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PT IA Chemistry > The Shapes and Structures of Molecules Pt. 1 > Flashcards

The Shapes and Structures of Molecules Pt. 1 Flashcards

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1
Q

what is a good thing to check to ensure you are drawing tetrahedral molecules correctly

A

draw a flat line across the point where the two lines in the plane of the paper meet, split the molecule into quadrants, if it’s a successful tetrahedral shape, there should be one bond in each quadrant

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2
Q

what are two things to note generally when drawing skeletal formulae of molecules

A
  • your structure needs to be consistent with which bonds ‘face’ which way, i.e. which direction you are viewing the molecule from
  • if you write a carbon, you MUST write everything attached to the carbon as well
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3
Q

what do we need to remember when drawing triple bonds

A

they are linear, unlike double bonds

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4
Q

what are the common abbreviations for methyl, ethyl, propyl and butyl

A

Me = methyl = -CH3

Et = ethyl = -CH2CH3

nPr = normal propyl = -CH2CH2CH3

iPr = isopropyl = -CH(CH3)2

nBu = normal butyl = -CH2CH2CH2CH3

tBr = tertiary butyl = -C(CH3)3

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5
Q

What are the common abbreviations for acetyl, phenyl, alkyl and aromatic group (aryl)

A

Ac = acetyl = -CH3CO(X)

Ph = phenyl = -C6H5 (benzene group)

R = alkyl (usually)

Ar = any aromatic (aryl) group e.g. substituted benzene ring

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6
Q

Why do we need to be careful when using lines between molecules in inorganic chemistry

A

they usually don’t represent a shared pair of electrons, simply some connection between atoms

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7
Q

why is it the case that although O- does have a truly negative charge, the oxygen on OH- doesn’t

A
  • we sometimes assume O in OH- is negative because we assume the electrons are shared equally between atoms through covalent bonds, this is not always the case as O is more electronegative than H
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8
Q

what is the name for assuming charges on certain atoms even when this might not be the true case and how do we calculate it

A

Formal charges, when calculating these we omit the fact that atoms may have different electronegativities

to calculate formal charge:

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9
Q

what is the name for assuming charges on certain atoms even when this might not be the true case and how do we calculate it

A

Formal charges, when calculating these we omit the fact that atoms may have different electronegativities

to calculate formal charge:
1) count number of electrons Ne by counting one for each bonded pair and two for each lone pair
2) calculate valence electrons associated with the neutral atom through its group Nv
3) the formal charge is (Nv - Ne)

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10
Q

what are some of the trivial names of common solvents

A

Propanone = acetone
Ethanoic acid = acetic acid
diethyl ether = ether
methylbenzene = toluene
tricholoromethane = chloroform

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11
Q

what occurs when X-rays are focused at a crystal and why

A

they are diffracted because the interatomic spacing of the atoms in the crystal is comparable to the wavelength of the x-rays

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12
Q

what can be deduced by analysing the diffraction pattern

A
  • the positions of atoms within the crystal
  • bond lengths and angles
  • how molecules are arranged and pack together
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13
Q

what diffracts the x-rays and thus what is actually produced

A

the electrons in the molecules rather than the nucleons, so it actually produces an electron density map

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14
Q

what doesn’t show up well on x-ray diffraction and why

A

hydrogen atoms don’t show up on X-ray structures because they have very low electron density

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15
Q

what do greater contours suggest on an x-ray diffraction pattern

A
  • greater electron density
  • often implies greater electronegativities or more electrons
  • contours join regions of = electron density
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16
Q

what are the main advantages and disadvantages of x-ray diffraction

A

Advs:
- ultimate method for structural information

Disadvs:
- Need good quality crystals
- Sometimes hard to locate atoms

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17
Q

what is the crudest/original method to ionise molecules

A

fire high energy electrons at the vapourised sample, they ‘knock’ electrons off of the molecules

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18
Q

what is the more gentle/effective way to ionise molecules

A

electrospray,
- a sample is introduced as charged aerosol droplets
- the solvent evaporates leaving charged molecules
- what is detected is not the M+ ion but the molecule with the ion stuck to it

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19
Q

what is important to remember when looking at a mass spectrum

A
  • use the specific masses of the isotopes not the average molecular mass
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20
Q

how can we differentiate between molecules or fragments which have very similar/the same Mr using mass spec, what else can it be used for

A
  • take a very high resolution mass spec
  • use exact mass values and account for the electron loss/gain
  • e.g. N isn’t just 14, it’s 14.00307gmol^-1
  • this can also be useful to identify large molecules where there are a large range of possibilities of structure
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21
Q

what is fragmentation in mass spectrometry and what are the products of fragmentation

A
  • its where the ionised molecule becomes unstable and breaks apart
  • it will form a smaller ion and a radical
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22
Q

how can fragmentation be useful

A
  • it can give clues to the structure of the molecule
  • each molecule will have a unique fragmentation pattern, this means the patterns can be used for analysis and identifying compounds whose spectra have already been recorded
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23
Q

what is MS/MS

A
  • it is possible to isolate ions formed from the initial fragmentation in a mass spectrometer
  • these can then be ‘followed’ and it can be observed how these fragment and what they form on a mass spectrometer
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24
Q

Give the advantages and disadvantages of mass spectrometry

A

Advs:
- gives molecular formula
- excellent for analysis of mixtures, the components of the mixture can be identified through fragmentation patterns
- tiny sample needed, mass spectrometers are very sensitive

Disadv:
- often difficult to interpret

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25
Q

what features in a molecule are quantised and can be changed/analysed through certain frequencies of light

A
  • electron energy level
  • rotational energy
  • vibrational energy
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26
Q

what is the equation linking the difference in energy levels of the quantised properties of atoms and the energy of a photon

A

ΔE = hf

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27
Q

what do photons from each region of the EM spectrum cause transitions in

A

Radio waves - transitions in alignment of nuclear spins

Microwave - transitions in rotational energy

Infrared - vibrational transitions

Visible/ UV/ X-rays - Electronic transitions

Gamma - Nuclear Processes

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28
Q

what are the two main ways to present an IR spectrum

A

transmission percent against wavenumber

OR

absorbance units against wavenumber

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29
Q

what is wavenumber and what are the 2 main equations for it

A

wavenumber is the number of wavelengths per unit length (usually cm)

wavenumber (v^)(cm^-1) = 1/wavelength(cm)(lambda)

wavenumber (v^)(cm^-1) = frequency(v)(Hz) / c (ms^-1)

hence wavenumber is proportional to frequency

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30
Q

Given wavenumber is proportional to frequency, what does this tell us about absorptions on the left of the spectrum

A
  • they have a higher wavenumber, hence a higher frequency photon causes it, hence they require more energy
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31
Q

what is an important thing to remember about absorptions at particular frequencies

A

it is not directly caused by the vibration of one bond but can indicate a bond’s prescence

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32
Q

how can we model a vibration between two atoms and what is the graph that can indicate this, give the main features of a graph

A
  • We can consider a bond (length r) as a spring with a mass attached to each end
  • As the bond is stretched or compressed the energy rises and a force tries to restore it to an equilibrium position
  • the graph is E against r, it starts very high asymptotic to y, drops below the x axis, the lowest point is the equilibrium position, it then rises again and is asymptotic to the x axis
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33
Q

what is the equation for the frequency of the oscillation of the bond

A

v = sqrt(Kf / m)

v is frequency
Kf is force constant of bond
m is mass

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34
Q

what is the equation for the frequency of the oscillation of the bond

A

v = sqrt(Kf / m)

v is frequency
Kf is force constant of bond
m is mass

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35
Q

is there a correlation between bond strength and force constant

A

Yes, it’s a positive correlation but remember they are not the same thing

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36
Q

how is the model adapted for diatomic molecules/bonds where both atoms vibrate, how do we calculate this adaption

A

we need to now use the reduced mass for the system (μ)

μ = m1*m2 / (m1 + m2)

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37
Q

how does the reduced mass calculation change where one mass is much larger than the other

A

μ = m1*m2 / (m1 + m2)

if m1 is much greater than m2 then
μ = m2

because m2 on the bottom is negligible so the m1’s cancel

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38
Q

how do we calculate the frequency of vibration (wavenumber) for a vibrating diatomic

A

v^ = (1 / (2 x pi x c)) (sqrt(Kf / μ))

where
v^ = wavenumber (cm^-1)
c = speed of light (cm s^-1)
Kf = force constant (N m^-1)
μ = reduced mass of system (kg per molecule)

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39
Q

what are the two things that might cause a high wavenumber in a bond

A
  • a light atom attached to a heavy atom, this reduces mu, hence as v (wavenumber) is proportional to 1/sqrt(μ), v increases
  • bond strength, triple > double > single in bond strength and hence wavenumber as stronger bonds will have a greater force constant Kf, thus meaning v is greater as v is proportional to sqrt(Kf)
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40
Q

which types of bond typically vibrate at which regions in an IR spectrum, give values

A

4000 - 2500 cm^-1 = X-H single bonds

2500 - 2000 cm^-1 = C,C triple bonds and C,N triple bonds

2000 - 1500 cm^-1 = C,C and C,O double bonds

1500 - 500 cm^-1 = other single bonds (not with H) and other vibrational modes e.g. bending

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41
Q

what is the true way that molecules vibrate when interacting with IR and what are these called

A

regularly it involves all of the atoms in the molecule vibrating, these change depending on the molecule but are called Normal Modes

this shows that its not true to say one absorption is the presence of one bond in a molecule BUT vibrations may occur at a particular wavenumber in molecules containing certain groups

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42
Q

what are some examples of Normal Modes, use ethyne as an example

A

C-H symmetric stretch (in antiphase)

C-H antisymmetric stretch (in phase)

C,C triple bond stretch

Trans bend

Cis bend

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43
Q

what feature of a bond affects the size of the absorption on an IR spectrum

A

the dipole moment on the bond, when oscillating

bonds with no dipoles don’t cause an absorption but may still vibrate

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44
Q

when might a bond not cause an absorption in an IR spectrum, what can we use instead to analyse it

A

if the bond is completely symmetrical/has no dipole, Ramen Spectroscopy

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45
Q

How can we measure the vibrational frequencies/wavenumbers for bonds which have no dipole moment, what does it measure

A

using Ramen spectroscopy

IR spec observes frequencies of light absorbed by a sample

Ramen spectroscopy observes frequencies of light scattered by a sample

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46
Q

what are the wavenumbers for the C-H groups and the specific types on an IR spectrum

A

C-H = 2900 - 3200 cm^-1

tetrahedral (sp3) C-H = just less than 3000 cm^-1

trigonal (sp2) C-H = just over 3000cm^-1

C sp H = sharp peak at around 3300 cm^-1

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47
Q

where will an N-H peak occur and why may two peaks occur (if NH2), on an IR spectrum

A

N-H = around 3300cm^-1

two peaks may occur due to symmetric and antisymmetric stretching

symmetric = 3300cm^-1
antisymmetric = 3400 cm^-1

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48
Q

what is the most common shape and location for an OH peak on an IR spectrum, why?

A

O-H = approx. 3000-3500cm^-1

it is a very broad peak due to hydrogen bonding in OH groups, this is because it will cause many different strengths of bonds in a certain range

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49
Q

when might an OH group peak have a different shape to its usual on an IR spectrum

A

when it doesn’t contain hydrogen bonding it may show a sharp peak

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50
Q

where do carbon-carbon and carbon-nitrogen triple bond peaks occur on an IR spectrum

A

Carbon-Carbon triple = 2100 - 2250 cm^-1 (weak because small dipole)

Carbon-Nitrogen triple = 2250cm^-1 (strong absorption because large dipole)

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51
Q

where does a C=C double bond peak show up/different types, on an IR spectrum

A

C=C = 1635-1690 cm^-1

benzene = number of peaks at 1450-1625 cm^-1

both are fairly weak intensity

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52
Q

what peaks could we expect from an NO2 group and why, on an IR spectrum

A

two peaks due to an symmetric stretch and an antisymmetric stretch
(remember to draw NO2 with one double bond, one single and some formal charges)

symmetric stretch = 1350cm^-1

antisymmetric stretch = 1530cm^-1

both are strong peaks

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53
Q

where is the peak for a ketone group on an IR spectrum

A

ketone = 1715 cm^-1

(if nothing fancy is occuring)

54
Q

why might a carbonyl bond absorption occur at different frequencies on an IR spectrum and what causes this
(4)

A
  • if the carbonyl bond is stronger it absorbs at a higher frequency
  • if the carbonyl bond is weaker it absorbs at a lower frequency
  • any groups nearby which donates electrons will weaken the bond
  • any groups nearby which withdraw electrons will strengthen the bond
55
Q

where does the carbonyl bond absorption of an acyl chloride occur on an IR spectrum and why

A

acyl chloride C=O = 1750 - 1820cm^-1

  • this is higher than the standard 1715cm^-1 for ketone (+50)
  • because chlorine withdraws electrons and strengthens the bond
56
Q

where does the carbonyl bond absorption of an amide occur on an IR spectrum and why

A

amide C=O = 1640-1690cm^-1

  • this is lower than the standard 1715cm^-1 for a ketone (-50)
  • because nitrogen donates electrons and weakens the bond
57
Q

where does the carbonyl bond absorption of a carboxylic acid occur on an IR spectrum and why

A

carboxylic acid carbonyl = 1730cm^-1

  • this is slightly higher than the standard 1715cm^-1 for a ketone (+15)
  • because oxygen slightly withdraws electrons and strengthens the bond
58
Q

where do the carbonyl bond absorptions occur in an acid anhydride on an IR spectrum and why

A

acid anhydrides show two absorptions due to symmetric and antisymmetric stretches

antisymmetric stretch = 1820cm^-1

symmetric stretch = 1750cm^-1

59
Q

what is conjugation and what effect does it have on the frequency at which a carbonyl bond absorbs at on an IR spectrum

A
  • the carbonyl is said to be conjugated with a C=C double bond if the two double bonds are separated by one single bond
  • it lowers the base frequency by 20-30 cm^-1
60
Q

how does the size of a carbon ring affect the frequency at which a carbonyl bond on that ring absorbs at on an IR spectrum

A
  • the smaller the ring, the higher the frequency that the carbonyl bond absorbs at

moving from 6,5,4,3 carbons in a ring, frequency increases by 30-40cm^-1 each time

61
Q

why does the size of the carbon ring have an effect on the stretching frequency of the carbonyl bond

A

the smaller the angle in the ring, the more the C-C bonds have to compress during the vibration, this requires more energy and hence frequency increases

62
Q

What is the underlying principle behind NMR spec

A
  • nuclei posses spin
  • thus they have a small magnetic field
  • this interacts with a strong external magnetic field causing different energy levels
  • radio waves of a suitable frequency cause transitions between these energy levels
63
Q

what is the symbol for nuclear spin and how many energy levels exist for any given nuclear spin, what are these

A
  • Nuclear spin is specified by the quantum spin number I
  • there will be (2I + 1) energy levels
  • these will be from +I to -I in integer steps
64
Q

How can we predict the nuclear spin of a nucleus

A
  • nuclei with odd masses have half-integral spin, 1/2, 3/2 etc.
  • nuclei with odd numbers of protons and odd numbers of neutrons have integral spin 1,2,3 etc.
  • nuclei with even numbers of protons and even numbers of neutrons have 0 spin
65
Q

what does the exact difference in energy between the different spin states depend on

A
  • strength of external magnetic field, stronger field = bigger difference (directly proportional)
  • the specific nucleus
66
Q

What causes the differences in chemical shifts for different proton/carbon environments

A
  • the difference in energy between energy states (and hence frequency of absorption) depends on the magnetic field experienced by the nucleus
67
Q

What causes the differences in chemical shifts for different proton/carbon environments

A
  • the difference in energy between energy states (and hence frequency of absorption) depends on the magnetic field experienced by the nucleus
  • freq of radio waves for resonance directly proportional to B field experienced by nucleus
  • the greater the shielding around a nucleus the lower the B field hence the lower the chemical shift
  • if a nucleus is near electronegative groups, shielding decreases, chemical shift increases
68
Q

When are peaks in exactly the same place (NMR)

A
  • this occurs where the nucleus is in the exact same environment (due to symmetry)
69
Q

why do we use chemical shift and how is it calculated

A
  • the exact frequency that a nucleus resonates at changes depending on the magnetic field
  • we can standardise it so that the same compound gives peaks at the same chemical shifts

chemical shift, delta (ppm) = 10^6 * (frequency of

69
Q

why do we use chemical shift and how is it calculated

A
  • the exact frequency that a nucleus resonates at changes depending on the magnetic field
  • we can standardise it against a reference compound so that the same compound gives peaks at the same chemical shifts

chemical shift, delta (ppm) = 10^6 * (frequency of resonance - frequency of reference) / frequency of reference

  • this works because any changes in magnetic field affect all frequencies equally
70
Q

what do we use as a reference compound and why

A
  • TMS, Silicone with 4 methyl groups
  • its highly shielded so appears low on spectrum
  • its chemically inert and only has one peak
71
Q

Give the chemical shifts of
Si(CH3)4
CH3CH3
CH3Cl
CH3NH2
CH3OH
CH3NO2
CH3F
on a 13C NMR spectrum

A

Si(CH3)4 = 0 by definition
CH3CH3 = 7
CH3Cl = 26
CH3NH2 = 28
CH3OH = 50
CH3NO2 = 61
CH3F = 72

72
Q

What sort of Carbon environments occur at 0-50ppm

A

Tetrahedral carbons, sp3
CH3-CH3 or any just Carbon and hydrogen tetrahedral

CH2-NH2 at about 30ppm

73
Q

What sort of carbon environments occur at about 50-100ppm

A

tetrahedral/sp3 carbons with very electronegative atoms attached, and sp carbons e.g.

CH3-O-
C triple bond C

74
Q

what sort of carbon environments occur at 100-150ppm

A

sp2 carbons
benzene at about 123ppm
C=C double bond
C=C-O at about 150ppm

75
Q

what sort of carbon environments occur at 150-200ppm

A

sp2 carbons with very electronegative atoms attached
Carbonyl groups

76
Q

where do Ketones/aldehydes generally occur on a 13C NMR spectrum

A

around 200ppm

Ketones - typically just over 200ppm (unless conjugated)

Aldehydes - typically just under 200ppm

77
Q

Where do carbonyl derivatives (other than aldehyde and ketone) occur on a 13C NMR spectrum

and why is this surprising?

A

carboxylic acids, ester, acyl chloride, amide, acid anhydride

all occur around 160-170ppm
they occur at a lower chemical shift even though they have more electronegative groups attached

78
Q

what is different about peaks of quaternary carbons (ones with no hydrogens attached) on a 13C NMR spectrum

A

the height of the peaks is lower

79
Q

what can happen in 13C NMR when two nuclei both have spin of 1/2

A

the two nuclei can couple forming a doublet rather than a singlet

80
Q

what causes the doublet to form from the 1/2 spin nuclei

A
  • the two spin states can either reinforce, increasing the experienced magnetic field, forming a peak at a higher chemical shift
  • or the two spin states can reduce the magnetic field experienced and form a peak at a lower frequency
81
Q

where do the peaks of a doublet (due to coupling) in 13C NMR form relative to the non-coupled peak

A

they are symmetrically arranged either side of where the peak would occur had there not been any coupling

82
Q

what does the distance between the peaks of a doublet (due to coupling) in a 13C NMR spectrum represent

A
  • the value of their seperation IN HERTZ (Hz) is equal to the coupling constant, J
83
Q

what does coupling (in NMR) occur through and what notation can we use to represent it

A
  • coupling occurs through bonds
  • hence we rarely see coupling (in 13C NMR) over more than one or two bonds
  • we can write nJx-y to represent the coupling where n is the number of bonds it’s coupling over, and x-y are the atoms involved in the coupling
84
Q

why might we not see C-H or C-C coupling (in a 13C NMR spectrum) given they both have 1/2 spins

A
  • C-H won’t appear because 13C NMR spectra are usually proton decoupled
  • C-C won’t appear because it would only appear if both carbons were C13, this is only likely to occur naturally in 0.01% of cases and these will get lost in the noise
85
Q

why might we not see coupling between two atoms that are able to in an NMR spectrum

A
  • if the two nuclei are in equivalent environments then no coupling can be seen (even if it still occurs)
86
Q

What can we use to help predict number/location of peaks where multiple couplings are occuring

A

tree diagrams:

  • easiest to do highest J first then smaller J’s but gives same result
87
Q

why might the IR spectrum for a carboxylate ion appear weird

A
  • it does the same ‘bond and a half’ thing that NO2 does and has symmetric and antisymmetric stretches
88
Q

what occurs on an NMR spectrum when an atom couples to 2 identical nuclei (all spin 1/2)

A
  • a triplet forms with the intensities in a ratio 1:2:1
  • the highest peak occurs where both of the coupled atoms have spin up
  • the lowest peak occurs where both of the coupled atoms have spin down
  • the middle peak occurs where there is no net spin i.e. 1 up, 1 down, the shift is the same as without coupling
89
Q

how can we visualise the coupling that occurs in NMR when an atom couples with 2 identical nuclei (all spin 1/2)

A
  • use a tree diagram, the first coupling occurs as normal, the second coupling occurs such that the middle two peaks overlap giving it a greater intensity
  • this only works because the coupling constants for each coupling are identical
90
Q

what occurs on an NMR spectrum when an atom couples to 3 identical nuclei (all spin 1/2)

A
  • it appears as a quartet with intensities 1:3:3:1
  • the highest peak occurs where all 3 nuclei are spin up
  • the slightly higher than standard peak occurs where there’s a net spin of 1 up
  • the slightly lower than standard peak occurs where there’s a net spin of 1 down
  • the lowest peak occurs where all 3 nuclei are spin down
91
Q

how can we visualise the coupling that occurs in NMR when an atom couples with 3 identical nuclei (all spin 1/2)

A
  • draw a tree diagram
  • coupling to first forms doublet
  • coupling to second forms triplet like as before
  • coupling to third forms quartet
92
Q

what is the general rule for the number of peaks that occur (in NMR) where a nucleus couples to n equivalent nuclei with spin I

A

its resonance signal will be split into (2nI +1)

where spin = 1/2 it is
(n+1)

93
Q

What is the best way to analyse coupling where the nuclei its coupling to are not equivalent

A
  • draw a tree diagram where each coupling is considered separately
  • consider the coupling with the greater coupling constant first
94
Q

How are 13C NMR spectra proton decoupled and why is this useful

A
  • 13C NMR spectra are recorded such that coupling to protons does not occur
  • when the spectra are being recorded, the protons are irradiated in such a way as to cause their nuclear spins to move rapidly between up and down
  • if this happens fast enough, the couplings average to 0 and don’t show up
  • this is called broadband proton decoupling
95
Q

what additional information can we tell about carbon environments from a proton coupled spectrum

A
  • you can see whether carbons are CH3, CH2, CH or quaternary

CH3 = quartets
CH2 = triplets
CH = doublets
quarternary = singlets and shorter

96
Q

what are the disadvantages of proton coupled 13C NMR spectra

A
  • the signal strength is weaker
  • other signals can be lost in background noise
  • multiplets can overlap and make it difficult to see different environments
97
Q

what is APT and what can we find out from it

A
  • APT = attached proton test
  • the peaks from the carbons with an even number (0,2) of protons attached point one way, the peaks from the carbons with an odd number of protons attached point the other way (1,3)
98
Q

what is a consequence on 1H NMR spectra of the naturally low abundance of 13C

A
  • very little coupling to 13C from protons occurs
  • it usually gets lost in background noise unless the compound is 13C enriched
99
Q

what can happen in NMR spectra where there’s a middle proportion of a half spin isotope (e.g. 30%) attached to the atom being observed

A
  • some of the atoms will couple to the half spin isotope forming a doublet
  • some of the atoms won’t couple as they will be attached to the non-half spin isotope, which will form a singlet
  • this will appear as a triplet even though it isn’t
100
Q

what are the typical shifts that occur in 1H NMR spectroscopy and what do we need to remember about it

A

approx 0-14ppm

  • this is NOT just the bit at the start of a 13C NMR spectrum, it occurs at a whole different set of frequencies
101
Q

what is a key difference between 13C NMR and 1H NMR regarding peak height

A
  • we need to be very careful observing the heights of 13C NMR peaks, only quaternary carbons can tell us anything important
  • 1H NMR peaks can be integrated to give relative numbers of protons in certain environments
102
Q

how can we analyse the integration values in a 1H NMR

A
  • the integral is usually shown as a snake-like grey line
  • measure the height of each grey line
  • the ratio of the heights of the lines is the simplest whole number ratio of the number of protons in each environment
103
Q

over how many bonds do we usually see coupling in 1H NMR and what are their more common names

A
  • two bonds = geminal coupling
  • three bonds = vicinal coupling
104
Q

which sorts of coupling on 1H NMR spectra occur and which is the biggest

A
  • 3 bond coupling where single bonds/rotation = approx. 7Hz
  • 2 bond coupling where single bonds/rotation = approx. 13Hz
  • 3 bond trans coupling where double bonds/no rotation = approx. 15Hz
  • 3 bond cis coupling where double bonds/no rotation = approx 8Hz
  • 2 bond coupling where no rotation/double bonds = 0-3Hz
105
Q

what is roofing

A

“in H1 NMR for two signals that are coupling to each other, it is found that the intensity of the outermost lines decreases while the intensity of the innermost lines increase, the strength of the effect increases where the difference in frequency between the two hydrogen environments is lower”

106
Q

how can we calculate the chemical shift of a 1H NMR peak that has undergone roofing

A
  • its the weighted average of the two peaks, NOT just halfway between the two peaks
  • this is usually different because the inner peak will be bigger so the weighted average will sit closer to the centre
107
Q

why can proton NMR spectra become very complicated very quickly

A

if there are many protons with almost the same environment then their chemical shifts will be similar and they will overlap

108
Q

what peaks are likely to occur at 0.5, 1.0, 1.5ppm respectively on a 1H NMR spectrum

A
  • protons on three membered rings have an unusually low and complicated peak at about 0.5ppm
  • protons on a methyl (CH3) group occur at about 1ppm
  • protons on a CH2 group occur at about 1.5ppm
109
Q

what peaks occur at around 2.5ppm on a 1H NMR spectrum

A
  • protons on a terminal alkyne
  • protons on a carbon adjacent to a C=C or C=O double bond
  • protons on a carbon adjacent to a benzene ring
  • protons on a carbon adjacent to an amine group
110
Q

what peaks occur at around 3.5ppm on a 1H NMR spectrum

A
  • protons on a carbon bonded to a Cl
  • protons on a carbon as part of a C-O ether bond
  • protons on a carbon adjacent to an amide group
111
Q

what peak occurs at around 4ppm on a 1H NMR spectrum

A
  • protons on a carbon adjacent to an ester group
112
Q

what peak occurs at around 5-5.5ppm on a 1H NMR spectrum

A
  • protons directly attached to a C=C double bond
113
Q

what peak(s) occur at about 7.5ppm on a 1H NMR spectrum

A
  • protons on a benzene ring
114
Q

what peak occurs at 8-8.5ppm on a 1H NMR spectrum

A
  • a proton on a methanoate ester
115
Q

what peak occurs at 9.5-10.5ppm on a 1H NMR spectrum

A
  • protons on an aldehyde
116
Q

what are some useful things to remember when analysing peaks in a 1H NMR spectrum

A
  • the shift of protons in methyl groups is around 0.5ppm less than the shift of alkyl chains
  • the shift of protons on a 3 membered ring is unusually low
  • the shifts of protons directly attached to aromatic rings is higher than expected/than alkenes
  • the shifts of protons on terminal alkynes is lower than expected
117
Q

why are the shifts of protons on aromatic rings higher than alkenes

A
  • the local magnetic field set up by the pi system reinforces the applied magnetic field for protons outside of the ring
  • this increases the magnetic field ‘felt’ by the protons
  • hence they have a higher chemical shift
118
Q

why are the shifts of protons on terminal alkynes lower than expected

A
  • the triple bond sets up a ‘cylinder’ of electron density
  • this sets up a local magnetic field which counteracts the applied magnetic field for a proton on the end of the alkyne
  • hence it experiences a lower local magnetic field and has a lower chemical shift value
119
Q

where can protons on a water molecule appear on a 1H NMR spectrum

A

around 1.5ppm for standard water

around 4.5ppm for deuterated water

120
Q

where can protons on an alcohol group appear on a 1H NMR spectrum

A

usually <5ppm for alcohol groups on carbon chains

around 7ppm for alcohol groups on an aromatic ring

121
Q

where can protons on an amine group appear on a 1H NMR spectrum

A

around 2.5ppm for a standard amine
around 4.5ppm for an amine on an aromatic ring

122
Q

where do protons on an amide group and protons on a carboxylic acid group appear on a 1H NMR spectrum

A
  • around 8.5ppm for amide
  • around 10.5ppm for carbox acid
123
Q

what features do peaks for protons on NH and OH groups usually have and why

A
  • broad peaks and can appear over a larger range due to H bonding
  • they can be made to ‘dissapear’ by shaking with heavy water, they don’t actually dissapear but undergo proton exchange from H bonding with the D2O, the resonance frequencies for D are totally different than for 1H, so they don’t show up
  • they usually appear as singlets because the protons are exchangeable protons are rapidly swapping in solution, so some are spin up and some are spin down and nearby protons ‘see the average’ so no coupling occurs
124
Q

what is a way that coupling of NH and OH protons can actually be observed

A
  • by preparing a very dilute solution in a non-H bonding solvent and making the solution extremely dry
125
Q

why don’t we usually observe coupling between nuclei where there’s a spin of greater than 1/2

A
  • a process called relaxation occurs where the nuclei changes spin states, this occurs much quicker in nuclei where the spin is greater than 1/2 so all the couplings to that nucleus go to zero
126
Q

what is a common example of an atom with a spin greater than 1/2 which relaxes sufficiently slowly for coupling to occur

A
  • Deuterium
  • spin 1
  • using (2I + 1) there are 3 different spin states (1,0,-1)
  • this means that the coupling to it forms a triplet with equally intense peaks
  • where there’s coupling to multiple peaks, use (2nI + 1) to determine how many peaks will be present and draw a tree diagram to determine intensities
127
Q

why can’t we use CH2Cl2 as a solvent in NMR

A
  • we would only obtain peaks from it as it is in a large excess compared to the target compound
128
Q

what occurs to a carbonyl bond when it is 1 bond away from a C=C double bond or a species with a lone pair, what affect can this have on the 13C and 1H NMR values

A
  • it delocalises, this can give the carbon of the double bond a formal positive charge and the oxygen of the carbonyl a formal negative charge
  • this may make the carbon that is formally positive and the protons attach to it resonate at a higher frequency because they are less shielded so experience a greater B field
129
Q

what can occur on the IR spectrum of a carbonyl bond, if it delocalises with a nearby C=C double bond or species with a lone pair

A
  • it will absorb at a lower frequency because it will form a sort of ‘bond and a half’ which is weaker than the double bond
  • it’s weaker because it gains electron density
130
Q

what should we be careful of when considering 1H NMR couplings to hydrogens in different environments

A
  • it is possible that even if the hydrogens that you are coupling to are in different environments, they may have the same coupling constant to the hydrogen of interest
    (think vicinal, geminal coupling etc.)

thus it may form a different multiplicity to what is expected e.g. sextet not td

PT IA Chemistry (9 decks)

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