The final
• Thursday, December 12, 8:00 – 10:45 AM• Probably 18 multiple choice @ 5 pts, about 12 short answer @ 5 – 10• Rooms to be announced• ONE index card• Cover page at http://www.ic.sunysb.edu/Class/orgchem/che321/Exam_Cover.pdf
• Synthesis library• pKa table• Chemical shift table• IR peak table
Please remember to fill out…
• The undergraduate TA feedback survey• accessible from an email in your *@stonybrook.edu inbox• Closes Thursday, December 12, at 7:00 AM
• The official course evaluation• Accessible at https://stonybrook.campuslabs.com/courseeval/• Closes Monday, December 9, at 11:59 PM
Your presenters this evening are…
• Chapter 9: Jonathan Khan• Chapter 10: Shawn Raghunandan• Chapter 11: Pradeep Ravindra• Chapter 12/Synthesis: Arthur Makarenko
Nuclear Magnetic Resonance SpectroscopyJonathan Khan©2013, Jonathan Khan. Under no circumstances may any part of this presentation be duplicated or used for commercial purposes. This material may not be reproduced for any use without explicit permission of the author.
Solomons, T. W. G., Organic chemistry.
Brief Review of Spectroscopy
Spectroscopy is defined as the interaction of matter with electromagnetic radiation (light).
The technique is always the same, expose a sample of your molecule to a specific type of electromagnetic radiation, collect data, the make a conclusion based off of the absorption/emission pattern of the molecule.
Spectroscopic Assays of CHE 321
Different forms of spectroscopy allow us to probe a molecule in various ways to gain different data about the molecule in question. IR spectroscopy allows us to qualitatively analyze a
molecule for the presence of functional groups
NMR spectroscopy allows us to quantitatively and qualitatively analyze a molecule for the presence of specific functional groups and the way in which they are associated.
NMR Spectroscopy
Certain atomic nuclei such as 1H and 13C behave as if they were bar magnets spinning around an axis.
When placed into a strong magnetic field and subsequently irradiated with electromagnetic radiation (usually in the radio range), these nuclei absorb some of the energy in a process called magnetic resonance.
Each chemically inequivalent proton will resonate at a different frequency characteristic of the functional group to which it belongs.
Normally, the spin axes of the spin-active nuclei in the molecule are disordered.
When placed into a strong magneticfield, the spin axis of the spin-activenuclei align with the appliedmagnetic field.
Applied Magnetic Fields
Solomons, T. W. G., Organic chemistry.
Magnetic Resonance
When a pulse of electromagneticradiation (specifically radio waves)is applied to the molecules in themagnetic field, it causes some ofthe spin-active nuclei to “flip.”
This causes the spin of the “flipped” nucleus to be aligned against the magnetic field.
As the strength of the applied magnetic field increases, the energy required to “flip” the nuclei also increases
Solomons, T. W. G., Organic chemistry.
Chemical Shifts
For each nuclei resonating at a particular energy value, we see one signal on our spectrum corresponding to that energy value.
An increase in resonance energy results in a higher ppm. The position of the signal on the spectrum allows us to
determine the chemical environment that the proton is sitting in.
For a standard proton spectrum, the value of the chemical shift will almost never go beyond 13ppm.
Upfield and downfield are terms used to describe the position of the signal on the spectrum. Downfield means at a higher ppm
Upfield means at a lower ppm
Shielding
Shielding is a term used to describe the relative strength of the induced magnetic field felt by the proton.
Caused by circulating electrons
The magnetic field felt by a particular proton can be either: Stronger than the applied field
Deshielded proton, caused by a decrease in electron density.
Electronegative functional groups such as carbonyls and halogens draw electrons away from other groups
Weaker than the applied fieldShielded proton, caused by an increase in electron density.
sp3 carbons are a good example, many σ-bonds increase the electron density around a particular set of protons.
NMR SpectrumOjima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Clicker QuestionHow many 1H signals
will this molecule give?
A. 9
B. 10
C. 11
D. 12
E. 13
F. 14
N
O
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Clicker QuestionHow many 1H signals
will this molecule give?
A. 1
B. 3
C. 4
D. 5
E. 6
F. 7
OO
O
N
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Integration Values
The height of a signal is called the integral
The height of each peak represents the relative number of protons in the molecule which resonate at a particular chemical shift.
OO
O
N
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Splitting is only seen in 1H spectra.
Splitting arises due to changes in magnetic field caused by adjacent inequivalent protons.
Equivalent protons will not split each other.
Signal Splitting
OO
O
N
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Splitting
R
HAHB
HC
Splitting patterns can be predicted via the n+1 rule N is the number of adjacent protons which are chemically
inequivalent from our proton of interest. Addition of one to n, affords the splitting pattern.
Protons used for n have to be chemically equivalent to each other, but inequivalent from the proton of interest.
If there is more than one type of adjacent proton, such as on a double bond, the n+1 rule is done for as many proton types present.
A≠B≠C because there is no free rotation around a double bond due to the pi bond.
HA “feels” R and HB, HB feels HA and HC, while HC feels HB and R. therefore these protons are all in different chemical environments.
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Splitting for a particular proton can also be determined by using Pascal’s Triangle.
Each row (n) represents the splitting pattern of a proton with (n) number of adjacent equivalent protons.
If a particular proton has more than one type of adjacent proton, use Pascal’s triangle for each type of proton.
(0)
(1)
(2)
(3)
(4)
(5)
(6)
Clicker QuestionHow many peaks will
the proton on the β-carbon be split into?
A. 1, singlet
B. 2, doublet
C. 3, triplet
D. 4, quartet
E. 4, doublet of doublets
F. 6, triplet of doublets
N
O
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
13C Spectroscopy
Theoretically identical to proton spectroscopy.
Rules for13C spectroscopy
No splitting
No integration
1 peak per type of carbon
To interpret 13C spectra: Find the peak
Go to the provided chart and match the chemical shift values to the corresponding functional group.
N
O
O
O
Si
Clicker QuestionHow many 13C signals
will this molecule give?
A. 11
B. 12
C. 13
D. 14
E. 15
F. 16 Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Clicker QuestionHow many 13C signals
will this molecule give?
A. 1
B. 3
C. 4
D. 5
E. 6
F. 7
OO
O
N
O
O
Si
Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Since we are given the spectra of C, we will start with C. Doublet at 9.75, 1H
Aldehyde
Multiplet at 2.75, 1Hα-proton
Doublet at 1.0, 6HDimethyl
Move onto A. Ozonolysis oxidizes olefins
to carbonyls
Ozonolysis of A affords only C.
Therefore A is symmetrical
Oxidation with OsO4 affords a meso compound B.
Identifies the substitution pattern on olefin.
O
C
A
HO OH
B
And finally D
Reaction of A with CH2I2 and Zn(Cu) affords a hydrocarbon.
CH2I2 and Zn(Cu) convert olefins into cyclopropanes via addition of a carbene.
D
Citations
All exam questions were taken from old exams posted on: http://www.ic.sunysb.edu/Class/orgchem/che321/oldexams.php
All other images and figures were taken from:Solomons, T. W. G., Organic chemistry.
Structures for the clicker questions were taken from:Ojima, I.; Chen, J.; Sun, L.; Borella, C. P.; Wang, T.; Miller, M. L.; Lin, S. N.; Geng, X. D.; Kuznetsova, L. R.; Qu, C. X.; Gallager, D.; Zhao, X. R.; Zanardi, I.; Xia, S. J.; Horwitz, S. B.; Mallen-St Clair, J.; Guerriero, J. L.; Bar-Sagi, D.; Veith, J. M.; Pera, P.; Bernacki, R. J., Design, synthesis, and biological evaluation of new-generation taxoids. J Med Chem 2008, 51 (11), 3203-3221.
Acknowledgements
Teaching me NMR
Joshua Seitz
Jacob Vineberg
Prof. Isaac Carrico
Farhan Ahmed
Assistance with this presentation
Eman Kazi
Some kid named Peter Giattini
Formation & Reaction of Radicals
• Heterolysis
A : B A+ + B:-
Reaction is heterolytic meaning we cleaved the bond giving the electrons to B creating an anion plus A a cation.
Formation & Reaction of Radicals
• Homolysis
A : B A. + B.
Reaction is homolytic meaning we dissociated the bond and created two equal pieces. (two free radicals)
• Curved arrows should be half arrows• Useful with alkanes with no functional group
Conditions for radical reaction
R O O Rheat
R O2
Cl Clh
Cl2
• In order for radical reaction to occur we need conditions of heat or light.
Radical Stability
• We consider carbon radicals to be electron deficient.
Allylic > Benzylic >3o > 2o > 1o
CH3
CCH3 CH3
H
CH H
H
CCH3 H
H
CCH3 CH3> > >
(positive inductive effect of alkyl groups stabilize radical)
Benzylic Radical
• Resonance still holds true here with Benzylic radicals. The radical can be delocalized by the double bonds present in the benzene ring.
Radically Hideous
Requires more energy to create and is less stable thanAllylic, tertiary, secondary, and primary.
Equivalent to
Mechanism of Radical Reactions
• There are three steps in a radical reaction.– Initiation – Propagation – Termination
Initiation
• In this step you create your two radical species. – Possible (common) initiation reactants• X:X (halogens)
– Br, Cl,
• RO:OR• H:X
– Br, Cl, F, etc..,
Propagation With CH4
• There are multiple possible propagation step which is determined by the number of reactants.
• Note that that the propagation step can occur more than twice
Termination • There are several termination steps.– Basically you need to radicals to react in order to
cancel the reaction out.
Excess amounts of reactants
CH4 (large excess) + Cl2
CH3Cl (mainly)hn
CH4 + Cl2 (large excess)
CCl4 (mainly)hn
Selectivity of Br and Cl
• Bromine is more selective than Chlorine– Bromine tends to go for the more stable Radical – Chlorine tends to go for the more available positions
Radical addition of Alkenes
• Generally HBr follows Markovnikov’s rule
• With peroxide– Anti-Markovnikov’s rule
H HHBr
Br Brnot
RO ORheat
Ch. 10 - 55
HBr
RO-ORheat
HBr
Br
Br
(via more stable
2o carbocation)
(via more stable
2o radical)
RO:ORHeat
RO:ORHeat
Which is the correct Product?
A
B
Bond Energies and Delta H
• Bond Energies– The amount in kJ it takes to break a bond
• Observation of Delta H– Negative Delta H (exothermic)– Positve Delta H (endothermic)
Calculating Delta H
• Never base your calculations off of – Reactants - Products– This may work sometimes but not always.
• To calculate Delta H – Use: bonds broke – bonds formed
Physical Properties of Alcohols/Ethers
Boiling Point• Ethers – similar to hydrocarbons of same molecular weight• Alcohols – higher boiling points than ethers and hydrocarbons
Solubility• Ethers and alcohols of same molecular weight have similar solubility• Alcohol solubility decreases as length of hydrocarbon chain attached
increases (i.e. Butyl Alcohol > Pentyl Alcohol)• Branching increases solubility (i.e. Isobutyl Alcohol > Butyl Alcohol)
Alcohol CapabilitiesLone pairs on oxygen atom make it both basic and nucleophilic. Hydroxyl (OH) groups are also very poor leaving groups. Let’s examine what different scenarios there are.
• Basicity: The alcohol, in the presence of a strong acid (H-Cl, H-Br, H+), will become protonated, converting it into a great leaving group (H2O)
• Alcohol as a Leaving Group: With the presence of a nucleophile, we have a substitution reaction, with the leaving group being water.
• Therefore, by converting an alcohol to a group that departs as a weak base (i.e. H2O, CH3OH, CH3CH2OH), we form a good leaving group. This allows elimination and substitution reactions to occur.
Conversion of Alcohols into Alkyl Halides
Goal: Substitute an alcohol with a halogenMethod: Treat the Alcohol with either a strong acid or a reagent that will convert it into a suitable leaving groupReagents:
Hydrogen Halides (HCl, HBr, HI) – Substitute a halide (SN2 for primary alcohols, SN1 for Secondary, Tertiary, Allylic, and Benzylic Alcohols).
Phosphorus Tribromide (PBr3) – Substitute a Br (SN2)
Thionyl Chloride (SOCl2/Pyridine) – Substitute a Cl for primary or secondary alcohols (SN2 Mechanism for both)
Zinc Chloride (ZnCl2) – Substitute a Cl for a primary (SN2) or secondary (SN1) Alcohol
Hydrogen Halides
Order of ReactivityDegree of alcohol: 3 > 2 > 1< methylBy acid: HI > HBr > HCl (HF is generally unreactive)
Mechanism: SN1 for Secondary, Tertiary, Allylic, and Benzylic Alcohols
SN2 for Primary Alcohols
Remember for SN1, since a carbocation forms, hydride and methyl shifts will occur to stabilize it if able. Will produce both front and back attack products due to planar transition state of carbocation.
For SN2, the reaction is concerted meaning that the nucleophile comes in from the back side of the leaving group as it leaves.
Other Ways to Make a Hydroxyl Group of an Alcohol a Good Leaving Group
Conversion to a sulfonate ester derivative:• Common sulfonate esters: mesylates (OMs), tosylates (OTs), triflates (OTf)Reagents:1. OMs, OTs, or OTf with pyridine2. R-OH Group • These reactions retain the configuration of the OH (no affect on
stereochemistry)• These sulfonate esters are very good leaving groups because the sulfonate
anions that they become are very weak bases• Useful for setting up a substitution reaction or elimination reaction
Forming Ethers by Intermolecular Dehydration of Alcohols• Earlier, it was mentioned that oxygen’s lone pairs make it both basic and nucleophilic• Steps:
1. For the formation of diethyl ether, H2SO4 at a warm temperature is used to protonate the OH group, forming H2O+ in one molecule of ethanol.
2. Then, another molecule of ethanol by SN2 attacks the carbon bonded to the H2O+, forcing it to leave as water.
3. Finally, the H2O that left the first molecule of ethanol acts as a base and deprotonates the alcohol, forming diethyl ether
Williamson Synthesis of Ethers• SN2 Reaction of a sodium alkoxide with an alkyl halide, sulfonate, or
sulfate• Use: to synthesize an Ether from an Alcohol• Method: Use a Hydride to pull off a proton from the Hydroxide to
form a nucleophile, then have it react with an alkyl halide (halogens are good leaving groups)• Reagents: R-OH, NaH, R-X (Alkyl Halide)
Protecting Primary Alcohols• Using a tert-butyl ether, you can protect a hydroxyl group of a primary
alcohol in order to carry out a reaction in different part of the molecule
• Can be removed with a diluted acid
Silyl Ether Protecting Groups for Alcohols
• Protecting groups allow you to use reagents in a synthesis that would otherwise unintentionally interact with the Alcohol group.• Stable between pH 4 and 12
TBS-Cl
DMF
You can just write this as OTBS
How to make an Epoxide
• Alkene Epoxidation (turns an Alkene into an Epoxide)
NOTE: MCPBA is a specific type of peroxy acid. RCO3H is the generic formula for a peroxy acid. They are equivalent reagents, for your purposes.
Example of use:
(Notice how you can abbreviate it over an arrow. Much simpler!)
Stereochemistry of Epoxidation• Recall that double bonds can be Cis or Trans• Epoxidation is a SYN Addition• The oxygen atom can add to either face of an Alkene.
• Therefore, if the molecule you form doesn’t have a plane of symmetry, you end up with a Racemic (Equal amounts of R and S) Mixture.
• In the example below, since cis-2-Butene has a plane of symmetry, you actually form a MESO compound, meaning that it has no enantiomer (it’s “suspected” enantiomer is actually identical to it)
• If I had cis-2-pentene, would I form a meso compound after Epoxidation? Draw it out!
Uses of an EpoxideSo we’ve made an Epoxide. What’s so special about it?• As it turns out, since three-membered rings are highly strained,
epoxides are much more reactive towards Nucleophilic Substitution (SN) than other ethers
Say I wanted to undergo Nucleophilic Substitution. How do we break this highly strained ring? Choose your poison:1. Acid – Nucleophile forms on more stable carbocation (usually higher
degree carbon)2. Base - Nucleophile forms on least sterically hindered carbon (usually
lower degree carbon)Remember that the nucleophile comes in from the backside of the epoxide, so the nucleophile and the OH group will be anti with each other
OXIDATION-REDUCTION
Oxidation- increasing the oxygen content or decreasing the hydrogen content of an organic moleculeReduction- opposite of oxidation, increasing the hydrogen content or decreasing the oxygen content of an organic molecule
REDUCING AGENTS
-lithium aluminum hydride reduces aldehydes, ketones, esters, and carboxylic acids to alcohols.-sodium borohydride only reduces aldehydes and ketones to alcohols.
PRACTICE PROBLEMS
What reagents are needed to carry out the following reactions?
PCC
KMnO4
H2CrO4
H2CrO4 and KMnO4
All of them
A
B
C
D
E
PRACTICE PROBLEMS
What reagents are needed to carry out the following reactions?
PCC
KMnO4
H2CrO4
H2CrO4 or KMnO4
Any of them
A
B
C
D
E
A
D
E
GRIGNARDS REAGENTS Exactly the same as the organolithium reaction!
Preparation of Grignard reagents:
RESTRICTIONS OF ORGANOMETALLICS
Organometallic reagentscannot be prepared in thepresence of these groups
In the presence of an alcohol, a protecting group may be implemented to carry out a organometallic reaction. To apply the TMS protecting group, react the alcohol with TMSCl. To remove it, react with fluoride.
ORGANOMETALLIC REAGENTS AS BASES
competition between a nucleophilicity and a basicity is the reason why organometallic reactions cannot be performed with certain functional groups.
PROPOSE A SYNTHESIS OF THE FOLLOWING COMPOUND FROM REACTANTS CONTAINING FOUR CARBON ATOMS OR LESS
?O O
PROPOSE A SYNTHESIS OF THE FOLLOWING COMPOUND FROM REACTANTS CONTAINING FOUR CARBON ATOMS OR LESS
?
OH
PROPOSE A SYNTHESIS OF THE FOLLOWING COMPOUND FROM REACTANTS CONTAINING FOUR CARBON ATOMS OR LESS
?OH
O
OH
O
OTMS
O
F-
1. NaH
2.
Br
OTMS
OH
2. H3O+
1. Li
OTMS
O
?
PCC
OTMS
OHO
1.
2. H3O+
Li
OTMS LiBr
OTMS
Top Related