Chapter 9 Study Guide.pdf

7/25/2019 Chapter 9 Study Guide.pdf

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Infrared spectroscopy

Infrared light is of the exact frequency to resonate with the covalent bonds of organicmolecules.

Plancks theory of quantization shows us that only the exact energy will resonate withthe bond. Too much or too little energy will not cause it to resonate(too little may causerotation(microwaves) or too much may cause fragmentation(UV)).

The resonance of IR radiation with organic molecules is seen as stretching and bendingof the covalent bonds. Stretching can be symmetric or asymmetric. Bending can be

scissoring or twisting. The actual types do not concern us much. Lighter atoms absorbat a higher frequency. Stiffer atoms absorb at a higher frequency.

IR spectra are incredibly complicated; however certain regions are more important.

2500-3600

Single bond stretches of hydrogen appear in this region

OH stretches occur either 3590-3650 if non-hydrogen bonded or 3200-3550 if hydrogen

bonded or 2500-3000 if part of a carboxylic acid.

NH appears between 3300-3500. NH2 would appear as a pair of peaks.

Alkane CH stretches appear at 2853-2962. Alkene CH stretches appear at 3010-3095.Alkyne CH stretches appear at 3300.

2100-2300

Triple bond stretches appear in this region.

Alkyne CC stretches appear at 2100-2260. Nitrile CN stretches appear at 2220-2260.

1400-1800

Double bond stretches appear in this region.

Carbonyl CO stretches appear at 1630-1780. Esters and Acids appear at the high end of thisregion. Ketones and aldehydes in the middle and Amides at the low end.

Alkene CC stretches appear at 1620-1680. Aromatic CC stretches appear at 1400-1500.

1300-1400

Single bond stretches of carbon appear in this region.

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8. Match the spectra to the compounds. Also circle the peak on the spectra that led you to the identification.

B

C

D

A

A. Butanenitrile B. Butanol C. Butanone (ketone) D. Butene

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SpectroscopyChapter 9

Light spectrum

PlanckEnergy is quantized.

Light can behave as both a particle and a wavedual nature of lightPhotoelectric effectEinstein Nobel Prize

Electrons can behave both as a wave and a particleMatter waves

DeBroglieNobel Prize

SchrodingerQuantum mechanics to solve the equations for the movement and location

of electrons. What is life? GenesBiochemistry. Nobel prize.

Light is a source of energy. How the energy of light interacts with matter is calledSpectroscopy.

Types of light energy.

E = h; = c/;

c =3.00 x 108

m/sh = 6.63 x 10

-34Js

= frequency(Hz or s-1

)

= wavelength(nm or m)

X-rays UV visible(VBGYOR) IR Radio( , TV and radio)

wavelength increases in this direction

frequency and strength increases in this direction

1 nm 400 nm 700nm 4000nm cm km

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HISTORY OF NUCLEUS

Maxwell(1831-1879)

Invented color photography. Rings of saturn are many small particles. Created thesingle comprehensive theory of electricity and magnetism. Died in 1879(same year as

Einstein born). Coincidence Newton born same year Galileo died.

Hertz(1857-1894)

Created the first radiowave in 1887.

Roentgen

1895Discovered X-rays accidently. Cathode ray tube on his desk cause fluorescentscreen to flash. Nothing short of lead blocked the invisible, unknown rays. Forerunnerof modern X-ray.

Marconi

Transmitted radiowaves over the English channel and the Atlantic ocean, 1899 and1901. Discovered the ionosphere.

Becquerel/Curie

Discovered radiation and the elements Radium and Polonium(Poland). Marie won 2nobel prizes(1903 and 1912). Pierres nobel prize speech.

Planck

Came up with the theory of quantization(1900).

Einstein

1905Quantum Theory of Light, Photoelectric effect, Dual Nature of Light, Brownian

Movement and the Theory of Relativity. 1916General Theory of Relativity. Einsteincame up with the theory and the mathematics on how to turn radioactive elements intoenergy. E = mc2. He also furthered our understanding of light/electromagneticradiation.

Rutherford

1911Discovered the nucleus. Showed that it was mostly empty space.

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DeBroglie

1924Theorized the existence of matter waves. Matter waves are particles of matterbehaving as a wave. Einstein said light could be a wave or a particle. In other words

light could behave like a particle(matter). De Broglie said electrons could behave like

light. The electron microscope.

Schrodinger

1926Invented quantum mechanics to solve the mathematical calculations of howatoms behave. Quantum mechanics revolutionized physics and chemistry. It produced

the highest level of understanding of the physical world. After solving the mostcomplicated problem in math, chemistry and physics, Schrodinger set out to solve themost complicated problem in biology, What is Life? Which is the title of his book

which gave the blueprint to solve the mystery of life. He hypothesized the existence ofthe genetic code and of molecules whose different structure carried vital

information(genes). He laid out a blueprint on how to break the genetic code. JamesWatson and Francis Crick were inspired by his ideas and in 1953 discovered thestructure of DNA.

Heisenberg

1927Heisenbergs Uncertainty Principle. You can not know both the speed and the

momentum of a particle at the same time.

Meitner/Hahn/Strassman/Frisch

Hahn and Strassman converted Uranium into Barium. They could not explain it

however. Meitner(and her nephew Frisch) were sent copies of the results and Meitnerfigured out that the Uranium had undergone fission(her word) and been turned into the

Barium. Although, Meitners paper(1939) on the topic is only one page long it isconsidered one of the great papers of all time. Niels Bohr(the unofficial head of thescientific community) brought the paper over to the United States by boat to make sure

it got here before it was published. Why????

Fermi

Italian scientist. Discovered chain reaction in 1942. 1stNuclear reactor at University ofChicago. This leads to ??? in 1945.

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Chapter 9NMR and MS

NMR

NMR stands for Nuclear Magnetic Resonance. Basically the nuclei of atoms are made toresonate with an energy source. In this case magnets help create the resonance.

There are two types of magnets that can be usedelectromagnets and superconducting

magnets. Electromagnets are artificial magnets created by using an electrical source.Electromagnets usually consist of a coil of wire surrounding an iron(or metal) core.

Electricity is passed through the coil. This creates a magnetic field around the wire that isamplified by the number of coils and by the iron core. Electromagnets are utilized in mostdevices. Any device with a motor usually involves an electromagnet. Telephones,

electronics, computers, turbines and powerplantsall use electromagnets(a simpleelectromagnet can be made by coiling a wire around a screwdriver and hooking it to a

battery).

Superconducting magnets are naturally occurring magnets whose power is greatly increased

by cooling it to very low temperatures(4.3 K). Superconductors have no electrical/heatresistance(conductors have low electrical/heat resistance; semiconductors have limited

electrical/heat resistance; insulators have high electrical/heat resistance). A loop of ordinarywire has so much resistance that an electron will stop moving around the loop within asecond of the power being turned off(i.e. flip off the light switch and the light goes off

immediately it doesnt linger). In a superconducting loop the electron would go onforever(theoretically it would never stop; there was an experiment where an electron wasplaced in a loop and let go. It went for years without ever stopping). So if the wiring in your

house was superconducting wire. You would need to start the electrons moving and then noadded electricity would be needed to keep it going(eventually the resistance of the light bulb

itself would stop the electron moving).

The magnets are kept cool by the use of a Dewar chamber(see page 367).

NMR magnets have to have incredible field strength to function properly. Most operate at a

level 70,000 to 140,000 times the strength of the earths magnetic field(MS, which does notrequire the same strength as NMR often, uses permanent magnets or even quadrupoles tooperate). MRIs use a much smaller magnetic field also usually 10,000 to 20,000 times the

strength of the earths magnetic field.

Magnetic quenchingthe bane of NMR. As the temperature in the coil heats up resistancebuilds up in the wire causing heat to be generated which boils away the liquid helium. Thisquenches the magnetic field. A simple NMR costs from 250,000 to multiple millions of

dollars. Quenching a magnet can lead to repair costs in the tens of thousands of dollars.At UNO we spend $7,500-10,000/year on the cryogens to keep the NMR cool.

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Operation

When nuclei of atoms are placed in a magnetic field they will become aligned with oragainst the magnetic field of the NMR. When an electromagnetic pulse(Rf) is applied to

the compound, the nuclei will absorb energy from the pulse. Different types of nuclei

absorb energy at different energy levels. This differentiation can be used to identifydifferent nuclei on NMR spectra.

Types of NMR

Continuous wave(CW) vs. Fourier Transform(FT)

CW was predominant when NMR was in its infancy in the 70s and 80s. CW use

electromagnets instead of superconducting magnets. They do not need to be cooled. Sothere is no cost for cryogens. Spectra take 2-5 minutes to develop. In CW, the sample is

irradiated with a constant value electrical impulse while the magnetic field strength isvaried. Different nuclei absorb at different field strength. When nuclei absorb, theygenerate a current in the wire surrounding the sample. This current is amplified and

graphed on the corresponding spectra. CW only work on proton NMR. Other moderntechniques do not work with CW. Signal to noise ratio is very low. Most magnets were

60 or 90 MHz.

FT is the standard today. FT-NMR use superconducting magnets. So there is the need

to keep the magnet cool. Spectra develop in seconds instead of minutes. Multiplespectra are taken and analyzed and combined too greatly increase the signal to noise

ratio. Therefore, FT-NMR can be done with much less concentration of sample. Nuclei

other than hydrogen can be analyzed. In FT-NMR, an electromagnetic pulse irradiatesand excites all the nuclei at the same time. This generates a complex current pattern for

all the excited nuclei that is amplified and detected. The data is then analyzed by acomputer program(Fourier Transform) that provides spectra indicating the different

types of nuclei.

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Nuclear Spin

Different nuclei will have a different spin value in magnetic fields. If you rememberfrom general chemistry, electrons were given a quantum number value for msof + or

. The different nuclei that organic chemists deal with have different spin quantum

numbers, I. Different isotopes of each nuclei have different spin numbers also. 1H,ordinary hydrogen; 13C, isotopic carbon, 19F and 31P, normal fluorine and phosphorousall have spin quantum numbers of +/. Therefore these are all observable by NMR.12C, 16O, 32S and 2H all have spin quantum numbers of +/1 and are not observable by

NMR. Remember electrons have spin = +/. Electricity is a flow of electrons.Therefore if a nuclei also has a spin of +/ then it can resonate with the magnetic field

to generate a current(similar to the flow of electrons). As stated previously when theprotons are placed in a magnetic field they will either align with(lowerenergy/stable/ground state) or against(higher energy/unstable/excited state) the magnetic

field. In NMR we want all the protons to be aligned. So energy is added to flip all theprotons aligned with the field to be aligned against the field(excited state). The amount

of energy needed to do this varies depending on the strength of the magnet. Thestronger the magnet the larger the separation will be between ground and excited states.The energy is provided by an electromagnetic pulse. Once the protons absorb the

energy they are aligned(resonating) with the magnetic field(called magnetic resonance).The value of the pulse corresponds to the magnets strength. In an older magnet(1.41

tesla), the pulse needed to flip all the protons is 60,000,000 Hz(cycles/second) or 60MHz. For modern NMR(7.04 tesla) the pulse needed to flip all the protons is300,000,000 Hz or 300 MHz(recall if you have a cordless phone at home they operate at

900 MHz; AM radio is measured in kHz1280 on the AM dial is 1,280,000 Hz or 1.28MHz; 99.5 on the FM dial is 99,500,000 Hz or 99.5 MHz).

Solvent

To run a NMR a sample must be dissolved in a liquid. Since the sample is organic an

organic liquid is usually used. But it has to be an isotopic solvent. Ordinarydichloromethane can not be used because the Hs on the dichloromethane would showup in the spectra. Since the solvent is always much higher in concentration the Hs for

dichloromethane would swamp out the spectra of the compound. Instead CD2Cl2orCDCl3or CD3COCD3or D2Oare used. These are called deuterated solvents.

However carbon tetrachloride can be used nondeuterated. Why???

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Chemical Shifts

In IR we saw that different functional groups absorb IR at different frequencies. InNMR different nuclei absorb magnetic radiation at different frequencies also. In NMR

this is called chemical shifts. Chemical shifts of different hydrogens are given on page

378. Chemical shifts are classified as downfield(deshielded/less magnetic field strength)or upfield(shielded/larger magnetic field strength). Most NMR read out in ppm scale

from 0-10. If the hydrogen lies closer to 10, then it is called downfield. If it lies closerto 0 it is called upfield. A sample of tetramethylsilane is added to the NMR solvent to

set the 0 line. Two factors, electron density and electron circulation can determine theposition of the proton in the spectra. The movement of electrons in a magnetic fieldgenerates their own magnetic fields. The magnetic fields of moving electrons shield the

proton from the magnetic field of the NMR itself(think about two south poles of amagnet repulsing each other; or all those star trek movies where a magnetic pulse

actually moves the space ship). The induced magnetic field of the electrons surroundingthe proton shields it from the external magnetic field of the NMR. Basically they feel

less the presence of the external magnetic field. This is called shielding. Since theproton is shielded it does not absorb as much magnetic radiation as it normally shouldand stronger field strength is needed to make the proton absorb and resonate. Thus its

signal is seen upfield(close to 0 ppm). The more electron density around the proton themore it will be shielded and the larger the magnetic field to which it must be subjectedin order for the proton to resonate. If a proton is attached to a carbon with an

electronegative group or attached directly to an electronegative group it will bedeshielded as the electron density is drawn away. The other factor is circulation of

delocalized(from conjugation) electrons. As seen on page 377, the circulation ofelectrons around pi bonds causes a field to be developed. In terminal alkynes that fieldshields the protons and move them upfield. In aromatic rings such as benzene, the

protons are often deshielded and moved downfield.

TMS has 12 protons. So a small concentration yields a large signal. The Hs attached tothe silicon are more shielded since Si is less electronegative than C(why???). Thereforethey appear way upfield where no other ordinary organic compounds normally appear.

The chemical shift is a ratio of the shift away from TMS/frequency of the instrument:

= (obs. shift from TMS(Hz)) x 106/(freq. of the instrument(Hz))

The units are ppm.

Integration

Each peak on the spectra takes up a certain area of the paper. Taking the integral of the

peak gives us the relative number of hydrogens in that peak. By comparing integrationswe can determine how many hydrogens are represented by each peak. I.E. if we

integrate the area under the peaks we may get an integral of 2. This could correspond toa CH2.

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Signal Splitting

In addition to chemical shift data and integration to help us understand NMR, there is a

third important feature called signal splitting. Hydrogens within in 3 sigma() bonds ofeach other will split each others signal into doublets, triplets, quartets and multiplets. If

hydrogens are equivalent or enantiotopic they will not split each other. If they are in adifferent environment or diastereotopic they will split each other. This is the N+1 rule.If a hydrogen is next to 3 other hydrogens its signal will be split into 4(3+1) peaksa

quartet. The 3 hydrogens(assuming they are equivalent) will be split into 2(1+1)peaksa doublet. Therefore on the spectra, the peak that corresponds to a CH next to aCH3, will appear as a quartet. The CH3peak will appear as a doublet. The integral of

the quartet peak will be 1(remember it represents 1 H). The integral of the doublet peakwill be 3(it represents 3 H). So integral and splitting are not related.

Why doesnt the carbons split the hydrogens? If a methyl hydrogen is split by ahydrogen on a carbon next to it, why not by the carbon itself?

Equivalent/Nonequivalent protons

Homotopic protons are protons with the same chemical shift. Homotopic protons must be in

the exact same environment as each other. Therefore 2 sets of CH3s or 2 sets of CHSeeexample page 379. If you have 2 sets of protons that you think are the same replace one

hydrogen in one set with another group(Cl). Then redraw the molecule and replace onehydrogen in the other set with another group(Cl). If the two molecules you have drawn arethe same new compound then they are homotopic.

Enantiotopic and diastereotopic are hydrogens on the same atom that may have a differentenvironment. For example, a methylene carbon with 2 Hs on it may have differentenvironments and give different shifts. To determine if they are enantiotopic ordiastereotopic switch the hydrogens for new groups like in the homotopic case. If the new

pictures are enantiomers of each other then the Hs will have the same chemical shift. If thenew pictures are diastereomers of each other then the Hs will have different chemical shifts.

Spin-spin Coupling

Spin-spin coupling gives rise to signal splitting we talked about earlier. We are not

going to go into the details of the coupling other than to talk about field strength of themagnets. The splitting of the proton peaks into quartetsis the same in all field

strength. The coupling constant(which determines how far apart the peaks are in thequartet) is the same in all magnetic field strengths. However high field strength NMRsallow for greater separation of the chemical shifts of the protons. In other words, 2 sets

of protons that may overlap in a 60 MHz instrument can be resolved in a 300 MHzinstrument.

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Process Rate

Because of the speed of the relaxation of protons after being bombarded by em pulse,NMR can not usually detect rapidly interconverting processes. I.e. a chair flip from

equatorial to axial hydrogen should yield two separate chemical shifts. The NMRs

processing is so slow however that it will not observe the interconversion and give onlyone shift. Lowering the temperature can slow the interconversion and allow theinstrument to detect it.

Carbon NMR

Because carbon has a spin quantum number of 1 it can not ordinarily be detected byNMR. But the isotope of carbon has a spin quantum number of so it can be detected.Since only 1.1% of all carbon is isotopic 13-Carbon the signal is much weaker. It can

only be done on a FT-NMR. Carbon-carbon splitting is possible. However the chance,of having an isotopic carbon next to each other is 1 in 10,000. Therefore carbons

usually show up as one unsplit peak. Each carbon will have its own unique chemicalshifts. Isotopic carbons can be split by hydrogen. But a special type of probe is usedthat cuts out the proton coupling(called a broadband proton decoupled probe). If the

decoupling is turned off, the splitting pattern of carbons can be used to determine thenature of the carbons. The chemical shifts of the different types of carbon are found on

page 394. These carbons are all considerable downfield from the hydrogens. The sameprinciples of shielding/deshielding apply. On the spectra on page 396 explain why thesolvent shows up here and not on proton NMR.

Special Types of NMR

DEPT 13C NMR

DEPT spectra yields information concerning the protons attached to the carbon. InDEPT, four different spectra are taken. The first shows only CH3s. The second shows

only CH2s. The third shows only CHs and the fourth shows only Cs. The fourspectra are then converted into one spectra as shown on page 397.

2D and 3D NMR(HETCOR/COSY)

COSY

COSY(1H-1H correlation spectroscopy) is a NMR technique where the protons of a

compound are plotted against themselves. Protons that are coupled to protons on adifferent carbon show up in the spectra as a high-density area in a contour-style map.

HETCOR

HETCOR(Heterogenous correlation spectroscopy) is similar to COSY except itcorrelates the protons to the carbon it is attached to.

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H3C

H2C

HC

C

C

C

H3C

H2C

HC

CH

CH

CH

H3C CH2

H2C CH2

HC CH2

H3C CH3

H2C CH3

HC CH3

singlet integration of 3



doublet integration of 3



triplet integration of 3



quartet integration of 3



H3C

H2

C

C

H2

appears as quartet due to CH3 and triplet because of CH2with integration of 2

H2C

CH

H2C

Cl

appears as triplet due to CH2 to the left and triplet due to CH2 on the right.

with integration of 1.

C

H

H2C

C

H2

appears as doublet

of triplet with int.

of 2

H

C

CH

Br

CH3

doublet due to CH, quartet due to CH3,

integration of 1.

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NMR examples

a

b d

e f

a

c c

O

a

c

b

f

de

0.51.02.05.06.0

a The CH3's labeled A have only one neighbor that

being the CH at B. Since 1 + 1 = 2. A appears

as a doublet. Since the carbons at A are not next

to anything other than sp3carbons, they appear

between 0 and 1 with an integration of 6.

b The CH at B has 6 neighbors, 3 each from Carbons

A. This could appear as a group of 7 peaks or as

a quartet of quartets with integration of 1.

c The CH3's at C have no neighbors so they appear as

a singlet with integration of 6.

d This is a CH attached to a double bond. This will

appear in the 5-6 region. It has one neighbor from

the CH on E so it will be a doublet, integration of 1.

e This is a CH attached to a double bond. This will

appear in the 5-6 region. It has one neighbor from

the CH on D so it will be a doublet, integration of 1.

f This is a CH3 next to a ketone. It will appear around

2.0. It has no neighbors so it will be a singlet with

an integration of 3.

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5 4 3 2 1

T

M

S

4 singlets at 0.8 integration of three, represents hydrogens on carbon 13, 14, 15 and 16 with no neighbors

triplet at 0.9, integration of three, represents hydrogens on carbon 11 with 2 neighbors from carbon 10

doublet at 0.9, integration of three, represents hydrogens on carbon 12 with 1 neighbor from carbon 8doublet at 0.9, integration of three, represents hydrogens on carbon 17 with 1 neighbor from carbon 5

quartet at 1.7, integration of one, represents hydrogens on carbon 5 with 3 neighbors from carbon 17

quartet at 2.4, integration of two, represents hydrogens on carbon 10 with 3 neighbors from carbon 11quartet at 2.5, integration of one, represents hydrogens on carbon 8 with 3 neighbors from carbon 12

singlet at 3.1 integration of one, represents hydrogens on carbon 1 with no neighborssinglet at 3.8 integration of one, represents hydrogens on carbon 3 with no neighbors

Proton NMR

2

3

4

5

6

7

8

12

1

O

9

10

O

11

H

H Cl

17

16 15

1314

HH

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MORE NMR PROBLEMS

1

2

3

4

6

7

8

9

12

13

O

14

15

O

O

510 11

6 5 4 3 2 1

6 7 8

2 15

5 10 11

1214 1

6, 7 and 8 have integral of 1. 2, 12 and 14 have integral of 2 and 1, 5, 10, 11 and 15 have integral

of 3

1, 3 and 9 have integral of 1. 4, 6, 8 and 10 have integral of 2.

1

2

3

4

5

6

7

O

8

9

10

OOCl

6 5 4 3 2 1

13 4

6

810 9

5 and 9 have integral of 1. 2, 3, 6 and 10 have integral of 2. 1, 7 and 11 have integral of 3.

1

2

3

4

5

8

9

10

11

O

6

7

Br

5 4 3 2 1.5 1 0.8

117 12 6109 35

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4

5 7

3

O

6

2 1

1

2

3 5

6 9

1 4 7

8

O

1

6

45 72

83

2

7 3.5 2.92.5

3

1.0

1

9

2.15.0 .81.01.9

4

1.6

5

2.6

6

1.2

7

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87

7 3.5 2.6 1.0

7

8

9

10

5

6

14

12

13

O 11

4

3

2

1

O

H

2.0

11

2

10

1 3

4 14

12 13

2.15.0 .81.01.9 1.62.6

1

2

4

5

6

7

8

10

O

11

12

14

3

139O

O

145 6

3.5

11

10 28

13 9 1 3

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H-NMR

0-2 CH3, CH2 and CH attached to other R groups.1.5-2 CH3, CH2 and CH allylic to a double bond.

2-2.5 CH3, CH2 and CH next to benzene or ketone.

2.5-3.5 H attached directly to a triple bond.3-3.5 CH3, CH2 and CH attached to F, Br, I.

3.5-4 CH3, CH2 and CH attached to Cl.3.3-4 CH3, CH2 and CH attached to O(alcohol or ether).

4-6 H attached directly to double bond6-9 H attached directly to benzene ring.9-11 H attached to aldehyde or to O of carboxylic acid.

C-NMR

0-50 Carbon attached to R group.50-100 Carbon attached to heteroatom(O, N, Cl, Br) or carbon as part of a triple

bond.100-150 Carbon is part of a double bond(alkene, benzene).

150-200 Carbon is part of a carbonyl(ketone, aldehyde, acid, ester).

IR

3400-3600 OH

3300 Alkyne Hs3030 Benzene/aromatic Hs2900 CHs

1620-1780 Carbonyls1480-1600 Alkene

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MS Datam/z % intensity

202(M+) 89.45

203(M++1) 13.78

204(M++2) 87.69

IR DataSharp peak at 3030 Strong multiplet at 2900

Strong peak at 1720 Multiple, Medium peaks from 1480-1550

NMR Data

8 7 6 5 4 3.5 3 2 1 0

Peaks doublet at 0.6 has integral of 3

doublet at 2.0 has integral of 2 doublet at 2.2 has integral of 2

doublet at 2.4 has integral of 2

tripled triplet at 3.1 has integral of 1

singlet at 3.5 has integral of 3

tripled quartet at 3.9 has integral of 1

5 singlets at 7.1, 7.2, 7.3, 7.4, 7.5 all with integral of 1

DEPT spectra indicates that there are 14 carbon peaks. At 180 is a Carbon with no

Hydrogens. At 145, 143, 142, 141 and 140 are Carbons with 1 hydrogen. At 150 is a

Carbon with no Hydrogens. At 80 is a carbon with 1 Hyrogen. At 75 is a carbon with 3hydrogens. At 70 is a carbon with 1 hydrogen. At 35, 30 and 25 are carbons with 2

hydrogens. At 28 is a carbon with 3 hydrogens.

What is the structure?

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MS Datam/z % intensity

299(M+) 89.45 (13.78/89.45)*100 = 15.4/1.1 = 14

203(M++1) 13.78 (87.69/89.45)*100 = 98 = Br

204(M++2) 87.69 MF = C14H19BrO2

IR DataSharp peak at 3030--Aromatic Ring Strong multiplet at 2900CHs

Strong peak at 1720--Ketone Multiple, Medium peaks from 1480-1550--Alkene

NMR Data

8 7 6 5 4 3.5 3 2 1 0

Peaks doublet at 0.6 has integral of 3 CH3 next to CH

doublet at 2.0 has integral of 2 CH2 next to CH and ketone

doublet at 2.2 has integral of 2 CH2 next to CH and ketone

doublet at 2.4 has integral of 2 CH2 next to CH and benzene

tripled triplet at 3.1 has integral of 1 CHBrbetween 2 CH2 singlet at 3.5 has integral of 3 OCH3 tripled quartet at 3.9 has integral of 1 CHO between CH3 and CH2

5 singlets at 7.1, 7.2, 7.3, 7.4, 7.5 all with integral of 1 Benzene CH's


Hydrogens. At 145, 143, 142, 141 and 140 are Carbons with 1 hydrogen. At 150 is a

Carbon with no Hydrogens. At 80 is a carbon with 1 Hyrogen. At 75 is a carbon with 3hydrogens. At 70 is a carbon with 1 hydrogen. At 35, 30 and 25 are carbons with 2

hydrogens. At 28 is a carbon with 3 hydrogens.

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FRAGME TS FROM MR

HC CH3HC

H2C

O

HC

H2

C

O

H2C

CH

H2C

CH

CH2

Br

O CH3

H2C

CH

O

CH3 H

H

H

H

H

H2C

CH

O

CH3

CH3HC

H2C

O

H2C

C

H

H

H

H

H2

C

H

H

CH

0.6 2.0 2.2 2.4

3.13.5

3.9

7.1-7.

5

7.1-7.5 with 2.40.6 to 3.9 and 3.5 to 3.9

2.0 to 2.2

H

H

H

H2C

H

H

C

H

AB C

A to 3.1

CH2

Br

HC

H2

C

O

H2

C

CH

CH3

O

CH3

B to C

HC

H2C

O

H2C

CH

CH3

O

CH3

Br

H2C

H

HH

H

H

22


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MS Data

m/z % intensity

122(M+) 84.20

123(M

+

+1) 3.70124(M++2 27.79

IR Data

Strong multiple peaks at 2900

Strong Peak at 1740

NMR Data


10 9 8 7 6 5 4 3 2 1

0

T

M

S

Peaks 2.2 has integral of 1.8,

3.4 has integral of 1.8


DEPT spectra indicates that there are 4 carbon peaks. At 185 is a C with no Hydrogens. At 80 is a carbon with 3 Hydrogens. At 65 is a

carbon with 2 Hydrogens and at 15 is a carbon with 2 Hydrogens.

23


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MS Data

m/z % intensity

122(M+

) 84.20 (3.70/84.20)*100 = 4.4/1.1 = 4 Carbons123(M

++1) 3.70 (27.70/84.20)*100 = 33% = 1 Chlorine

124(M++2) 27.79 Based on DEPT = 7 Hydrogens

C4H7Cl = 90122 90 = 32 = 2 Oxygens

IR Data C4H7O2Cl

Strong multiple peaks at 2900CHs

Strong Peak at 1740 Carbonyl(C double bond O) Ketone or ester due tolack of aldehyde or carboxylic acid peak in proton NMR.

NMR Data

10 9 8 7 6 5 4 32 1 0

TMS

Peaks 2.2 has integral of 1.8,



DEPT spectra indicates that there are 4 carbon peaks. At 185 is a C with no Hydrogens. At 80 is a carbon with 3 Hydrogens. At 65 is a

carbon with 2 Hydrogens and at 15 is a carbon with 2 Hydrogens.

Peak at 2.2 = 2 Hydrogens with 2 neighbors next to ketone(CH2)

Peak at 3.4 = 2 Hydrogenswith 2 neighbors next to chlorine(CH2Cl)

Peak at 3.8 = 3 Hydrogens with no neighbors but next to Oxygen(OCH3)

185 =C

O

80 = C next to O(OCH3) 65 = C next to Cl(CH2Cl) 15 = CH2

4

O

3

2

1

Cl

O

4

12

24


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Problem #1

MS Data

m/z % intensity86(M

+) 54.30

87(M++1) 2.99

88(M++2 0.05

IR Data

Strong peak at 2900

Strong peak at 1725

NMR Data


10 9 8 7 6 5 4 3 2 1

0

TM

S

Peaks triplet at 0.8 has integral of 7.8,

triplet of quartet at 1.3 has integral of 5.2 singlet at 2.1 has integral of 7.8 triplet at 2.4 has integral of 5.2

DEPT spectra indicates that there are 5 carbon peaks. At 200 is a C with no Hydrogens. At

50 is a carbon with 2 Hydrogens. At 40 is a carbon with 3 Hydrogens. At 35 is a carbon with

2 Hydrogens and at 15 is a carbon with 3 Hydrogens.

25


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MS Data

m/z % intensity

86(M+) 54.30 (2.99/54.30)*100 = 5.5/1.1 = 5 Carbons

87(M

+

+1) 2.99 (0.05/54.30)*100 = 0.09% = Nothing88(M++2 0.05Based on DEPT = 10 Hydrogens

C5H10= 70IR Data 86 70 = 16 = 1 Oxygen

C5H10O

Strong peak at 2900 CHsStrong peak at 1725 Carbonyl(C double bond O) Ketone or ester due to lack

of aldehyde or carboxylic acid peak in proton NMR.NMR DataP

10 9 8 7 6 5 4 3 2 1

0

TM

S

Peaks triplet at 0.8 has integral of 7.8,

triplet of quartet at 1.3 has integral of 5.2 singlet at 2.1 has integral of 7.8

triplet at 2.4 has integral of 5.2


50 is a carbon with 2 Hydrogens. At 40 is a carbon with 3 Hydrogens. At 35 is a carbon with

2 Hydrogens and at 15 is a carbon with 3 Hydrogens.

Peak at 2.4 is next to ketone with 2 neighbors.

Peak at 2.1 is next to ketone with no neighbors.Peak at 1.3 is nex t to 2 and 3 hydrogens.

Peak at 0.8 is next to 2 hydrogens

200 = carbonyl, 50 = CH2next to ketone, 40 = CH3 next to ketone, 35 = CH2and 15 = CH3

1

2

3

4

5

O

3

1

4

5

26


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MS Data

m/z % intensity

271(M+

) 75.80272(M

++1) 10.01

273(M++2 74.28

IR Data

Small, sharp peak at 3030Strong multiplet at 2900

Strong peak at 1680Medium set of peaks in 1500 and 1600s

NMR Data


10 9 8 7 6 5 4 3 2 1

0

TMS

Peaks triplet at 1.0 has integral of 9.75

triplet at 1.2 has integral of 9.75 singlet at 2.2 has integral of 9.75 quartet at 2.5 has integral of 6.5 quartet at 3.8 has integral of 6.5

singlet at 8.2 has integral of 3.25 singlet at 8.8 has integral of 3.25

DEPT spectra indicates that there are 12 carbon peaks. At 180 is a Carbon with noHydrogens. At 165 and 160 are Carbons with 1 hydrogen each. At 150, 152, 156 and 158

are Carbons with no Hydrogens. At 50 is a carbon with 2 Hyrogens. At 45 is a carbon with 2hydrogens. At 40 is a carbon with 3 Hydrogens. At 35 is a At carbon with3 Hydrogens andat 30 is a carbon with 3 Hydrogens.

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MS Data

m/z % intensity271(M+) 75.80 (10.01/75.80)*100 = 13.2/1.1 = 12 Carbons

272(M++1) 10.01 (74.28/75.80)*100 = 98% = 1 Bromine

273(M+

+2 74.28 Based on DEPT = 15 Hydrogens C12H15Br= 239

IR Data 271 239 = 32 = 2 Oxygens C12H15O2Br

Small, sharp peak at 3030 Aromatic Hs

Strong multiplet at 2900 CHsStrong peak at 1680 Ester or Ketone(lack of aldehyde/acid in H-NMR)Medium set of peaks in 1500 and 1600sAromatic double bonds

NMR Data

10 9 8 7 6 5 4 3 2 1

0

T M

S

Peak s triplet at 1.0 h as integral of 9.75 3 prontons w ith 2 neighbors(C H 3CH 2 )

triplet at 1.2 has integral of 9.75 3 protons w ith 2 neighbors(C H 3CH 2 )

singlet at 2.2 has integral of 9.75 3 protons next to ketone or aromatic ring(no neighbors)

quartet at 2.5 has integral of 6.5 2 protons next to ketone or arom atic ring(3 neighbors)

qua rtet at 3.8 has integral of 6.5 2 protons next to ether(ester) (3 neighbors) singlet at 8.2 ha s integr al of 3.25 1 arom atic proto n

singlet at 8.8 ha s integr al of 3.25 1 arom atic proto n

DE PT spectra indicates that there are 12 carbon peaks. A t 180 is a Carbon w ith no H ydrogen

At 165 and 160 are Carbons with 1 hydrogen each. A t 150 , 152, 156 and 158 are Carbonswith no H ydrogens. A t 50 is a carbon with 2 Hyrogens. A t 45 is a carbon with 2 hydrogens.

At 40 is a carbon with 3 H ydrog ens. A t 35 is a At carbon w ith 3 H ydrogens and at 30 is acarbon with 3 H ydrogens.

180 = ketone or ester, 165, 160, 158, 156, 152, 150 = benzene ring, 50 = CH 2next to ether, 45 = CH 2 next to ring, 40 = CH 3next to ring,

3 5 = C H 3and 30 = CH 3

1

2

H

H

Br

7

5

6

O

O

3

4 1

2

H

H

Br

7

O 5

O

3

4

6

Both are po ssible. Also the groups

could be arranged differently around

the ring.

1 2

3 45 6

7

28


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MS Data

m/z % intensity

172(M+) 69.95

173(M

+

+1) 6.93174(M++2) 0.56

IR Data


Strong Peak at 1740Strong Peak at 1600

NMR Data

Peaks triplet at 0.8 has integral of 3,triplet of triplet at 1.3 has integral of 2

triplet at 2.4 has integral of 2


quartet at 3.5 has integral of 2

doublet at 3.6 has integral of 2 triplet of triplet at 5.3 has integral of 1



120 is a carbon with 2 Hydrogens and at 110 is a carbon with 1 Hydrogen. At 95, 85 and 75

are carbons with 2 Hydrogens. At 45 is a carbon with 2 Hydrogens. At 25 is a carbon with 3Hydrogens and at 15 is a carbon with 2 hydrogens.

6 5 4 3 2

1

29


30/39

MS Data

m/z % intensity (6.93/69.95)*100 = 9.9/1.1 = 9 Carbons

172(M+) 69.95 (0.56/69.95)*100 = 0.8% = No Cl, Br or S

173(M

+

+1) 6.93 Based on DEPT = 16 Hydrogens174(M++2) 0.56 C9H16= 124

172 - 124 = 48 = 3 OxygensIR Data C9H16O3


Strong Peak at 1740 Carbonyl(C double bond O) Ketone or ester due to lackaldehyde or carboxylic acid peak in proton NMR.

Strong Peak at 1600 ALKENENMR Data

Peaks triplet at 0.8 has integral of 3,triplet of triplet at 1.3 has integral of 2



quartet at 3.5 has integral of 2

doublet at 3.6 has integral of 2 triplet of triplet at 5.3 has integral of 1



120 is a carbon with 2 Hydrogens and at 110 is a carbon with 1 Hydrogen. At 95, 85 and 75are carbons with 2 Hydrogens. At 45 is a carbon with 2 Hydrogens. At 25 is a carbon with 3

Hydrogens and at 15 is a carbon with 2 hydrogens.

Peak at 0.8 is next to CH2.

Peak at 1.3 has CH2's on each side.

Peak at 2.4 is next to ketone and CH2.

Peak at 3.3 is next to oxygen and CH2.

Peak at 3.5 is next to oxygen and CH3

Peak at 3.6 is next to oxygen and CHPeak at 5.3 is alkene and has CH2's on each side.

Peak at 5.8 is alkene and next to CH

6 5 4 3 2

1

1

2

O

3

4

5

6

O

7

8

9

O

9 8 7 2 6 4 5 1

30


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MS Data

m/z % intensity

150(M+) 88.69

151(M

+

+1) 5.85152(M++2) 29.27

IR Data

Strong multiplet at 2900

Strong peak at 1680

NMR

Peaks doublet at 0.8 with integral of 3

doublet at 2.5 with integral of 2

singlet at 3.3 with integral of 3 singlet at 3.5 with integral of 2

triplet of quartet at 3.8 with integral of 1

DEPT spectra indicates that there are 6 carbon peaks. At 200 is a C with no hydrogens. At90 is a carbon with 3 hydrogens and at 85 is a carbon with 2 hydrogens. At 50 is a carbon

with 1 hydrogen. At 45 is a carbon with 2 hydrogens and at 15 is a carbon with 3 hydrogens.

3.8 3.5 2.5 0.

8

3.3

31


32/39

MS Data

m/z % intensity

150(M+) 88.69 (5.85/88.69)*100 = 6.6/1.1 = 6 Carbons

151(M

+

+1) 5.85 (29.27/88.69)*100 = 33% = 1 Cl152(M++2) 29.27 Based on DEPT = 11 Hydrogens

C6H11Cl = 118150 118 = 32 = 2 Oxygens

C6H11ClO2IR Data

Strong multiplet at 2900 CHs

Strong peak at 1680 Carbonyl

NMR

Peaks doublet at 0.8 with integral of 3

doublet at 2.5 with integral of 2

singlet at 3.3 with integral of 3

singlet at 3.5 with integral of 2triplet of quartet at 3.8 with integral of 1

DEPT spectra indicates that there are 6 carbon peaks. At 200 is a C with no hydrogens. At

90 is a carbon with 3 hydrogens and at 85 is a carbon with 2 hydrogens. At 50 is a carbon

with 1 hydrogen. At 45 is a carbon with 2 hydrogens and at 15 is a carbon with 3 hydrogens.

3.8 3.5 2.5 0.

8

3.3

Peak at 0.8 is next to CH.

Peak at 2.5 is between ketone and CH.

Peak at 3.3 is CH3 next to O.

Peak at 3.5 is CH2 next to O.

Peak at 3.8 is a CHCl between a CH2 and a CH3.

1

O

2 3

4

5

O Cl

4

2 1

3

5

32


33/39

MS Data

m/z % intensity

210(M+) 91.56

211(M

+

+1) 12.09212(M++2) 30.21

IR Data


Strong peak at 1680Medium set of peaks in 1500 and 1600s

NMR

Peaks singlet at 2.1 with integral of 3



triplet at 2.5 with integral of 2triplet at 3.7 with integral of 2




170, 165, 160 and 155 there are carbons with 0 Hydrogens and at 150 and 145 are carbonswith 1 Hydrogen. At 50 is a carbon with 2 Hydrogens. At 45 is a carbon with 2 Hydrogens.

At 40, 35 and 30 are carbons with 3 hydrogens.

7 6 5 4 3 2

33


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MS Data

m/z % intensity

210(M+) 91.56 (12.09/91.56)*100 = 13.2/1.1 = 12 Carbons

211(M

+

+1) 12.09 (30.21/91.56)*100 = 33% = 1 Cl212(M++2) 30.21 Based on DEPT = 15 Hydrogens

C12H15Cl = 194210 - 194 = 16 = 1 Oxygen

IR Data C12H15ClOSmall, sharp peak at 3030 Aromatic Protons = Benzene

Strong multiplet at 2900 CHsStrong peak at 1680 Carbonyl

Medium set of peaks in 1500 and 1600s Double bonds = Benzene

NMR

Peak s singlet at 2.1 with integral of 3

singlet at 2.2 with integral of 3 singlet at 2.3 with integral of 3

triplet at 2.5 with integral of 2

triplet at 3.7 w ith integral of 2 singlet at 7.0 with integral of 1 singlet at 7.5 with integral of 1

D EPT spectra indicates that there are 12 carbon peaks. At 200 is a C with no Hydrogens. At

170, 165, 160 and 155 there are carbons with 0 Hydrogens and at 150 and 145 are carbons

w ith 1 Hydrogen. At 50 is a carbon with 2 Hydrogens. At 45 is a carbon with 2 Hydrogens.At 40, 35 and 30 are carbons with 3 hydrogens.

7 6 5 4 3 2

7 6

5 3

4

HH

12

ClO

7 6 2 1

3 4 5

Peak at 2.1 is CH3 next to ring.Peak at 2.2 is CH3 next to ring.

Peak at 2.3 is CH3 next to ring.Peak at 2.5 is in between keto ne and CH2

Peak at 3.7 is in between CH 2 and Cl.Peak at 7.0 is CH of benzene ring.Peak at 7.5 is CH of benzene ring.

Peaks at 7.0 and 7.5 are singlets. This means

they are not next to each o ther.

34


35/39

MS Data

m/z % intensity

202(M+) 65.60

203(M

+

+1) 7.94204(M++2) 21.60

IR Data

Strong multiplet at 2900

Strong peak at 1680Medium peak at 1580

NMR Data

7 6 5 4 3 2 1 0

Peaks doublet at 0.6 has integral of 3 triplet at 0.7 has integral of 3 tripled quartet at 0.8 has integral of 2 doublet at 0.9 has integral of 3

singlet at 1.5 has integral of 3 doubled quartet at 2.3 has integral of 1 triplet at 2.4 has integral of 2 quartet at 3.6 has integral of 1



Hydrogens. At 145 is a Carbon with 1 hydrogen. At 140 is a Carbon with noHydrogens. At 80 is a carbon with 1 Hyrogen. At 45 is a carbon with 1 hydrogen. At40 and 35 are carbons with 2 hydrogens. At 30, 25, 20 and 15 are carbons with 3hydrogens.


35


36/39

m/z % intensity (7.94/65.6)*100 = 12.1/1.1 =11 Cs

202(M+) 65.60 (21.60/65.6)*100 = 33% = Cl

203(M++1) 7.94 Dept indicates 19 Hs

204(M++2) 21.60 C11H19Cl = 186

202 - 186 = 16 = 1 Oxygen

C11H19OCl

IR DATAStrong multiplet at 2900CHs bondsStrong peak at 1680Ketone (only 1 oxygen so not ester and no Hs at 9-

11 onNMR so not Aldehyde or Acid)

Medium peak at 1580Alkene

NMR Data

7 6 5 4 3 2 1 0

d o u b l e t a t 0 . 6 h a s i n te g r a l o f 3

t r i p l e t a t 0 .7 h a s i n t e g r a l o f 3

t r i p l e d q u a r t e t a t 0 . 8 h a s i n t e g r a l o f 2 d o u b l e t a t 0 . 9 h a s i n t e g r a l o f 3

s in g l e t a t 1 .5 h a s in t e g ra l o f 3

d o u b l e d q u a r te t a t 2 .3 h a s i n t e g r a l o f 1

t r i p l e t a t 2 .4 h a s i n t e g r a l o f 2 q u a r te t a t 3 .6 h a s in t e g r a l o f 1

d o u b l e t a t 5 . 7 h a s i n t e g r a l o f 1

D E P T s p e c t r a i n d i c a te s t h a t t h e r e a r e

1 1 c a r b o n p e a k s . A t 1 8 0 i s a C a r b o nw i t h n o H y d r o g e n s . A t 1 4 5 i s a

C a r b o n w i t h 1 h y d r o g e n . A t 1 4 0 i s a

C a r b o n w i t h n o H y d r o g e n s . A t 8 0 isa c a r b o n w i t h 1 H y r o g e n . A t 4 5 is a

c a rb o n w i th 1 h y d r o g e n . A t 4 0 a n d3 5 a r e c a r b o n s w i t h 2 h y d r o g e n s . A t

3 0 , 2 5 , 2 0 a n d 1 5 a r e c a r b o n s w i t h 3h y d r o g e n s .

P e a k s0 . 6 = C H 3 n e x t t o C H , n o f u n c t i o n a l g r o u p s c lo s e

0 . 7 = C H 3 n e x t t o C H 2 , n o f u n c t i o n a l g r o u p s c lo s e

0 . 8 = C H 2 b e t w e e n C H 2 a n d C H 3 , n o f u n c t io n a l g r o u p s c lo s e0 . 9 = C H 3 n e x t t o C H , n o f u n c t i o n a l g r o u p s c lo s e

1 . 5 = C H 3 n e x t t o C H o f a d o u b le b o n d2 . 3 = C H b e t w e e n C H , C H 3 a n d k e to n e

2 . 4 = C H 2 n e x t t o C H 2 a n d k e t o n e

3 . 6 = C H n e x t t o C H 3 a n d C l a n d C

7

6 5

4 3

2

1

8

C l

9

O

36


37/39

Test #5NMR Name:

MS Data

m/z % intensity164(M+) 44.86

165(M++1) 5.43

166(M++2) 0.18

IR Data


Multiple, Medium, sharp peaks at 1400-1500

NMR Data


37


38/39

KEY

MS Data

m/z % intensity164(M

+) 44.86 (5.43/44.86)x100 = 12.1/1.1 = 11 carbons

165(M++1) 5.43 (0.18/44.86)x100 = 0.4% = no chlorine, bromine, sulfur166(M

++2) 0.18 C11H16= 148; 164 148 = 16; = 1 oxygen

C11H16OIR Data

Small, sharp peak at 3030 Aromatic hydrogen peaks

Strong multiplet at 2900 CH peaks(no carbonyls detected)

Multiple, Medium, sharp peaks at 1400-1500 Aromatic double bond peaks

NMR Data

38


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NMR data

H-NMR

0-2 CH3, CH2 and CH attached to other R groups.

1.5-2 CH3, CH2 and CH allylic to a double bond.

2-2.5 CH3, CH2 and CH next to benzene or ketone.

2.5-3.5 H attached directly to a triple bond.

3-3.5 CH3, CH2 and CH attached to F, Br, I.

3.5-4 CH3, CH2 and CH attached to Cl.

3.3-4 CH3, CH2 and CH attached to O(alcohol or ether).

4-6 H attached directly to double bond

6-9 H attached directly to benzene ring.

9-11 H attached to aldehyde or to O of carboxylic acid.

C-NMR

0-50 Carbon attached to R group.

50-100 Carbon attached to heteroatom(O, N, Cl, Br) or carbon as part of a triple bond.

100-150 Carbon is part of a double bond(alkene, benzene).

150-200 Carbon is part of a carbonyl(ketone, aldehyde, acid, ester).

IR

3400-3600 OH 3300-3500 NH

3300 Alkyne Hs 3000-3100 Alkene H2

3030 Benzene/aromatic Hs 2900 CHs

1620-1780 Carbonyls 1620-1680 Alkene

39

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