Textbook: Loudon 5th Ed. (2009)

Chapter 13: Nuclear Magnetic Resonance Spectroscopy

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Individual Problems

Individual Problems

Problem # 660

Using your knowledge of ¹H NMR, predict the NMR spectrum for the compound below. (draw out the spectrum you would expect to see). Be sure to include:

peak integrations
peak multiplicities
chemical shifts (approximate)

Solution

There are four types of protons (hydrogens) in the compound below, so we would expect to see four peaks (A, B, C, and D).

A is not adjacent to any other types of protons, so its multiplicity is 1, also called singlet (s). ( n=0, so n + 1 = 0 + 1 = 1). A also is a CH3, so it will have an integration of 3. Finally, A's proton are on a carbon adjacent to an oxygen, so its chemical shit will be about ~4 ppm.

B is a CH2 so its integration is 2. It's adjacent to a 2 hydrogens (a CH₂) so its multiplicty is 3, also called a triplet (t). (n=2, so n + 1 = 3). B is also adjacent to oxygen, so it will be close to ~4 ppm, but a little further downfield than A because it's more substituted than A.

C is also a CH2 so its integration is 2. It's also adjacent to 2 hydrogens so it's a triplet (n+1 rule). C's hydrogens are on a carbon adjacent to a carbonyl (C=O), so its chemical shift will be around ~2 ppm.

D is an aldehyde proton, which is always a singlet with near ~10 ppm. It's only one proton, so its integration is 1.

Note: notice that the aldehyde proton doesn't couple to another other protons! So even through C is surrounded by a CH₂ on its left and an aldehyde proton on its right, when determining multiplicity we only count the CH₂, and not the aldehyde proton.

Problem Details

MendelSet practice problem # 660 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): proton NMR

Problem # 661

The proton NMR of cyclohexane gives only one peak when the NMR is run at room temperature.

But when the temperature is lowered to -100 ºC the proton NMR spectrum shows two peaks. Explain.

Solution

Every proton on cyclohexane appears identical, but remember than cyclohexane is usually in a chair form, so there are really two types of protons: axial and equatorial.

In theory each of these two protons should give its own NMR peak. But at room temperature the molecule is undergoing chair flip so rapidly the two peaks converge into one.

But at low temperature the rate of chair flip slows down enough that two distinct peaks emerge.

Problem Details

MendelSet practice problem # 661 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): cyclohexane chair forms, proton NMR

Problem # 662

The ¹H and ¹³C NMR spectra of a compound with chemical formula C₁₀H₁₄O are shown below. The compound's IR spectrum shows a broad peak at 3,300 cm^-1. Determine the structure of this compound.

Solution

Let's go through the steps you should take to solve any NMR structure elucidation problem.

1. Are there any hints?

Yes. A broad IR peak around 3,300 cm^-1 tells you this compound contains an alcohol.

2. How many IHD are there?

(indices of hydrogen deficiency are also called degrees of unsaturation or double bond equivalents, depending on the textbook.)

The formula is C₁₀H₁₄O. That's the same as C₁₀H₁₄ (oxygens don't change IHD count). A fully saturated compound has formula C_nH_2n+2, so a C₁₀ molecule should have 22 hydrogens (2 x 10 + 2). The difference between C₁₀H₂₂ and C₁₀H₁₄ is 8 hydrogens, which corresponds to 4 IHD.

3. Draw some C₁₀H₁₄O structures with 4 IHD and eliminate, learn, repeat.

This isn't shown on the image below, but if you don't have an idea of the structure of the molecule, just start drawing structures! Never look at a blank page- just start drawing structures with the correct number of IHD (in this case, 4 IHD).

The peaks around ~7 ppm on the ¹H NMR tell you its probably aromatic (also, benzene has 4 IHD) so that's a good place to start. Draw a few benzene candidate structures with formula C₁₀H₁₄O, but then eliminate structures that don't fit the data. How? I like to go through this check list:

Eliminate by number of signals: Do the candidate structures you drew give the proper number of ¹H and ¹³C NMR signals? If not, eliminate!
Eliminate by multiplicity: Do your structures have the correct ¹H NMR multiplicities? If not, eliminate!
Eliminate by integration: Do your structures have the correct ¹H NMR integrations?If not, eliminate!
Eliminate by chemical shifts: Would the structures you drew have chemical shifts that are about the same as the chemical shifts in the ¹H spectrum given in the problem?

That's the order in which I usually eliminate candidate structures- fastest method (determining the number of expected NMR signals) to slowest method (predicting chemical shifts).

Some things we know so far about this molecule:
Form the IR we know It contains an alcohol.
The gigantic singlet with integration of 9 screams tert-butyl
The total integration of the aromatic region (~7 ppm) is 4, which means this molecule is disubstituted benzene. They're both doublets with integration of 2 which points to a para-substitution pattern.

As you elminate incorrect structures you will learn what fits the data, and be able to draw better candidate structures. Then you can repeat this process. Once you have a structure that passes these four problems, you probably have the correct structure.

Notice how we didn't even really need the ¹³C NMR to answer this problem. The ¹H NMR is usually enough.

I can't stress how important it is to just draw something! Even if you have no idea what the answer might be, don't even leave a blank page for an NMR problem. Just starting drawing out structures with the proper formula and IHD count!

Problem Details

MendelSet practice problem # 662 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): proton NMR, carbon NMR

Problem # 663

The 1H and 13C NMR spectra of a compound with chemical formula C₄H₆O₂ are shown below. The compound's IR spectrum shows a sharp peak at 1,700 cm^-1. Determine the structure of this compound.

Solution

For a detailed explanation of the general strategy for solving NMR structure elucidation problems, see problem 662.

Here are the steps I would take to solve this problem:

1. Are there any any hints?

Yes. The sharp IR peak at 1,700 cm^-1 tells you this molecule contains a carbonyl (C=O).

2. How many IHD are there? (also known as DBE or degrees of unsaturation)

C₄H₆O₂ is the same as C₄H₆ (oxygens can be ignored) which if fully saturated would be C₄H₁₀ (from C_nH2_n+2).

So this compound is missing 4 H's (C₄H₁₀ - C₄H₆), which corresponds to 2 IHD.

3. Draw some C₄H₆O₂ structures with 2 IHD and eliminate, learn, repeat.

Some things we know so far:

From the IR we know this molecule must have a carbonyl group. This "uses" 1C, 1O, and 1 IHD so we have 3C, 1O and 1 IHD remaining to work with).
What is the the remaining 1 IHD? It's probably not another carbonyl since the ¹³C NMR shows only one carbonyl peak (~170 ppm). It's probably not an alkene since we don't see any vinyl hydrogen peaks on the ¹H NMR (~5-6 ppm). That leaves a ring.
Another clue that this molecule contains a ring is that we only see CH₂'s in the ¹H NMR spectrum (all peaks have an integration of 2).
The carbonyl peak on the ¹³C NMR spectrum is around ~170 ppm, which means it's an oxidation state III carbonyl (ester) rather than an oxidation state II carbonyl (aldehyde or ketone).

Draw a few structures based on these clues and you'll arrive at the correct answer.

Problem Details

MendelSet practice problem # 663 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): proton NMR, carbon NMR

Problem # 730

N,N-dimethylformamide (DMF) is shown below. Based on its structure, you might expect to see only one -CH₃ signal in the ¹H NMR spectrum. But instead DMF shows two different -CH₃ signals. Explain.

Solution

DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its ¹H NMR should only show one methyl signal (singlet, 6H).

So why is that the real life the ¹H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).

Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.

For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.

But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.

In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).

Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the ¹H NMR.

Problem Details

MendelSet practice problem # 730 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amino Acids, Peptides, and Proteins

Question Type: Concept (explain why this is so..)

Keyword(s): proton NMR, amides

Problem # 665

The ¹H and ¹³C NMR spectra of an unknown compound are shown below. The compound's mass spectrum shows a molecular ion with m/z ratio of 86. Determine the structure of this compound.

Solution

This is a difficult problem because we are never given the compound's molecular formula, only its mass (the molecular ion is the mass of the compound).

So first, let's make a guess of this compound's molecular formula.

Do we have any clues about what atoms are present in this molecule? Yes. The 13C NMR peak at ~210 ppm indicates a carbonyl (specifically an aldehyde or ketone). So there must be at least one oxygen (and 1 IHD.)

So let's do some math:

1O is 16. Mass is 86. 86-16 - 70. That's the total mass of C's and H's we have to work with.

How many C's can we "fit" in 70? 6 C's would be 72, so let's try 5 C's.

5 C's is 60, so we need 10 H's to make 70.

So a formula of C₅H₁₀O has the correct mass. It also has the correct number of IHD (1). So let's go with it.

Now we follow the series of steps laid out in problem 662:

1. Are there any hints?

We sort of did this step already, but yes- this molecule contains an aldehyde or ketone. Since we don't see a singlet at ~10 ppm on the 1H NMR, it can't be an aldehyde. So it must be a ketone.

2. How many IHD are there? (also known as DBE or degrees of unsaturation).

C₅H₁₀O is the same as C₅H₁₀ (ignore oxygen). C5 would be C₅H₁₂ if fully saturated (because of C_nH_2n+2), so this molecule is missing 2 hydrogens, which corresponds to 1 IHD.

3. Draw some C₅H₁₀O structures with 1 IHD and eliminate, learn, repeat.

Some things we know from the NMR spectra:

There's a doublet with an integration of 6, and a multiplet with an integration of 1. This screams isopropyl group.
That singlet with an integration of 3 is probably a methyl group.
As mentioned above, since we don't see the aldehyde proton anywhere, the carbonyl must be a ketone.

Now draw a few candidate structures based on these clues and eliminate those structures that don't fit the data. If anything doesn't fit, you must elminate the entire structure.

These (no molecular formula, only MS data) are the hardest kinds of NMR problems you'll get in sophomore organic chemistry, so don't worry if you find it challenging- you're supposed to.

Problem Details

MendelSet practice problem # 665 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): mass spec, proton NMR, carbon NMR

Problem # 666

The ¹H and ¹³C NMR spectra of an unknown compound are shown below. The compound's mass spectrum shows a molecular ion with m/z ratio of 122. Determine the structure of this compound.

Solution

This is similar to problem 665- a very difficult NMR problem because we're not given the unknown's molecular formula, only its molecular mass (122 from the molecular ion peak on the mass spec).

So the first thing we have to do is guess the molecular formula based on this mass. Here's my thought process:

There are signals in the aromatic region on the ¹H NMR (~7 ppm). So we probably have at least 6 carbon's with at least 4 IHD's (a benzene ring has 4 IHD).
There are two other signals on the ¹H NMR, so there are probably at least 2 more carbons, for 8 carbons in total.
8 carbons with 4 IHD would give a formula of C₈H₁₀ (C₈H₁₈ from C_nH_2n+2 minus 8 H's for the 4 IHDs). which only gives a mass of 106. But we need a mass of 122- we're 16 mass units short. That's exactly one oxygen!
Based on this, C₈H₁₀O is a good guess for the formula.

Now we can follow the steps laid out in problem 662:

1. Are there any hints? & 2. How many IHD are there?

Mass spec NMR problems require us to have done these two steps by now.

3. Draw some C₈H₁₀O structures with 4 IHD and eliminate, learn, repeat.

Some things we know form the NMR spectra:

Because there are only four protons in the aromatic region (total integration of 4), the benzene ring is disubstituted. The two peaks in the aromatic region (~7 ppm) are doublets with integrations of 2, so this ring is probably para substituted.
The benzene ring "uses" 6 carbons which leaves 2 carbons for the other peaks. They're probably both methyls because they both have integrations of 3. They're also both singlets, so there's not adjacent to any other carbons that have a hydrogen. The peak that's further downfield (¹H ~3.8 ppm) is probably near the oxygen.

Draw a few structures based on these clues, and eventually you will come to the correct structure.

Problem Details

MendelSet practice problem # 666 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): mass spec, proton NMR, carbon NMR

Problem # 672

Compound A has molecular formula C₆H₁₂O and shows a sharp peak at 1,710 cm^-1 in its IR spectrum.

Treatment with 1 equivalent of phenyl Grignard yields compound B, which has formula C₁₂H₁₈O and whose IR shows a broad peak at 3,350 cm^-1.

Compound B's ^₁H NMR spectrum is shown below. Determine the structures of compounds A and B.

Solution

Let's use steps similar those outlined in problem 662 to solve this NMR structure elucidation problem.

1. Are there any hints?

Yes, from the IR peaks. The starting material is a carbonyl (sharp IR peak at ~1,700 cm^-1) and the product is an alcohol (broad IR peak ~3,300 cm^-1).

2. How many IHD are in each compound? (also known as degrees of unsaturation or DBE).

C₆H₁₂O is the same as C₆H₁₂ (ignore oxygen) which should be C₆H₁₄ if fully saturated (C_nH_2n+2). It's missing 2 hydrogens so C₆H₁₂O has 1 IHD. This 1 IHD must be the carbonyl.

C12H18O is the smae as C12H18 which should be C12H26 if fully saturated. It's missing 8 hydrogens so it has 4 IHD. The benzene from the Grignard reagent must account for all 4 IHD.

3. Draw some C₆H₁₂O and C₁₂H₁₈O structures and elminate those that don't fit the data, then learn and repeat.

Things we know from the problem and NMR:

The starting material is a carbonyl and we're adding a Grignard, so we expect the product to be an alcohol. The NMR shows a peak that disappears with D₂O addition, which also confirms that the product is an alcohol.
The product must have a benzene ring because the reagent was phenyl Grignard. That accounts for the aromatic signals (~7 ppm) and the 4 IHD in the product.
the NMR shows a doublet with an integration of 6 and a multiplet with an integration of 1. This is the splitting pattern on an isopropyl group.
The NMR also shows a quartet with an integration of 2 and a triplet with an integration of 3. This is the splitting pattern of an ethyl group.

Once you have a few clues from the NMR, start drawing structures! And then elminiate those that do not fit the data (too many signals, wrong multiplicity/integrations etc.).

Problem Details

MendelSet practice problem # 672 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Grignard and Masked Carbanions (alcohols from carbonyls)

Question Type: Concept (explain why this is so..)

Keyword(s): Grignard, proton NMR

Problem # 679

Compound A (C₅H₁₂O) is oxidized using aqueous chromium (Jones reagent) to compound B (C₅H₁₀O₂), which is then treated with methanol under acidic conditions to yield compound C (C₆H₁₂O₂) and water.

The ¹H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.

Solution

Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.

1. Are there any hints?

Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.

Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.

2. How many IHD are there?

Compound A: C₅H₁₂O = C₅H₁₂ should be C₅H₁₂ (C_nH_2n+2) so 0 IHD.

Compound B: C₅H₁₀O₂ = C₅H₁₀ should be C₅H₁₂. Missing 2H, so 1 IHD.

Compound C: C₆H₁₂O₂ = C₆H₁₂ should be C₆H₁₄. Missing 2H, so 1 IHD.

These IHD counts fit our assumptions from part 1).

3. Draw some structures and eliminate, learn, repeat.

Some clues from the NMR:

The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
We know from before we have one IHD, and it's probably an ester.

So start drawing structures and eliminate those that don't fit the data!