Orgo Exam 3 LOs PDF

SN2 ❖Predict the product(s) for an SN2 reaction and explain the stereochemical outcome (chirality,number of stereoisomers formed) Product: inversion of configuration due to backside attack (opposite of LG) **LG also drawn for products ○ SN2 can only be attacked from the back which c...

SN2 ❖Predict the product(s) for an SN2 reaction and explain the stereochemical outcome (chirality,number of stereoisomers formed) Product: inversion of configuration due to backside attack (opposite of LG) **LG also drawn for products ○ SN2 can only be attacked from the back which causes the inversion No space ○ High electron density and a node in the front The HOMO and LUMO wouldn’t be able to overlap if it bonded on the front Chirality flips ○ R -> S ○ Wedge to dash ❖Draw the curved arrow mechanism to show electron movement for an SN2 reaction Concerted No rearrangements ❖Identify all components involved in an SN2 reaction (substrate, nucleophile, solvent), explain the factors that influence the effectiveness of each, and predict the resulting outcome Good LG Substrate: methyl, then primary, then secondary are favored ○ As the number of alkyl halide groups increase, sterics in the transition state increase Which increases the Ea and slows down the reaction. Also, the more sterics, the harder it is for the nucleophile to backside attack Solvent: polar aprotic because it wont solvate the nucleophile- so it can still backside attack Nucleophile: SN2 favors fast nucleophiles to increase the reaction rate and therefore decrease the Ea Nucleophiles, solvents, LGs ❖Classify nucleophiles as fast or slow Fast nucleophiles: ○ Favor SN2 ○ Negative charge ○ Highly polarizable compounds (big atoms)  Because electrons are further from nucleus making them polarizable Nucleophilicity Increases down a periodic table and to the left Slow nucleophiles: ○ No charge ○ Unwilling to share electrons ○ High electronegativity ❖Classify leaving groups as excellent, good, and poor and use Classification to predict reactivity of substrate Poor LGs: ○ Strong bases ○ Alcohol (must be converted) Good LGs: ○ water, positive alcohol, ether Very good LGs: ○ Stable ○ Weak bases Increased size Increased electronegativity Low pKa ○ Halogens ○ If they form resonance structures when they leave ❖Classify solvents as nonpolar, polar protic, or polar aprotic and understand the effects of on rate and reaction pathway Nonpolar: no dipole moments- no difference in EN Polar: have dipole moments caused by a difference in EN Polar protic: polar solvents that have at least one hydrogen on an EN heteroatom ○ Capable of Hydrogen bonding with the solvent ○ SN1 reactions occur faster in polar protic solvents The solvent stabilizes the ionic intermediates and transition states, which lowers the activation energy required, therefore speeding up the reaction rate Polar aprotic: polar solvents that only have hydrogens on carbons ○ Not capable of hydrogen bonding with the solvent ○ SN2 reactions occur faster in polar aprotic solvents The nucleophile is less stable in a polar aprotic solvent because the solvent Can't stabilize an anion, so the nucleophiles energy is raised which Lowers activation energy and speeds up reaction rate SN1 ❖Predict the product(s) for an SN1 reaction and explain the stereochemical outcome (Chirality, number of stereoisomers formed) Stepwise, so can have carbocation or methyl rearrangements The intermediate carbocation can be attacked from both sides, so SN1 yields a racemic mixture ○ Both chiral centers form- 2 stereoisomers ○ racemic ❖Draw the curved arrow mechanism to show electron movement for an SN1 reaction ❖Identify all components involved in an SN1 Reaction (substrate, nucleophile, solvent), explain the factors that influence the effectiveness of each, and predict the resulting outcome Excellent LG Substrate: tertiary is favored-> carbocation intermediate will be more stable with more alkyl groups ○ Secondary will occur It’ll be even slower Nucleophile must be weak for SN1 to occur over SN2  Solvent: polar protic is favorable because it stabilizes the ionic intermediates and transition states Nucleophile: a weak nucleophile is favored because it attacks slower, allowing the carbocation intermediate to form ○ Nucleophile doesn’t participate in the RDS ❖Predict the favored reaction between SN1, and SN2 for a given substrate 3- ALWAYS SN1 ○ Too much steric hindrance for SN2 ○ Alkyl groups stabilize the intermediate 2- both, depends on speed of nucleophile methyl/1- ALWAYS SN2 ○ unless resonance stabilized-> SN1 E2 ❖Predict the product(s) for an E2 reaction and explain the stereochemical outcome (E/Z, number of stereoisomers formed) Zaitsev and Hoffman products form ○ Zaitsev (markovnikov) is the major product unless the base is sterically hindered- then the Hoffman (anti-markovnikov) product is major Stereoselective outcome ○ Expect both trans and cis products to form if the Beta position has 2 protons Trans is major product because it’s lower in energy Still have to make sure the LG and protons are antiperiplanar Stereospecific outcome- if the beta position only has 1 proton ○ The hydrogen and LG must be antiperiplanar (staggered) because it has a lower energy TS than synperiplanar (eclipsed) ○ The LG and it’s proton must be rotated so that they are in plane (not on a wedge or dash)  In chair conformations, the LG and H must be axial and in opposite planes so that it is antiperiplanar with the neighboring Hydrogen In cyclohexanes, the beta protons must be on a dash next to the wedged LG to yield a product ❖ Draw the curved arrow mechanism to show electron movement for an E2 reaction Concerted mechanism No rearrangements ❖ Identify all components involved in an E2 reaction (substrate, base, solvent), explain the factors that influence the effectiveness of each, and predict the resulting outcome *opposite of SN2 Substrate: 3>2>1>methyl ○ Bases aren’t affected by sterics ○ Explained by electronics. The TS is stabilized by alkyl substituents Base:  ○ A strong base is favored because E2 is concerted ○ A sterically hindered base will favor the Hoffman product Solvent: not important E1 ❖ Predict the product(s) for an E1 reaction and explain the stereochemical outcome (E/Z, number of stereoisomers formed) Products: Zaitsev will always be major and Hoffman minor no matter how sterically hindered the base is E and Z (cis and trans) isomers can form and the isomer with the least steric hindrance will be favored because it’ll have a lower energy- usually E ❖Draw the curved arrow mechanism to show electron movement for an E1 reaction Stepwise Carbocation intermediate ○ Rearrangements can occur ❖Identify all components involved in an E1 Reaction (substrate, base, solvent) explain the factors that influence the effectiveness of each, and predict the resulting outcome Substrate: 3> rarely 2> rarely 1> rarely methyl ○ Resonance stabilized intermediates that aren’t 3 CAN form Base: weak base is favored because it’ll attack the positive charge slower and give the carbocation intermediate time to form Solvent: polar protic ❖Predict the favored reaction between E1 and E2 for a given substrate Tertiary: E2 favored for a strong base and E1 favored for a weak base Secondary: E2 favored for a strong base, weak base won’t react Primary: E2 favored for a strong base, weak base won’t react ❖Draw and accurately use curved arrows to show electron pushing mechanisms for E2 and E1 reactions Elimination v substitution ❖Identify all components involved in each reaction, what factors influence the effectiveness of each, and how each component affects the type of reaction that occurs and use this information to determine which reaction pathway will occur (substitution vs. elimination) Heat is used for elimination substrate slow nucleophile fast nucleophile fast nucleophile slow nucleophile strong base strong base weak base weak base 1 E2 Major: SN2 SN2 X Minor: E2 2 E2 major : E2 SN2 X Minor: SN2 3 E2 E2 SN1 SN1 or E1 SN2: 1>2 ○ Doesn’t occur in 3 SN1: 3>2 ○ Rarely occurs in 2, doesn’t occur in 1 E2: 3>2>1 E1: 3>2 ○ Rarely occurs in 2, doesn’t occur in 1 ❖Describe a reaction as stereospecific, Stereoselective, or neither Stereospecific: The configuration of the product is dependent on the configuration of the starting material. Only one product forms ○ SN2- inversion product due to backside attack  ○ E2- antiperiplanar LG and Hydrogen proton are opposite and in the same plane Stereoselective: multiple products are formed and some stereoisomers are favored ○ E2- Zaitsev and Hoffman, E and Z ○ E1- E or Z can be favored Neither ○ Racemic SN1 reaction Stereospecific stereoselective SN1 no no SN2 yes no E2 yes yes E1 no yes Radicals ❖Predict and rank the stabilities of radicals benzylic>allylic>3>2>1>methyl>vinylic Alkyl groups stabilize unpaired electrons via delocalization effect called hyperconjugation ❖ Given a synthetic transformation, account for the flow of electrons via radical mechanisms using curved arrow notation and label each step (initiation, propagation, termination) Initiation: homolytic cleavage Propagation: hydrogen abstraction, halogen abstraction, addition to a pi bond, elimination  Termination: coupling Radicals don’t undergo rearrangement ❖Given the starting material and reagents, predict the major organic product for radical reactions, including selectivity Chlorination: Cl2, hv ○ Products: major is Cl in primary position, minor is Cl in tertiary ○ Racemic mixture if chiral center Bromination: Br2, hv ○ Products: only Br in tertiary position ○ Racemic mixture if chiral center Allylic bromination: NBS ○ Must use NBS ○ Br Installed at allylic position (draw all resonance structures) ○ Mixture of products can form if the radical is resonance stabilized Radical addition of HBr with peroxides ○ Racemic mixture ○ Anti Markovnikov addition ○ It is only a radical rxn when ROOR is used. Just HBr is an ionic reaction ❖Supply reagents necessary for the halogenation of alkanes and alkenes Chlorination: Cl2 and initiator: hv (light) or delta (heat) Bromination: Br2 and hv or delta Allylic bromination NBS Radical addition of HBr: ROOR and HBr ❖Explain the radical selectivity of halogenation reactions using the Hammond Postulate Bromination is more selective than chlorination because the RDS of chlorination is exothermic and the RDS of bromination is endothermic. ○ Therefore the TS in chlorination more closely resembles the reactants and bromination the products. This leads to the C-H bond in bromination being almost completely broken (to form a radical), so it has more radical character. Because it has more radical character, the TS is more sensitive to changes in the substrate, therefore bromination is more selective ❖Predict the products of allylic bromination and account for their formation using mechanism arrows Br will be installed in the tertiary position because it is more stable for the radical in the TS ❖Draw the complete mechanism for radical addition of HBr in the presence of peroxides ❖Draw the polymerized chain from a monomer and vice versa IR spectra ❖ Analyze and describe the three characteristics of signals (wavenumber, intensity, and shape) Wavenumber ○ Conjugated, unsaturated (have C=C) tend to have lower wavenumbers because the compounds can participate in resonance and therefore have more single bond character (which has less s character) ○ Hooke’s law states that wavenumber depends on… Bond strength Increasing bond strength (therefore bond order) increases wavenumber because the amount of s character increases with bond strength  ○ Ex: triple bond carbons are sp hybridized which is 50% s character Masses of the atoms Intensity ○ Intensity increases with an increases dipole moment ○ Signals just below 3000 tend to be strong when an abundance of C-H bonds are present Shape ○ When alcohols have broad signals they typically participate in H bonding ○ Alcohols sharp signals are “free” from H bonding ○ This effect is more pronounced (broader) in alcohols that are in carboxylic acids Because these molecules can H bond twice- “dimer” forms Asymmetric compounds tend to create more than one peak because molecules are vibrating in different ways. This causes the shape to change ○ Ex- in methyl multiple C-H peaks show up ○ A fully symmetrical compound will have one peak ❖ Interpret IR spectra to identify functional groups, match compounds to their spectra, and predict the appearance of a spectrum based on a given compound ❖ Predict and explain location (wavenumber) of signal for organic bonds, including all types of carbonyls Carbonyl frequencies in functional groups ○ Anhydride peak 1: 1820 ○ Acid chloride: 1790 ○ Anhydride peak 2: 1760 ○ Ester: 1735 ○ Aldehyde: 1730 ○ Ketone: 1720- BASE VALUE ○ Carboxylic acid: 1715 ○ Amide: 1650  Carbonyls have different wavenumbers due to electronics ○ Check inductive and resonance effects If Induction is major contributor it’ll have a dipole moment pulling electron density into the carbonyl bond Increases frequency If resonance is the major contributor then it’ll cause the carbonyl bond to have more single bond character (less s) Lowers frequency ❖Explain and predict the effects of H-bonding in an organic sample H bonding will cause the shape of the peak to broaden ❖Propose structure given a molecular formula and IR spectrum ❖Explain how to monitor the progress of a reaction using IR Synthesis ❖Plan syntheses involving reactions of alkenes and alkynes, substitutions and eliminations, acid/base, and radical reactions Is there a change in the carbon skeleton? Is there a change in functional group or location of functional group? Use retrosynthetic analysis ❖Given a synthetic target, devise multistep syntheses via retrosynthetic analysis using all reagents learned to this point ❖Understand how all reactions learned fit together in synthesis ❖ Utilize carbon-carbon bond forming reactions to increase the size of the carbon skeleton and utilize carbon-carbon bond breaking reactions to decrease the size of the carbon skeleton in synthesis Remove carbons through ozonolysis Add carbons through substitution- make the nucleophile the chain of carbons that are added with a positive compound ❖Plan a retrosynthetic analysis for a given target compound ❖Determine the most efficient synthesis possible