IR/NMR Step-by-Step Approach

[DieTrying93]

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Hi! I'm currently studying for the MCAT, but I've taken organic chemistry two years ago. Not only that, but I had a terrible professor and I barely learned anything in class. The professor who taught it was an adjunct who was quickly (and haphazardly, if I say so myself) bc our original professor had to take a leave of absence.

Nonetheless, we over the course of the year, we did not cover IR/NMR spectroscopy. And I'm trying to teach it to myself but I'm very confused. I was wondering if anyone can give me a step-by-step approach on how they solve IR and NMR problems. I'd really appreciate it! Thanks!

 

[polyatomic]

At first, IR & NMR can seem very overwhelming with the crazy peaks, but once you get the hang of it you'll find it relatively easy. I'll give you a rundown on how to read these in general, but keep in mind that the MCAT probably won't be as in-depth.

What you need to know for IR:
1. First, you'll have to memorize some typical IR peaks, which you can find here: http://www.chem.ucla.edu/~webspectra/irtable.html.
Focus on main/important peaks like alcohols, carboxylic acids, & carbonyls (you can group ketones, esters, amides, etc into a big carbonyl group. You won't have to differentiate between those on the MCAT.)
2. Nomenclature:
Broad (b) peak means the peak looks like a big, wide U shape. Sharp (s) peak means it looks like a V or a needle. Those are the main characteristics of peaks.
Don't worry about bending or stretching. That's likely beyond the MCAT.

How to read IR:
1. Peaks below 1300 cm-1 should be ignored. This is the footprint region and is relatively unreliable
2. Look for the most common/obvious peaks:
Alcohols - always around 3300 and crazy broad. Can't miss it. Super common IR peak
Carboxylic acids - center around 3000 and even crazier broad.
Carbonyls - around 1600-1700 and always very sharp and obvious.
3. Look at the multiple choices you have. Eliminate those that didn't have the obvious peaks you spotted. Again, the MCAT won't go very in-depth into IR, so you should be good by just memorizing some big peaks. If you find a practice question, I can show you step by step how to solve it.

What you need to know for NMR:
There's two main types of NMR: Hydrogen (H) NMR and Carbon (C) NMR.

<---Ethanol

1. CNMR has single, sharp peaks that represent every carbon unless there's symmetry in the molecule. That means if two carbons are mirror images of each other in a molecule, the two peaks will show as one. In the case of ethanol, you would have two carbon peaks.
2. HNMR has splitting. This is indicative of how many hydrogens are next the the hydrogen you're looking at The formula for how many peaks it splits into is (1 + n). "n" is the number of hydrogens next to the hydrogen you're looking at.
For instance, there are 2 hydrogens on C2, which is directly adjacent to C1. This means C1 will have (1 + 2) = 3 split peaks. This makes it a "triplet."
Hydrogens are ONLY split by hydrogens on carbons. So in ethanol, C2 would NOT be split by the hydrogen on Oxygen. Therefore it would only be split by the 3 hydrogens on C1, making it (1 + 3) = 4 --> a quartet. Also, hydrogens on non-carbon atoms will not show any splitting. So the hydrogen on oxygen here will be a singlet.
Because of its splitting, HNMR is hugely useful for finding out a structure.

NMR
3. Where a peak appears along the x-axis depends on how "deshielded" that carbon or hydrogen is. Deshielded means the electrons have been stripped from hydrogens, due to the presence of an electronegative atom nearby. Below is an HNMR of ethanol. Oxygen is very electronegative, so this "deshields" C2, the carbon adjacent to it. However, it leaves C1 fairly untouched because it is much farther away from oxygen. Thus below, you see that C2 has a higher numerical shift than C1.

<---H

From here, you just need to try more and more practice problems and carefully study the solutions to understand the reasoning for problem solving.

Again -- if you have any specific examples to show me, I'd be happy to work through it step by step.

Let me know if anything's is unclear or if you have more specific questions! 🙂

 

[Czarcasm]

At first, IR & NMR can seem very overwhelming with the crazy peaks, but once you get the hang of it you'll find it relatively easy. I'll give you a rundown on how to read these in general, but keep in mind that the MCAT probably won't be as in-depth.

What you need to know for IR:
1. First, you'll have to memorize some typical IR peaks, which you can find here: http://www.chem.ucla.edu/~webspectra/irtable.html.
Focus on main/important peaks like alcohols, carboxylic acids, & carbonyls (you can group ketones, esters, amides, etc into a big carbonyl group. You won't have to differentiate between those on the MCAT.)
2. Nomenclature:
Broad (b) peak means the peak looks like a big, wide U shape. Sharp (s) peak means it looks like a V or a needle. Those are the main characteristics of peaks.
Don't worry about bending or stretching. That's likely beyond the MCAT.

How to read IR:
1. Peaks below 1300 cm-1 should be ignored. This is the footprint region and is relatively unreliable
2. Look for the most common/obvious peaks:
Alcohols - always around 3300 and crazy broad. Can't miss it. Super common IR peak
Carboxylic acids - center around 3000 and even crazier broad.
Carbonyls - around 1600-1700 and always very sharp and obvious.
3. Look at the multiple choices you have. Eliminate those that didn't have the obvious peaks you spotted. Again, the MCAT won't go very in-depth into IR, so you should be good by just memorizing some big peaks. If you find a practice question, I can show you step by step how to solve it.

What you need to know for NMR:
There's two main types of NMR: Hydrogen (H) NMR and Carbon (C) NMR.

View attachment 177890<---Ethanol

1. CNMR has single, sharp peaks that represent every carbon unless there's symmetry in the molecule. That means if two carbons are mirror images of each other in a molecule, the two peaks will show as one. In the case of ethanol, you would have two carbon peaks.
2. HNMR has splitting. This is indicative of how many hydrogens are next the the hydrogen you're looking at The formula for how many peaks it splits into is (1 + n). "n" is the number of hydrogens next to the hydrogen you're looking at.
For instance, there are 2 hydrogens on C2, which is directly adjacent to C1. This means C1 will have (1 + 2) = 3 split peaks. This makes it a "triplet."
Hydrogens are ONLY split by hydrogens on carbons. So in ethanol, C2 would NOT be split by the hydrogen on Oxygen. Therefore it would only be split by the 3 hydrogens on C1, making it (1 + 3) = 4 --> a quartet. Also, hydrogens on non-carbon atoms will not show any splitting. So the hydrogen on oxygen here will be a singlet.
Because of its splitting, HNMR is hugely useful for finding out a structure.
View attachment 177893
NMR
3. Where a peak appears along the x-axis depends on how "deshielded" that carbon or hydrogen is. Deshielded means the electrons have been stripped from hydrogens, due to the presence of an electronegative atom nearby. Below is an HNMR of ethanol. Oxygen is very electronegative, so this "deshields" C2, the carbon adjacent to it. However, it leaves C1 fairly untouched because it is much farther away from oxygen. Thus below, you see that C2 has a higher numerical shift than C1.

View attachment 177892<---H

From here, you just need to try more and more practice problems and carefully study the solutions to understand the reasoning for problem solving.

Again -- if you have any specific examples to show me, I'd be happy to work through it step by step.

Let me know if anything's is unclear or if you have more specific questions! 🙂

Excellent, excellent explanation! You summarized it so perfectly.

@OP: The only thing I would add to this is to briefly mention that in NMR, you have to be careful for Hydrogen environments. As mentioned above, the neighboring hydrogens determine the splitting pattern. So in the example, the 3 methyl hydrogens influence the splitting pattern of the neighboring hydrogens on the middle carbon (the -CH2 group). Thus we expect it to have 4 peaks, as was just explained. But in this scenario, all 3 hydrogens on -CH3 were equivalent. Why? Because the Hydrogens can rotate about the single bond and orient to every position in space -- thus they are equivalent. And usually, for most easy NMR questions, that's the case.

But keyword is usually. You may get a question really testing you. I mention this because you'll encounter a brief section about complex splitting and this is specifically due to non-equivalent hydrogen environments, (typically) due to a chiral center. Those -CH3 hydrogens will always be equivalent (atleast in every problem I ever solved). However, in the presence of a chiral center, you really want to pay attention to those -CH2 hydrogens as in some instances those hydrogens can be unequivalent due to a neighboring chiral center and therefore, the lack of symmetry in the molecule as a whole. In such a scenario, when the splitting is due to non-equivalent hydrogens you can simply refer to the splitting as a multiplet. I really won't dive into explaining this as a review book would probably explain this more clearly, but I mention it just to give you an idea and so that you're aware.

I hope that makes sense and if not, I'm sorry lol.

 

[polyatomic]

Excellent, excellent explanation! You summarized it so perfectly.

@OP: The only thing I would add to this is to briefly mention that in NMR, you have to be careful for Hydrogen environments. As mentioned above, the neighboring hydrogens determine the splitting pattern. So in the example, the 3 methyl hydrogens influence the splitting pattern of the neighboring hydrogens on the middle carbon (the -CH2 group). Thus we expect it to have 4 peaks, as was just explained. But in this scenario, all 3 hydrogens on -CH3 were equivalent. Why? Because the Hydrogens can rotate about the single bond and orient to every position in space -- thus they are equivalent. And usually, for most easy NMR questions, that's the case.

But keyword is usually. You may get a question really testing you. I mention this because you'll encounter a brief section about complex splitting and this is specifically due to non-equivalent hydrogen environments, (typically) due to a chiral center. Those -CH3 hydrogens will always be equivalent (atleast in every problem I ever solved). However, in the presence of a chiral center, you really want to pay attention to those -CH2 hydrogens as in some instances those hydrogens can be unequivalent due to a neighboring chiral center and therefore, the lack of symmetry in the molecule as a whole. In such a scenario, when the splitting is due to non-equivalent hydrogens you can simply refer to the splitting as a multiplet. I really won't dive into explaining this as a review book would probably explain this more clearly, but I mention it just to give you an idea and so that you're aware.

I hope that makes sense and if not, I'm sorry lol.

Thanks for the compliment & add on! Forgot to mention that.

 

[DieTrying93]

Thank you both so much!! Your explanations are perfect, I really appreciate your time and effort 🙂

 

[polyatomic]

Thank you both so much!! Your explanations are perfect, I really appreciate your time and effort 🙂

Finally, the spectroscopic methods class came in handy.

 

[1TB4RKSB4CK]

oohhh kill emmm
great explanation for anyone who has a tough time on this

 

[stickywebz]

hey why isnt the hydrogen attached to the oxygen more deshielded and further down stream?

I understand that C2 will be more deshielded than C1 but isnt the hydrogen much closer to the oxygen so it should have a great affect?

thanks!