NMR + chromatography Flashcards

(85 cards)

1
Q

2 types of NMR

A
  1. 13C NMR
  2. 1H (proton) NMR
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2
Q

number of signals = methylpropene

A

2

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3
Q

number of signals in methylamine

A

2

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4
Q

number of signals in propene

CH2=CH-CH3

A

3

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5
Q

calibration

A

-small quantity of TMS (tetramethylsilane) is added to samples as it produces a single providing internal standard to which other peaks are compared

-only givies one single, non-toxic, inert, low boiling point

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6
Q

NMR

A

-number of peaks in an NMR spectrum refers to Hydrogen atoms

-area under the peak = proportional to the number of hydrogen atoms

Chemical shift = closer the hydrogen is to an electronegative atom the greater the shift

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7
Q

why compounds absorb radiowaves

A

-the nucleus of some atoms have a nuclear spins e.g 1H, 13C, 19F etc)

-a nucleus with spin generates a small magnetic field

-there is a small difference in energy ebtween these 2 alignments that corresponds to the energy of radiowaves

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8
Q

how NMR is done

A

-the sample is usually dissolved in a solvent in a long thin tube and placed inside the machine where it is bombarded with radiowaves in a strong magnetic field

-the stronger the magnetic field the better the resolution of the signals

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9
Q

solvents

A

-if the solvent contains any 1H atoms then these will appear in the spectrum

-common solvents used = CCl4

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10
Q

chemical shifts

A

-the chemical shift depends on what other atoms/groups are near the H. The closer the H to electronegative atoms the greater the shift.

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11
Q

scales

A

-horizontal scale = chemical shift

-vertical axis = intensity of the absorption

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12
Q

signals

A

-same distance of CH3 = 1 signal. Difference = multiple signals

-in 1 H NMR the area of the signal is proportional to the number of hydrogen atoms is represents

-the height of the signal doesn’t matter but rather the area which the signal covers.

At A-level to indicate the relative intensity of the signals from which the simplest whole number ratio can be calculated:

e.g 1.2 : 1.2 : 1.8 = 2:2: 3

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13
Q

calculating signals

A

1) Divide all values with the smallest number

2) Find whole number

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14
Q

splitting patterns

A

-the number of lines in a signal is split into and gives information about neighbouring H atoms

-usually only H atoms on the neighbouring C atom cause splitting

-number of lines = 1 + number of H atoms on adjacent carbon atoms (n+1)

-H atom on OH group in alcohols rarely causes splitting or is split itself.

Doublet = 1:1

Triplet = 1:2:1

Quartllet = 1: 3: 3: 1

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15
Q

outline the steps needed to locate positions of the amino acid on TLC plate and determine their RF values

A

-spray with developing agent
-measure distances from initial pencil line to spots of pigment (x)
-measure distance from initial pencil line to solvent front (Y)
-Rf value = X / Y

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16
Q

why do amino acids have different Rf values

A

amino acids have different polarities
-different solubilities in the developing solvent (mobile phase) and retention values in the stationary phase

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17
Q

solid vs liquid stationary phase

A

A solid stationary phase separates by adsorption,

A liquid stationary phase separates by relative solubility

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18
Q

polar vs non-polar stationary phase

A

If the stationary phase was polar and the moving
phase was non- polar e.g. hexane. Then non-polar
compounds would pass through the column more
quickly than polar compounds as they would have
a greater solubility in the non-polar moving phase.

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19
Q

Precuations for chromotaography

A

Wear plastic gloves to prevent contamination
from the hands to the plate
pencil line –will not dissolve in the solvent
tiny drop – too big a drop will cause different
spots to merge
Depth of solvent– if the solvent is too deep it
will dissolve the sample spots from the plate
Will get more accurate results if the solvent is
allowed to rise to near the top of the plate but
the Rf value can be calculated if the solvent
front does not reach the top of the plate
lid– to prevent evaporation of toxic solvent
dry in a fume cupboard as the solvent is toxic
UV lamp used if the spots are colourless and
not visible

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20
Q

Practical procedure for TLC

A

a) Wearing gloves, draw a pencil line 1 cm above the
bottom of a TLC plate and mark spots for each sample,
equally spaced along line. b) Use a capillary tube to add a tiny drop of each solution to a
different spot and allow the plate to air dry.
c) Add solvent to a chamber or large beaker with a lid so that
is no more than 1cm in depth
d) Place the TLC plate into the chamber, making sure that
the level of the solvent is below the pencil line. Replace
the lid to get a tight seal.
e) When the level of the solvent reaches about 1 cm from
the top of the plate, remove the plate and mark the solvent
level with a pencil. Allow the plate to dry in the fume
cupboard.
f) Place the plate under a UV lamp in order to see the spots.
Draw around them lightly in pencil.
g) Calculate the Rf values of the observed spots.

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21
Q

procedure for column chromatography

A
  • A glass tube is filled with the stationary phase usually silica
    or alumina in powder form to increase the surface area. * A filter or plug is used to retain the solid in the tube.
    Solvent is added to cover all the powder. * The mixture to be analysed is dissolved in a minimum of a
    solvent and added to the column. * A solvent or mixture of solvents is then run through the
    column. * The time for each component in the mixture to reach the
    end of the column is recorded (retention time)
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22
Q

types of chromatography

A

-Thin layer chromatography = plate is coated with a solid and a solvent moves up the plate

-Column chromatography –> column is placed with a solid and solvent moves down the column

-Gas chromotography –> gas is passed through column under pressure at high temperature

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23
Q

separation dependency

A

Separation depends on a balance between solubility in moving phase and retention in stationary phase

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24
Q

mobile phase

A

Mobile phase = carries soluble component (more soluble = moves faster)

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25
stationary phase
Stationary phase = phase which holds back the component (more affinity = moves slower)
26
retention time
Retention time = how long it takes to travel through the chromatogram. Depends on affinity for mobile and stationary phase Rf value = between 0 and 1
27
equation for retention values
Rf = distance moved by component / distance moved by solvent
28
most polar =
ionic substances = NaCl
29
what does retention time depend on
Retention time depends on the polarity of a substance The stationary phase attracts polar substances = longer retention time as the substance moves slower
30
detection
Amino acids can be stained using ninhydrin Colourless samples can be viewed using UV light
31
phases
-Mobile phase = solvent that moves up the plate = solvent soaks up the TLC plate and dissolves the compounds at the baseline. -Separation occurs based on different attractions to the stationary and mobile phase -Adsorption = compound temporarily binds to silica gel surface -more strongly absorbed = less distance it will travel
32
factors affecting separation
-higher solubility = faster movement up the plate -stronger adsorption = slower movement e.g compound forming hydrogen bonds with silica gel will move slower than van der waals
33
column chromatography
-tube is usually filled with silica or aluminia in powder form to increase surface area -used for separating mixtures on a larger scale -stationary phase = silica gel -mobile phase = liquid solvent -chromatography column is filled with silica gel -concentrated sample solution is added at the top -solvent is added continiously to wash compounds through the column -more polar compounds interact more with the stationary phase = move slower
34
Gas chromatography
-uses an inert gas e.g He,H2,N2 -long coiled tube placed in powder -samples analysed using mass spec -mobile phase = gas solvent -stationary phase = high boiling point liquid adsorbed on a solid -used for analysing small volatile compounds 1) Tiny sample is injected with a syringe 2) sample vaporises instantly in a heated injector 3) sample travels through a coiled column coated with stationary phase liquid 4) separation depends on boiling point and solubility in the stationary phase 5) results appear as peaks on a chromatogram
35
separation is too slow
-if separation is too slow, use more polar solvent not compound
36
factors affecting retention time
-higher boiling point = longer retention time -solubility in the stationary phase = more soluble = longer retention times -higher temperature = shorter retention time
37
gas chromatogram
-peak position = retention time to identify the compound -peak area = proportional to the amount of compound present
38
when calculating Rf values
measure up to the middle (midpoint) of the spot not the top
39
ethanol forms hydrogen bond with stationary phase =
adsorbs strongly = longer retention time
40
wider + tall =
high abundance = chromatogram
41
each peak =
each environment that absorbs specific amount of energy 1 peak = 1 environment e.g ethene
42
maximum environments in carbon 13 NMR
depends on the carbon atoms present
43
butane =
2 peaks in C13 due to equivalent
44
C13 for chlorobenzene
4 environments
45
1H for chlorobenzene
3 environments
46
polar sample =
dissolve in a polar solvent
47
non-polar solvent
tetrachloromethane CCl4
48
polar solvent
CDCl3
49
tetramethylsilane Si(CH3)4
-only gives a single peak -far away from peaks produced by the sample -inert = wont react with sample -non-toxic -highly volatile = remove from sample easily
50
larger chemical shift =
closer to an electronegative atom
51
turn decimal ratio into whole number ratio =
multiply all ratios by the smallest value
52
integer ratio =
adds up to show all hydrogen atoms in the molecule
53
splitting pattern in high resolution NMR
singlet = single line/spike doublet = 2 lines/spike split into 2 etc triplet, quartet, multiplet
54
n + 1
n = number of non-equivlaent hydrogen atoms n =1 hydrogen = doublet forms
55
splitting only occurs
for hydrogen atoms on carbon atoms e.g hydrogen in OH group = singlet + NH2
56
example = propanoic acid
3 peaks = 3 environments = types of non-equivalent 1H integration ratio = 1:2:3 O-H = singlet CH3 = triplet as split by adjacent CH2 CH2 = quartlet
57
no splitting =
singlet
58
triplet next to a quartet =
CH3 - CH2
59
two triplets =
CH2-CH2
60
number of environments =
number of peaks
61
tetramethylsilane formula
Si(CH3)4
62
CHCl4 not used for 1H
would show hydrogen peak
63
any ratio with 9 =
3 CH3 groups surrounding a carbon atom
64
ratio based on number of hydrogens
e.g 3 : 3 : 2 3 hydrogens , 3 hydrogens, 2 hydrogens
65
largest energy gap =
fewest electrons
66
carbon 12 nuclei
dont have spin
67
stronger magnets =
higher frequencies interact with the same sample
68
higher peak
more molecules in the same environment never use peak heights in C13 NMR analysis
69
3 rules for hydrogen NMR
-hydrogens bonded to the same carbon are in the same environment -hydrogens bonded to carbons in the same environment are in the same environment -
70
hydrogens on double bond
not necessarily the same environment due to restricted rotation need to look at groups on the opposite side of the double bond
71
polar vs non-polar solvents
2a (Both) have no H (atoms so give no signals in spectrum) tied to either CDCl3 or CCl4 2b CCl4 non polar (- good solvent for non-polar organic molecules) 2c CDCl3 polar covalent molecule (– good solvent for polar organic compounds)
72
TMS
3a (Lots (12) of equivalent H to) give one signal / single environment 3b Signal in an area away from other typical H signals / peak upfield from others OR (Low electronegativity of Si shifts) signal right 3c Easy to remove / volatile / low bp
73
molecular formula of tetramethyl silane
C4H12Si
74
why are there 3 peaks in the carbon NMR spectrum
there are 3 pairs of equivalent carbon atoms
75
structuring long NMR response
-state number of peaks = number of environments -find integration ratio -look at graph for singlets/doublets etc e.g singlet = ratio 1 = due to H in OH alcohol e.g singlet (3) due to CH3 with no adjacent hydrogen
76
C6H3Br3
benzene structure
77
splitting pattern for cyclohexane
singlet 1.2-1.4
78
alkenes
may be in different envrionments if the side of double bonds are not equal
79
ratio and signals for methylcyclohexane
5 peaks 3:1:4:4:2
80
3:2 ratio
ethyl group CH3-CH2 triplet and quartet
81
2:2 ratio
triplet and triplet CH2-CH2
82
3:1 ratio
doublet and quartet CH3-CH
83
ratio of 3
CH3 = methyl
84
ratio of 1
singlet OH group
85
no absorption for C=O in its infrared spectrum =
try drawing OH instead