doublet instead of triplet? If you look #209 of Destroyer 2009, you will see what I mean. Please confirm this with me 🙂

Okay, then the proton at the Cl would just be a doublet, and the proton attached to the ring at the blue methyl would be a sextet (3 from the blue methyl, and then only two from the green CH2 groups since they are symmetric). I think we have it sorted it out now!

 

[whawha]

Okay, then the proton at the Cl would just be a doublet, and the proton attached to the ring at the blue methyl would be a sextet (3 from the blue methyl, and then only two from the green CH2 groups since they are symmetric). I think we have it sorted it out now!

Thank you for confirming this with me 🙂

 

[Extirpator]

I think nze just meant to draw a benzene instead of cyclohexane...in that case all his statements would be correct

No, it's still be wrong even if it was a benzene. The majority of the bonds he was referring are still 4 bonds away, not 3. He just forgot to count the c-h bonds is all (if it was benzene).

 

[whawha]

No, it's still be wrong even if it was a benzene. The majority of the bonds he was referring are still 4 bonds away, not 3. He just forgot to count the c-h bonds is all (if it was benzene).

If it was a benzene, it would have been 3 singlets, right?

 

[Extirpator]

yes, with the largest peak the furthest downfield.

 

[whawha]

yes, with the largest peak the furthest downfield.

That is due to the presence of chlorine which is electronegative, right?

Thanks.

 

[bryantje]

I had like 3 questions on this on my DAT today. Nothing too alarming, but it was my "If this is on there I just gonna guess" topic for Orgo (I had one of those for each section haha)