M8 ACI 2

IQ2 How is information about the reactivity and structure of organic compounds obtained?

8.2.1 conduct qualitative investigations to test for the presence in organic molecules of the following functional groups:

a) carbon–carbon double bonds

b) hydroxyl groups

c) carboxylic acids

8.2.2 investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to:

a) proton and carbon-13 NMR

b) mass spectroscopy

c) infrared spectroscopy

View videos:

• ACI#10 https://www.youtube.com/watch?v=CHE-5iPZ5jY&list=PLeFSFSJ9WqSC36UhgaA0FsiqNtiSo8I6H&index=10 [5.38 mins]
• Alkane and alkene https://www.youtube.com/watch?v=PE1CDR1S5pk [0.59 mins]

2.1 conduct qualitative investigations to test for the presence in organic molecules of the following functional groups:

b) OH (hydroxyl) groups

Identify the presence of a primary or secondary alcohol using oxidation - K2Cr2O7/H+ (or KMnO4/H+) test.

If potassium dichromate with a small amount of acid (acidified potassium dichromate - K2Cr2O7/H+) is added to a primary or secondary alcohol, the orange dichromate turns blue-green. This test will not work for tertiary alcohols, there will be no colour change.

2.1 conduct qualitative investigations to test for the presence in organic molecules of the following functional groups:

c) carboxylic acids

Identify the presence of a carboxylic acid using one or more of:

• pH meter (pH under 7)
• litmus paper - blue litmus turns red, red litmus unchanged
• calcium carbonate - bubbles [review acid + carbonate reaction from ABR]

Of possible use:

• formula for finding the degree of unsaturation 1/2 (2n - [2n + 2])

eg: ethene 1/2 [(2x2) - {(2x2) + 2}]

= 1/2 [4 - (4 + 2)]

= 1/2 [4 - 6]

= 1/2 x 2 = 1

1 is a double bond, or cycloalkane

OVERVIEW

A small bag containing a mysterious white powder has been found by investigators in a local park. We have been tasked with identifying this compound. Is it an illicit drug? sugar? medication?

Molecules are too small to see with microscopy, so how do we identify the molecular-level differences between substances? We can identify molecules making up substances based on signal patterns and absorption peaks from instruments like spectrometers. Techniques such as Mass Spectrometry (MS) can help us to work out the formula of substances, Infrared (IR) Spectroscopy can tell us what functional groups are present in a molecule, and Nuclear Magnetic Resonance (NMR) spectroscopy can tell us about molecular structure and connectivity for molecules in solution.

Below are images of the instruments and read-outs of forms of spectroscopy and spectrometry.

Knowledge of functional group is particularly important to spectroscopy. Functional groups control the behaviour of molecules; they are where chemical reactions take place, making identifying them in a molecule very important.

Simple molecules that contain the same functional group in their structure can be expected to react in similar ways. More complicated chemical molecules may contain more than one functional group within their structure, which can sometimes affect the chemical reactions they undergo. The names of organic molecules are systematic references to the functional groups within the molecule, and so can be used to identify them.

So, what is actually going on in the molecule that allows IR spectroscopy to give us useful information?

Infrared frequencies make up a part of the electromagnetic spectrum. The IR machine shines the range of frequencies in the infrared range through the organic compound sample. Some of those frequencies match particular vibrational frequencies (bending, stretching, compressing and twisting) of the covalent bonds in the sample molecule. Different bonds absorb at different and characteristic frequencies.

For the information that follows - explanation plus spectra - you do not need to know/remember details. You will be given data sheets with the bond and wavenumbers, but you will need to be able to apply these to identify functional groups.

MASS SPECTROMETRY

2.2 investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to:

b) Mass Spectrometry

Infrared spectroscopy can help us identify a compound based on functional groups. But unless you are given a limited number of options with different functional groups (eg alcohol and alkene), IR spectroscopy can only limit likely identities to a group of compounds that share functional groups eg C=O could be aldehyde, ketone, acid, ester, etc.

Even if you do have a limited number of options, interpreting IR spectra can sometimes be challenging and you will want to confirm your identity by some other means (not to mention that science is at its strongest when multiple lines of evidence are pursued).

Mass spectrometry can help by giving the molecular mass of a molecule which can be matched to known masses, or used to predict a molecular formula. Some structural information can also be deduced from patterns created during mass spectrometry, and these patterns can be used like a ‘fingerprint’ to search databases or compare samples with known compounds. Mass spectrometry (MS) can even help us identify the elements present in a molecule due to isotope patterns in the spectra.

How does the mass spectrometer achieve this?

There are three key steps involved in mass spectrometry:

1. Ionisation of the compound
2. Separation of the resulting ions
3. Detection of the ions

Step One: Ionisation

MS only detects ions (charged species), so first the sample is bombarded with high energy electrons which can knock electrons out of some molecules, giving a molecular ion (M)+• (a radical cation).

In most cases, the molecule will then break into fragments (‘fragmentation’), with the charged pieces appearing in a spectrum. These fragments tend to form in characteristic and predictable ways. A fragmentation pattern can be used like a ‘fingerprint’ to help identify an unknown compound by matching its spectrum to the spectrum of a known compound.

Step Two: Separation of ions

The ions are accelerated through a magnetic field, which affects their path and so separates the ions depending on their mass-to-charge ratio or ‘m/z’. It is a little like throwing a light ping pong ball and solid golf ball of similar size into the downward draft from a fan. The path of the ping pong ball (which will be greatly deflected by the stream of air) will be markedly different from that of the golf ball, despite having the same external forces acting on each.

Step Three: Detection of Ions

The separated ions are now recorded by a detector which takes into account the path of an ion and uses this information to calculate m/z. The detector also records the relative number of ions hitting it.

Together, these give a spectrum of mass-to-charge (m/z, along the horizontal or x-axis) plotted against abundance (=amount, along the vertical or y-axis). If ionisation does not break a molecule apart (fragmentation) then the highest m/z peak corresponds to the whole molecular ion’s mass or M+ and so is equal to the molecular mass of the compound.

NMR SPECTROSCOPY

Nuclear magnetic resonance (NMR) spectroscopy is a powerful tool for compound elucidation. The most commonly used types of NMR are carbon-NMR (13C-NMR) which looks at the carbons in a molecule, and proton-NMR (1H-NMR) which looks at a molecule’s hydrogens (specifically 1H, i.e., protons). In this section we will focus on 13C-NMR and introduce 1H-NMR.

How does NMR work?

The short answer is: ‘spin.’

An atom’s nucleus displays ‘spin’ if it has an odd mass number and/or an odd atomic number. As the isotopes 1H and 13C meet this criterion, they display spin (with this spin randomly oriented), whereas 12C and 16O do not.

Nuclei act like small bar magnets because they have charge and movement, and so each nucleus has its own magnetic moment. If an external field is applied to a nucleus, its magnetic moment will either align (line up) with the field or oppose it. You get a mixture of aligning and opposing.

When aligned, the nucleus is in a lower energy state and when opposed it is in a higher energy state.

View videos:

• How NMR works https://www.youtube.com/watch?v=sfiQFQYgJuQ [1.18]
• NMR Animation https://www.youtube.com/watch?v=ywR6aLpfjl0 [3.01]

A sample is first prepared in a thin glass tube and then subjected to a very powerful magnet in the NMR. Inside this magnetic field, the nuclei in the sample molecules must either align or oppose their spin to the external field and there is a difference in energy between the aligned and opposed nuclei.

The size of this gap is dependent on the strength of the magnetic field that the nuclei are experiencing.

Nuclei can ‘flip’ between spin states by absorbing radiofrequency energy or emitting energy corresponding to the gap between spin states. It is this energy that we measure in NMR spectroscopy.

When there is no external magnetic field then there will be no difference in energy between the two spin states. The stronger the external magnetic field applied, the greater the difference between the lower (aligned) and the higher (opposed) energy states; this difference is termed ‘∆E’.

E = hv, (h is a constant, but v is frequency), if we apply electromagnetic radiation at the correct frequency, then we can cause the nucleus to ‘flip’ its spin from the lower energy state to the higher one. Because ∆E depends on the strength of the magnetic field, the exact frequency necessary for this flipping also depends on the strength of the external magnetic field applied. The energy gap between the two spin states of a nucleus (∆E) grows as the magnetic field becomes stronger.

13-C NMR (CARBON NMR)

2.2 investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to:

c) carbon-13 NMR

13C-NMR helps map the carbon framework of molecules, revealing the number and types of carbon atoms present.

A 13C-NMR spectrum for a molecule consists of sharp peaks each usually corresponding to one* carbon in the molecule (*we will introduce an exception: equivalent carbons). The peaks are spread across the spectrum rather than all piled on top of each other because of their ‘molecular environment’ (eg if they’re near a functional group, if there are many or few other atoms nearby, etc).

The ‘0’ point on the spectrum (which is on the right) is arbitrarily set in both 13C-NMR and 1H-NMR as the point that gives a peak for the compound tetramethylsilane (TMS). TMS is thus added to samples to give a reference point.

How does the magnetic field produce a spectrum?

If all the carbons in a molecule are exposed to the same external magnet in 13C-NMR, why don’t the peaks for the carbons just pile up on top of each other at the same frequency (ppm)?

Remember that ∆E (the difference in energy between spin states in the presence of a magnetic field) depends on the strength of the magnetic field experienced? In addition to the NMR’s magnets, electrons in molecules create their own magnetic fields which interact with the field applied by the NMR, generally shielding the nucleus from external magnetic forces. That is, even with a constant magnetic field coming from the NMR spectrometer (B-applied), the actual magnetic field ‘felt’ by a nucleus (B-effective) varies depending on the electrons nearby (the ‘local environment’ - neighbours, attachments - of nuclei). When magnetic fields vary, so too does the difference in energy between the two spin states, meaning they absorb at different radiofrequencies.

The carbon nuclei’s environment affects the energy required to cause flipping, so different carbons within a molecule have signals at different frequencies (ppm) within the spectrum.

The actual magnetic field ‘felt’ by a nucleus:

Carbon nuclei experiencing such variations in effective magnetic fields are said to be ‘magnetically inequivalent,’ and their differences in energy absorption leads to ‘chemical shifts’ on a spectrum (the peaks for different carbons in a molecule are spread across a spectrum, not all in one place).

The 13C-NMR spectrum of ethanol demonstrates chemical shift.

For the carbon next to an oxygen (C-O, circled in red below): the oxygen is electron withdrawing (electronegative), pulling electrons away and causing less electron density around the carbon to which it is attached. That carbon’s nucleus has less ‘shielding’ and feels a higher B-effective which means a larger energy gap between the spin states causing it to resonate at at a higher frequency (= higher ppm signal). It has shifted downfield to around 60 ppm.

For the other carbon (-CH3, indicated in blue): is further from the oxygen and so doesn’t really get much deshielding (and so feels a lower B-effective which means a smaller energy gap and thus the lower frequency or ppm signal around 20 ppm).

Note the three peaks (there is also a negligible TMS peak at 0 ppm): the peak around 80 ppm is a solvent peak (deuterated-chloroform) and the two remaining peaks are from ethanol. Ethanol produces two peaks because there are two different ‘types’ of carbon in ethanol (one attached to oxygen and another which is not).

In highly symmetrical molecules, multiple carbons will experience identical environments and are said to be ‘equivalent’. All the carbons experiencing the same environment will appear as just one peak.

See pentane (below).

The two outer carbons (blue) are experiencing exactly the same environment as there is symmetry in the molecule (they are both attached to one carbon and three hydrogens). The second carbon in from each side (orange) is also experiencing equivalence to each other (each is attached to an outside carbon, the central carbon and two hydrogens). The central carbon (red) is the only carbon to experience being attached to two other carbons each attached to outside carbons. In total, three peaks will arise in a 13C-NMR spectrum.

Although all three middle carbons are -CH2, the very central carbon is in a different environment because it has two chains of two carbons attached, whereas the carbons circled orange each have a single outer carbon attached on one side and a chain of three carbons attached to the other. Therefore they show a separate peak.

Pentane and equivalent carbons. In the case of ethanol presented earlier there were two peaks in the carbon spectrum for the two carbons in the molecule. We don’t see five peaks for pentane because some carbons are equivalent: they experience the same environment.

How do we interpret NMR spectra?

Reference tables have been made to assist in interpreting NMR spectra. This is possible because different chemical environments cause the shifting of peaks in predictable ways. An example of a table (from the NSW HSC Examination Sample Questions) is below right.

1-H NMR (Proton NMR)

2.2 investigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to:

d) proton NMR

It is important to note that 1H nuclei contain a single proton and so in the context of 1H-NMR we often use the terms hydrogen, proton and 1H nuclei interchangeably.

The physics behind 1H-NMR (proton NMR) is essentially the same as for 13C-NMR, but the spectra provide us with more information.

1. Number of peaks (signals)

The number of signals reflects the number of non-equivalent hydrogens, ie the number of magnetically different protons (= usually chemically different protons, eg the proton on -OH is different to that on -CH.)

In -CH3 all 3 protons (hydrogens) would usually share the same peak as they are all attached to the same carbon.

We see proton equivalence in 1H-NMR with symmetrical molecules like cyclohexane, but not in cyclohexanone.

2. Chemical shift

As for 13C-NMR, the location of a signal in a spectrum provides information about the chemical environment of the proton. Because these shifts are predictable, tables have been produced to assist interpreting spectra:

Note:

In the 1H-NMR spectra the chemical shifts (in ppm) fall in the range of around 0-13 ppm (for 13C-NMR spectra it is more like 0-250 ppm).

Peaks are much broader for 1H-NMR.

3. Number of protons

A major departure from 13C-NMR is that in 1H-NMR the area under a signal’s peaks tells us about the relative number of protons that are involved in that signal. This is called ‘intensity’.

Notice the numbers under the peaks (4,4,2). These are the results of integrating the area under the signals (calculated by software) and indicate the relative numbers of protons producing the signal. Note that peak height does not necessarily correlate with peak area.

4. Signal splitting

Signals for 1H-NMR are often (but not always) split into multiple peaks. The signal for a proton will split into n+1 peaks, where n = number of neighbour Hs, ie the number of protons attached to adjacent carbons. Peaks are often referred to as singlets (1 peak = 0 neighbour Hs), doublets (2 peaks = CH neighbour), triplets (3 peaks = CH2 neighbour), quartets (4 peaks = CH3 neighbour).

Peak splitting: n + 1

In this example of 1,1,2-trichloroethane there are 3 protons.

• CH2: Two of these protons are equivalent (purple) and so make 1 peak at around 4 ppm. The area underneath the peak corresponds to 2 protons. The peak has then been split into 2. From n+1 we know this means there is 1 proton on the neighbouring carbon.
• CH: The peak at around 5.8 ppm corresponds to a single proton. The area under the peak would integrate to 1. The peak is split into 3 due to the 2 protons on the neighbouring proton (2 + 1 = 3).

Another way to work on this would be

# H on next carbon = # peaks - 1

View videos:

• Proton NMR https://youtu.be/uNM801B9Y84 (ignore 2.49-6.12 on how the instrument works [8.42])
• ACI#13 https://www.youtube.com/watch?v=rAMqhWp4PkY&list=PLeFSFSJ9WqSC36UhgaA0FsiqNtiSo8I6H&index=13 [19.32 mins]

Clues for Bottle 5

MS:

– MM=88.

IR:

– Notice there are no broad OH peaks.

– Notice there are no amine peaks.

– What does a 1750 cm-1 peak suggest?

– The fingerprint region (>1500 cm-1) is hard to interpret.

13C-NMR:

– There are 4 peaks, which suggests how many carbons? (Let’s start off assuming there is no equivalence, we can adjust this assumption if the other data does not support it).

– A peak around 170 ppm would suggest what options?

– The peak just above 60 ppm suggests what options?

– The two peaks between 0 and 40 ppm suggest what options?