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year 2 spec > redman nmr > Flashcards

redman nmr Flashcards

1
Q

when assigning peaks in nmr,, what. should we look at the molecule for

A

we should see if there are any resonance forms of itttt.

aka if theres double bonds, carbonyls, edg or ewg with lone pairs etc

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2
Q

what alters a groups chem shift

A

its proximity to a fg

resonance structures

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3
Q

how does resonance affect ppm

A

when we draw resonance theres normally a redistribution of e- density.

they place the e- came from will normally be (+) ,, means its less shielded and will have a higher chem shift

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4
Q

okay so if we have butanal with a double bond between 2and 3 ,, what will be seen in spec

A

we should identify that resonance can occur and draw out the structures!!

we should then see which carbonds are most deshielded and will have a higehr chem shift.

this is normally the carbons bonded to the ewg (no2, cor) and thennnnn the carbon that is (+) due to resonance.

then u look at the other carbonds and their proximity to the fg (the closer they are = the larger the effect of the fg on it)

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5
Q

okay so resonance and substituted benzene

A

the ortho position is unpredictable bc resonance occurs on it,, but its also very close to the substituent so other factors effect its chem shift

meta: no resonance so it has the same chem shift as unsubstituted benzene

para: this is the special position!!!! this is always affected by resonance!! edg = more shielded,, lower ppm ,,, ewg = deshielded = larger ppm

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6
Q

para posoition and ewg

A

(+) : deshielded = larger ppm than unsubstituted benzene values

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7
Q

para position and edg

A

(-) = more shielded = smaller ppm value than unsubstituted benzene.

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8
Q

diff forms of coupling

A

scalar dipolar

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9
Q

scalar coupling

A

between bonds

bc e- also have a magnetic field

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10
Q

dipolar coupling

A

between space

they dont need to be connected via bond

doesnt rlly matter for in solution nmr

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11
Q

when is coupling to d observed

A

when u use deuterated solvent : so basically for everything

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12
Q

d i value =

A

1

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13
Q

why would u switch h for d

A

bc in methane: u cant calculate the coupling between the h’s bc. its not oberserved bc theyre said rto be in the same env.

if u switch some for d,, there will be coupling

J.HD /J.HH = yd / yh

yd/yh = look up
J.HD = can find using spec

use this to find J.HH

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14
Q

what is gyromagnetic constant ,, y

A

constant that shows the strength of the magnetic field of the nucleus

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15
Q

the number to the lhs of J is what

A

the number of bonds between the 2 things that are coupling

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16
Q

what effects J

A
  • distance // number of bonds
  • gyromagnetic constant
  • hybridisation
  • bond angles (orbital overlap is effected)
  • electronegativity
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17
Q

J value is Hz can beeee

A

positive or negative values

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18
Q

Hz of 1J

A

280Hz

very large bc close proximity

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19
Q

2J Hz between a carbonyl

A

0 - 3 Hz

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20
Q

2J geminal (attached to the same C)

A

-20 — 5 Hz

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21
Q

3J vicinal (on adjacent C’s with a good bond rotation between them)

A

2 – 12Hz or 7Hz

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22
Q

3J trans alkene

A

12 – 18 Hz

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23
Q

3J cis alkene

A

7 - 12Hz

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24
Q

how can we find out if smt is a cis or trans alkene

A

we look at the coupling constant value,,,

7–12Hz = cis

12 - 18 Hz = trans

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25
Q

is 4J ever seen ??

A

yes girllll

4 j can be seen when theres pi bonding (double bonds) in alkenes or benzene

or in rigid structures where the 2Hs coupled via 4J are in a W position,, aka in cyclohexane or when there are 2 cyclic structures fused together

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26
Q

why is 4J normally not seen and why can it sometimes be seen in structures with pi bonding or W geometry

A

4J normally has such a small Hz coupling constant that the peaks merge together and are seen as one peak.

however,, in structures with pi bonding or W geometries,, the coupling constant values in Hz is larger than usual,, so the peaks can be resolved to see 2 peaks!!!

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27
Q

so can coupling between 2H’s on benzene (1,3 subs) be seen

A

yessss

bc its got pi bonding so the 4J coupling constant in Hz is larger thsn usual,, allowing us to resolve the 2 different peaks

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28
Q

what should we thino of when wer talking about 4J and Hz and peaks being resolved etccc

A

we should think of the splitting diagram tree!!!

and how the different peaksks in the tree are separated with a certain J Hz value.

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29
Q

whats roofing

A

when the intensity of the peaks doesnt match what we think we should get based on pascals triangle

its when u can draw a roof above the peaks on a spectra

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30
Q

4J allene Hz value

A

6hz

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31
Q

4J trans alkene hz

A

-2Hz

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32
Q

1-3 benzene subs with Hs’ 4J hz

A

2Hz

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33
Q

cyclohexane chair W geometry 4J hz value

A

2Hz

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34
Q

2 cyclic structures fused 4J hz value

A

7hz

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35
Q

normally when we do 13C nmr,, what do we do

A

we run it on proton decoupled mode so u dont see coupling to H ,,, only D from the solvent

or singlets most of the time

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36
Q

values of J. CH coupling : 1J sp3

A

125 hz (more p character = nose in the middle of p orbital : scalar coupling depends is via e- in bonds so if theres less e-,,, less hz

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37
Q

values of 1JCH sp2

A

155Hz

more s character bc less p character

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38
Q

values of 1J CH sp

A

250hz

lots of s character which has no nodes so lots of e-!!!

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39
Q

2J and 3J

A

-4 - 6 hz

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40
Q

when we say bond angle affects J value,, what do we means by bond angle

A

we mean the torsion angle,, aka we should think of a newman projection

eclipsed groups = 0 torsion angle

antconformation = 180 angle

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41
Q

eclipsed Jhh torsion angle and J value in hz

A

torsion angle = 0
12Hz

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42
Q

90* torsion angle in newman projection J value in hz

A

0 Hz

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43
Q

anti conformation, torsion angle and J value in hz

A

torsion angle = 180
14Hz

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44
Q

all the torsion angles and theyre effcect on J is wjatttt

A

its averaged out lowkey,, bc the bond is spinning freely most of the time so the groups keep rotationg so theyre always a mix of all the torsion angles

it gets averaged out to about 7Hz

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45
Q

what is the karplus relationship

A

we can find torsial angles based on J hz values!!! when a molecules rotation is hindered ( pi bonding or cyclic structures)

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46
Q

karplus relationship : cyclohexane: axial axial coupling in hz

A

180* angle
larger J or 8-13Hz

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47
Q

karplus relationship
cyclohexane: equitorial equitorial // equitorial axial ,, angle j value in Hz

A

60*
small J 2-5Hz

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48
Q

how can the karplus relationship be used

A

okay so we can lowkey find out if we have a a cis or trans cyclohexane!!

we do this by looking at the H attached to the same C as one of the substituents.

we look on the spec and we see that different peaks have different splitting J values.

we draw the 2 possible cis and trans structures and the H.

if the H is axial,, u get a a mix of small Js for the equitorial groups but u also get large J’s due to aial axial interactions!!!

if the H is equitorial,, u only get small J values bc theres no 180* interaction bc its equitorial. so. the peak with small J hz will be this one.

then u find the integration for each peak (this tells u how many H’s (aka molecules) correspond to abs of that ppm)

% of cis isomer =
integration of cis / total integration x 100

% of trans isomer =
integration of trans isomer / total integration x 100

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49
Q

whats second order coupling

A

when things coupling to eachother have very similar chemical shifts (like in this lecture when we keep moving the chem shift difference closer and closer until the chem shift difference is 0 and a singlet is visible)

so theres no observed coupling between the 2 groups,, but they are still coupled.

this gives peaks that have roofing ,, and stuff that look like a quartet when its acc a 2 doublets with roofing

aka complex splitting patters

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50
Q

how do we know if we’re going to get a complex splitting pattern

A

look if the molecule has any magnetic inequivalence (alkenes or benzene usuallyyyy) other structures too probs

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51
Q

what leads to second order effects in spec

A

multiplets caused by magnetic inequivalence

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52
Q

when do we see a nice doublet

A

if the chem shift difference between 2 different peaks IN HZ is much larger than J

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53
Q

if we decrease the chem shift value between 2 different peaks,, and reduce the chem shift difference,, what happens

A

the change in chem shift value is reduced,, its closer to the J value in Hz.

roofing occurs
where the inner peaks are larger in intensity,, and the end peaks reduce in intensity

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54
Q

if we have a doublet,, then roofing occurs where the outer peak is reduced in intensity,, what happens to the chem shift value

A

the chem shift value shifts towards the line with the higher intensity and awak from the centre

the centre being the centre when the 2 peaks were of the same intensity

when they have different intensities,, the chem shift goes closer to the more intense peak.

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55
Q

can roofing be helpful

A

yes,, if we draw the roofs,, we can find out what peaks are coupled in spec.

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56
Q

how to find normal chem shift values

A

u add the 2 ppm values of the peaks and divide by 2

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57
Q

how to find the chem shift values when u have roofing

A

(v x i) ( v x i) // i + i

where v = freq = ppm
i = intensity

u do this but obvs for each peak. to get the averaged chem shift when intensity differes and chem shift shifts towards the peak with higher intensity.

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58
Q

ppm =

A

ppm = Hz/MHz

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59
Q

when do we see roofing

A

when change in chem shift is similar to J

we dont see roofing when change in chem shift in Hz is up to x10 larger than J.

imagine 2 doublets
change in chem shift is from one of their chem shifts to another. J is the gap between the 2 dohblet peaks of 1 doublet.

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60
Q

the smaller u make change in chem shift what happens

A

the more u change in intensity of the lines

the more intense roofing u get

bc the closer lines are getting more intense,, and the shorter lines are getting less intense. the chem shift values are also being altered.

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61
Q

how can we tell if smt is rlly a quartet,, quintet etc and not just second order coupling

A

u measure the distance between the peaks

aka all the J values of the observed // apparent quartet.

if theyre all equal then they are a quartet

if theyre not then its not a quartet girl - its a second order coupling // second order coupling.

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62
Q

2 ways we can tell if smt is acc a quartet or a second order coupling this

A
  • check the J values between all the peaks,, if theyre equal then its a quartet,, if not them its a second order coupling thing
  • also chem=ck if the intensities match the theoretical intensities, aka if its a quartet,, do the ‘quartet’ peaks have a 1:3:3:1 ratio
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63
Q

okay what if we go crazy and we make the chem shift difference value 0 🫢

A

the 2 peaks will become one!! bc they both have the same chem shift,, meaning theyre both on the same ppm!!!

it looks like a singlet bc the outer peaks now have an intensity of 0,, it makes it look as if theres no coupling but there acc is,, theres just no observable coupling,, bc it looks like a singlet 2ni+1.

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64
Q

if second order coupling occurs,, what happens and why is this a problem and how can we resolve

A
  • the different peaks have the same // similar chem shift values which makes it hard to find the coupling patters as u get weird multiplets that overlap instead of nice peaks.
  • we can resolve this,, as we learnt in the previous lecture,, to switch one of the atoms in the molecule for D,, which will couple to the H’s // other atoms, then we can use the JHD value and the magneticgyro constant ratio of H and D to find the JHH value!!

JHH = JHD x hyromagnetic constant value. this is kinda long tho bc u need to add a D in a very specific specific place which is lowkey tricky

or we can make use of 13C,, which is nmr active,, so the H’s will couple to themselves ,, and the 13C (the 13C peaks will be much smaller tho!!! bc their abundance is so low)

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65
Q

what effects how spec is observed and describe everything about it

A

the magnetic field strength of the spectrum!!

when we have a spec with a higher magnetic field strength,,, aka 500MHz vc 200MHz the roofing and second order effects are minimised,, due to the chem shift difference between 2 peaks being much much larger than the J value.

more ‘predictable peaks’ from tree diagrams are observed rather than the second order peaks.

find J in Hz
find chem shift diff in hz
see what number must be multipied to get them both

then u see that the chem shift difference = XJ Hz

where x is the number than links the chem shift and J

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66
Q

how does a larger MHz effect spectra and peaks

A

less second order effects as chem shift difference,, delta delta,, is increased

and the peaks of the doublet get closer together.

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67
Q

how can we check for magnetic // chemical inequivalence

A

chemical - yk this girl, u can litch just do this via observation

magnetic: do the 2 things have the same J values to the same things. to check : draw a newman projection,,, 180* torsial angle = larger J value,,, 60* = smaller J value,, so they wouldnt have the same J values for the same atom,, so theyre not magnetically equivalent. remmeber to pick one group and compare the 2 things to the one u picked.

they could have the same nJ value,, but diff acc J values in Hz.

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68
Q

magnetic inequivalence normally leads to whattttt

A

it leads to multipletsssss!!!
weird looking peak,, dont even try to draw the tree diagram just write ‘m’

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69
Q

when is magnetic inequivalence present

A

CH2 CH2 molecules!!
aromatic molecules (1-4 substituted ones esp)

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70
Q

how can we tell if smt is chemically equivalent

A

find a mirror plane,,, spin the molecule round,, if they take eachothers place, theyre in the same chemical environment

71
Q

CH2 CH2 magnetic inequivalence cekcer

A

draw newman projection and look if the 2 things ur comparing are separated from the same things with the same angles

karplus method

72
Q

aromatic molecule magnetic inequivalence checker

A

draw the aromatic molecule and see if the 2 chemically equivalent things are the same amount of bonds away from a certain group.

nJ method

73
Q

when we’re looking to see if in spec things are averaging,, by bond rotation for example,, what should we look for

A

we should look for lone pairs and double bonds to look for any resonance effects.

we can resonate stuff onto C,, but C(-) obvs isnt very favourable so this resonance wont be very favourable,, so we can kinda forget it.

74
Q

double bonds cant rotate so if theres resonance structures what should we think

A

if theres resonance structures where youve got a new double bond.

the double bond stops rotation // more energy is needed to overcome this double bond for it to rotate

so the groups attached to the double bonds may no longer be averaged together via bond rotation,, and may give 2 different peaks, instead of one averaged peak

75
Q

when drawing resonance structures,, what should we compare between the structures

A

which resonance form has the most influecne.

we do this by finding out which one is more stable // where the (-) is more favoured.

76
Q

number of observed resonances aka number of observed peaks is affectted by what

A

rate of exchange,, aka rate of bond rotation

77
Q

diff types of exchange limits

A

fast exchange limit
slow exchange limit

78
Q

describe slow exchange limit

A

occurs at lower temp
system has less energy so bond rotates slower as theres less energy to overcome the Ea of bond rotation

2 peaks are seen,, the groups are not averageddes

79
Q

desribe a fast exchange limit

A

occurs at a higher temperature

the system has more energy,, more likely to overcpme the Ea of bond rotation,, bond rotates faster,, groups are averaged

u get 1 peak,, an average chem shift of both the separate resonances

(aka the 2 peaks u get from slow limit exchanged are averaged to give one peak at an averaged chemical shift)

80
Q

describe dmf and its different forms

A

amide with Ch3 attached to N’s

dont think of resonance forms,, think of the different structures u get from rotating bonds.

in DMF,, both the groups attached to the N are CH3,, meaning theyre both the same

so theyre equally likey to be present. one conformation is not more favoured than the other

81
Q

if both conformations are equally likely to be present,, what does this mean

A

the integration of both peaks for both conformations will be the same.

the peaks will be of the same length

82
Q

if both the conformations are equally likely to be present,, what is the Keq value

A

1!!!

bc its the same for both

83
Q

describe a slow exchange limit,, what is seen in the spec,, what is integration

A

in the spec u see 2 different resonances

due to seeing more peaks as they are not averaged out

the integration of the peaks depends on what conformation is more favoured.

84
Q

id Keq is 1,, what do we see in slow exchange limit

A

the v.cis and v.trans (aka the freq of both conformations) is both seen!!!

the integration will be equal as molecules are equally favoured.

85
Q

what if the Keq is not 1,, for a slow exchange limit

A

if Keq isnt one,, one conformation is favoured over the other

the peaks will have different intensities,, wirth the more common//favoured conformation having a larger intensity.

more abundant conformation = larger peak

lesss abundant conformation = smaller peak

86
Q

wait so when we have 2 conformations,, do to bond rotation,, what is the difference between them

A

theyre different energy conformations,, aka rotating the bond can change what intewractions are occuring between the substituents.

some conformations will have higehr energy,, some will have lower energy.

87
Q

how do we find the equailibrium constant of the reaction where ur rotating the bond to get different conformations of the same molecule,, when we have the % abundance of each conformation

A

product / reactant x100

bc K = [pro]//[rea]

so ur doing that but with the abundance as a decimal number.

88
Q

describe fast exchange limit

A
  • done at hotter temp
  • average peak is seen in an average chem shift (average but remember it depends on the abundance of each one)
  • aka average chem shift is weighted by the mole fraction of each species.
89
Q

in fast limit exchange,, if theres a 1:1 ratio of both conformations,, how do we find the averaged chem shift (v.average)

A

1:1 ratio so each conformations v (ppm x MHz) has an equal contribution to the V.average.

V.cis + V.trans / 2

90
Q

okay what about if,, in fast limit exchange,, the ratio between the conformations was 8:92,, how would u find the V.average

A

V.average is weighted by the mole fraction of each conformation

so the conformation with 92 will have a larger contribution to the v.average

V.av = 0.08V.x + 0.92V.y

91
Q

okay wait explain fast limit exchange slowly: how is the V.average weighted and what do u see and how do u see that if its only one peak

A

so imagine u lowkey have slow limit exchange so yk where the V.cis and V.trans are.

more abundant conformation = more influence on V.average

u find v.av by doing 0.08V.x + 0.92V.y

aka u multiply abundance by the V value and add them to get V.average.

92
Q

as u increase T from slow limit exchange to fast limit exchange,, what happens to the peaks

A

the peaks get broader,, they merge,, and then they sharpen into 1 peak

93
Q

what is coalescence

A

where the 2 peaks from slow limit exchange are super flat and they become one.

94
Q

what is Kex ^c

A

exchange rate constant at coalescence !!!

95
Q

whats does Keq ^c equal to

A

🔺vo (n/ root 2)

where vo = resonance separation (difference between 2 peaks) in slow exchange limit in Hz

n = pi

root 2 = square root of 2

units are s-1 bc its

96
Q

when does fast limit exchange occur

A

when Kex > Kex^c

97
Q

when does slow limit exchange occur

A

when Kex < Kex^c

where Kex^c = 🔺Vo (n / root 2)

where 🔺v = separation, in Hz between the 2 resonance peaks in slow limit exchange

98
Q

what can be altered to make smt slow // fast exchange limit

A

temp!! high temp = fast exchange limit
low temp = slow exchange limit (watch so ur solvent doesnt dissolve tho!!)

spectrometer frequency!! higher frequency = higher frequency needed for coalescence to occur bc its harder for Kex> Kex^c to occur !! bc Kex^c will be such a larger freq value. spec effects 🔺o

99
Q

what is Kex^c

A

its the exchange rate we needddd for coalescence!!!!!!!!

100
Q

how to find Kex^c

A

find the ppm difference between 2 resonance peaks

x spec frequency to get it into Hz.

do Hz(n/ root 2)

this is the freq u need for coalescence to occur,, in Hz

101
Q

what processes can move H’s

A

bond rotation

NH and OH exchange (protonating OH using H3O+,, then deprotonating the other H, so ur swapping the molecules original H with H3O+ H,, giving an average resonance of the water and the alcohol in fast exchange limit (OH exchange = fast ,, u get a single line,,, but NH exchange may be slow)

cyclohexane ring flips : axial to equitorial

organometallics and coordination chem

102
Q

organometallics and coordination chem H movement

A

ring wizzing : the C bonded to the metal keeps swapping

rotation of rings : the ring spins round and roundd so u get an average for all corners of the cyclic molecule.

substituents on double bonds in an cyclic structure hopping onto other double bonds in the structure

103
Q

when we say we pulse smt by 90* what do we mean

A

think of the rotary diagram,, the net magnetisation flips 90* to the xy plane when its excited,, this is what we mean

104
Q

13C nmr is usually alsoooo

A

1H decoupled nmr,,, so we can get singlets instead of multiplets.

so H doesnt effect splitting by giving multiple peaks.

105
Q

to decouple smt,, what must we do

A

we must irradiate it using its radio frequency to prevent there from being a population difference.

106
Q

describe 13C [H] nmr using the fid graphs

A

okay so for 13C nmr we would irradiate the molecule to excite 13C, so we draw a tall peak,, and draw the FID the 13C nucs give out after this.

then under this graph,, we draw the 1H bit, we basically draw a wide block with the word ‘decoupled’ in it,, and we make sure to line this up with the fid of 13C ,, the nuc we want to observe.

107
Q

when we decouple H,, what frequency do we use

A

we use a range of frequencies bc the Hs in a molecule all have different shifts

this is called broadband heteronuclear //homonuclear decoupling

108
Q

why is the decoupling block so wide

A

the H’s must be decoupled for the whole FID of the 13C.

its also broadband heteronuclear decoupling bc u need a large range of frequencies to irridate Hs as they all have dfferent chemical shifts.

109
Q

difference between broadband heteronuclear decoupling and broadband homonuclear decoupling

A

heteronuclear: ur observing and decoupling different types of nucleus

homonuclear : ur observing and decoupling the same type of nuc.

110
Q

irridate meansssss

A

excite

111
Q

what is selective homonuclear decoupling used for + descibe the graph for it

A

its used to work out coupling patters

we irridate and observe the same type of nucleus.

we get the irridation peak,, then the FID but superimposed onto the FID is the decoupled block.

so everything is on one graph.

112
Q

in selective homonuclear decoupling,, what is done

A

so u have ur spec, and youll have a peak that u dk what H it corresponds to.

so u irradiate the peak at its own frequency.

and irridating this ALTERS the peaks of the H’s near it in the molecule.

irridate the peak we dk
assign the other peaks based on ppm
figure out the tricky one.

bc irridating one nuc affects the others

113
Q

nuclear overhauser effect description and simplification

A

nuclear overhauser effect = NOE!!!

when we irridate one nuc,, nucs close to it via space (dipolar coupling) will also be affected. there will be a change in their intensity depending on their proximity to the irridated nucleus.

closer together = arger NOE = more intense peak

further apart = weaker NOE = weaker increase in intensity.

114
Q

when does NOE occur

A

in molecules less than 5A away

it can occur between different molecules!!

115
Q

why is NOE useful,, name the experiment and its graph

A

1D NOE difference experiment!!!

decouple the H of interest at its own radiofrequency (zap the peak that doesnt make sense),,, irridate the sample,,, measure the fid.

116
Q

1D NOE difference experiment explained

A

so u have ur spec and dk what one of the peaks correspond to .

so u zap that peak,, making its intensity 0,, and record the FID of the remaining sample to give an nmr spec. where the peak we irridated will be 0 intensity.

then u make a control spec. where u irridate the wholw molecule with a radiofrequency that no H’s absorb (look at spec and pic a ppm –> Hz where there are no peaks)

then u do irridated - control and compare the 2.

the intensity of the irridated//unknown peak will be -100. and bc ur doing irridated (where the H’s close in proximity would have experience NOE and increased their intensity) - control (no NOE effect) the irridated - control spec,, will show how much the intesnity increased by!!!!!!!

the highest intensity peak will therefore have experienced the larger NOE and would have been closest to the H we didnt know!! if we find what H this is using ppm,, we can. figure out the one we didnt know.

the small wiggles i nthe spec = artefacts and we can ignore these.

117
Q

can NOE work on heteronuclear bonds aswell

A

yes!!!

118
Q

whats a benefit of heteronuclear NOE

A

it can increase the intensity of atoms with a normally low intensity (13C) if theyre bonded to a H (that we can irridate)

we would irridate the H at the start,, then irridate the moelcule at 13C freq (bc thats the fid and spec we want) and collect the 13C fid

119
Q

when we do NOE,, when do we irridate the peak we dk

A

at the beginning,,, during relaxation decay ( at the beginning of the graph)

120
Q

max enhancement (aka increase in intensity) we get from NOE

A

1+ 1/2(ya/yb)
where a = the nuc ur irridation and b = nuc u want the fid for,, the one ur observing.

121
Q

yH / y 13C ,, and therefore what will the max enhancement be

A

4!!

max enhancement = 1+ 1/2( yh/yc)

so 3!! intensity increade of 3.

122
Q

DEPT nmr

A

C with odd H’s = upp
C with even H’s = down
quaternary carbons = gone
135DEPT

90 DEPT = CH up only,, everything else is gone. so it helps us differentiate CH3 and CH from DEPT 135.

123
Q

what does the number in dept refer to

A

the angle of theta. aka the angle of the last H irridation//pulse.

124
Q

dept: spacing begtween the H irridatioons =

A

time frame

1/2J

J = coupling constant.

125
Q

in dept,, what is the first nuc that is irridated

A

the more sensitive one,, between H and C its H. H is irridated first at 90,, then 180,, then theta (any angle)

then C is pulsed,, alongside the H’s,, but slightly after bc the most sensitive nuc bust first be irridated.

126
Q

NOE graph exp when its homonucleaer

A

decoupled block, irridation, fid.

bc we need to decouple // irridate the peak we dk.

then find the fid for the rest. to give the irridated spec.

think of the steps for NOE and then think of the graph girl.

also its homonuclear so itll be on the same line,, youve got this.

127
Q

okay so until the last 2 lectures of redman,, what dimension were the nmr

A

all other nmr = 1D

these ones = 2D

128
Q

how do we know we have a 2D spec

A

theres 2 spectrums, one on the x axis and one on the y axis

129
Q

okay so when we first start thinking about nmr and the different dimensions,, what do we need to think about + explain them

A

the graphs // fid diagrams

the more dimensions = the more time periods

u have t1, t2, t3 for 3D
just t1 for 1D
t1 and t2 for 2D.

the last time period is the acquisition time period and this has the FID on it.

between each time period theres a ‘mixing peak’ which looks like the irridation peak.

apart from the last acquisition peak,, the rest are evolution time periods

before the mixing // irridation peaks theres the preparation stage.

so we have preparation,, irridation, t1 (evolution), mixing, t2 (fid) for 2D.

130
Q

what do we do for every 2D or 1D experiment

A

for every experiment,, we need to very the length of T1 by slightly increasing it. we do this by letting it occur for longer.

131
Q

why does multidirectional nmr spec take a lot of time

A

bc u need to record many T1,, increasing its duration every time.

in order to get more data.

132
Q

for 2D ,, how many fourier transformations do we need

A

2

bc u always start with T1 and for 2D, T2 is the final one,, the one with the FID.

and u need to trnasform the T’s from seconds to frequency, cm-1.

133
Q

describe a 2D, spec

A

1D spectrum 1 on the x axis above

another 1D spectrum on the y axis

134
Q

in 2D nmr,, the type of correlation associated with 2 peaks // resonance (one on each axis) depends on what

A

depends on the pulse sequence

135
Q

whats on the x axis in 2D spec and how can we increase the resolution of it

A

on the 2D spec,, x axis = T2 = FID acquisition

the longer u spend recording the FID,, the more resolved it will be.

unless the FID has returned to 0,, then theres no point continuing the recording.

136
Q

whats the y axis on 2D spec and how do we increase its resolution

A

on the y axis theres another 1D spectrum,, but this one depends on T1,, and the more T1 recordings u make (aka the amounttt of T1 recordings) ,, the more resolved the y axis will be.

we can control the amount of T1 recordings we do. every time we do them, we must increase the length of how long it was recorded for.

137
Q

y axis in 2d

A

1d spectrum
resolution depends on the amount of T1 recordings u make

138
Q

x axis in 2D

A

another 1d spectrum
resolution depends on how long u record T2 aka FID for. the longer the more resolved,, unless the FId has already got to 0.

139
Q

when we say resolution in 2D spec what do we mean

A

think of biology - the less resolved it is,, the closer the 2 contours are together - u cant tell if the contour is made up of 2 or if its made up of 1.

140
Q

what does 2D spec help us do

A

helps us assign peaks in spec

141
Q

when u extend 2 identical peaks in 2D nmr (one from each axis) and they meet at a contour what will this contour be called.

A

autocorrelation!!

contours that appear between 2 identical peaks are called autocorrelations.

bc they automatically correlate bc theyre the same thing!!!

142
Q

whats it called when two different peaks (one from each axis) in 2D spec correlate

A

cross peaksssss

cross peaks occur when 2 different peaks correlate.

they also show what 2 peaks couple!!

the relationship // correlation between them depends on the nature of the pulse sequence.

143
Q

2D nmr when ur looking at 2 1H nmr spectra,, trying to see which H’s couple to eachother

A

1H 1H COSY

when 2 H’s couple to eachother

shows 1J, 2J, 3J and strong 4J coupling aswell

144
Q

what if theres no corss peaks in 1H 1H COSY

A

it means there are no Hs coupling

145
Q

what is 1H 1H COSY - brief

A

its a correlation experiment.

146
Q

COSY SHOWS

A

coupling between H’s!!! they couple if there are cross peaks

147
Q

what 2d experiment is the one where u see 13C and 1H coupling

A

the 1H - 13C HMQC

the homonuclear multiple quantum correlation

148
Q

describe 1H - 13C HMQC

homonuclear multiple quantum correlation

A

1J CH SEEN ONLY!!!

only 13C and 1H bonded directly to eachother are seen.

no 2JCH or 3JCH seen!!!

which C is bound to which H.

helps us assign C with similar chem shifts to H’s and seeing if they’re coupled to the same thing or not.

u will still see coupling to deuterium.

149
Q

what isnt seen in 1H 13C HMQC

A

quaternary 13C arent seen - obvssss!!!

bc its about 1H - 13C coupling,, if theyre not bonded to H, obvs they wont be there!!! womp wompw

150
Q

how to work out 1H 13C - HMQC

A

u find which H is bound to which 13C. by seeing which peaks have crosspeaks between the H and C nmr on the axis.
u need to assign the H peaks on the 1D spec and then youll see which C correlate aka couple aka are directly bonded to eachother.

151
Q

COSY CROSS PEAKS SHOW WHATTTTT

A

H and H coupling

1,2,3 and 4J coupling

152
Q

HMQC CROSS PEAKS SHOW

A

H C 1J bond coupling!!

aka H and C that are directly bonded to eachother.

153
Q

CDCl3 peak in 13C nmr and 1H nmr

A

CDCl3 = 13C nmr = 77ppm

CHCl3 = 1H nmr = 7ppm

154
Q

how close do nucs need to be to experience the NOE effect

A

5A in proximity

155
Q

drop off effect for NOE

A

1/r^6

where r = distance between nucs

156
Q

cross peaks in 1H 1H NOESY are due to what

A

cross peaks in NOESY are seen when H’s are in close proximity to eachother.

157
Q

describe the points in a 1H 1H NOESY spec

A

first slice of peaks (the ones near the top) = normal NOESY spec (1D)

second slice of peaks (the ones near the middle) = difference NOESY spec(1D)

158
Q

whats the good thing about using a 1H 1H NOESY spec

A

u get the normal NOESY and the NOESY difference spec in one!!

so u get all ur info on one spectrum.

159
Q

whats a negative about NOESY spec

A

there cant be any paramagnetic nuclei in the sample.

so we must use a vaccum to prevent O2 (which has a triplet state) from dissolving from gaseous phase in the atmosphere into the sample.

160
Q

whats another problem with NOESY spec

A

the NOE occurs via space,, and when the analyte is in a liquid,, the moelcule tumbles, which changes the magnetic field felt.

so we must measure the r.c (the rotational correlation time)

161
Q

what is r.c in 1H 1H NOESY

A

its the rotational correlation time

aka the average time it takes for a molecule to move through an angle of 1 radian.

162
Q

the larger the r.c in 1H 1H NOESY

A

the larger the r.c = the longer it takes for a molecule to rotate an angle of 1 radian.

therefore the molecule must have a larger molecular weight

or the liquid must be more viscous.

163
Q

describe 1H 1H NOESY and the r.c graph

A

so u have r.c on the x axis
and numbers on the y axis.
its a cross shaped graph.

theres a NOE and an ROE line.

they both start in the upper left box,, NOE dips below the x axis before reaching the y axis and continues being negative y axis until the end of the x axis.

the ROE remains positive throughout the whole x axis.

the LHS = smaller r.c value = small molecule + less viscous liquid

RHS = larger r.c value = larger molecule + more viscous liquid.

164
Q

what hapoens in the 1H 1H NOESY when the NOE line drops negative (below the x axis)

A

nothing is seen in the spec here, which is bad and we lowkey use ROE to avoid this bc ROE remains positive.

NOE dips past 0 if the molecule is above 1000 in mw.

165
Q

what happens when the line for NOE in 1H 1H NOESY is above the x axis (aka positive)

A

theres a high peak intensity

166
Q

what happens when the NOE dips below the x axis and becomes negative

A

u see a low peak intensity.

167
Q

describe ROESY

A

rotating frame overhauser effect

its always positive no matter the r.c

unline NOESY: which is positive for low r.c,,, but negative for larger r.c

168
Q

whats a negative of ROESY

A

it gives many artefacts ue to J coupling

169
Q

when to choose between using NOESY/ROESY or the KArplus relationship

A

NOESY//ROESY: when there are H’s that are in close proximity to eachother!! aka H’s on 2 different aromatics that are fused together etc.
- the closer they are = the more enhanced they are. the further they are,, the less enhanced they are. (helps u find the conformation of the isomer it is,, if the H is axial and near the other H’s for eg or if its equitorial and further away for eg. but tbh karplus can also be used here.

karplus is usually used when the H’s are not close to any other H’s!!
we use it to find which conformation is correct: the one with H axial (larger J value) or if the H is equitorial (small J value)

170
Q

COSEY: what spec is observed + what its used for

A

1H spec

assigning peaks in complicated spec

171
Q

HMQC: what spec is seen and what its used for

A

1H and 13C spec observed

assigning H and C peaks in H and C spec starting from known signals

172
Q

NOESY : what spec is seen and what its used for

A

1H observed

evidence for spatial proximity using NOE and enhancementssss

closer = larger nehancement
further = less enhancement.

173
Q

on the r.c graph thing,, what is on the x and y axis and what is the line

A

rc = x axis
max noe on y axis

the line shows how noe changes as r.c increases.

small molecules and non viscous solvents have a small r.c

larger molecules and more viscous liquids have a larger r.c

year 2 spec (10 decks)

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