Elucidation of Unknown Compound. The molecule that is the solution to this problem has a boiling...


Elucidation of Unknown Compound. The molecule that is the solution to this problem has a boiling point of 84 degrees C and is a commonly employed local anesthetic. It was initially obtained by hot water extraction of the leaves and female flowers of the willow (Salix).

Use the following information to deduce the structure of the unknown terrestrial natural product. Interpret each step (i- iv) of your elucidation for credit.

i) The mass spectrum, M.S. of this unknown material shows a parent ion {eq}(M^+) {/eq} at m/z = 124 and a base peak at m/z = 123.

ii) It's infrared, IR spectrum exhibits a strong, broad absorption centered at 3200 {eq}cm^{-1} {/eq} with a noticeable shoulder at 3600 {eq}cm^{-1} {/eq} . It also contains a strong absorption at 3020 {eq}cm^{-1} {/eq} , a medium and sharp absorption at 1600 {eq}cm^{-1} {/eq} , and a characteristic absorption at 750 {eq}cm^{-1} {/eq} .

iii) The{eq}^1H {/eq} NMR of this compound consists of a multiplet at 6.9 ppm (4H), a singlet at 4.8 ppm (2H), and a broad singlet (2H) at 4.0 ppm.

iv) The{eq}^{13}C {/eq} NMR exhibits absorptions at the following ppm's: 155.6 (s), 133.5 (s), 129.1 (d), 121.3 (d), 120.7 (d), 115.4 (d) and 62.6 (t).

Elucidation of Unknown Compound:

Elucidation of an unknown compound requires a combination of techniques to determine the presence of atoms, types of bonds and functional groups. Each of the four spectra, mass spectroscopy, infrared spectrometry, H1 and C13 nuclear magnetic resonance, provide insight to a specific structure of our unknown compound.

Mass spectrometry is most useful for determining the chemical formula of an unknown compound. The spectra can be produced via a multitude of ionizing techniques and passing the compound through a magnetic field. The amount that the compound is deflected by that magnetic field indicates its mass over charge ratio (m/z). A low spec mass spectrometer, as found in this question, will only provide an integer value for its m/z. The base peak represents the most stable ion in the fragmentation process beginning from the parent peak.

Infrared spectroscopy functions by emitting infrared light through the unknown compound over a broad energy range and observing which frequencies are absorbed by the compound. Adsorptions a specific energy levels indicate different types of bonds and functional groups. This technique is the least useful and reliable structure determination technique of the four available to us. It provides a good starting point of what functional groups we may expect in our compound.

NMR, or nuclear magnetic resonance, is by far the most useful structure determination technique available. It is a non-destructive technique that uses magnetic fields to determine the presence of chemically unique atoms in a compound. Each of these chemically unique atoms interact with their neighboring atoms and provide a greater insight into the atom's chemical and electronic environment. More specifically, in the spectrum provided in this question, we can determine whether that atom is adjacent to, and how many hydrogens. The atom's proximity to electron withdrawing groups, EWG, or electron donating groups, EDG, also greatly affect where the chemical shift appears in the spectra.

Answer and Explanation: 1

Elucidation of an unknown compound requires first reviewing all available information to determine the building blocks of the compound. Once we are reasonably confident in the chemical formula and components of our unknown compound, we can start to propose structures that align with our data. This process often involves many revisions.

In this answer, we will:

  • Determine the chemical formula
  • Review the data provided from MS, IR, H1 NMR, and C13 NMR
  • Confirm proposed structure with additional data in the question
  • Discuss potential issues with the structure

Determine the Chemical Formula

The mass spectrum provided is low resolution. We cannot determine the exact chemical formula for our compound only using the mass spectrum. Using the information from our mass spectrum in tandem with our other spectra we can deduce the chemical formula and the degrees of unsaturation.

We know from our mass spectrum that the molecular weight of our compound is 124m/z. The most stable fragmented ion of this compound is 123m/z. That would imply the loss of a hydrogen.

From the infrared spectrum, a large broad absorption around 3200 cm^{-} usually means the presence of an alcohol group. This means we have at least one oxygen in our compound.

In the proton NMR, three peaks are integrated for a total of 8 protons. Keep in mind that this integration is only a ratio between the peaks. Therefore the three peaks for 4, 2, and 2 protons could be any multiple of that.

The carbon NMR exhibits 7 individual peaks which is usually a reliable indication of the number of carbons present in a compound. If the molecule is symmetrical, we must be cautious as we would only see one peak for each chemically equivalent carbon.

Totally the mass so far:

7 Carbon = 7*12.01 = 84.07

8 Hydrogen = 8*1.01 = 8.06

1 Oxygen = 1*16.00 = 16.00

Total = 108.14

Notice that there is a difference of approximately 16 between our first attempt at a chemical formula and the parent peak in the mass spectrum. That is a good indication that there is one more oxygen present in our compound. The chemical formula is C_{7}H_{8}O_{2}

We can calculate the degrees of unsaturation by the following formula

$$DOU=\frac{2+2C+N-H-X}{2} $$

Where DOU is degrees of unsaturation, C is the number of carbons, N is the number of nitrogens, H is the number of hydrogens, and X is the number of halogens.

For our chemical formula, it has a 4 degrees of unsaturation. Each degree of unsaturation represents a ring, double bond, or one half of a triple bond (triple bonds account for two DOU).

Mass Spectrum

We have already discussed the majority of the useful information about the mass spectrum in the determination of the chemical formula section. We can confirm a few additional details from the limited information given.

Since the parent peak is an even number that means that there must be either an even number of nitrogen or none at all. Since we have already accounted for the 7 carbon, 8 hydrogen, and at minimum 1 oxygen, it is a safe assumption that no nitrogen is present.

The absence of a M+2 peak with a ratio of 1:1 or 3:1, strongly implies the absence of any Br or Cl in our compound respectively. Both of these atoms have stable isotopes and would produce the same molecule with two distinctive parent peaks.

Infrared Spectrum

Recall that infrared spectra does not indicate the amount of any functional group but rather it's presence. We are using this spectrum to identify the type of bonds and functional groups to help map out our structure. It is often useful to associate an adsorption from our IR spectrum and map it to certain peaks in our NMR spectrum.

There are 5 mentioned absorption:

(1) 3200 cm, strong and broad

(2) 3600 cm, shoulder of (1)

(3) 3020 cm, strong

(4) 1600 cm, medium and sharp

(5) 750 cm, characteristic absorption

Broad adsorptions in infrared spectroscopy indicates hydrogen bonding. Alcohols are typically found in the range of 3400 - 3700 cm-1 which would explain the shoulder (2). Peak (1) most likely is the carbon-hydrogen stretching apart of an alkene.

Peak (3) is consistent with carbon-hydrogen stretching in an alkane. C-H stretching typically has an energy level of 3000 cm-1 with medium to strong intensity.

Peak (4) with an energy level of 1600 cm-1, would indicate the presence of a Carbonyl group. The peak corresponds to the C=O bond.

All peaks with energy levels lower than 1500 cm-1 are considered in the finger print region and are difficult to discern any useful information. Absorptions in this section of a spectrum can be used for verification of the correct compound.

Proton NMR

Three chemical shifts appear in the proton NMR. Let's label them:

1) 6.9 ppm, multiplet, integrates for 4H

2) 4.8 ppm, singlet, integrates for 2H

3) 4.0 ppm, broad singlet, integrates for 2H

Chemical shift (1) appears downfield in the aromatic region. A multiplet integrating for 4H with no specific splitting pattern makes it difficult to assume any type of coupling with neighboring protons. Since it is in the aromatic region, it is a safe assumption at this point, that the protons are part of aromatic benzene ring.

From our analysis of the chemical formula and DOU, we can work off the assumption that there is a benzene ring present. With a single multiplet, we would expect a 1,2-di-subsituted benzene ring. That would explain why we observe a single aromatic shift.

Chemical shift (2) does not appear in the aromatic region but it relatively far downfield. This implies that these two protons are adjacent to an electron withdrawing group (EWG). Fortunately, benzene is a strong EWG due to the electrons orbiting the benzene ring in the conjugated system. We know that our chemical formula is C7H8O2, and 6 of those carbons are present in the benzene ring. That only leaves one carbon left. A great candidate for chemical shift (2) would be a CH2 group.

Chemical shift (3) appears as a broad singlet. This almost always means the presence of a OH or NH2 group since these protons are subject to replacement during the NMR scan. No nitrogen is present from our analysis of our MS. From our discussion of chemical shift (1) we assume that our benzene ring is ortho-substitued. A great fit would therefore be an alcohol substituent adjacent to the CH2 group on the benzene.

From our analysis of the proton NMR, we have accounted for C7H7O, remaining is OH. We have accounted for all the DOU. We can be confident that the remaining oxygen does not have a double bond but is rather another alcohol group. The best fit for the alcohol is to be attached to the CH2 substituent discussed in (2). From this point we have a proposed structure solely from MS, IR, and H1 NMR. Let's now confirm our proposal with C13 NMR.

Carbon NMR

Recall the N+1 rule for C13 NMR. The splitting pattern for chemical shifts in C13 NMR shows how many protons are bonded to the carbon atom, where N is the number of hydrogens. A singlet has no hydrogen attached. A doublet has one proton attached, and so on.

C13 NMR has seven chemical shifts. Let's label them:

(1) 155.6 ppm, singlet

(2) 133.5 ppm, singlet

(3) 129.1 ppm, doublet

(4) 121.3 ppm, doublet

(5) 120.7 ppm, doublet

(6) 115.4 ppm, doublet

(7) 62.6 ppm, triplet

We know that the carbon associated with chemical shift (1) does not have any protons bonded to it. It is also the chemical shift that appears the furthest downfield. The only carbon in our proposed structure so far that meets these criteria is carbon 1 (figure 1).

Chemical shift (2) also represents a carbon with no protons bonded to it. Since we have already assigned carbon 1 (figure 1) the only remaining option is carbon 2. This is consistent with the chemical shift value of 133.5 ppm which is in the aromatic region.

Chemical shifts (3), (4), (5), and (6) are all doublets implying that they have 1 proton bonded to them. This is what we would expect with a disubstitued benzene. With these last 4 chemical shifts, we have accounted for all 6 aromatic carbons. It is difficult to assign these last 4 chemical shifts to specific carbons in the ring since their shift values are so close together. We can make some strong assumptions based on their proximity to the two substituents on carbon 1 and 2.

The chemical shift (7) is the only triplet in the spectra. We know from our N + 1 rule that two protons are bonded to this carbon. It is present in the alkane region and would be very consistent with carbon 7.

Verification with Additional Information

The introduction to this question mentions that this compound can be extracted from the leaves and female flowers of the willow (Salix). https://drugs.ncats.io/substance/FA1N0842KB

Concerns With Proposed Structure

After the analysis of four different spectrum, we can be reasonably confident in our proposed structure. 2-(Hydroxymethyl)phenol, otherwise known as salicyl alcohol, has largely fit all the information from our spectra. The largest exceptions are from the MS and from the proton NMR.

With our proposed structure, it would be unlikely that the parent peak is the deprotonated version of our compound. However, this spectra does not inherently disprove our proposed structure.

The largest area of concern with the proton NMR is the integration pattern of the three peaks. We have successfully identified peak (1) as representing the 4 aromatic protons in our benzene. Peak (2) represents the 2 protons on carbon 7. At this point of our analysis, 6 of the 8 protons have been identified. The remaining two hydorgens in our proposed structure are apart of an alcohol group. Since these two alcohols are not chemically equivalent their peaks would not overlap in the proton NMR. We can have confidence that peak (3) does represent an alcohol group since it is a broad peak and the hydrogen is subject to replacement. We would therefore expect peak (3) to integrate for 1H rather than 2H.

If this was an error in integration, than we would expect to see an additional peak representing the second alcohol group. Since no actual spectra is provided, we can assume this to be an oversight of the question due to the overwhelming evidence pointing towards our proposed structure.

Other information about this compound was also given in the question. The boiling point of this compound is found to be 84 degrees celcius. The melting point of salicyl alcohol is is approximately 87 degrees Celsius. This appears to be an additional oversight of the question. The author of this question seems to have mixed up boiling point and melting point.

Learn more about this topic:

The Different Types of Chemistry


Chapter 1 / Lesson 12

Learn about different types of chemistry. Explore the five branches of chemistry and the differences between them, including what they are and what they study.

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