spectrum problem

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Spectrum Problem : Hint 1

Mass Spectrum

1. Look at the molecular ion peak. Since it is 72, the molecular weight is 72. Since there is no M+2 peak, there are no halogens in the molecule. Also, because the mass is an even number, there are no nitrogen atoms in the molecule.

2. Divide the mass by twelve to find the maximum number of carbons that could be in the compound. This gives six, but there is no remainder so the compound must have fewer than six carbons (some mass must be left over for H's). First try compounds with five carbons. The only possible compound is C5H12. Next try four carbons. The only possible compound is C4H8O. Finally, a possible compound with three carbons is C3H4O2.

At this point it would be useful to determine the degrees of unsaturation for the various formulae

The Probable Molecular Formulae Are Thus:C5H12 = zero degrees of unsaturation (no double bonds or rings)C4H8O = one degree of unsaturation (one double bond or ring)

C3H4O2 = two degrees of unsaturation (double bonds and/or rings)


IR Spectrum

1. There is a strong peak at 1614 cm-1, and it seems to overlap another peak at 1649 cm-1. These could be two distinct groups, or one of the peaks could be " noise." There also clearly are some peaks just above 3000 cm-1, indicating (hydrogens attached to) a carbon with a double bond. This rules out the possibility of C5H12, which is a saturated alkane.

2. Next, compare the groups that could be formed in C4H8O or C3H4O2 with the groups that could result in IR peaks around 1615 cm-1 to 1650 cm-1. There are only two possibilities: an alkene (peak should be at 1625 cm-1) or a ketone attached to an alkene (peak should be at 1680 cm-

1). Of these two, the only one that is really likely is the alkene, since the peak ends quite a ways from 1680 cm-1.

The probable molecular formulae are now:C4H8O and C3H4O2

and the compound is probably an alkene

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Carbon NMR

1. There are four carbon signals, which obviously rules out C3H4O2. This means that the compound is C4H8O. Since there are four distinct carbon signals, no two carbons in the molecule are equivalent magnetically.

Also, the signals around 85 and 65 ppm are consistent with being next to oxygen or part of a double bond. The signal at ca. 150 ppm is VERY downfield. Perhaps this is a carbon that is both part of a double bond and next to an oxygen.

The molecular formula is C4H8O, the compound is an alkene, all carbons are different.


Proton NMR

1. There are eight hydrogen signals, which matches the formula C4H8O. Since there is one degree of unsaturation, there is either a double bond or a ring. Recalling the IR spectrum, there must be a double bond, and it

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must be an alkene. This means that the oxygen is not double bonded, so it must be in an ether or an alcohol group. Since there was clearly not an alcohol stretch on the IR spectrum, there must be an ether group.

2. (As we already decided) There are only three C4H8O molecules that can be drawn with an ether group and an alkene group. Two of these include three carbons on one side of the oxygen, and one carbon on the other side. These two can immediately be ruled out because they would create 3-H singlets (from the methyl groups on the ether)there are no 3H singlets in the spectrum.

Also, the signals around 85 and 65 ppm are consistent with being next to oxygen or part of a double bond. The signal at ca. 150 ppm is VERY downfield. Perhaps this is a carbon that is both part of a double bond and next to an oxygen,

This leaves only one possible compound: ethyl vinyl ether.......

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NMR Problems

Self-Teach Spectrum Problem

by Liam Bisson, CHM 331 Spring 2002

Sketch the NMR spectrum you would expect for the following compound:

Liam suggests first drawing the various peaks you expect to find, then redrawing then in the correct positions (with correct chemical shifts) according to electronegativity, proximity to double bonds ...


First identify the methyl groups in the compound. These groups are (often) the easiest to identify, and will (often) occur at the far right of the proton NMR spectrum

Are any of the methyl groups magnetically equivalent?

Instructor NoteMethyl groups by definition are at

the ends of chains, therefore are often away

from electronegative

groups and unsaturation

Count up the hydrogens on the carbons adjacent to the methyl groups. use the (N + 1) rule to determine the number of peaks you would expect to see in the spectrum. E.g. if there were 2 hydrogens on the adjacent carbon, then (2 + 1) = 3. There would be 3 peaks in the NMR spectrum.

We conclude that there are 2 magnetically equivalent methyl groups. There is one adjacent carbon carrying one hydrogen.

Self-Teach Spectrum Problem, Answer

Sketch the NMR spectrum you would expect for the following compound:

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Some information about double bonds:

There is free rotation about a carbon-carbon single bond. As a result, any hydrogens attached to a single bonded carbon are considered to be identical, or "magnetically equivalent". The reason for this is that due to the free rotation around the carbon-carbon single bonds, the hydrogens rotate so fast that they are impossible to distinguish from each other. (On the timescale of the NMR experiment, the rotation of the hydrogens is such that they all "see" and occupy the same area of space in the molecule, and consequently experience an identical local magnetic field).

With a double bond, the carbon atoms can not rotate freely and therefore are "stuck" in place. As a result the hydrogens attached to these double bonds (and indeed to the same carbons on these double bonds) do not rotate and can be distinguished and therefore considered as individuals. (They occupy different areas of space in the molecule and experience different local magnetic fields).

Free rotation around the single

bonds means that Ha is

magnetically equivalent toHb. There is no free rotation around

the double bond, and thus Hc is not equivalent

to Hd

The bottom line is that the three hydrogens are non-equivalent. Do you need to be told how

many signals to expect so far?

There will be a doublet from the methyl groups and three signals from the three alkene hydrogens


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Some information about signal sizes:

The signal sizes are related to the number of magnetically equivalent protons, and are refereed to as (relative) integrals.

For example, if six hydrogens from two different carbons are magnetically equivalent (as in the present case) then they will add together to make a peak that is twice as large as one from a single methyl group. In other words, more hydrogens contributing to the same peak will push it higher.

The four signals we have identified so far will have different integrals. In this course we usually make it easy for you at give the integral in terms of the number of hydrogens contributing.

Do you need to be told the

integrals for the various signals?

6H doublet from the methyl groups and three 1H signals from the three alkene hydrogens


Some information about splitting:

The three hydrogens will experience a "complex splitting pattern". Each one is split by the other hydrogen on the same carbon, and also by the hydrogen on the adjacent carbon.

Instructor NoteDo you want to

know more about complex splitting

The term complex splitting usually refers to the situation where a proton is split by protons with different J-coupling constants. The simple (N + 1) rule works because on alkyl chains the various adjacent protons have the same J-coupling constants, whatever carbon they are on.

Instructor NoteDo you want to

know more about J-coupling constants

The splitting of each alkene hydrogen will be the similar and complex. Do you need to be told the

splitting pattern for the signals?

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Organic Structure Determination

Analytical Chemistry

Instrument-based methods for determination of structure of organic molecules

1) Mass Spectrometry - yields molecular weight/elements/possible molecular formulas

2) Infrared Spectroscopy - yields functional groups

3) NMR Spectroscopy - very important, yields structure

1 Mass Spectrometry• Gives molecular weight, some elemental information, possible molecular formulas (and degrees of unsaturation)

• schematic overview of the spectrometer...

Example: Butane

• various positively charged ions are formed in the mass spectrometer

• the ion corresponding to the structure of interest, the "molecular ion", is a radical cation, and its mass is equal to the molecular weight of the structure

• the parent radical cation will also fragment via bond breaking to form cations of lower masses

• radical cation and cations are deflected in magnetic field and the ion mass is "selected" by variable magnet

An Idealized mass spectrum of butane...

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• mass spectrum is plot of ion current (i.e. # of ions) versus mass/charge ratio (m/z), where z is charge

because the charge is almost always positive one (multi-charged species are rarely encountered in mass spectra), m/z is basically the same as mass

• largest m/z value corresponds to "molecular ion", i.e. the unfragmented molecule - molecular weight

• molecular ion peak not necessarily the most "abundant" ("tallest"), but has largest m/z

1.1 Important Information from Molecular Ion Peaks

Example of: A compound containing C, H and O only (the molecular ion mass is an even number)

• M+1 peak due to molecules containing D, 13C, etc.

• an even numbered molecular weight means that the molecular contains carbon and hydrogen, and possibly (but not necessarily) oxygen. There is no specific "signature" for oxygen in mass spectrometry

Example of: A compound containing Br

• has TWO molecular ions, one for the molecules that contain 79Br, and one for the molecules that contain 81Br

• these two molecular ion peaks are separated by 2 mass units and have ca. 1:1 size ratio

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Example of: A compound containing Cl

• has TWO molecular ions, one for the molecules that contain 35Cl, and one for the molecules that contain 37Cl

• these two molecular ion peaks are separated by 2 mass units and have ca. 3:1 size ratio

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Example of: A compound containing N (the molecular ion mass is an odd number for a single N atom)

• odd molecular weight means a single nitrogen (the M.W. for 2, 4, 6 etc. nitrogens would be even)

• strictly speaking, any odd number of N atoms, but usually one in this class

1.2 Important Information From Molecular Weight

• Determine Possible Molecular Formula from Mass Spectrum, for example.....

what are POSSIBLE and also REASONABLE molecular formulas for this mass spectrum?

• reasonable molecular formulas are not determined by the degree of unsaturation, but usually (but not always!) by having the number of H atoms to be greater than or equal to the number of C atoms

• divide MW by 12 to determine maximum possible number of H atoms, the remainder (R) is equal to the mass of the H atoms

divide M.W. by 12 = 100/12 = 8 (R4) - maximum number of carbons = 8

C8H4 is possible (7 degrees of unsaturation, BUT not reasonable (too few H atoms))

• replace 1 C atom by 12 H atoms, maintains the correct MW

C7H16 is possible AND reasonable (0 degrees of unsaturation)

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• replace 1 C atom and 4 H atoms by 1 O atom, maintains correct MW

C6H12O (1 degree)

and C5H8O2 (2 degrees)

and C4H4O3 (3 degrees)

these 3 are both possible AND reasonable

Thus, there are FOUR possible AND reasonable molecular formulae for M.W. = 100

2 Infrared SpectroscopyElectromagnetic Radiation (light!) oscillating electric and magnetic field vectors

The Electromagnetic Spectrum

• the range of frequencies of the electromagnetic radiation in the infrared region matches the range of frequencies of vibrations of bonds in molecules

• if the frequency of the radiation exactly matches that of a particular bond vibration in a molecule, then the electric field vector of the radiation can interact with the dipole moment of the vibrating

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bond, the radiation can be absorbed by the molecule and used to increase the bond vibration amplitude

• the electromagnetic frequency must match the frequency of VIBRATION of the bond (IR absorption is quantized)

if the frequency is higher or lower than the bond vibration frequency it will not be absorbed

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2.1 A Real Infrared Spectrum

Infrared Vibrations are:

• can be localized on bonds or groups of atoms (the more useful vibrations are localized on specific bonds)

• different bonds vibrate with different frequencies

• need to know how many, what frequencies and how strong!!

1) How Many Vibrations?

• for n atoms there are MANY POSSIBLE vibrations: (3n + 6)

• but many are complex, i.e. are not localized on individual bonds, occur with lower frequencies (less than 1500 cm-1) in the "fingerprint region", which is the area that has signals that are specific to a particular molecule, not to a specific functional group, in this class we ignore this region

2) What are the frequencies (energies)?

• vibrational frequencies are determined by bond strength and atomic mass

• a bond as a "spring" analogy is useful

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• a strong bond (spring) vibrates with a higher frequency

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• a bond (spring) with small/light atoms attached will vibrate with a higher frequency

Bond Dissociation Energies, Vibrational Frequencies and Vibrational Energies (in wavenumbers) for selected simple bonds

Visualize the how stronger bonds and lighter atoms result in higher vibrational frequencies

Approximate Regions in the Infrared Spectrum

• bonds to the very light atom H have the highest vibrational frequencies, stronger bonds to H having the highest frequencies

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http://chm233.asu.edu/notes/spectroscopy/visualize/vibrations.html



• triple bonds to heavier atoms come next

• double bonds to heavier elements come next, with higher frequencies for stronger bonds

• single bonds to heavier elements have the lowest frequencies, usually in the fingerprint region where identifying functional groups is difficult, and is not included in this course

3) How strong are the absorptions (how big are the peaks)?

• electric field vector interacts with bond dipole moment

• large (change in) dipole moment results in strong IR absorptions

2.2 Real Absorption Bands

Vibrations of bonds to the light element Hydrogen around 3000 cm - 1

• H atoms are LIGHT, bonds to H atoms tend to be high frequency (large nu), large cm-1

• peaks due to C-H vibrations that are found at frequencies less than 3000 cm-1 are due to H atoms that are attached to sp3 hybridized carbons. Bonds from hydrogen to sp3 carbons are somewhat weaker than bonds to, e.g. sp2 hybridized carbons, and thus are found at somewhat lower frequencies

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• peaks due to C-H vibrations that are found at frequencies just above 3000 cm-1 are due to H atoms that are attached to sp2 hybridized carbons. Bonds from hydrogen to sp2 carbons are somewhat stronger than bonds to sp3 hybridized carbons, and thus are found at somewhat higher frequencies

• peaks due to C-H vibrations that are found at frequencies less than 3000 cm-1 are due to H atoms that are attached to slp3 hybridized carbons. Bonds from hydrogen to sp3 carbons are somewhat weaker than bonds to, e.g. sp2 hybridized carbons, and thus are found at somewhat lower frequencies

• C-H vibrations around 3300 cm-1 are due to H atoms that are attached to sp hybridized carbons. These are strong bonds with high vibrational frequencies. They are distinguished from O-H and N-H bonds by the fact that they are not broad

Vibrations greater than 3000 cm - 1 are for the light Hydrogen atom when it is bonded to more electronegative elements than carbon, these bonds are stronger and vibrate with a higher frequency

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• the alcohol O-H stretching vibration is broad due to hydrogen-bonding that results in a distribution of O-H bonding situations that results in a distribution of vibration frequencies. The absorption is centered at ca. 3300 cm-1, which distinguished alcohols from carboxylic acids (see later). One O-H absorption is equivalent to several C-H absorptions together, the larger dipole moment of the O-H bond results in stronger absorption

• the amine N-H stretching vibration is also broad due to hydrogen-bonding, but N-H hydrogen bonding is weaker than O-H, and some non-hydrogen bonded N-H can be observed as small sharp peaks on top of the broad absorption

• there are usually 2 small peaks for a primary amine that has two N-H bonds

• there is usually 1 small peak for a secondary amine that has one N-H bond

• of course, a tertiary amine has no N-H bonds and no signals at all are observed in this region in this case.

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Vibrations of the C-H bond in an aldehyde just below 3000 cm-1

• aldehydes have 2 small peaks around 2730 and 2820 for the single C-H bond that is attached to the C=O. These are sometimes difficult to distinguish, and can range between ca. 2720 - 2740 and ca. 2810 - 2830, but the aldehyde also has the strong C=O stretching vibration at ca. 1700 cm-1 (see further below). Observations of BOTH vibrational features helps to identify an aldehyde

• carboxylic acids have a broad O-H peak for the same reason that alcohols do, hydrogen bonding, however, the broad absorption band in this case is distinguished from the alcohol O-H in that it is centered around 3000 cm-1, basically right in top of the usual C-H region. Carboxylic acids are also distinguished from alcohols by having the C=O stretching vibration at ca. 1700 cm-1 that is very strong (see later).

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Vibrations around 2500 - 1700 cm -1 The TRIPLE bond region

• There are only two kinds of vibrations observed in this region, the C-N triple bond of the nitrile functional group and the C-C triple bond of the alkyne functional group

• C-N triple bond absorptions due to bitrile tend to be string becaue the dipole moment associated with this bond is also large

• carbon-carbon triple bond absorptions tend to be somewhat weak (the bonds have very small dipole moments) and are only really observed for the asymmetrical terminal alkynes (alkynes in which one carbon is attached to hydrogen, the other top an alkyl or aryl group). For terminal alkynes of course, the H-C(sp) vibration is also observed at ca. 3300 cm-1.

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Vibrations Around 1750 – 1600 cm -1 The DOUBLE bond region

• important region of the IR spectrum, as usual, stringer bond vibrate at higher frequencies, which means that C=O double bonds have higher vibrational frequencies than C=C double bonds

• The second electronegative oxygen that is connected to the C=O bond in the ester makes all of the bonds stronger, thus, ester C=O vibrations occur at relatively high frequencies, around 1720 - 1730 cm-1

• Vibrations of the C=O bond in aldehydes and ketone occur closer to 1700 cm-1, often ca. 1710 cm-1. The small frequency difference between alkdehydes/ketones and esters is reliable and is very diagnostic

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• "Conjugated" ketones have a C=C bond adjacent to the C=O bond, the minor resonance contributor illustrates that the C=O bond has some single bond character in these cases. The more important the minor resonance contributor, the more single bond character (the C=O is less of a pure double bond), the weaker the bond, the lower the vibrational frequency

• conjugated aldehydes and ketones have vibration frequencies around 1680 cm-1, the difference compared to non-conjugated aldehydes and ketones (ca. 1710 cm-1) is reliable enough to distinguish these cases

• The minor resonance contributor for the amide also shows that the C=O bond has some single bond character

• In the case of an amide the minor resonance contributor is even more important than the minor contributor in the conjugated system above, because the electrons that are involved in resonance "start": as non-bonding and thus higher energy than those in the double bond above, and are thus more "available" for resonance

• the vibration frequency is thus further decreased to ca. 1640 cm-1. Just like amines, the amide will have N-H vibrations (with peaks) in the 3300 cm-1 region

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• C=C double bonds tend to have small dipole moments and are usually have weak (small) absorptions, the bonds are also weaker than C=O bonds and vibrate with lower frequencies, ca. 1620 cm-1

• The vibration associated with the benzene ring is not of a single bond, but of the entire ring. The peak is often not very strong (small dipole moment again), but is usually sharp and very close to 1600 cm-1, and is thus often readily identified.

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Return to Real Spectra: Example: Acetophenone

• note that you will need to be able to distinguish "real" peaks from peaks due to impurities or other artifacts

• note that a benzene ring adjacent to a C=O bond represents a common example of a the more generic conjugated C=C adjacent to C=O

Example: a hydroxy ketone

• the strong peak at 1710 cm-1 must be an aldehyde or a ketone, in this case it must be a ketone because the two C-H aldehyde peaks at ca. 2700 and ca. 2800 are not observed

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• the chart below is what you are provided with on a test to help you assign peaks:

3 NMR Spectroscopy

3.1 Basic Principles: Nuclear Spin and Radio Frequency Absorption (more)

• NMR relies on a property that several (magnetic) nuclei possess: nuclear spin

• Organic chemists are mainly interested in spectra of two particular nuclei, that of hydrogen (a proton) and the 13C isotope of carbon.

• nuclear spin is a quantum phenomenon, but can be understood as a classical magnetic moment

• the nuclear magnetic moments will align either with, or opposed to, an external magnetic field

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http://chm233.asu.edu/notes/spectroscopy/S1.html

• spins aligned with the field (alpha) are lower in energy than those opposed to the field (beta)

• this energy difference, Delta E, depends upon the strength of the external field (H0)

• The alpha spins can be converted into beta spins by absorption of electromagnetic radiation of the appropriate energy (Delta E). For NMR this is in the radio frequency region.

• radiation of the appropriate frequency, nu(abs), is absorbed, see schematic spectrum above

3.2 Shielding and Deshielding (more)

• NMR works because although all the nuclei in a molecule experience the same external field, nuclei in different parts of the molecule experience slightly different local magnetic fields. The electrons around a nucleus generate their own field that opposes the external field, i.e. electron motion magnetically SHIELDS the nuclei

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• electronegative atoms very effectively "pull" electron density away from proximal hydrogens, decreasing the shielding of its nucleus, increasing the effective magnetic field. Hydrogens close to increasingly electronegative atoms require increasingly larger frequencies to flip their nuclear spin.

3.3 Chemical Shifts (more)

• different frequencies for different protons are measured in terms of a chemical shift, i.e. relative to a standard, invariably tetramethylsilane, TMS.

• chemical shifts are measured in delta, which is unitless, and range from ca. 0 - 10 ppm in proton NMR spectra and from ca. 0 - 200 ppm in 13C NMR spectra. The delta value for TMS is always zero, by definition.

3.4 Which Direction is Which in the Spectrum? (more)

• MORE DESHIELDING means LARGER DELTA means LARGER CHEMICAL SHIFT

3.5 Which Groups Deshield and by How Much? (more)

• 2 main factors influence deshielding

(1) the inductive effect of substituents, i.e mainly electronegativity

(2) the magnetic effect of substituents, which is mainly a property of unsaturated groups.

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• thus, we concern ourselves with, ELECTRONEGATIVITY, UNSATURATION and COMBINATIONS OF THESE TWO

Electronegativity. Let's look at a real spectra to illustrate...

• proximity to the oxygen deshields, increases delta

• the magnetically equivalent signals

• similarity between proton and 13C spectra. 13C spectrum gives number of equivalent carbons in the molecule

Unsaturation For example, the effect of proximity to pi bonds

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• deshielding by the benzene ring even larger than for "simple" double bonds

• note the characteristic frequencies of the aromatic protons, i.e ca. 7 ppm, and 130 ppm.

• in these spectra, the aromatic 1H signals overlap extensively and the magnetically inequivalent protons can not be distinguished, although they can in the 13C spectrum

Combined Unsaturation/Electronegativity For example, proximity to pi- bonds

• unsaturation and electronegativity combine to give largest chemical shifts

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• there are ranges of chemical shifts even for similar kinds of nuclei, but the ranges are fairly well defined.

Approximate regions are shown below

Finally, Extent of Substitution in Carbon Spectra

• 13C spectra are more sensitive to deshielding factors than proton spectra because there are more electrons around each C nucleus compared to H. Even the number of alkyl groups around a carbon can be determined in carbon spectra. Even though the electronegativity difference between C and H is small, replacing H with the slightly more electronegative C results in slightly more deshielding. Specifically, going from a primary carbon to secondary to a tertiary carbon results in signals that increasingly deshielded, let's revisit a spectrum from above....

even though both of the circled C atoms are directly attached to oxygen, the "central" carbon of the t-butyl group, C2, is significantly more deshielded than the carbon of the methyl group C2 BECAUSE it is tertiary (has more C atoms attached to it) compared to C1, which is primary.

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Chemical Shift Summary

• in a test you will be given the schematic summary of chemical shifts shown below, but you should know the "crude" correlation chart shown above. Don't set out to memorize it, you will learn most of it by default by working problems.

3.6 Signal Splitting: (N + 1) Rule and Coupling Constants (more)

• the presence of proton HB on an adjacent carbon results in TWO local magnetic fields at the proton HA due to the two magnetic moments of HB, the peak for Ha is split into TWO (it becomes a doublet)

• the presence of the TWO hydrogens HB result in THREE local magnetic fields at proton Ha, the peak for HA I split into THREE, it appears as a triplet.

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Splitting "Rules"

• N+1. The number of peaks into which a signal is split is equal to the number of non-magnetically equivalent protons with which it interacts + 1 (assumes all equal J value, which is explained below)

• in our courses, splitting only occurs with protons on ADJACENT carbon atoms

• the area ratio's of the peaks in a splitting pattern is given by Pascal's triangle

• coupling constants (J) have to be identical for interacting protons (see below)

The coupling constant is the extent to which it is split, and is given the symbol J

• in the example below, the 3 H(a)'s "split" the two H(b)'s (and vica versa).

• the splitting of both signals are identical because they are "coupled"

• J is usually ca. 7 Hz for protons on alkyl chains with free rotation

• the Table summarizes other coupling constants (you don't have to memorize these)

Coupling Constants, J (Hz)

Here is the carbon nmr spectrum of the same molecule as in the proton spectrum above

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IMPORTANT POINTS

1) All of the 13C spectra in OUR organic chemistry courses are "PROTON DECOUPLED", i.e. no splitting is observed in our carbon spectra peaks, only singlets are observed

2) The same factors that determine chemical shift in the proton spectrum determine chemical shifts in the carbon spectrum

3) There are more peaks in the carbon spectrum compared to the proton spectrum, which means that one of the carbon atoms does not have any hydrogens attached (the C=O carbon in this case)

Example Problem: which spectrum belongs to which molecule?

• how are the benzene protons C and D distinguished in this second structure?

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3.7 Signal Sizes: Integration

• In 1H spectra, the signal size relates to the number of contributing protons

• spectrometers provide relative signal sizes, historically as an integration trace, in most spectral problems you receive, however, the ABSOLUTE number of hydrogens that contribute to a signal are given (which is a lot easier)

• it is the INTEGRATED AREAS under the peaks that determine signal "size", not peak height

• signal size is related to # of equivalent H's in proton spectra

• signal size is NOT related to # of equivalent C's in 13C spectra due to spin relaxation effects

3.8 Complex Splitting (more)

• Breakdown of the (N + 1) rule occurs when protons are split with different coupling constants (J)

• Splitting by different protons with different J value results in complex splitting patterns, usually best understood with the aid of a "splitting tree" (note the "order" of the splitting does not matter)

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• the protons Ha and Hc are also doublets-of-doublets, although one of the splittings is so small that they look like simple doublets at the resolution of this spectrum

• the individual aromatic signals are not resolved in this spectrum

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3.9 Exchangeable Hydrogens

• under normal conditions, N-H and O-H protons exchange on the NMR timescale, thus....

• they appear as singlets, possibly broad (depending upon conditions)

• they do not split and are not split

• their chemical shifts depend upon conditions, hard to predict

• they are exchangeable, at any concentration, O-H signal disappears upon shaking with D2O

3.10 Real Spectra (more)

• real spectra are often not as simple as those provided in this lecture course!

• you are usually given the absolute number of hydrogens for each peak in 1H NMR, however, real integrations give only relative numbers of hydrogens, see above

• spectra are run in solvents in which the H's have been exchanged for D's, however, exchange is never complete and peaks due to solvent are often observed

• real spectra often contain peaks from impurities and other contaminants, for example the acetone used to clean the NMR tube! You will have to sort this out yourself.

• small peaks, e.g. on tertiary carbons, may get "lost" in the noise in 13C NMR spectra

Examples:

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4 Solving Spectral Problems; A Suggested Procedure (more)1. Get molecular weight and elemental information from the mass spectrum and possible molecular formulas and degrees of unsaturation

2. Get functional group information from the IR

3. Get the number of chemically inequivalent carbons, and confirm functionalities from 13C NMR

4. Compare # of signals in proton and 13C NMR's to determine whether there are carbons without hydrogens

5. Build molecular fragments and put them together to make tentative structures

6. See if the structures fit the proton NMRImportant! Try to get as much information from the other spectra before going to the proton NMR

Important! After you have your structure, predict a proton spectrum and see if it matches the provided spectrum. You should not be in any doubt as to whether you have the correct structure

Example Problem 1 Look at the spectra in this order: 1) mass 2) IR 3) 13C nmr 4) 1H nmr

The MW is 116, there are no signs of Br, Cl. The even MW suggests no nitrogen

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The IR spectrum clearly shows no C=C unsaturation (no C-H stretch above 3000 cm-1), and clearly shows a carbonyl group. This is around 1740 cm-1, and so must be an ester

• look for the methyl groups in the proton NMR, they come in integral multiples of 3, they tend to be les deshielded because they have to be at the end of a chain, and from their splitting patterns you can figure out what they are connected to

Example Problem 2

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• FOUR degrees of unsaturation immediately makes us think of the possibility of a benzene ring

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Example problem 3

• questions of this style will be on the exams

* five degrees of unsaturation, almost certainly a substituted benzene ring

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spectrum problem

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