135B- organic spectroscopy Flashcards

1
Q

DBE equation

A

nitrogen could also be boron

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2
Q

DBE above 4 generally looking at aromatic substance

A
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3
Q

DBE for this

A

1

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4
Q

equation for wavenumber

A

length is usually in cm-1

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5
Q

E=hf equation units

A
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6
Q

equation for c speed of light

A

this is in ms-1, but if you want cm-1 use 3x10^10

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7
Q
A
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8
Q

which electrons are higher in E/ HOMO

A

lone pairs as they are non bonding

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9
Q

Why do we only usually see electrons promoted to π* not σ*

A

π* is lower in energy as worse overlap so it is easier to be promoted to something of lower E

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10
Q

This will have two absorption bands as there is a choice of 2 electrons to be promoted. Which promotion is higher in E

A

pi–> pi*
n–>pi* lone pair is higher E than the pi electron so there is a lower jump in energy

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11
Q

On a UV spec where will the higher energy photons be on the frequency axis

A

higher energy is closer to 0
lower wavelength= higher frequency higher frequency is higher energy E=hf

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12
Q

which has lower HOMO/LUMO gap

A

red/ blue peak is lower than red E=hf

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13
Q

Why is graphite black but diamond white (HOMO/LUMO gap)

A

in graphite carbons are sp2 hybridised and have a lone electron in p orbitals/ delocalised so has lots of conjugation. Very small HOMO/LUMO gap so can absorb all frequencies/wavelengths of light and is black

diamond is white as has high HOMO/LUMO gap

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14
Q

What makes a smaller HOMO/LUMO gap

A

conjugation

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15
Q

will a higher conjugation molecule have a lower/higher molar extinction co-efficient

A

higher as smaller HOMO/LUMO gap so can absorb more frequencies of light

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16
Q

what colour would this be

A

colourless/ absorbing nothing in the visible light region

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17
Q

units for molar coefficient ε

A

M-1 cm-1

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18
Q

what is the Beer-Lambert Law (ε)

A

A = ε·c·l
c= M
l= cm

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19
Q

what is CD

A

difference in absorbance between left and right-handed circularly polarized light

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19
Q

enantiomer CD spectral should be mirror images

A
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20
Q

what is c speed of light in cm-1 instead of m

A
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21
Q

what are the equations for working out absorbance and transmittance

A
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22
Q

How to plot a graph using Beer-Lambert Law A = ε·c·l

A
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23
Q

What are needed for fast/ slow vibrations

A
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24
Q

what does a higher wavenumber mean in terms of energy

A

higher wavenumber= high energy

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25
Q

What is the only type of vibration where the bond length changes

A

stretching

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26
Q

Why will molecules never not vibrate

A

we can never get to 0K

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27
Q

axis for IR spec (transmittance)

A

wavenumber from highest to lowest
and transmission

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28
Q

axis for UV spec (absorbance)

A

wavelength (nm) vs absorbance

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29
Q

Where is the fingerprint region

A

below 1500 cm-1

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30
Q

C=O stretch on IR

A

1700 cm-1
ester is 1740

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31
Q

C-O stretch on IR

A

1100 cm-1

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32
Q

C-H stretch on IR

A

3000 cm-1

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33
Q

O-H stretch on IR

A

3300 cm-1

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34
Q

CO2 stretch on IR

A

2350 cm-1

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35
Q

Why is the acid OH stretch broader than the alcohol stretch

A
  1. acid OH vibrating at slightly lower E (lower WN)
  2. forming hydrogen bonds slightly stretching O-H in HOH bond making it weaker than the other one. The one that’s H bonding is going to give you broad/ the one that wouldn’t would give steep (if O would only form 1H bond (one H not both)
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36
Q

How will an IR spec be skewed if the same is wet

A

will have an OH stretch

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37
Q

Why do arene C-H bonds and alkyl C-H bonds have different wavenumbers

A

In arene C is sp2; bonds are slightly stronger
In alkyl C is sp3 so bonds are weaker

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38
Q

How do you tell the difference in IR spec between an aldehyde and its conjugate ketone (ketone and aldehyde between A and B graphs)

A

Aldehyde= A
Ketone= B
only an aldehyde will have the alkl and arene peak/ the two fingers at about 3000
aldehyde will have the twin peaks and ketones will just have the one

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39
Q

explain why a ketone needs more energy to vibrate the C=O bond than an ester

A
  1. C-O bond is polarised in the ester as O is more electronegative/ draws e- towards/ C becomes δ+
  2. In C=O the electrons usually lie more towards the O, but because C-O is drawing electrons away from C, the e- in C=O are pulled more towards an equal sharing
  3. So needs less energy to vibrate
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40
Q

Which plate of the detector will get the heavier fragment

A

higher mass= less travel so will be the first plate

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41
Q

What type of solutions would you see the sodium cation

A

buffers

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42
Q

What are the tell tale signs on a spectra that the compound contains a bromine atom

A

Two lines 2 mass units apart due to two isotopes in a 1:1 ratio

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43
Q

what would a diatomic bromine mass spec look like

A
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44
Q

which peak is the most stable species

A

the biggest peak- so the 43 fragment

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45
Q

What would M-15 have lost

A

a methyl group

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46
Q

What is the M+1 peak

A

M+. peak is due to the C13 isotope presence

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47
Q

What does the difference in 1da vs 0.5 da mean

A

peaks are closer together as Z increases
so z=1 at 1da difference and z=2 at 0.5 da difference

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48
Q

boron useful to know

A
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49
Q

Why would we use high resolution mass spec when investigated a complex compound

A

can measure to many decimal places to distinguish between groups and some fragments/molecules can have the same mass number

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50
Q

What are the tells of chlorine in a mass spec

A

two isotopes of 35 and 37. 35 is more abundant

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51
Q

What is the difference in the peaks seen between ESI+ and ESI- spectroscopy

A
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52
Q

boron mass spec

A

two isotopes 10 and 11. 10 is 23 and 11 is 100

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53
Q

What effects how far the peaks move along the scale on a NMR diagram

A

how much electron density there is around a carbon/ this is related to wether they are connected to an electron withdrawing group and deshielded

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54
Q

If the carbon has an electronwithdrawing group next to it will it be more/less deshielded and therefore will the peak be higher/lower on an NMR scale

A

if the carbon is more positive it is more deshielded and therefore higher on the scale

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55
Q

which carbon resonates at a higher frequency

A

1- higher on the scale as it is more positive and deshielded

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56
Q

what makes a carbon more/less shielded

A

more electrons around the carbon= more shielded/ has a δ- charge
less electrons around the carbon= less shielded

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57
Q

Why does CH3F resonate at a higher frequency than the other compounds

A

F Is the most electronegative out of the others. Electron withdrawing from C making it δ+, therefore will resonate at higher frequency than the carbons in the other molecules (most positive C)

58
Q

why does t-BuOH only have 2 peaks but butanol has 4

A

carbon closest to O resonates at higher F
1- t-BuOH only has two carbon environments/ oxygen has same relationship to other carbons that are- same magnetic environment so only give one signal for the 3 carbons
2- butanol- 1 is the least shielded and shielding increases across carbons so each have their own peak

59
Q

Why does CH3Li have the lowest chemical shift

A

C is more electronegative than Li so pulls electrons towards its giving it a δ-. Meaning C is more shielded and therefore resonates at lower frequency

60
Q
A
61
Q
A

C-O can also be other electron withdrawing groups

62
Q
A
63
Q

Rank the carbons from least to most shielded/ resonating at higher to lower frequency

A
  1. C=O Is closest to the strongest electron withdrawing group
  2. Purple is closest to weaker electron withdrawing group
  3. Pink is the adjacent carbon to the C=O (stronger withdrawing)
  4. Green is adjacent to C-O (weaker withdrawing)
64
Q

Will sp, sp2 or sp3 carbons resonate at higher frequencies

A

sp and sp2 carbons will have higher frequencies/higher shifts

65
Q

How many carbon environments does this molecule have

A

the last carbon on the benzene ring will always have its own environment

66
Q

will carbons will similar resonate have their own peaks

A

no- the peak will just be a thicker line

67
Q

Rank the carbons from highest chemical shift to lowest/ predict NMR spectra

A
68
Q

Rank the carbons from highest chemical shift to lowest/ predict NMR spectra

A

A most shift- is attached to an electronegative group/ least shielded as has the least amount of electrons around it

69
Q

Rank the carbons from highest chemical shift to lowest/ predict NMR spectra

A

the last carbon on the benzene ring is the least shielded after the one that is next to the NO2 (?)

70
Q

Rank the carbons from highest chemical shift to lowest

A

1 will have the highest chemical shift as it is the least shielded
- methyl is always last as it is sp3 hybridised

71
Q

Rank the carbons from highest chemical shift to lowest

A
72
Q

What is the solvent used for C-13 DEPT spectra

A

CDCl3

73
Q

How is a C-13 DEPT spectra different from a normal one

A

and in C13 spectra the3 protons are decoupled so there’s no C-H splitting- all peaks are singlets

74
Q

Which is the C-13 DEPT spectra and which is the C-13 one

A

DEPT is the one on the bottom

75
Q

What is the difference between APT (attached proton test) spectra and DEPT spectra

A

They are the same except you do see quandary carbons in APT which you don’t in DEPT
quandary carbon- carbon attached to 4 other carbons

76
Q

In HNMR which side of the spectrum will shielded protons be and which side are desheilded ones

A
77
Q

What makes a proton more deshielded/ resonate with a higher frequency

A

being close to electronegative atoms/ withdrawing groups

78
Q

What are the most desheilded protons and why

A

aldehyde, because you can draw a resonance structure to put the positive charge on the carbon
desheilded= less electrons around it

79
Q

Which protons have the most electron density around them and which have the least

A

left-least
right-most

80
Q

The H-NMR spec for acetic acid has 2 peaks. Which H has the highest chemical shift and which peak will be the highest

A

red has highest shift as it is the H closest to the O
green highest integral height as there are greater ratio of hydrogens with that environment
-you need to reference that 6mm represents 1 proton

81
Q

What does the integral (blue lines) represent on a H-NMR spec

A

the number of protons e.g. green integral would represent 4 protons and the red would be 1

82
Q

rank the protons from least to most shielded

A

purple- least shielded
then pink, then green

83
Q

HNMR- why do we have doublets

A

protons influence each other, we allow protons to couple,
-the applied MF on A will be influenced by the minute MF that X has (X has magnetic field due to electrons being around it which have their own spin and MF) and visa versa
- the two magnetic fields will interact
diagram- top line is where the protons resonate at without influence/ second line-A will resonate at higher F when the two magnetic fields are aligned

84
Q

HNMR-why is the distance between the two green peaks the same distance as between the two red ones (bottom spectrum) and what is the distance called

A

the influence of HA on HX is the same as the influence of HX on HA
-distance called the coupling constant (J) measured in Hz) and tells you have strong the influence on each proton is due to the other

85
Q

*****THIS CALCULATION DOESNT WORK
The coupling constant/splitting value (distance between the two green/ two red peaks is called J. How do you calculate J and what would its value be if the splitting distance was 0.2ppm in a 400MHz field

A
86
Q

What is the HNMR spec of an ethyl group

A

quartet and a triplet

87
Q

HNMR- Why will the CH3 peak be a triplet

A

the three protons are influenced by the neighbouring 2 protons. If both protons in the CH2 are aligned with the externally applied MF, the CH3 peak will be pushed to a higher chemical shift
- If both are opposite aligned will be pushed lower
-the two lines should be roughly equal in height because influence should be the same

-the peak in the middle is where the CH3 would be if there wasn’t any influence

88
Q

HNMR- Why is there a ratio of 1:2:1 of the triplet

A

-one alignment of the magnetic fields between CH2 and CH3 push the peak to the right, and one pushes to the left
-two sets of influences and because of that the peak in the middle is double the intensity of the other ones

89
Q

HNMR- Why will the CH2 peak be a quartet

A

the two protons are influenced by the neighbouring 3.
-[two outside peaks] when all 3 proton magnetic fields are aligned with the applied one/ when all 3 protons are have opposite MF
-[peaks in the middle] you have 2 possibilities, where 2 up and 1 down, or 1 up and 2 down and there are 3 different arrangements of each
- so will have the same size peak in a ratio of 3:3

90
Q

what is the rule relating the number of peaks on an HNMR spec and the number of neighbouring hydrogens on the H environment you’re measuring

A

no. peaks = neighbouring H + 1

91
Q

HNMR- This relates the number of peaks to number of neighbouring protons- describe the pattern to the triangle

A

the number in the middle is the sum of the numbers on either side

92
Q

154 ppm into A

A

1.54 A (/100)

93
Q

HNMR- What is the usual J value for alkyl chains and what can be implied about a molecule if the value is less than this. An alkyl J value implies it has flexibility

A

there is rigidity or the bond could be shorter in the molecule for some unknown reason

94
Q

what changes the value of the coupling constant J

A

angle between the two hydrogens

95
Q

HNMR- Why do Z and E isomers have different coupling constant J values even though they are the same hydrogens (both attached to sp2 carbons around a double bond)

A

J is dependant on the angle between the two hydrogens and the angle is different between the two structures

96
Q

HNMR- What is the coupling constant for terminal double bonds, Z alkenes and E alkenes

A

E alkene (on the end) is more like 14-18 for X-X not XY

97
Q

HNMR- Why does an E alkene have a 14-18 (higher range) of J values when it has two X groups (groups that are the same) either side compared to when they have different groups (Y and X)

A

Y and X will influence the chemical shift because they will add or withdraw electron density with different abilities (as X and Y are different), therefore changing the geometry and chemical shift and electron density of the molecule

98
Q

What is the Benzene resonance in ppm and why is it a singlet in HNMR

A

7.3 ppm, all the hydrogens are the same and no reason for multiplicity
cannot have splitting (HA influence on HX) if the two protons are identical

99
Q

In HNMR Other double bonds are around 4.5 to 6 ppm but benzene is at 7.3ppm. This means its hydrogens have less electron density around them, why is this

A

higher ppm= the protons must be more desheilded
-carbon is sp2 hybridised so has one p orbital with electron which forms the ring
-each e- has a spin and so forms a ring current which generates a magnetic field which will influence the applied magnetic field

exam- each H is deshielded by local magnetic field due to spinning of π electrons and resonate at high frequencies

100
Q

are protons in a benzene ring or Z/ E alkene more desheilded

A

benzene protons are more desheilded

101
Q

In CNMR why are the carbons in a benzene ring more desheilded than the carbons in a different double bond (E/Z) and will therefore resonate at a higher frequency

A

-carbon is sp2 hybridised so has one p orbital with electron which forms the ring
-each e- has a spin and so forms a ring current which generates a magnetic field which will interact with the applied magnetic field

exam- each C is deshielded by local magnetic field due to spinning of π electrons and resonate at high frequencies

102
Q

In HNMR- benzene resonates at 6.3 ppm
If there was an electron donating group or electron withdrawing on the ring, where would the new peak shift to

A

electron donating- slightly lower ppm
electron withdrawing- higher ppm (more desheilded)

103
Q

HNMR- Why do the hydrogens on the bottom diagram come out at a higher chemical shift than the top one, and how do you explain this with diagrams

A

top- electron donating group so hydrogens are more sheilded
bottom- electron withdrawing group means less electrons around the H and therefore more desheilded

104
Q

is OMe electron withdrawing or donating

A

donating

105
Q

Which configuration is ortho, meta and para

A
106
Q

Is this the H-NMR spec for para, meta or ortho

A

para

107
Q

Is this the H-NMR spec for para, meta or ortho

A

meta

108
Q

Is this the H-NMR spec for para, meta or ortho

A

ortho
doublet, triplet, triplet, doublet

109
Q

Why will these protons be different on an HNMR spec and how many peaks overall will there be.
-the peaks for the outlines H’s.

A

one is by the Y and one is by the X, both will have different electron withdrawing/donating abilities
4 peaks- the H’s one removed from both X and Y are also different

110
Q

Which hydrogens will be doublets and which will be triplets

A

orange and blue- doublets
greens and yellow- triplets (coupled to two protons)
note- triplets depend on coupling constant; green is coupled to blue and yellow. If the coupling constants are very close then it will be a triplet. If they are different it will be a doublet

111
Q

will the H’s be doublet, singlet, triplet

A

blue-singlet (no H neighbours)
yellow and green-doublet
red-triplet (if coupling constant of nearby protons is similar)

112
Q

Why are the blue H’s the same and the yellow H’s the same. What HNMR graph will this show

A

because of the line of symmetry through X and Y. Two doublets

113
Q

Why is this a J3 coupling

A

coupling of hydrogens 3 bonds away (blue to yellow is 3 bonds difference)
short range coupling

114
Q

what is a W/ J4 coupling of hydrogens on an inorganic molecule and how can the protons interact if they are not next to each other

A

hydrogens are 4 bonds away
the lobes of the C-H bonds are coming into the centre of the benzene ring so you can have magnetisation transfer between this proton and that proton
large range coupling

115
Q

C-H bond is sp2s

A

sp2 orbital from carbon and s orbital from H

116
Q

What is true about the two hydrogens coupling constants because they are so far away

A

they will be very small

117
Q

What is a normal coupling constant for J3 hydrogens

A

6,8,10 Hz

118
Q

What is a normal coupling constant for J4/ W hydrogens

A

1.5 Hz

119
Q

HNMR- Which colour is the W coupling constant and which is the J3
-this diagram is called a doublet of doublets

A

green- J3
blue- W (1.5 Hz)
the fact the normal peaks are split into smaller peaks (smaller doublet) is due to W proton interaction

120
Q

Why is the splitting of the CH3 peak in a 1:2:1 ratio due to coupling constants in this HNMR spec

A

1/ CH3 group is influenced by the pink and purple H
2/the first split is due to the pink H, by the CC Ja-x
3/the purple H will split the new CH3 singles again by the same amount (same coupling constant)
4/ 2 in the middle because 2 lines from the splitting diagram

121
Q

Why is the splitting of the CH2 in a 1:3:3:1 ratio for this HNMR diagram

A

all splits are by the same amount as all have the same coupling constant

122
Q

These diagrams are both doublet of doublets. What is the difference between the two HNMR

A

diagram 1- normal splitting (due to J3 hydrogens next to each other), then second smaller splitting (1.5Hz CC from a W/J4 splitting)
-one coupling constant is large for one pair of hydrogens and the other coupling constant is smaller for the other coupling of hydrogens

diagram 2- both couples of hydrogens have similar coupling constants

123
Q

What’s the difference between a doublet of doublets and a triplet (in both senarios something is being influenced by 2 protons) in HNMR

A

doublet of doublet- one H coupling constant is different from the other
triplet- both H’s have the same coupling constant

124
Q

HNMR- this is a doublet of triplets. The red line is the H being split. How many hydrogens it it interacting with/ being split by and what are their environments

A

being split by 3 protons. On one side, 1 proton is in an environment, and on the other side 2 protons are in a different one
There are 2 identical protons in the blue environment

125
Q

How many proton environments does this have

A

3

126
Q

Which signal is which HNMR

A

yellow- OH as H is the most desheilded
blue- triplet because split by 2 identical protons (their coupling constant is the same. Stops this from being a doublet of doublets)
red- doublet as only influenced by blue proton
pink- singlet

127
Q

What halogen is NMR active

A

F, has a half spin like a protons so protons will couple with fluorine
Br, Cl are not NMR active

128
Q

How will the highlighted peaks be different on a HNMR spec for each molecule

A

7.15- will be singlet as no neighbouring interactions
6.98- will be doublet [flourine is NMR active and so interacts as a proton]

129
Q

This is a hypothetical diagram for the yellow highlighted proton. This has J3 and W coupling. What is it J3 coupled with and what is it W coupled with.
-Which lines on the graph correlate to which coupling. [the W coupling hydrogens have different CC’s]

A

J3=flourine. Is closest so will have the largest coupling constant
W coupling- with the blue and orange hydrogens
- the orange H has a lower CC than the blue one

130
Q

What hydrogen coupling does the green proton have

A

J3 coupling with brown and pink- making it a triplet
W coupling with flourine- making it a messy triplet

131
Q

If pink has a W coupling constant with mustard, mustard will also have a W coupling with pink
- W couplings arnt unidirectional

A
132
Q

do they easy ones for this website chances are you’ll do well in the exam http://nmr-challenge.uochb.cas.cz/

A
133
Q

what are the two bromine isotopes Br= 79.9

A

79 and 81

134
Q

A UV spectra peak at 264nm usually means what is true about a compound

A

aromatic

135
Q

where are peaks on an HNMR spec that show he compound is aromatic

A

7-8 ppm

136
Q

What is benzene mass number on a mass spec

A

78

137
Q

Which peak corrosponds to what carbon environment

A

red- most desuhielded
yellow- most shielded
blue- is the second most deshielded

138
Q

where do aldehydes and ketones fall on this scale

A

above 190ppm about 200ppm. Anything below this is usually an ester

139
Q

what functional group could these labeled peaks be

A

C=C

140
Q

What does a DBE of 5 mean when we know the molecule is an aromatic

A

benzene is DBE=4, so there must be another double bond in the molecule still (perhaps in a side chain)

141
Q
A
142
Q
A