New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Organic Chemistry 241 Exam 1 Notes

by: Denice Arnold

Organic Chemistry 241 Exam 1 Notes Chem 241

Marketplace > University of Pennsylvania > Chemistry > Chem 241 > Organic Chemistry 241 Exam 1 Notes
Denice Arnold

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Chapters 12, 13, and 4 of Wade's Organic Chemistry textbook. These notes cover all the material needed for Exam 1 in Organic Chemistry I.
Organic Chemistry I
Dr. Winkler
Chemistry, Organic Chemistry, orgo, Chem241
75 ?




Popular in Organic Chemistry I

Popular in Chemistry

This 10 page Bundle was uploaded by Denice Arnold on Saturday January 23, 2016. The Bundle belongs to Chem 241 at University of Pennsylvania taught by Dr. Winkler in Fall 2015. Since its upload, it has received 29 views. For similar materials see Organic Chemistry I in Chemistry at University of Pennsylvania.


Reviews for Organic Chemistry 241 Exam 1 Notes


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 01/23/16
Orgo Chapter 12 Sunday, October 4, 2015 10:31 AM Chapter 12 Infrared Spectroscopy and Mass Spectrometry (p. 513-552) Absorption spectroscopy - measurement of the amount of light absorbed by a compound as a function of the wavelength of light PowerpointNotes: On spectrometer graph, dips are where absorbed most (so transmits least) IR Spectroscopy - observes the vibrationsof bonds and provides evidence of the functional groups present Higher frequencies mean "absorbs at greater frequencies" - needs more energy Mass Spectrometry - bombards molecules with electrons and breaks the molecules into fragments, givingthe molecular Stronger bonds require more energy, thus higher frequency to be stretched or compressed weight Conjugation makes bonds more stable by decreasing the electron density in the double bond Nuclear Magnetic Resonance - observesthe chemical environments of H or C atoms and provides evidence for the Less stiff bonds, vibrate slower (lower stretching frequency) structure of the alkyl groups and cluesto the functional groups Less stiff bonds = lower stretching frequency UV Spectroscopy - observes electrical transitions and provides information on the electrical bonding in the sample Electromagnetic spectrum The more s character in the orbital, the closer the electrons will be to the nucleus and the lower in energy (more stable) they will be X-rays excite electrons past all energy levels,causing ionization UV waves excite electrons to higher energy levels within molecules More s character, stronger the bond, higher the frequency IR waves excite molecular vibrations Microwave waves excite rotations For N--H, # spikes corresponds to # hydrogens (not primary, secondary, tertiary) Radio waves excite the nuclear spintransitions observed in NMR spectroscopy Infrared Region In carbonyls, electrons move away from C (to O) thus carbon is electron deficient Anything that contributes electron density to carbon, it is stabilizing More stable compounds have lower stretching frequencies Wavenumber is the reciprocal of the wavelength (wavenumber) (wavelength) = 10,000 Frequency decreases with increasing atomic weight More stable = lower stretching frequency Heavier atoms vibrate more slowly than lighter ones Frequency increases with bond energy Normally, carbonyls are around 1700. Stretching frequency decreases when Stronger bonds usually vibrate faster than weaker bonds conjugation is present Triple bonds (stronger) vibrate at higher frequencies than double bonds Infrared spectrum - graphs of the energy absorbed by a molecule as a function of the frequency or wavelength of light Conjugation lowers stretching frequencies in carbonyls A nonlinear molecule with n atoms generally has 3n - 6 fundamental vibrational modes Amide - nitrogen next to carbonyl Symmetric stretching, antisymmetric stretching, scissoring Other types of vibrational modes include twisting, rocking, etc Amine - nitrogen next to regular carbon -1 Fingerprint region (600 - 1400 cm ) Contains most of these complex vibrations Generally ignored in our study of IR spectra -1 Simple stretching vibrations in the 1600 - 3500 cm region are the most characteristic IR-active - alternate stretching and compressingof bond produces a changing dipole moment so that energy may be absorbed IR-inactive - vibration produces no change in dipole moment, therefore no absorption of energy Zero dipole moments sometimes produce absorptions (weak) because molecular collisions, rotations, and vibrations make them unsymmetrical temporarily Overtone peaks - small peaks at a multiple of the fundamental vibration frequency due to strongly polar bonds Infrared spectrometer Sample beam passes through the sample cell while the reference beam passes through the reference cell that contains only the solvent Record the difference in light transmittance between the two beams Higher frequencies (shorter wavelengths) appear toward the left of the chart 100% transmittance (no absorption) at the top 0% transmittance (absorption of all the light) at the bottom Characteristic carbon bond stretching frequencies C--C 1200 cm -1 C==C 1660 cm -1 C-=-=C <2200 cm -1 Conjugated double bonds - two double bonds one bond apart Slightly more stable than isolated double bonds due to slight pi bonding Less electron density in the double bonds Less stiff, so vibrates a little more slowly than isolated Aromatic C==C bonds result in even lower stretching frequencies sp hybrid carbons absorb higher frequencies than sp hybrid carbons The more s character, the stronger and stiffer the bonds, so the higher the vibrational frequency Characteristic C==C stretching frequencies Isolated C==C 1640-1680 cm -1 Conjugated C==C 1620-1640 cm -1 Aromatic C==C ~1600 cm -1 O--H and N--H stretching frequencies Alcohol O--H 3300 cm -1broad Acid O--H 3000 cm -1broad Amine N--H 3300 cm -1 broad with spikes Ketones, aldehydes, and acids Conjugation of a C==C double bond lowers its stretching frequency Orgo Page 1 Conjugation of a C==C double bond lowers its stretching frequency Delocalization of pi electrons reduces electron density of carbonyl double bond Weakened double bond so lower stretching frequency Presence of a C==C double bond can be inferred from its effect on the C==O frequency Resonance generally lowers carbonyl frequencies Angle strain on the carbonyl group forces more electron density into C==O bond, making it stiffer Increased angle strain raises stretching frequencies C--N single bond stretch rarely useful for structure determination C--N similarto C--C and C--O absorptionfrequencies (around 1200 cm ) -1 -1 C==N similarto C==C double bonds (around 1660 cm ) C==N stronger absorption due to greater dipole moment Often resembles a carbonyl absorption intensity C--N bond stretching frequencies C--N 1200 cm -1 -1 C==N 1660 cm -1 C-=-=N >2200 cm C-=-=C <2200 cm -1 Limitations and Advantages of IR Spectroscopy Ambiguities between functional groups, but can provide conclusiveproof that two compounds are either the same or different 1. Indicates functional groups in a compound 2. Shows the absence of other functional groups 3. Can confirm the identify of a compound by comparison with a known sample Mass Spectrometry Sample is struck by high energy electrons, breaking the molecules apart Masses of fragments are then measured Mass spectrometer ionizes molecules and sorts them according to their masses Records the abundance of ions of each mass Mass spectrum plots mass against the relative number of ions of each mass Electron Impact Ionization Neutral molecule ionized when struck by an electron, knocking out an additional electron Positive charge and one unpaired electron yields a radical cation Regular carbocation: three-bonded carbon atom with six paired electrons in valence shell +. Radical cation: seven electrons and bonded to four other atoms [CH ] 4 Fragmentation from the impact of the electron gives a characteristic mixture of ions M+ is the molecular ion, the cation corresponding to the mass of the original molecule Only the positively charged fragments are detected by the mass spec Separation of Ions of Different Masses After ionization and fragmentation, ions are separated and detected (magnetic deflection) Heavier atoms bend too little and lighter ions bend too much Mass-to-charge ratio (m/z),where z isgenerally +1, determines the radius of curvature Since charge is generally standardized, curve depends only on mass On a massspectra, base peak (strongest peak), isdefined to be 100% M+1 peak is one mass unit heavier than the M+ peak M+2 peak… etc. Ex: Br has a M+2 that is as large as M+ meaning that 79Br and 81Br are of equal abundances Two peaks can suggest one element - depends on relative abundances Fragmentation patterns Bond breaking tends to form the most stable fragments Most molecular ions have even mass numbers while most fragments have odd mass number Compounds containing aromatic ringstend to fragment at the carbon (benzyliccarbon) next to the ring 1. Alkanes - cleavage to give the most stable carbocations 2. Alcohols - loss of a water 3. Alkenes and aromatics - cleavage to give allylic and benzylic carbocations 4. Amines - alpha cleavage next to the carbon bearing the nitrogen 5. Ethers - loss of an alkyl group 6. Ketones and aldehydes - loss of alkyl groupsnext to the carbon bearing the oxygen Orgo Page 2 Orgo Chapter 13 Sunday, October 11, 2015 12:34 PM Chapter 13 Nuclear Magnetic Resonance Spectroscopy (p. 563-613) Nuclear Magnetic Resonance NMR limited to odd numbered nuclei 1H (major), 13C, 17O, 15N, 19F (major), 31P A spinning proton induces an electric current which generates a magnetic field called the magnetic moment When a small bar magnet is placed in the field of a larger magnet, it twists to align Has to be aligned either with the external field or against the field Lower energy state is with the field (alpha-spin) Higher energy state is against the field (beta-spin) The energy it takes to move a proton from an alpha state to a beta state is proportional to the strength of the magnetic field A nucleus is "in resonance" when it is irradiated with radio-frequency photons having an energy equal to this energy Magnetic Shield The electrons circulating the proton generate a small induced magnetic field that opposes the externally applied field Proton bathed in electron density - shielded Energy required to flip spin increases; must overcome shielding More shielded, higher upfield on NMR spectrum Proton denuded of electron density - deshielded Less energy required to flip spin More deshielded, lower downfield on NMR spectrum 1. The number of different absorptions (also called signals or peaks) implies how many different types of protons are present a. Number of protons 2. The amount of shielding shown by these absorptions implies the electronic structure of the molecule close to each type of proton; location of signals a. Chemical shifts 3. The intensities of the signals imply how many protons of each type are present a. Integration 4. The splitting of the signals gives information about other nearby protons a. Splitting Chemical Shifts Chemical shift - the difference in parts per million (ppm) between the resonance frequency being observed and that of tetramethylsilane (TMS, (CH3)4Si) TMS used as reference bc silicon is less electronegative than carbon Protons on methyl groups are well shielded: way upfield Defined as 0.00 ppm on the delta scale Chemical shift is the same regardless of operating field and frequency Frequency shift (in Hz) is proportional to the operating frequency Effect of an electronegative group on the chemical shift depends on its electronegativity Also depends on distance from protons Effect decreases with increasing distance Usually negligible if separated by four or more bonds Vinyl and Aromatic Protons Vinyl and Aromatic Protons Double bonds and aromatic rings produce large deshielding effects At the center of the ring, the induced field opposes the external field On the outside (where the protons are) the field lines curve, so the induced field enhances the external field, thus deshielding the protons Similar argument for double bonds, but effect not as large Acetylenic Hydrogens Expect acetylenic hydrogens (--C-=-=C--H) to be even more deshielded due to pi bonds But acetylenic proton lies along axis of induced field, which is shielded Acetylenic Protons Aldehyde Protons More deshielding due to C==O double bond and by inductive effect of O (--CHO) Hydrogen-Bonded Protons For O--H and N--H, chemical shifts depend on concentration Hydrogen bonding contributes to more deshielding When diluted, H bonding becomes less important, so less (?) shielding occurs Carboxylic Acid Protons Strongly deshielded due to oxygen next to carbonyl group Shifts greater than delta 10 Number of Signals Generally, number of NMR signals corresponds to the number of different kinds of protons Protons in identical chemical environments with the same shielding have the same chemical shift; said to be chemicallyequivalent In some cases, there may be fewer signals in NMR spectrum than there are different protons Think protons on an aromatic ring with one or two odd substituents Although there could be two or three different types, but only one shift Accidentallyequivalent- not chemically equivalent, but happen to have same shift Areas of the Peaks The area under the peak is proportional to the number of hydrogens contributing to the peak Spin-spinsplitting of 1,1,2-tribromoethane Integrals tell us the ratio of different types of H's, not the actual number of H's Spin-SpinSplitting Orgo Page 1 Areas of the Peaks The area under the peak is proportional to the number of hydrogens contributing to the peak Spin-spinsplitting of 1,1,2-tribromoethane Integrals tell us the ratio of different types of H's, not the actual number of H's Spin-SpinSplitting When two different types of protons are close enough that their magnetic fields influence each other, they are said to be magneticallycoupled Spin-spinsplittingis the splitting of signals into multiplets When alpha-H is aligned with the field, beta-H protons feel a slightly stronger total field and absorb at a lower field (deshielded) and vice versa N+1 Rule: If a signal is splitby N neighboring equivalent protons, it will be split into N+1 peaks 0 neighboring protons corresponds with a singlet 1 neighboring proton corresponds with a doublet (1:1 ratio) 2 neighboring protons corresponds with a triplet (1:2:1 ratio) Continues following the ratios set by Pascal's triangle Most spin-spin splitting is between protons on adjacent carbons - vicinalprotons Protons bonded to the same carbon atom (geminalprotons) can only split each other if they are nonequivalent Peaks within a multiplet are not quite symmetrical; tend to lean toward the signal of the proton responsible for the splitting When drawing an NMR spectrum 1. Determine the types of protons present, together with their proportions 2. Estimate the chemical shifts of the protons 3. Determine the splitting patterns 4. Summarize each absorption in order, from the lowest field to the highest 5. Draw the spectrum, using the information from the summary p. 585 (Skip coupling constants and complex splitting) StereochemicalNonequivalenceof Protons To determine whether similar-appearing protons are equivalent, substitute another atom for each of the protons in question Diastereotopic- when the two imaginary products are diastereomers, nonsuperimposable mirror images - think double bonds or cyclicsystems or saturated, acyclic compounds Appear in the NMR at different chemical shifts,in any environment Able to split each other On a ring, protons that are cis and trans to another atom are diastereotopic to each other When a molecule contains an asymmetric carbon atom, the protons on any methylene group are usually diasteretopic Enantiotopic- imaginary replacement of either two protons forms enantiomers NMR cannot distinguish between enantiotopic protons - equivalent by NMR However, if in a chiral environment, NMR is able to differentiate the protons Homotopic- identical, in any environment InterpretingProton NMR Spectra 1. Find index of unsaturation, if possible. i. Consider rings, double bonds, triple bonds… ii. Match integrated peak areas with number of protons in the formula 2. Broad singlet past 10 ppm, think --OH (or --NH if less than 10 ppm) 3. Delta 3-4 suggests a carbon bearing an electronegative element(O, halogen) i. Lower deltas are probably protons more distant from electronegative atom 4. Delta 7-8 suggests presence of aromatic ring i. If farther downfield than 7.2, possibleelectron-withdrawing substituent 5. Delta 5-6 suggests vinyl protons (across a double bond, either cis or trans) 6. Splitting patterns indicate number of adjacent protons i. Learn ethyl groups and isopropyl groups 7. Delta 2.1 to 2.5 suggests protons adjacent to a carbonyl group or next to an aromatic ring i. Singlet at delta 2.1 often from a methyl group bonded to a carbonyl group 8. Delta 9 to 10 suggests an aldehyde 9. Singlet around delta 2.5 suggests a terminal alkyne Orgo Page 2 Orgo Chapter 4 Sunday, October 18, 2015 11:22 AM The Study of Chemical Reactions (p. 132 - 167) Halogenationof Alkanes Simple substitution of a halogen for a hydrogen that can occur in the gas phase, without a solvent to complicate the reaction Mechanism - step by step pathway from reactants to products Thermodynamics - energetics of the reaction at equilibrium Allows us to compare stability of reactants and products Predicts which compounds favored by equilibrium Kinetics - variation of reaction rates with different conditions and concentrations Predicts how the rate will change if we change reaction conditions Chlorination of Methane 1. Does not occur at room temperature in the absence of light 2. Most effective wavelength of lightis a blue color 3. Light initiated reaction has a high quantum yield Many molecules of product are formed for every photon of light absorbed InitiationStep Generates a reactive intermediate Chlorine molecule absorbs a photon Cl-dot is a reactive intermediate- short-lived species that reacts as quickly as it is formed Species with unpaired electrons are called radicals or free radicals Electron deficient because they lack an octet PropagationSteps Formation of products with regeneration of reactive intermediates Reaction intermediate reacts with a stable molecule to form a product and another reactive intermediate Allows chain to continue until supplyof reactants is exhausted or the reactive intermediate is destroyed Final product - hydrogen chloride Free radical product - methyl radical Free radical product - chlorine atom TerminationSteps Destruction of reactive intermediates Side reactions that destroy reactive intermediates and tend to slowor stop the reaction Combination of any two free radicals decreases number of free radicals Reaction of free radicals with wall also decreases number of free radicals Production of chloromethane (product) consumes free radicals All these steps break the chain and are thus terminating End of reaction: When relatively few molecules of reactants are available, free radicals are less likely to encounter a molecule of reactant than they are to encounter each other - chain reaction quickly stops Equilibrium Constantsand Energy Gibb's Free Energy - difference between the free energy of the products and the free energy of the reactants Measure of the amount of energy available to do work Change in enthalpy - heat of reaction Exothermic- weaker bonds broken, stronger bonds formed Exothermic- weaker bonds broken, stronger bonds formed Endothermic- stronger bonds broken, weaker bonds formed Largest factor in the driving force for chlorination Similar case in most organic reactions Entropy - randomness, disorder, freedom of motion Usually small in relation to the enthalpy term Assume that Delta G = Delta H in organic reactions Bond DissociationEnthalpies Homolyticcleavage - each bond retains one of the bond's two electrons Forms free radicals Heterolyticcleavage - one of the atoms retains both electrons Forms ions Sum of values of enthalpy changes for the individual propagation steps gives the overall enthalpy change for the reaction Do not need to include initiation step Propagation steps come in two's Alternative mechanisms - alternate propagation steps are possible Most plausible mechanism is the one that is the lowest energy (skipped two sections) TransitionStates Transitionstate - highest energy state in a molecular collisionthat leads to a reaction Not an intermediate,which exists for some finite length of time Rate-determiningstep - highest energy step Highest point in the energy diagram is the TS for the rate-limiting step Greatest Ea, slowest reaction rate Selectivityin Halogenation Halogenation - substitution where halogen atom replaces a hydrogen Methane reacts differently with different halogen radicals Fluorination has a very high rate - explosive Chlorination difficult to control at higher temperatures Bromination is very slow Iodination is exceedingly slow, even at high temperatures - unreactive Different types of hydrogens (primary, secondary, tertiary) have different reactivities Preference for reaction at the secondary position results from the greater stability of the secondary free radical and the transition state leading to it Tertiary carbon is most stable because the electron deficient carbon gets electron density donated from the surrounding carbons - closest to an octet Chlorination of isobutane Nine primary hydrogens and one tertiary hydrogen Multiply the number of each hydrogen by its respective reactivity to calculate the probability of its free radical formation Even though primary hydrogens are less reactive, there are so many of them so that the primary product is the major product To calculate the percentage of product, divide the probability by the total Chlorination of propane Reaction-energy diagram for the first propagation step in the chlorination of propane Formation of the secondary radical has a lower activation energy than does formation of the primary radical Exothermic - energy maxima closer to the reactants Small difference in Ea Bromination of propane Bromination is much more selective than chlorination because the major reaction is favored Bromination is much more selective than chlorination because the major reaction is favored by a larger amount Reaction-energy diagram for the first propagation step in the bromination of propane Energy difference in the transition states is nearly as large as the energy difference in the products Endothermic - energy maxima closer to products Large difference in Ea HammondPostulate In an endothermic reaction, TS closer to products in energy and structure. In an exothermic reaction, TS closer to the reactants in energy and structure. HammondPostulate - Related species that are closer in energy are also closer in structure. The structure of a TS resembles the structure of the closest stable species Exothermic processes are less selective because the products are closely related to the transition state. Endothermic processes are more selective because the products have more distinctions between each other Free Radical Reactions 1. Initiation: Draw a step that breaks the weak bond Homolytic cleavage to give two radicals 2. Propagation: Draw a reaction of the initiator with one of the starting materials Yields a free radical version of the starting molecule Look for most energetically favorable position (usually tertiary) 3. Propagation: Draw a reaction of the free-radical version of the starting material with another starting-material molecule to generate a new radical intermediate 4. Termination: Draw sidereaction termination steps Reaction of any two free radicals, collisionof a free radical with the reaction vessel (skipped radical inhibitors) Reactive Intermediates


Buy Material

Are you sure you want to buy this material for

75 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Kyle Maynard Purdue

"When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the I made $280 on my first study guide!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.