Spectroscopic Detection of a Benzene Ring in Organic Compounds

Learning Objectives

To interpret a mass spectrum of a compound which may contain a benzene ring.
To know characteristic IR bands benzene rings.
To associate characteristic oop frequencies with substitution patterns on the benzene ring.
To be able to associate ¹³C and ¹H NMR chemical information with the presence of benzene rings in a compound.

Few people have contributed as much to science as Michael Faraday. Albert Einstein had Faraday's portrait on his wall along with Newton's and Maxwell's. One of Faraday's many discoveries was the substance we know as benzene.

Figure 6.1 (Spec) Michael Faraday (1791-1867) and his wife Sarah (1800-1879)

See page for author [CC BY 4.0 (http://creativecommons.org/licenses/by/4.0)], via Wikimedia Commons

But on this page we're looking not especially at the compound benzene. We're looking at compounds that contain a benzene ring. All three of the compounds we used for the acid-base extraction activity contained a benzene ring. Literally thousands upon thousands of compounds contain this structural moeity. Figuring out the structure of benzene took many decades of research after Faraday's initial isolation and discovery. Figuring out the reactivity took still more decades.

IR (FTIR)

C-H stretches

Like any C=C-H, there is a very distinctive sharp absorption band just above 3000 cm-1.

C=C stretches

The conjugated double bonds in the benzene ring stetch in the plane of the ring and usually give two intense absorption bands around 1600 and 1500 cm^-1.

OOP C-H bends

Most informative, though most troublesome to remember, is the information revealed in the bending region. In particular, the out-of-plane (oop, pronounced oh-oh-pea) bending region from 650 cm^-1 to 900 cm^-1. Table 6.1 (below) is titled "Classic." The "classic" refers to the requisite that the "R" groups are nonpolar groups like ethyl or isobutyl. If however the groups are polar, that is they involve a very electronegative atom, the frequencies and even the number of peaks indicated do not apply.

Table 6.1(Spec) "Classic" oop Bending Frequencies

Another way to consider the characteristic bands present in substituted benzenes is in terms of the number of adjacent hydrogens on the ring.

Table 6.2(Spec) "Classic" Aromatic Bending Frequencies in Terms of Hydrogens on the Ring

In reconciling Table 6.1(Spec) and Table 6.2(Spec) there can be some complications, especially with regards to the meta substitution. Table 6.1 says to look for 3 bands but table 6.2 indicates just two. One from the isolated H and one from 3 H's but where'd that third band in table 6.1(Spec) at 670 to 710 cm^-1 come from? There's an explanation, but it's not worth diving into at this point. For now, memorize table 6.2(Spec) and remember there's a complication for the meta case.

Aromatic Overtones

These are three or four very weak, evenly spaced bands around 2000 cm^-1 that arise from more intense absorptions at lower frequencies.

¹H-NMR

Proton NMR makes it particularly easy to spot a benzene ring. The hydrogens attached to the benzene ring have a distinctive chemical shift. See figure 6.3 (Spec). The number "7" is easy to remember.

Table 6.3 (Spec) Some ¹H chemical shifts adapted from http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/NMR%3A_Interpretation

Besides just for spotting the presence (or absence) of a benzene ring, a proton NMR spectrum can tell you a lot more about the protons on the ring. How many peak groups are in the aromatic region? What is the integration?

¹³C-NMR

The ¹³C-NMR spectrum is not always useful for distinguishing an arene from an alkene. Both groups have carbons with chemical shifts around 130 ppm. With the aid of the ¹³C spectra, you might be able to pick out how many carbons around 130 have one attached hydrogen and how many carbons with no attached hydrogens. If you run across a carbon with two attached hydrogens at 130, it must be from a vinyl group and not a benzene carbon.

Example 6.1 (Spec)

Identify the unknown compound using the following four spectra.

Figure 6.2 (Spec) Mass Spectrum

Figure 6.3 (Spec) IR Spectrum

Figure 6.4 (Spec) ¹³C NMR Spectrum

Spectrum from SDBSWeb : http://sdbs.db.aist.go.jp (National Institute ofAdvanced Industrial Science and Technology, Accessed September 10, 2017)

Figure 6.5 (Spec) ¹H NMR Spectrum

proton NMR spectrum, three peak groups, 7.0, 2.6 and 1.2 ppm

Spectrum from SDBSWeb : http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, Accessed September 10, 2017)

Answer

Step 1: Identify the probable M⁺ ion in the mass spectrum (Figure 6.2 (Spec)). The most likely candidate is at 134 m/z. The peak at 135 is probably the ¹³C isotopic peak. Normalize the peaks by dividing 100 by the intensity at 134.

$\frac{100}{51.39} = 1.946$

Now apply the normalization factor to the peak intensity at 135.

1.946 × 5.99 = 11.66

Divide the normalized abundance by the natural abundance of ¹³C (1.1%).

$\frac{11.66}{1.1} = 10.60 \approx 10 or 11$

So, it looks like 10 or 11 carbons in the M⁺.

To get to a mass of 134 with 10 carbons would require 14 hydrogens. To get a mass of 134 with 11 carbons would allow for only 2 hydrogens. C11H2 would be most improbable. Let's go with C₁₀H₁₄.

Step 2: From the MS analysis in Step 1, there's no reason to look for heteroatoms; however, a quick scan of the IR shows no evidence of oxygen. No beard or belly, no carbonyl stretch, no C-O stretches. Running the rings + double bonds on C₁₀H₁₄ gives 4 rings +db. This is consistent with the presence of a single benzene ring with no other rings or double bonds.

Step 3: The IR confirms the presence of a benzene ring.

Three peaks between 3100 and 3000 cm^-1 could be =C-H stretches.
Aromatic overtones appear at 1900, 1800 and 1700 cm^-1 . They look like fingers.
Curiously, there's no C=C stretch at 1600 cm^-1; however, there is a peak that looks like a C=C stretch at 1520 cm^-1 .
There is clearly an OOP at 830 cm^-1 . This is consistent with para substitution. When similar hydrocarbon substituents are present in a para arrangement, the expected C=C at 1600 cm^-1 will be absent.

Step 4: The 1H NMR clearly shows the hydrogens attached to the benzene ring are identical. The "aromatic" hydrogens are at 7.1 ppm. The integration for the 1HNMR totals only 7 hydrogens. We know there are 14 hydrogens. There must be 2-fold symmetry in the molecule. This is consistent with para substitution where the substituents are identical. The integration for the peak at 2.6 ppm is 2 and it's chemical shift is consistent with a group adjacent to a benzene ring. So this is likely a CH₂ adjacent to the benzene ring. The integration for the peak at 1.2 ppm is 3 and it's chemical shift is consistent with a group more removed from the benzene ring. This is likely a CH₃ attached to the CH₂ on the ring mentioned earlier. We suspect two ethyl groups para substituted on the benzene ring.

Step 5: The ¹³C NMR shows only 4 types of carbons. The peak at 142 ppm and the peak at 128 ppm are likely aromatic carbons. From the chemical shift, the peak at 28 is likely a CH₂ and the peak at 17 is likely a CH₃ . The ¹³C confirms our hunch that we have two ethyl groups para substituted on a benzene ring.

Figure 6.6 (Spec) Structure of 1,4-diethylbenzene

Image: NIST (Public Domain)