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A2 Chem Unit 4.9: Spectroscopy & chromatography

Eedxcel A2 Unit 4: Infrared, Mass Spec, NMR, chromatography
by

J Amuah-Fuster

on 8 July 2016

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Transcript of A2 Chem Unit 4.9: Spectroscopy & chromatography

Number of peaks = number of chemically different H’s on adjacent atoms + 1


1 neighbouring H 2 peaks “doublet” 1:1


2 neighbouring H’s 3 peaks “triplet” 1:2:1




3 neighbouring H’s 4 peaks “quartet” 1:3:3:1




4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1 Each proton type is said to be chemically shifted relative to a standard (usually TMS).
The chemical shift is the difference between the field strength at which it absorbs and the field strength at which TMS protons absorb the delta (d) scale is widely used as a means of reporting chemical shifts

Observed chemical shift (Hz) x 10
δ = ppm (parts per million)
Spectrometer frequency (Hz)

The chemical shift of a proton is constant under the same conditions (solvent, temperature)
The TMS peak is assigned a value of ZERO (δ = 0.00)
All peaks of a sample under study are related to it and reported in parts per million
H’s near to an electronegative species are shifted “downfield” to higher chemical shift values NMR spectra provide information about the structure of organic molecules from the ...

number of different signals in the spectrum
position of the signals (chemical shift)
intensity of the signals
splitting pattern of the signals


A liquid sample is placed in a tube which spins in a magnetic field
solids are dissolved in solvents which won’t affect the spectrum - CCl4, CDCl3
TMS, tetramethylsilane, (CH3)4Si, is added to provide a reference signal
when the spectrum has been run, it can be integrated to find the relative peak areas
spectrometers are now linked to computers to analyse data and store information NMR stands for Nuclear Magnetic Resonance, adn it involves causing the nuclei of certain atoms to spin in a magnetic field as a result of radiowaves.

Proton nuclear magnetic resonance spectroscopy provides information about the hydrogen atoms in molecules
Carbon nuclear magnetic resonance spectroscopy provides information about the carbon atoms in molecules

Proton NMR provides information by...

spinning a sample of the compound in a magnetic field

hydrogen atoms in different environments responding differently to the field

each different environment of hydrogen producing a signal in a different position

the area under each peak / signal being proportional to the number of hydrogens

the splitting of signals according to how many H’s are on adjacent atoms When a whole molecule, M, is bombarded with high-energy electrons, it becomes ionised:


This molecular ion then break up into a smaller positive ion, X+ and a radical, Y·:


The peak with the largest m/z value is the molecular ion; the molecular formula can be deduced by comparing the m/z value of the highest peak with the empirical formula of the substance. Nowadays, mass spectrometry is used to identify unknown or new compounds.

When a molecule is ionised it forms a MOLECULAR ION which can also undergo FRAGMENTATION or RE-ARRANGEMENT to produce particles of smaller mass.

Only particles with a positive charge will be deflected and detected.

The resulting spectrum has many peaks.

The final peak (M+) shows the molecular ion (highest m/z value) and indicates the molecular mass. The rest of the spectrum provides information about the structure. Spectroscopy & Chromatography Gas Chromatography & High Performance Liquid Chromatography Unit 4 overview Chapter 9:
Spectroscopy & chromatography
Specification Unit 4 exam worth 120 UMS points
40% of A2 grade Exam dates:
TBC Nuclear Magnetic Resonance Infrared Spectroscopy Explain the effect of different types of radiation on molecules and how the principles of this are used in chemical analysis and in reactions, limited to:
i.infrared in analysis
ii.microwaves for heating
iii.radio waves in nmr
iv.ultraviolet in initiation of reactions

Explain the use of high resolution nmr spectra to identify the structure of a molecule:
i.based on the different types of proton present from chemical shift values
ii.by using the spin-spin coupling pattern to identify the number of protons adjacent to a given proton
iii.the effect of radio waves on proton spin in nmr, limited to 1H nuclei
iv.the use of magnetic resonance imaging as a non-invasive technique, eg scanning for brain disorders, or the use of nmr to check the purity of a compound in the pharmaceutical industry

Demonstrate an understanding of the use of IR spectra to follow the progress of a reaction involving change of functional groups, eg in the chemical industry to determine the extent of the reaction

Interpret simple mass spectra to suggest possible structures of a simple compound from the m/e of the molecular ion and fragmentation patterns

Describe the principles of gas chromatography and HPLC as used as methods of separation of mixtures, prior to further analysis (theory of Rf values not required), and also to determine if substances are present in industrial chemical processes. Mass Spectroscopy Molecular ions and fragmentation detection of small amounts of chemicals,
distinguishing isomers (molecules with the same molecular formula but different structural formulae) of chemicals,
linked with gas chromatography (GC) and high performance liquid chromatography (HPLC) to identify individual chemicals in a mixture.


the isotopic composition of an element can be used to determine the source of the element, e.g., the radioactive polonium used to poison the Russian Alexander Litvinenko was traced to its source by analysing the isotopic composition of the polonium found in his body.
The proportion of radioactive C-14 in a dead sample of tissue is used to determine how long ago the tissue died.


trace amounts of chemicals can be detected in body fluids.


a mass spectrometer can be used on a probe or a space shuttle to analyse elements and compounds found out in space and on planets (probes only so far). Uses of the mass spectrometer Modern Analytical Chemistry Mass spectrometer Modern Analytical Chemistry Example IR spectra (6) Example IR spectra (5) Example IR spectra (4) Example IR spectra (3) Modern Analytical Chemistry Example IR spectra (2) Example IR spectra (1) Modern Analytical Chemistry An infrared spectrometer measures the amount of radiation absorbed.
The infrared light absorbed by the organic sample (organic compound in solvent) and a reference sample (solvent) is compared;
Different wavelengths of infrared light is passed through the samples;
Different bonds in a covalent molecule absorb radiation of different frequencies, which are normally measured as wavenumbers;
Infrared spectra usually has the range 4000 cm-1 to 600 cm-1 How the Infra Red Spectrometer works Modern Analytical Chemistry Infrared spectroscopy relates to the absorption of energy, that increases the vibration (stretching of the bonds) in a molecule.
The frequency of the infrared radiation absorbed depends on the nature of the bond that is vibrating.
A molecule must have a polar bond to absorb infrared (the greenhouse gases, H2O, CO2, CH4 and NO all have polar bonds). This is why O2 and N2 are not greenhouse gases. Infrared spectra the electron gun produces a beam of electrons which ionises the gaseous sample and breaks up molecules into charged fragments.


an electric accelerates the positively charged ions.


powerful electromagnets deflect the charged particles. Ions of smaller mass are deflected the most (track D in diagram). The magnetic field is increased gradually to deflect ions of greater mass.


a detector calculates the m/e ratio (mass/charge) of each positive ion and its relative abundance is compared with the most abundant particle (the parent ion).
A graph is generated. Mass spectrometer INFRA RED SPECTRA - INTERPRETATION Infra-red spectra are complex due to the many different vibrations taking place in each molecule.

Total characterisation of a substance based only on its IR spectrum is almost impossible unless one has computerised data handling facilities for comparison of the obtained spectrum with one in memory.

The technique is useful when used in conjunction with other methods:
nuclear magnetic resonance spectroscopy
mass spectroscopy

Peak position depends on:
bond strength
masses of the atoms joined by the bond Modern Analytical Chemistry The Infra-red Spectrophotometer • a beam of infra red radiation is passed through the sample
• a similar beam is passed through the reference cell
• the frequency of radiation is varied
• bonds vibrating with a similar frequency absorb the radiation
• the amount of radiation absorbed by the sample is compared with the reference
• the results are collected, stored and plotted Interpretation of thousands of spectra has shown that many classes of organic compound show characteristic fragmentation patterns due to their functional groups.
It is possible to identify the type of compound from its spectrum by looking at the ...

position of peaks
difference in values between major peaks THE MASS SPECTRUM Spectra obtained for organic molecules have many peaks.
Each peak is due to a particular fragment with a certain m/z value.

Highest m/z value usually corresponds to the molecular ion
its position provides information about the molecular mass of a substance

The tallest peaks come from the most stable species CHARACTERISTIC ABSORPTION FREQUENCIES
C-H Alkane 2965 - 2850 strong
C-C Alkane 1200 - 700 weak
C=C Alkene 1680 - 1620 variable

C=O Ketone 1725 - 1705 strong
Aldehyde 1740 - 1720 strong
Carboxylic acid 1725 - 1700 strong
Ester 1750 - 1730 strong
Amide 1700 - 1630 strong

C-O Alcohol, ester, acid, ether 1300 - 1000 strong

O-H Alcohol (monomer) 3650 - 3590 variable, sharp
Alcohol (H-bonded) 3420 - 3200 strong, broad
Carboxylic acid (H-bonded) 3300 - 3250 variable, broad

N-H Amine, Amide 3500 (approx) medium
CN Nitrile 2260 - 2240 medium

C-X Chloride 800 - 600 strong
Bromide 600 - 500 strong
Iodide 500 (approx) strong Modern Analytical Chemistry SYMMETRIC BENDING ASYMMETRIC
STRETCHING STRETCH INFRA RED SPECTROSCOPY Different covalent bonds have different strengths due to the masses of different atoms at either end of the bond.

As a result, the bonds vibrate at different frequencies

The frequency of vibration can be found by detecting when the molecules absorb electro-magnetic radiation.

Various types of vibration are possible.

Examples include... STRETCHING and BENDING INFRA RED SPECTRA - INTERPRETATION measured in (waves per centimetre) / cm-1 The only way to completely identify a compound using IR is to compare its spectrum with a known sample.
The part of the spectrum known as the ‘Fingerprint Region’ is unique to each compound. IDENTIFICATION OF COMPOUNDS BY DIRECT COMPARISON OF SPECTRA The presence of bonds such as O-H and C=O within a molecule can be confirmed because they have characteristic peaks in identifiable parts of the spectrum. INFRA RED SPECTRA - USES IDENTIFICATION OF PARTICULAR BONDS IN A MOLECULE • esters show a strong absorption between 1750 cm-1 and 1730 cm-1
• this is due to the presence of the C=O bond IR SPECTRUM OF AN ESTER • alcohols show a broad absorption between 3200 and 3600 cm-1
• this is due to the presence of the O-H bond IR SPECTRUM OF AN ALCOHOL • carboxylic acids show a broad absorption between 3200 and 3600 cm-1
• this is due to the presence of the O-H bond
• they also show a strong absorption around 1700 cm-1
• this is due to the presence of the C=O bond IR SPECTRUM OF A CARBOXYLIC ACID • organic molecules have a lot of C-C and C-H bonds within their structure
• spectra obtained will have peaks in the 1400 cm-1 to 800 cm-1 range
• this is referred to as the “fingerprint” region
• the pattern obtained is characteristic of a particular compound the frequency
of any absorption is also affected by adjoining atoms or groups. FINGERPRINT REGION • carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm-1
• this is due to the presence of the C=O bond IR SPECTRUM OF A CARBONYL COMPOUND All identified species must be written as positively charged ions. m/e = 65: 37Cl-CHCH3+
m/e = 63: 35Cl-CHCH3+
m/e = 43: CH3CHCH3+
m/e = 41: CH2CHCH2+ Identification of molecules:
Answers Identify the species responsible for the peaks at m/e = 65, 63, 43 and 41 This parent peak is of 2-chloropropane containing the 35Cl isotope. This parent peak is of 2-chloropropane containing the 37Cl isotope. One can tell the difference between alcohols, aldehydes and carboxylic acids by comparison of their spectra. CARBOXYLIC ACID ALDEHYDE or KETONE ALCOHOL AND C=O STRETCH O-H STRETCH C=O STRETCH O-H STRETCH WHAT IS IT! Why is the peak at a m/e=78 also a parent ion? You can check this by working out the molar mass of the molecule This last peak is the parent ion – in other words the ionised whole molecule This is the mass spectrum for 2-chloropropane Identification of molecules MOLECULAR MASS DETERMINATION USING MASS SPECTROMETRY FRAGMENTATION RE-ARRANGEMENT FRAGMENTATION MOLECULAR ION IONISATION CHARACTERISTIC FREQUENCIES C-Cl CN C-C alkanes C=C C-H Aromatic C-C N-H C-O C=O O-H The tiny peak is the M + 1 peak and is due to fragments containing carbon-13 atoms. This isotope makes up 1 % of all carbon atoms. Mass spectra (1) Mass spectra (2) ALDEHYDES AND KETONES FRAGMENTATION PATTERNS Cleavage of bonds next to the carbonyl group (C=O) is a characteristic fragmentation of aldehydes and ketones. A common fragment is carbon monoxide (CO) but as it is a molecule and thus uncharged it will not produce a peak of its own. However, it will produce an m/z drop of 28 somewhere in the spectrum.

The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. Multiple peaks occur in the molecular ion region due to different halogen isotopes.

There are two peaks for the molecular ion of C H Br, one for the molecule containing the isotope Br-79 and the other for the one with the Br-81 isotope. Because the two isotopes are of similar abundance, the peaks are of similar height. FRAGMENTATION PATTERNS HALOGENOALKANES IDENTIFY THE COMPOUND Breaking the bond between the butyl group and the carbonyl group produces two further ions, depending on how the bond breaks.

Two peaks at m/z values 43 and 57 will appear in the mass spectrum. Aldehydes and ketones FRAGMENTATION PATTERNS The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. The species due to the final signal is known as the molecular ion and is usually corresponds to the molecular mass of the compound. THE MASS SPECTRUM - THE MOLECULAR ION In the spectrum of octane, a signal occurs at 114 due to the species C H + The rest of the spectrum provides additional information of the molecule’s structure.
Peaks appear due to characteristic fragments (e.g. 29 due to C H +) and differences between two peaks also indicates the loss of certain units (18 for H O, 28 for CO). THE MASS SPECTRUM - FRAGMENTATION The small peak (M+1) at 115 due to the natural abundance (about 1%) of carbon-13. The height of this peak relative to that for the molecular ion depends on the number of carbon atoms in the molecule. The more carbons present, the larger the M+1 peak. THE MASS SPECTRUM - THE MOLECULAR ION A further peak occurs at m/z = 72 (100-28) due to loss of CO Aldehydes and ketones FRAGMENTATION PATTERNS The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. Modern Analytical Chemistry Horizontal scale shows the chemical shift (δ-ppm). It increases from right to left.

H-1 spectra show the different chemical environments of the hydrogen atoms in a compound.

There is always a small peak at 0 from the reference compound (TMS)
The area under the peaks tells you the ratios involved.

Splitting is caused by “coupling” with protons on adjacent carbon atoms which could either have the same or opposite spins.

The more hydrogens on the adjacent atom, the more splitting that occurs. High resolution NMR Put simply Nuclear magnetic resonance how many different sets of equivalent H atoms there are


information about chemical environment of H atom


gives ratio of H atoms for peaks



how many H atoms on adjacent C atoms Number of signals



Position of signals



Relative intensities



Splitting NMR: SUMMARY ethyl propanoate

1,2-dibromopropane

dimethylethyl propanoate

but-2-ene methylpropene

propene

2-chloropropane

propanone

methylamine For each of the following compounds, predict the number of signals, the relative intensity of the signals, and their spin-spin splitting patterns. MULTIPLICITY (Spin-spin splitting) 4 adjacent H’s
gives 5 peaks in the ratio 1 : 4 : 6 : 4 : 1 MULTIPLICITY (Spin-spin splitting) Splitting patterns are worked out by considering the effect adjacent, chemically different protons have on another signal in a given environment.

The spin of the proton producing the signal is affected by each of the two forms of the adjacent proton.

One orientation augments/enhances its field and the other opposes/reduces it. This is done by calculating the various possible combinations of alignment of adjacent protons.

HOWEVER:
Signals for the H in an O-H bond are not affected by hydrogens on adjacent atoms so are not split NMR SPECTROSCOPY MULTIPLICITY (Spin-spin splitting) Number of peaks = number of chemically different H’s on adjacent atoms + 1


0 neighbouring H’s signal isn’t split 1 peak “singlet”


1 neighbouring H signal split into 2 peaks “doublet” ratio = 1:1

2 neighbouring H’s signal split into 3 peaks “triplet” 1:2:1

3 neighbouring H’s signal split into 4 peaks “quartet” 1:3:3:1

4 neighbouring H’s signal split into 5 peaks “quintet” 1:4:6:4:1 SPIN-SPIN COUPLING (SPLITTING) Summary 4 sets of equivalent H’s: ratio 6:1:2:3 5 sets of equivalent H’s: ratio 3:1:2:2:3 4 sets of equivalent H’s: ratio 3:1:2:3 2 sets of equivalent H’s: ratio 6:2 (3:1) PROVIDES THE REFERENCE SIGNAL The molecule contains four methyl groups attached to a silicon atom in a tetrahedral arrangement.

All the hydrogen atoms are chemically equivalent. TETRAMETHYLSILANE - TMS non-toxic liquid - SAFE TO USE
inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED
has a low boiling point - CAN BE DISTILLED OFF AND USED AGAIN
all the hydrogen atoms are chemically equivalent - PRODUCES A SINGLE PEAK
twelve hydrogens so it produces an intense peak - DON’T NEED TO USE MUCH
signal is outside the range shown by most protons - WON’T OBSCURE MAIN SIGNALS
given the chemical shift of d = 0
the position of all other signals is measured relative to TMS 3 signals: ratio 3:2:9 2 signals: ratio 6:2 (3:1) 3 signals: ratio 2:1:3 Hydrogen nuclei in the same environment will show up as one peak on low resolution NMR. The chemical shift and the relative area of the peak helps identify the presence of three terminal –CH groups. 9 2 The integration trace gives the area under the peaks – i.e. The relative areas of the peaks. NMR for 1-bromo-2,2-dimethylpropane MULTIPLICITY (Spin-spin splitting) O adjacent H’s
There is no effect

1 adjacent H
can be aligned either with a or against b the field
there are only two equally probable possibilities
the signal is split into 2 peaks of equal intensity


2 adjacent H’s
more possible combinations
get 3 peaks in the ratio 1 : 2 : 1



3 adjacent H’s
even more possible combinations
get 4 peaks in the ratio 1 : 3 : 3 : 1 FOR AGAINST ANALOGY
Imagine you had an opinion on something. If nobody influenced you, your opinion would be the same. However if another person had a view on the topic, they would either agree or disagree with you. Their ideas would either enhance what you thought or diminish it. There would be two possibilities of equal chance.



If there were two people offering views they could either be both for it (1 possibility) , both against (1 possibility) or one could be in favour and the other against (2 possibilities). There would be three possibilities of relative chance 1:2:1 MULTIPLICITY (Spin-spin splitting) THE BASIC ELEMENTS OF AN NMR SPECTROMETER RADIOFREQUENCY
OSCILLATOR NMR SPECTROMETERS The sample is spun round in the field of a large electromagnet and a radio-frequency (RF) field is applied. The magnetic field is increased and the excitation or “flipping” of nuclei from one orientation to another is detected as an induced voltage resulting from the absorption of energy from the RF field.









An nmr spectrum is the plot of the induced voltage against the sweep of the field. The area under a peak is proportional to the number of nuclei “flipping”

Not all hydrogen nuclei absorb energy at the same field strength at a given frequency; the field strength required depends on the environment of the hydrogen.

By observing the field strength at which protons absorb energy, one can deduce something about the structure of a molecule. This is a quartet despite the fact that there are 4 H’s on adjacent atoms - the H on the OH doesn’t couple OH hydrogens are always seen as a singlet ... there is no splitting O-H bonds and splitting patterns The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting

H’s on OH groups do not couple with adjacent hydrogen atoms

Arises because the H on the OH, rapidly exchanges with protons on other molecules and is not attached to any particular oxygen long enough to register a splitting signal. Low resolution nmr gives 1 peak for each environmentally different group of protons

High resolution gives more complex signals - doublets, triplets, quartets, multiplets

The signal produced indicates the number of protons on adjacent carbon atoms LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE LOW RESOLUTION - HIGH RESOLUTION low resolution nmr gives 1 peak for each environmentally different group of protons

high resolution gives more complex signals - doublets, triplets, quartets, multiplets

the signal produced indicates the number of protons on adjacent carbon atoms MULTIPLICITY (Spin-spin splitting) 4 signals: ratio 3:2:2:3 2 signals: ratio 6:1 2 signals: ratio 3:2 3 signals: ratio 2:1:3 1 signal 2 signals: ratio 6:2 (3:1) The broad peaks are split into sharper signals HIGH RESOLUTION SPECTRUM OF 1-BROMOPROPANE LOW RESOLUTION - HIGH RESOLUTION A spinning nucleus such as H-1 behaves as a spinning charge and generates a magnetic field. It can be likened to a bar magnet.

When it is placed in an externally applied field it can align with, or against, the field.

The energy difference between the two states depends on the applied field. A nucleus without spin cannot be detected by nuclear magnetic resonance spectroscopy. All nuclei possess charge and mass. Those with either an odd mass number or an odd atomic number also possess spin. This means they have angular momentum. NMR SPECTROSCOPY – ORIGIN OF SPECTRA Measure the distance between the top and bottom lines.

Compare the heights from each signal and make them into a simple ratio. IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio HOW TO WORK OUT THE SIMPLE RATIOS
Measure how much each integration line rises as it goes of a set of signals

Compare the relative values and work out the simple ratio between them

In the above spectrum the rises are in the ratio... 1:2:3 INTEGRATION Approximate
chemical shifts

The actual values depend on the environment CHEMICAL SHIFT Scientists regularly perform chemical analysis on food samples to make sure they do not contain illegal and/or toxic substances.

One method of analysis is chromatography. it works by separating mixtures into their constituent parts for identification. Chromatogram The x-axis measures time taken for solute to pass through column (Retention time).

The stronger the interaction between the solute and the stationary phase, the longer the solute takes to travel through the column.

Substances that pass through quickly have a stronger attraction to the mobile phase. Vertical axis - Absorbance the stronger the absorbance the larger the peak Horizontal axis - Frequency (wavenumber) OR Wavelength measured in microns (m);
1 micron = 1000 nanometres Example IR spectra Bond Class of compound Range / cm-1 Intensity strong bonds and light atoms absorb at lower wavenumbers

weak bonds and heavy atoms absorb at high wavenumbers Pentane IDENTIFY THE FRAGMENT/MOLECULAR ION PEAKS (by formula) Mass Spectrometry Chlorine has several isotopes, therefore it can form several diatomic molecules Peak II is due to the positive ion of the Cl-37 isotope, while peak IV is due to a chlorine molecule (Cl ) made up of one Cl-35 atom bonded to a C-37 atom. 2 2 5 The more stable the acylium ion RCO+, the more abundant it will be and;

the more abundant the species the taller its peak in the mass spectrum 2 2 5 8 18 M (g) + e-  M+ (g) + 2e- M+ (g)  X+ (g) + Y· Pharmaceutical industry Isotopic analysis of elements and radioactive dating Drug testing in sports Space exploration Ionisation Acceleration Deflection Detection Examples of Mass Spectra WHAT IS NMR AND WHAT DOES AN NMR SPECTRUM TELL YOU? POSSESS SPIN
H H C F P DON’ T POSSESS SPIN
C 1 2 13 19 31 1 1 6 9 15 12 6 OBTAINING SPECTRA INTERPRETATION OF SPECTRA 6 DOWNFIELD - ‘deshielding’ the splitting pattern depends on the number of hydrogens on adjacent atoms Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet (n + 1 rule) EXPLAIN THE THEORY BEHIND THE SPLITTING PATTERN FOR 4 ADJACENT H’s INTEGRATION the area under a signal is proportional to the number of hydrogen atoms present
an integration device scans the area under the peaks
lines on the spectrum show the relative abundance of each hydrogen type

By measuring the distances between the integration lines one can
work out the simple ratio between the various types of hydrogen.


before integration after integration NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET 3 NMR Exercises s s s s s s d d d d d d qrt qrt qrt qrt qrt t t t sext sext
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