Unit OnePart 9nmr spectroscopyequivalent hydrogensintegral vs. number of hydrogens how do we know all this info about molecules shape, interactions etc?
Part 9Unit Onenmr spectroscopyequivalent hydrogensintegral vs. number of hydrogens well a lot of our knowledge comes from looking at molecules using various forms of spectroscopy. Today we’ll look at nmr and next lecture ir and mass spectrometry...
ENERGY electromagneticradiation describes a form of energy...
ENERGY it is a continuum of energy that is arbitrarily divided into a number of regions (the most obvious being visible light)
ENERGYlow energy EM radiation has a long wavelength and low frequency (askthe physicists) whilst high energy is small wavelength and high frequency
can electromagnetic radiation interact with molecules? a stupid question...you know what the answer must be otherwise why have I introduced this topic?
sunburn...simpleinteraction of energy and molecules... yes!
the fact you can see this slide means EM(visible light) must be able to interact with our eyes yes!
why canwe getsunburnt on acloudy day ? but not warm
differentmolecules interact with differentenergy
different molecules interact...water in clouds blocks infraredenergy (which makes us hot) but withdoes not block ultraviolet energy different (which burns us) energy
so how dowe use this?
E2if we take a molecule in itsresting state (low energy or happily sitting around chilling) and... E1 molecule in energy state E1
...irradiate the molecule with energy some will be absorbed causing the molecule... E2absorption of energy E1 molecule in energy state E1
...to become excited or to rise to a higher energy level... E2absorption of energy E1 molecule in energy state E1
molecule in energy state E2 or ‘excited’ the amount absorbed E2will depend on what wehave changed within the molecule...we can measure this... E1
molecule in energy state E2 or ‘excited’eventually the molecule will relax E2 and emit energy as it returns to its resting state...we can measure this energy as well... emission of energy E1
...so, by measuring the energy absorbed or emitted we can‘see’ changes in the molecule. So, what can we see? what part of the molecule is effected?
ultraviolet-visible (UV) UV excites an electron
ultraviolet-visible (UV) high energy UV radiation can excite an electron between orbitals ...this tells us about the conjugation within the molecule (multiple bonds separated UV by one single bond) excites an electron
infrared (IR)vibrates bonds O H
infrared (IR) IR has slightly less energy andalters the vibration of bonds within a molecule...this can tell us about the functional groups within that molecule vibrates bonds O H
nuclear magnetic resonance (NMR) C-H framework H H H H H C C H C O C H H H H
nuclear magnetic resonance (NMR) NMR uses low energy radio waves to alter the spin of the nucleus of certain atoms... this can give us a lot of useful information about the position of C and H within the molecule. It is by far the most useful form of spectroscopy we routinely use C-H framework H H H H H C C H C O C H H H H
this beast is an MRI...it isidentical to an NMR machine except it spins a magnetaround a patient and an NMR spins the sample...
nobody likes the I guess it’s called MRI and notword ‘nuclear’NMR as no patient would allow a doctor to insert them in amachine with the word “nuclear” in the title
so what is nmr? ...but seriously...what does an NMR (or MRI) do?
hydrogen a hydrogen atom comprises of one proton and one electron. for nmr we are primarily interested with the single proton of the nucleus... +ve –ve atom
hydrogenthe positively charged proton is spinning and this creates a magnetic field (is it the left or right-handed rule?) nucleus spins
sample in a molecule (or collection of molecules) we have lots of protons (or hydrogens) all spinning in various directions...but if we apply a magnetic field to the sample...random orientation
sample β-spin stateMAGNETIC FIELD ENERGY α-spin state aligned
sample β-spin stateMAGNETIC FIELD ENERGY ...then all the protons will align with the field... α-spin state aligned
sample β-spin stateMAGNETIC FIELD ENERGY if we give the protons energy we can excite them so that they oppose the field (these are the only two states they can exist in once in the field) α-spin state aligned
sample β-spin stateMAGNETIC FIELD ENERGY α-spin state excited
sample β-spin stateMAGNETIC FIELD the energy required to do this is in the radio wave frequency (so not much) ENERGY α-spin state excited
β-spin sample state take away the energy source and the protons will relax (becomeENERGY aligned with the field once more) α-spin state relaxes
β-spin sample stateENERGY record energy emitted when they do this they emit energy as a radio wave...which we can measure α-spin state relaxes
β-spin sample stateENERGY record energy emitted now, the cool bit is... the amount of energy they will emit depends on the strength of the magnetic field that they experience... α-spin state relaxes
hydrogen each proton has an electron spinning around it...this is a spinning charge so it too produces a magnetic field... +ve –ve atom
β-spin sample stateENERGY record energy emitted ...so, the amount of energy emitted is influenced by the electrons around the proton or the chemical environment. As bonds are electrons we now have information about bonds / structure of the molecule α-spin state relaxes
O CH3H3C N N H O N N CH3 caffeine CH3 H3C Si CH 3 H3C TMS NMR tells us about the individual H in a molecule...
O CH3H3C N N in caffeine there are H four different kinds of H N (or proton) depending O N on where they are... CH3 caffeine CH3 H3C Si CH 3 H3C TMS
O CH3H3C N N H O N N CH3 caffeine so the NMR spectrum has four CH3 peaks...one for each kind of H H3C Si CH 3 H3C TMS
O CH3H3C N N H O N N CH3 caffeine CH3 H3C Si CH 3 H3C TMS
deshielded shielded H H H H H C X C C C C C H alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld predict roughly we can high ﬁeld δ scale where a peak will be found in the H NMR spectrum as it is related to the high low electron electrons concentration of electron density H near each density
deshielded shielded if there are lots of H H H H H electrons near to an H X C then it C set to be is C C C C H shielded... alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld high ﬁeld δ scale high low electron electron density density
deshielded shielded if there aren’t many electrons near the H it is H H H deshielded...H H C X C C C C C H alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld high ﬁeld δ scale high low electron electron density density
electron withdrawing groups pull electrons away causing objects to be deshielded and once again...deshielded shielded H H H H H C X C C C C C H alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld high ﬁeld δ scale high low electron electron density density
deshielded shielded H H...proving it is useful to H H H C X know about polarisation C C C C C of H bonds etc alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld high ﬁeld δ scale high low electron electron density density
deshielded shielded H H H H H C X C C C C C H alkenyl X = O, N allylic alkyl aromatic or halide 8 6 4 2 0 low ﬁeld high ﬁeld δ scale high low electron electron density density
correlation table Type of Type of Type of δ (ppm) δ (ppm) δ (ppm)hydrogen hydrogen hydrogen H2 C CH3 0.70–1.30 C C Ph 2.60 C CH2 4.60–5.00 H2 H2 C C CC C C 1.20–1.35 C C I 3.10–3.30 H 5.20–5.70 C H2C C C 1.40–1.65 C C Br 3.40 CHCl2 5.80–5.90 H H2H3C C C 1.60–1.90 C C Cl 3.50 Ph H 6.60–8.00H3C C O O 2.10–2.60 H2C O 3.50–3.80 9.50–9.70 C C H OH3C N 2.10–3.00 H3C O 3.50–3.80 10.00–13.00 C C OHPh CH3 2.20–2.50 Ar OH 4.00–8.00 R OH 0.50–5.50 H2 C C H 2.40–2.70 C C F 4.30–4.40
correlation table Type of Type of Type of δ (ppm) δ (ppm) δ (ppm)hydrogen hydrogen hydrogen H2 C CH3 0.70–1.30 C C Ph 2.60 C CH2 4.60–5.00 H2 H2 C C CC C C 1.20–1.35 C C I 3.10–3.30 H 5.20–5.70 C H2C C C 1.40–1.65 C C Br 3.40 CHCl2 5.80–5.90 H H2H3C C C 1.60–1.90 C C Cl 3.50 Ph H 6.60–8.00H3C C O O 2.10–2.60 H2C O 3.50–3.80 9.50–9.70 C C H shows where we would OH3C N expect 2.10–3.00 to find peaks in our O H3C 3.50–3.80 10.00–13.00 C C OH NMR data...you will alwaysPh CH3 be given a table like this OH 2.20–2.50 Ar 4.00–8.00 R OH 0.50–5.50 for your exams... H2 C C H 2.40–2.70 C C F 4.30–4.40
chemically equivalent hydrogen OH S CH3H H this is DMSO (dimethylsulfoxide)...it has six hydrogen atoms...do we see six peaks?
chemically equivalent hydrogen O S H3C CH3 nope, just one peak...why?
chemically equivalent hydrogen O O OH S H S H S CH3 CH3 CH3H H H H H H all the hydrogens are chemically equivalent...if we rotate the C–C bond we see we can put each H in exactly the same place...so we can not tell them apart and neither can the nmr. They are identical
chemically equivalent hydrogen CH3 N H3C CH3 trimethylamine has 9 hydrogen but all chemically equivalent so one peak
chemically equivalent hydrogen CH3 N H3C CH3 the peak is slightly shifted compared to DMSO as N does not pull the electronstowards it as much as the S=O group
two chemical environments CH3 H H mesityl has two peaks as it has one set of H attached to the aromatic H3C CH3 ring and one set attached to methyl groups H
three chemical environments O δ 3.42 ppm δ 2.15 ppm O H3C CH3 H H δ 4.03 ppm three chemicalenvironments (methyl ether, methyl ketone and the one in the middle) so three peaks
three chemical environments Oδ 3.42 ppm δ 2.15 ppm O H3C CH3 H H this peak shifted down field (deshielded) as the ketone is electron δ 4.03 ppm withdrawing and the electronegative O is electron withdrawing so low electron density near the two H
integration (hydrogen counting) O when the NMR prints a spectrum in normally has a second bit of O information attached...this blue H3C CH3 line (although the colour is H H unimportant)
integration (hydrogen counting) O O H3C CH3 H H this is called theintegral and its height is proportional to the area of each peak
integration (hydrogen counting) O O H3C CH3 H Hit is proportional to the number of H responsible for the 3x 3x peak 2x
integration (hydrogen counting) O O so without any knowledge of polarity H3C CH3 (electron withdrawing groups) we know that this peak must be due to the central CH2 as H H it is smaller than the other two peaks (which are caused by 3 H each) 3x 3x 2x
integration (hydrogen counting) CH3 this peak must belong to H H the nine H of the three CH3 groups H3C CH3 H 9x 3x
integration (hydrogen counting) H3C O CH3 we have 6 x H on thetwo CH3 groups so they must be the peak near1 ppm whilst we have 4 x H near the O so theyare the deshielded H at 6x 3.5 4x
integration (hydrogen counting) H3C O CH3 6x 4x but note...it is only a ratio...ethanol(CH3CH2OH) would givea very similar spectrum as the ratio 3:2 is the same as 6:4
integration (hydrogen counting) H3C O CH3also note how broad the peaks are...this is because nmr is a lotmore complicated than I’ve shown... 6x 4x
integration (hydrogen counting) H H HH H HH H H H 6xbroad as two 4x different H
integration (hydrogen counting) H H HH H HH H H H Cheeky example... The H and H turn out to be similar so overlap or 6x come at the samebroad as two 4x point...we only see one different H broad peak
integration (hydrogen counting) H H H H HH HH H H 6x 4x
integration (hydrogen counting) H H H H HH HH H H see how much sharper 6x the peak is when the H are all identical 4x (chemically equivalent)
integration (hydrogen counting) HHH H H H H H H H O HH HH HH H H H H 18x H HHH 3x 2x 1x
integration (hydrogen counting) HHH H H H H H H H O HH HH HH H H H H 18x H HHH 3x 2x 1x Hopefully, you can see that there are four chemically equivalent hydrogens. (in real speak we say they have the ‘same environment’)
integration (hydrogen counting) H H H H N H 2x 2x 1x
integration (hydrogen counting) H H H H N H the green H is most deshielded (further down field) as the nitrogen is electronegative and is pulling the 2x 2x electrons away from the H and towards itself... 1x
integration (hydrogen counting) H H H H N H so why is the red H more shielded than the blue H even though it is closer to the 2x 2x electronegative N? 1x
integration (hydrogen counting) H H H the shielding effect tends to alternate around the ring. The C with H N H the green H is deshielded (δ+) so it dragselectrons towards itself meaning the C with the red H is now slightly shielded (δ–). As the red CH has had the electrons pulled onto it, they must have come from somewhere... so the blue H is δ+... 2x 2x 1x
integration (hydrogen counting) HH H H H H H H HH H H H H H H H H HH 11x 9x very broad (>1ppm) as 7 different H (& t-Bu group)
integration (hydrogen counting) 4 HH 1 H4 2 H H 5 H H6 H HH H H 2 H 3H H 7 H 5 H3 H HH there are only seven chemically equivalent hydrogens around the ring because it is symmetrical. The axial H are different from the 11x equatorial H. Hopefully the numbering shows this... 9x very broad (>1ppm) as 7 different H (& t-Bu group)
this is chemistry......it gets more complicated...
signal splittingH3C O CH3 if we look closer the spectrum actually shows seven lines (a quartet and a triplet) 6x 4x
signal splittingH3C O CH3 6x 4x this is due to peak splitting and it tells me a lot about the structure of the molecule...luckily you don’t need to know about it...yet...
a real 1H NMR spectrum H O looks nasty but S is very useful! Tol H