SBU CHE 321 Final F13 Review Slides

Final Exam ReviewCHE 321Fall 2013

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

http://www.ic.sunysb.edu/Class/orgchem/che321/Exam_Cover.pdf

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

https://stonybrook.campuslabs.com/courseeval/

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


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


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.

Common Chemical Shifts and Their Meaning


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


A. 1

B. 3

C. 4

D. 5

E. 6

F. 7

OO

O

N

O

O

Si


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


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


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


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


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


A. 1

B. 3

C. 4

D. 5

E. 6

F. 7

OO

O

N

O

O

Si


Exam Problems

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

Credit to Anthony Antonelli for this

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.

http://www.ic.sunysb.edu/Class/orgchem/che321/oldexams.php



Acknowledgements

Teaching me NMR

Joshua Seitz

Jacob Vineberg

Prof. Isaac Carrico

Farhan Ahmed

Assistance with this presentation

Eman Kazi

Some kid named Peter Giattini

CHE 321Final Review

Fall 2013

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.

Allylic Radical

• Allylic Radicals act similar to Benzylic radical in terms of resonance.

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..,

Initiation Mechanism with Cl:Cl

Cl Clh

(homolyticcleavage)

2 Cl

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

Propagation Continued

• This can be done until all the Hydrogens have been replaced by Cl.

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

Chlorine Selectivity

AB

CD

AB

AB

What is the major product?

Bromine & Chlorine

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

What are the products of the initiation step?

A

B

P1

P2

T

CH3 radical

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)

Bond Energies

Bond Energies

Which bond has a higher bond dissociation?

Bond Energies

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

Calculating Delta H

a. -101b. 117c. 101

Delta H

The Delta H for step 3 is +8kJ mol-1 and -109 kJ mol-1 for step 2.

A. TrueB. False

Chapter 11 Final Review

By Pradeep Ravindra

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

Sulfonate Esters

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)

Synthesis of Ethers by Alkoxymercuration-Demercuration

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

??????

Synthesis ExampleStarting with this reactant, make the product

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

Fill in the reagents

Answers

Review

Answers

Predict the product(s) formed when the following reactants are treated with excess H-Br:

Answers

Mechanism Practice

Mechanism Practice

Challenge your knowledge

CHAPTER 12 OVERVIEW

Arthur Makarenko

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 PROBLEMSWould you use LiAlH4 or NaBH4 for the following reactions?

A B

PRACTICE PROBLEMSWould you use LiAlH4 or NaBH4 for the following reactions?

A B

B

A

A

OXIDIZING AGENTS

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

ORGANOLITHIUM

GRIGNARDS REAGENTS Exactly the same as the organolithium reaction!

Preparation of Grignard reagents:

GRIGNARDS REAGENTS An ester can react with two equivalents of a Grignard reagent.

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.

SODIUM ALKYNIDES

Can also perform nucleophilic attack on carbonyls

ORGANOMETALLIC REAGENTS AS BASES

competition between a nucleophilicity and a basicity is the reason why organometallic reactions cannot be performed with certain functional groups.

SYNTHESIS REVIEW

Arthur Makarenko

PROPOSE A SYNTHESIS OF THE FOLLOWING COMPOUND FROM REACTANTS CONTAINING FOUR CARBON ATOMS OR LESS

?O O

mCPBA

O

O

ORA

B

O O

mCPBA

O

?

O O

mCPBA

O

O

H2

LindlarNaNH3

A B

O O

OR

mCPBA

O

O

H2

Lindlar

O O

?

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

?

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

O O

1. 1.

2. H3O+ 2. H3O+

A BOR

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

O

1. 2. H3O+

?

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

O

1. 2. H3O+

LDA

H2O

H3O+

HO-

A

B

C

D

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

O

1. 2. H3O+

LDA

?

mCPBA

O

O

H2

Lindlar

O O

PCC

HO

O

1. 2. H3O+

LDA

Br


?

OH

OH

A

B

C

Which bond is best to break?

O

1.

2. H3O+OH

Li

?

O

1.

2. H3O+OH

Li

Br

Li

?

O

1.

2. H3O+OH

Li

Br

Li

OH

HBr

PBr3

BOTHWORK

A

B

C

O

1.

2. H3O+OH

Li

Br

Li

OH

PBr3

?

O

1.

2. H3O+OH

Li

Br

Li

OH

PBr3

Li

O

1. 2. H3O+


?OH

O

OH

O

OTMS

O

F-

?

OH

O

OTMS

O

F-

1. NaH

2.

Br

OTMS

OH?

OH

O

OTMS

O

F-

1. NaH

2.

Br

OTMS

OH

2. H3O+

1. Li

OTMS

O

?

OH

O

OTMS

O

F-

1. NaH

2.

Br

OTMS

OH

2. H3O+

1. Li

OTMS

O

?

PCC

OTMS

OH

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

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

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

Br

OH

TMSCl

SBU CHE 321 Final F13 Review Slides

Documents

strong magnetic field

spin axis

magnetic fieldssolomons

ofthe spinactive nuclei

spin axes

resonance energy results

organic chemistry

themagnetic field