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.
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.
MendelSet practice problem # 661 submitted by Matt on July 17, 2011.
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.
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 CnH2n+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!
MendelSet practice problem # 662 submitted by Matt on July 17, 2011.
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.
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 CnH2n+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.
MendelSet practice problem # 663 submitted by Matt on July 17, 2011.
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.
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.
MendelSet practice problem # 730 submitted by Matt on July 24, 2011.
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.
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 CnH2n+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.
MendelSet practice problem # 665 submitted by Matt on July 18, 2011.
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.
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 CnH2n+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.
MendelSet practice problem # 666 submitted by Matt on July 18, 2011.
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.
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 (CnH2n+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.).
MendelSet practice problem # 672 submitted by Matt on July 18, 2011.
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.
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 (CnH2n+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!
MendelSet practice problem # 679 submitted by Matt on July 19, 2011.