NMR .........ALKENES

The NMR Spectra of Alkenes seem to have some very strange properties that defy explanation by the simple N+1 rule. In this article we will learn:

1. The Difference in NMR Spectra for cis and trans Alkenes.

2. What Coupling Constants (J Values) are.

3. Violations of the Simple N+1 Rule.

4. How to interpret the NMR Spectra of Alkenes.

NMR Spectra of Cis vs. Trans Alkenes

Based on what we’ve discussed about NMR spectra so far you would think that the cis and trans isomers of alkenes would have identical spectra.

In fact, based on the example above this turns out not to be true. The distance between the peaks, called the coupling constant J, is not the same for cis and trans alkenes.

Coupling Constants (J Values)

Coupling constants (J values) are defined as the interpeak distance in a particular peak in a NMR spectra.  They depend on the distance and angle between the coupled protons.

If you really want to be able to master this stuff it helps to know ballpark numbers for each of these J values in the table above.

Violations of the Simple N+1 Rule

Let’s say a proton is linked to 4 other protons. By the simple N+1 rule we would expect to see its spectra split into a quintet. However, it turns out that the simple N+1 rule only holds if the J values for each of the linked protons is the same.

Let’s see what happens if the J values aren’t the same:

The example on the right is what we’d expect to see if the J ab and J bc values were the same. However, because they are not we must apply the compound N+1 rule.

The compound N+1 rule says that when we have different coupling constants we apply the N+1 rule sequentially to each. So, for example our protons labeled as Hb are split into triplets by the Ha protons and then split again into triplets by the Hc protons, creating a triplet of triplets.

Guess where else we see the compound N+1 rule? That’s right, in cis / trans systems for alkenes.

NMR Spectra of Alkenes

Let’s analyze protons a, b and c.

Ha: Split by Hb (J geminal = 0.5-3 Hz from above) and by Hc (J trans = 12 -18 Hz from above). Because they’re different we apply the compound N+1 rule and get a doublet of doublets.

Hb: Split by Ha (J trans = 12-18 Hz) and Hc (J cis = 7-12 Hz). Of the three It looks most like a doublet of doublets since it has the largest J values (peaks are more spaced out)

Hc: Split by Ha (J geminal = 0.5-3 Hz) and Hb (J cis = 7-12 Hz). Also adoublet of doublets, and very similar to Ha except the peaks are closer since J cis < J trans.

This type of system is very common amongst alkenes, read more about it here.

Example 1

1. Calculate the DoU and Hypothesize:

DoU = 1

Most probably an alkene because of the signal at 5 ppm.

2. Write the pieces:

1.0 ppm (t, 3H) CH3

1.6 ppm (s, 3H) CH3

1.7 ppm (s, 3H) CH3

2.0 ppm (multiplet, 2H) CH2

5.3 ppm (triplet, 1H) CH

3. Structural units:

CH3-CH2-

CH3-

CH3-

CH=C

4. Subtract from Molecular Formula:

Everything’s accounted for.

5. Assemble:

Two possible ways to arrange the pieces from above -

Which one is it? The multiplet at 2 ppm tells us the answer. In the structure on the left the CH2 group should be split into a doublet of quartets. In the structure on the right it would be split into a doublet. It must be the structure on the left.

6. Verify:

The chemical shifts and splitting patterns look correct!

Example 2

1. Calculate DoU and Hypothesize:

DoU = 1

Probably an alcohol because the singlet at 3.5 ppm is most likely the OH proton.

2. Write the pieces:

3.5 ppm (s, 1H) OH

4.1 ppm (d, 2H) CH2

5.1 ppm (dd, 1H) CH

5.3 ppm (dd, 1H) CH

5.9 (multiplet, 1H) CH

3. Structural Units:

The last 3 peaks are all part of a double bond: CH=CH2

CH2

OH

4. Subtract from Molecular Formula:

We’ve accounted for everything.

5. Assemble:

6. Verify:

Looks good, the 2 Hydrogens on the end of the alkene would be a doublet of doublets and the internal hydrogen is a doublet of doublet of triplets.

Examples for Further Practice

Solutions

Example 3 -

Example 4 -

Conclusion

We’ve learned four things today:

1. The Difference in NMR Spectra for cis and trans Alkenes – Trans alkenes have larger interpeak distances because of greater J values.

2. What Coupling Constants (J Values) are – J values are the interpeak distance for a set of coupled protons and depend on the distance and angle between those protons in the molecule.

3. Violations of the Simple N+1 Rule – If a proton is coupled to 2 or more sets of protons with different J values, you have to apply the compound N+1 rule and treat each set of protons separately.

4. How to interpret the NMR Spectra of Alkenes – Alkenes generally have different J values, so we see the compound N+1 rule usually applied, creating patterns like doublets of doublets.

DACCA, BANGLADESH

Farmer Harvesting Jute, Tangail,Dhaka, Bangladesh Photographic Print

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Endo-tricyclo[6.2.1.02,7]undeca-4,9-diene-3,6-dione

Synthesis of endo-tricyclo[6.2.1.0^2,7]undeca-4,9-diene-3,6-dione (3)

To a solution containing 541 mg (5.0 mmol) in 18 mL of dry methanol at -78º C under nitrogen atmosphere, was added cyclopentadiene freshly distilled (344 mg, 5.2 mmol, in 4 mL of dry methanol) also cooled to -78ÚC. Then, the reaction mixture was allowed to reach 0 ÚC (approx. 1h). After that, the solvent was removed under reduced pressure and the product was crystallized by using hexane, yielding the yellow crystals (854 mg, 4.9 mmol, 98%). 

mp. 63-65º C. 

¹H NMR (CDCl₃, 400 MHz), δ (ppm) 1.44 (dt, 1H, J 8.4 Hz and J 1.7 Hz; 1.55 (dt, 1H, J 8.4 Hz, J 1.7 Hz), 3.20-3.26 (m, 2H), 3.53 - 3.58 (m, 2H), 6.07 (t, 2H, J 1.7 Hz), 6.58 (sl, 2H). 

¹³C NMR (100 MHz, CDCl₃), δ (ppm) 48.3, 48.7 (2 x CH and 1 x CH₂), 135.3, 142.0, 199.4.

NMR STUDY EXAMPLE 2

In this spectrum, the non-equivalent alkene hydrogens, labeled Ha and Hb, have distinctly different chemical shifts. Hb is downfield, mixed in among the absorptions for the benzene hydrogens (b), (c), and (d). [Can you draw a resonance structure that explains why Hb is downfield from Ha?] Ha is coupled to both Hb and to the aldehyde H (a), which can be seen as a doublet near 10 ppm.

Since the two hydrogens to which Ha is coupled are different, we must apply the couplings separately in order to predict the observed splitting pattern. This is shown in the sketch below:

The coupling to Hb, trans- across a double bond, is quite large, and splits the Ha absorption into a doublet. Then the smaller coupling to (a) splits each half of the doublet again, giving a quartet. The kind of diagram shown here is called a "splitting tree", and can be constructed for any case in which a group of nuclei is coupled to two distinctly different neighbors.

Let's construct the splitting tree for this one:

NMR STUDY EXAMPLE 1

The three groups, (a), (b), and (c) produce spectral features typical of propyl groups. A 3H methyl triplet for (a), split by the neighboring CH₂ (b) is straightforward. Likewise, a 2H triplet for (c), also split by (b). The (b) CH₂ actually has two different groups of neighbors.

Nonetheless, because rotation within the alkyl group averages all the coupling constants to the typical alkyl 7.5 - 8 Hz, we see a sextet for (b), as if the CH₂ and CH₃ neighbors were a single CH₅ group. Such behavior is quite common is acyclic structures.

The other feature of note is that although the benzene ring actually bears three different kinds of H (e, f, and g), the CH₂substituent on the ring is not electronegative enough to induce a significant chemical shift difference between the hydrogens near it and those further away.

Thus the five hydrogens accidentally have the same chemical shift in this spectrometer. An instrument with a more powerful magnet would spread out the spectrum still further, and show us the three expected absorptions.

DACCA, BANGLADESH

Farmer Harvesting Jute, Tangail,Dhaka, Bangladesh Photographic Print

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Cuminaldehyde or 4-isopropylbenzaldehyde, Spectral data

Cuminaldehyde or 4-isopropylbenzaldehyde,

The homotopic methyls at (a) and the unique H at (b) form an isopropyl (1-methylethyl) group, with an extremely characteristic pattern: doublet + septet, 6 : 1 integration ratio. As is often the case, we must expand the septet region to be able to see all seven peaks; if such an expansion is not available, the absorption may be referred to as a "multiplet", meaning "There's a bunch of lines there, but I can't quite count 'em!"

The pattern of the hydrogens on the benzene ring, (c) and (d), is also typical - typical for a 1,4-disubstituted benzene (para-disubstituted) with groups that differ distinctly, so that the hydrogens next to one substituent are in a significantly different environment than those next to the other. The two widely separated doublets (one neighbor) show clearly in the expansion.

The benzene chemical shift arises from an induced circulation of electrons:

The aldehyde H in R-CHO is usually found shifted downfield to the neighborhood of 9.5 ppm by an induced circulation of thep-electrons of the C=O. In Ar-CHO the additional circulation of the p-electrons of the benzene ring produces a further shift, to the vicinity of 10 ppm.

Cuminaldehyde
Synonym(s)	Cumaldehyde Cuminal Cuminic aldehyde p-Isopropylbenzaldehyde
Latest JECFA evaluation	2001 (Session 57)
Status of specification	Full
Chemical name	4-Isopropylbenzaldehye
JECFA number	868
CAS number	122-03-2
FEMA number	2341
COE number	111
FLAVIS number	-
Molecular weight	148.2
Chemical formula	C10H12O
Physical form/odour	colourless to pale yellow liquid with a strong, pungent, spicy, green, herbaceous odour
Solubility	insoluble in water; soluble in organic solvents, oils
Solubility in ethanol	miscible at room temperature
Boiling point (°C)	235-236°
Assay min %	95% (by chemical analysis)
Acid value max	5.0
Refractive index	1.527-1.534
Specific gravity	0.973-0.981
Other requirements	(equivalence factor for assay = 74.04)
ID Test	IR
Spectrum	http://www.fao.org/ag/agn/jecfa-flav/img/img/868.gif

Pentanedioic acid, diethyl ester

• Formula: C₉H₁₆O₄
• Molecular weight: 188.2209
• CAS Registry Number: 818-38-2

In this spectrum the two methyls labeled (a) are homotopic, as are the two methylenes labeled (b).

• The three Hs within each methyl are homotopic, and the two Hs within each methylene are enantiotopic.
• Hence we get single absorptions for these groups, displaying the typical ethyl pattern of methyl triplet and CH₂ quartet.
• Note the numerical values for the integration shown under the chemical shift scale, which fit the 3 : 2 ratio of CH₃ to CH₂.

The two CH₂ groups labeled (c) are homotopic (enantiotopic within the group) and therefore have the same chemical shift.

• They are split to a triplet by the CH₂ at (d).

The two Hs of (d) are enantiotopic and have the same chemical shift.

• With a total of four equivalent neighbors (both (c) groups), (d) should be a quintet. The enlarged view shows this clearly.
• The apparent small additional couplings seen are probably artifacts. The integration values fit the 2 : 1 ratio of (c) to (d).

Part of the use we make of nmr spectra is simple pattern recognition. An ethyl is always triplet + quartet, although the separation between the two groups will vary according to the rest of the structure. A chain of three CH2 with no other neighbors is always triplet + quintet, again with variation in the separation on the chemical shift axis.