CHE 321 Midterm 3 Notes - Dale Drueckhammer; Robert Grubbs; Zachary Katsamanis PDF

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Dale Drueckhammer; Robert Grubbs; Zachary Katsamanis...

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Midterm 3 Substitution and Elimination - SN1, SN2, E1, E2 - Numbers refer to rate law - 1 → first order rate = k [substrate]; rate depends only on concentration of substrate - 2→ second order rate = k [substrate][nucleophile or base] rate depends on substrate/base con - All reactions have a leaving group - Solvent = nucleophile, means that something is attacking - Generally, S reactions attack the leaving group/ associated carbon; E reactions attack adjacent H - SN1 (unimolecular) - carbocation mechanism, shifts possible - Step 1 - Rate determining / slow step (formation of carbocation intermediate) - Arrow from bond to leaving group → carbocation int and LG- ion - Same step 1 as in E1 reaction - Step 2 - solvent/nucleophile attacks carbocation - Results in OR group attaching to molecule w/ + charge on O - Step 3 - deprotonate H, removes + charge from O - Final product: nucleophile replaces leaving group - Retention (stereochemistry stays the same) and inversion (stereochemistry changes) products are equally likely - NOTE: ROH + HCl/HBr/HI are SN1 reactions - E1 (unimolecular) - carbocation mechanism, shifting possible - Step 1 - Rate determining / slow step (formation of carbocation intermediate) - Arrow from bond to leaving group - Step 2 - adjacent H is displaced, forms double bond - No bonds off carbon double bonds have stereochemistry → no wedges/dashes - Final product: double bond forms adjacent to where leaving group was - SN2 (bimolecular) - backside attack (anti), concerted - NaCN / DMSO - NaCN dissociate into Na+ and CN-, but ignore the Na+ - DMSO - polar aprotic solvent - Step 1 - CN- directly attacks C that leaving group is attached to from the back - Leading group is on chiral carbon w/ stereochemistry → product of SN2 always inversion - Final product: inversion, nucleophile replaces leaving group but has opposite stereochemistry - E2 (bimolecular) - concerted - NaOCH3 - Step 1: OCH3- displaces H, and arrow from bond to LG - NOTE: H that you are displacing must have opposite stchem than LG so be careful with bond rotations - If LG and H have same stchem no reaction even if it may result in the most substituted double bond - Final product: more substituted double bond formed as major product (Zaitsev’s Rule) Leaving Groups - Nu:- (nucleophile w/ lone pair and - charge) - General mechanism: Nu-: R-LG → Nu-R + LG- What makes a good leaving group? stable → even more stable - Better leaving group as you go down column of halogens

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Cl-, Br-, and I- are big → can spread e- over wide distance when they’re ions so they’re stable, and I- spreads out the most - F- is most electronegative and very small → cannot spread out much and has repulsion → poor leaving group and prefers being bound to R Alcohols (OH-) are bad leaving groups and don’t undergo elimination (unless acid added) - OH- (like in NaOH) is a very strong base → dissociates almost immediately and is very reactive = unstable - ROH is more stable than OH, so OH will not leave - Can make into a better leaving group when reacting with strong acid HX → CA = H2O

Nucleophiles and Bases - Nucleophile - donates electrons, basicity = subset of nucleophilicity - Charge - higher e- density = higher nucleophilicity - Neutral = weak base - Negative charge = more nucleophilic (CB is always better nucleophile b/c high e- density) - OH- > H2O - NH2- > NH3 - H2S- > HS - Electronegativity - Across PT → increase electronegativity, hold e- tightly → less likely to donate, less nucleophilic - Nucleophilicity increases with decreasing electronegativity - CB: H3C- > H2N- > OH- > F- (correlates with basicity) - H2C- and H2N- are very basic → quick to gain H - CA: H3N > H2O > HF (order of basicity more apparent here) - More basic if you less electronegative b/c you’re more likely to donate - Solvent - Polar protic - capable of H bonds - H2O, CH3OH (alcohol), EtOH (alcohol), carboxylic acids - Solvation in H2O surrounds ion in a shell → can’t react - Top of table has highest capacity for H bonding b/c smaller size → most hindered - Nucleophilicity increases down periodic table in polar protic solvent - I- > Br- > Cl- > F- SeH- > SH- > OH- Polar aprotic - no H bonds - DMF, DMS, DMSO, MeCN, Acetone - Increasing basicity increases nucleophilicity - HI is the strongest acid → I- is the weakest base → least nucleophilicity - Nucleophilicity increases up halogen column in apolar protic solvent - F- > Cl- > Br- > I- (F- strongest base here) - Steric hindrance - smaller the better - More branching → less nucleophilic (bulkiness hinders reactivity) - For charge and electronegativity, nucleophilicity increases with basicity - Exception is in polar protic solvent, when H bond hinders nucleophilicity and the reverse is true where nucleophilicity increases with decreasing basicity - CN- good nucleophile but bad base E1 vs SN1 - OH is a poor leaving group, unless you add strong acid HX to it

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X is Cl-, Br-, or I- (these CBs are decent nucleophiles) → favors SN1 - Using strong acid w/ good nucleophile CB - X is HSO4- (poor nucleophile CB b/c - charge distributed by resonance = stable) → favors E1 - Using strong acid w/ poor nucleophile CB - Idea: good nucleophiles tend to be unstable → very basic/quick to react - Heat favors elimination - If you see H2SO4, TsOH, or H3PO4 w/ heat → elimination and form double bond, carbocation intermediate so shift is possible Factors affecting the rate of SN1 and E1 reactions - Structure of alkyl group: reactions faster if leaving group attached to 3C than 2C b/c substitution (no react w/ 1C) - Better leaving group: I > Br > Cl >> F (much slower w/ F) - Halide ions as LG different than as nucleophile - H bonding and solvation stabilize halide ions → good for LG b/c product is more stable, but bad for nucleophile cus reactant is more stable (want reactions to go right) - Solvent: fastest in polar protic solvent - Solvent polarity stabilizes the carbocation intermediate (stabilizes product) - Solvent polarity and H bonds stabilize leaving group (stabilizes product) Factors that affect rate of SN2 reactions - Structure of R group - Faster with small alkyl halides, 1C faster than 2C - 3C does not react - Leaving group - RI > RBr > RCl > RF - Alkyl fluorides are too unreactive to be used in SN2 - Nucleophile/solvent - Protic solvent: I- > Br- > Cl- > F- H bonding decreases down column → less hindered by solvation → more nucleophilic - Basicity does not correlate with nucleophilicity - Aprotic solvent: F- > Cl- > Br- > I- More basic = more nucleophilic (correlates well) Summary of Substitution and Elimination - SN2 - Good nucleophile and weak base (like Cl-) - 1C: SN2 - 2C: SN2 - 3C: no reaction - Logic: good nucleophiles will attack the carbocation, but not the H b/c not basic - E2 - Good nucleophile and strong base (EtO-) - 1C: substitution = major product - 2C: elimination = major product - 3C: elimination only - Using bulky alkoxide → increases elimination product and less substituted alkene - Tert-butoxide → removes H+ from less subbed C → less subbed alkene forms - Logic: good nucleophiles will attack, but will attack the H instead of carbocation b/c basic - SN1/E1 - No base/nucleophile, only heating alcohol or water in acetone (EtOH / heat) - 1C: no reaction - 2C: both substitution and elimination

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3C: both substitution and elimination

SN1, SN2, E1, E2? 1. Good LG? a. Yes → will react b. No → no reaction 2. Is carbon where LG attached to sp3? a. No reactions can happen if not sp3 3. Does the alkyl halide have X on a methyl group, primary C, secondary C, or tertiary C? a. Methyl - always SN2 i. No elimination → can’t form double bond between Cs anywhere on a methyl w/ only 1 C ii. Carbocation on methyl = very unstable/won’t form→ no SN1 b/c requires carbocation iii. SN2 is the only option for methyl b. Primary - SN2 unless bulky then E2 i. Primary carbocation is not that stable → no SN1 or E1 ii. Regardless of basicity → primary always SN2 except... 1. If substrate or nucleophile is bulky → E2 a. Ex: not bulky OCH3b. Ex: bulky OC(CH3)3iii. SN2 affected by steric hindrance → that’s why prefer to grab at the end (primary C) than to go in where Cs are more subbed 1. NOTE: primary carbon is 2 Cs bonded to each other (both are primary) but halide is attached to the alpha carbon, not the beta carbon 2. Small nucleophile → can attack alpha C (primary C that X is attached to) fine 3. Large bulky nucleophile → has hard time squeeze in between alpha C and Br → goes to the side instead and grabs an H from beta C 4. Branching on beta carbon → bulky substrate → hinders SN2 → go E2 instead c. Secondary i. Relatively stable → can form carbocation → cannot rule out SN1 or E1 ii. Secondary more subbed → more steric hindrance iii. Do you have a good nucleophile or strong base? 1. Cl- good nucleophile but poor base 2. OCH3- or OH- good nucleophile and good base 3. CN- good nucleophile but poor base 4. Tert-butoxide poor nucleophile but strong and bulky base 5. Any of these criteria → yes a. Strong Nu, weak base (Cl-) → SN2 b. Strong Nu, strong base (OCH3-, OH-) → E2 c. Weak Nu, strong base (tert-butoxide) → E2, but less subbed alkene forms iv. No → weak Nu, weak base (H2O, CH3OH) 1. Carbocation can shift 2. Heat or no heat? a. Heat favors elimination → E1 b. No heat → SN1 d. Tertiary i. Very bulky → lot of steric hindrance → no SN2 ii. Stable → can form carbocation → cannot rule out SN1 or E1 iii. Do you have a strong base?

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1. Yes → E2 No → weak base (H2O, H3COH, Cl-) 1. Cl- may be a strong Nu, but doesn’t matter here with tertiary b/c no SN2 2. Heat? a. Yes → E1 b. No → SN1

Making OH into a good leaving group 1. React ROH with HX → ROH2+ (good LG) + Xa. Depending on carbocation stability, OH2+ can leave on its own (SN1) or needs help (SN2) b. 2C or 3C → SN1 (protonation → elimination → carbocation → SN1 w/ Br-) c. 1C → slower SN2 (protonation → 1C unstable carbocation, skip elimination → go SN2 w/ Br-) 2. React ROH with SX or SP bond then SN2 3. Sulfonate esters - 2 steps (formation of sulfonate ester, substitution) a. Sulfonate esters have RSO3-, which are better LG than OH b. OTF > OTS > OMS in terms of better leaving group c. RSO2Cl / pyridine i. Pyridine = base that deprotonates ii. Stronger inductive effects of Rmake the RSO3- a better LG 1. OTF (alkyl triflate) - CF3 2. OTS (alkyl tosylate) - benzene ring w/ a methyl 3. OMS (alkyl mesylate) - methyl d. Mechanism for formation of sulfonate ester - Cl leaves once i. ROH + RSO2Cl -- pyridine (base) → sulfonate ester ii. Addition - bond forms between ROH and RS  O2Cl w/ O(+), and RS=O → RS-O(-) iii. Deprotonation - pyridine removes H from O+ iv. Elimination - S-O(-) → RS=O and RS-Cl → RS + Clv. DOES NOT CHANGE STEREOCHEM (BEFORE NU ADDED/SN2) e. Mechanism for substitution (SN2) i. ROT -- Nu → Nu-R + OTii. Cl- can come back as a Nu, or a new Nu like CN- can be used iii. Nu backside attacks alpha carbon, and OT is released (elimination) iv. BACKSIDE ATTACK CHANGES STEREOCHEM → INVERSION f. Overall reaction (2 steps): i. 1) ROH --RSO2Cl/pyridine→ ROT (same stereochem as ROH) ii. 2) ROT --Nu→ Nu-R (inversion substitution of original ROH) + OT1. Also floating pyridine(+) 4. Direct halogenation alcohol w/ SOCl2 - 1 step (key step is SN2) a. Reaction of ROH w/ thionyl chloride (SOCl2) is SN2 → best w/ 1C OH b. Mechanism - Cl leaves twice then SN2 (all in one “step” but not like concerted) i. Addition, deprotonation, and elimination are the same as sulfonate ester formation ii. SN2: Cl- comes back as Nu and does a backside attack (inversion) → R-Cl and 2nd Clleaves in (-)O-SO-Cl → SO2 + Cliii. BACKSIDE ATTACK CHANGES STEREOCHEM → INVERSION c. Overall reaction: ROH + SOCl2 --pyridine (base) → RCl (inversion) + SO2 i. Also floating pyridine(+) and Cl5. Direct halogenation of ROH w/ PBr3 - 1 step (SN2)

a. Can react w/ 1C or 2C OH b. Mechanism - similar to direct halogenation w/ SOCl2 but a lot shorter b/c add/elim same time i. Addition and elimination of Br-: bond forms between ROH and PBr3 w/ O(+), Br- leaves ii. SN2: Br- comes back as Nu and backside attacks alpha C → RBr + HOPBr2 iii. HOPBr2 hydrolyzes to H3PO4 iv. BACKSIDE ATTACK CHANGES STEREOCHEM → INVERSION v. Differences from SOCl2: no O+ deprotonation, Br- leaves once, comes back immediately c. Overall reaction: RCH2OH + PBr3 --pyridine(base) → RCH2Br (inversion) + HOPBr2 6. Triphenylphosphine dibromide (Ph3PBr2) - 1 step a. Involve formation of strong P=O bond b. In Ph3PBr2, one Br- is ionically bonded w/ P and the other Br- is covalently bonded c. Mechanism: i. OP bond forms → O+, and ionic Br- dissociates ii. Br- grabs H on O+ → previous OH bond now btw OP and 2nd Br- dissociates → 2 possible resonance structures where O and P share + charge 1. R-(+)O=PPh3 (O still has + b/c bond just moved over) 2. R-O-P(+)Ph3 a. Br- backside attacks alpha C → (-)OP(+)Ph3 eliminated = O=PPh3 iii. BACKSIDE ATTACK CHANGES STEREOCHEM → INVERSION d. Overall reaction: ROH --Ph3PBr2→ RBr (inversion) + (-)O-P(+)Ph3 = O=PPh3 Alkyl halides vs alkanols - RX → ROH - SN1 - (-)OH attacks after Br- leaves to form carbocation - SN2 - (-)OH attacks alpha C directly and Br- leaves - ROH → RX by - SN1 - react with HX - SN2 - after S or P ester formation - RX → alkene - E1 dehydrohalogenation - Br good LG → comes back to grab H → lone pairs form double bond - Forms alkene and HX - ROH → alkene - E1 dehydration - acid catalyzed carbocation formation (ROH + HA) - (+)OH2 good LG → comes back to grab H → lone pairs form double bond - Forms alkene and H3O+ - Alkene formation favored by removing water (dehydration) - Alcohol formation favored by adding water (reverse reaction) E1 - more substituted carbocations (stable) form faster and most stable alkene preferred (more formed) How to make less substituted alkenes? - E1 - 1C OH → add H2SO4 and heat → most subbed = major product - 2C OH → add H2SO4 → most subbed = major product - E2 - Convert ROH to ROTs then treat w/ base - 1C OH → ROT → 1-alkene

Chart for nucleophiles and bases Strong Base

Weak Base

Good Nucleophile

(-)OH (-)OR R-C-triple bond-C(-)

I(-), Br(-), Cl(-) RS(-) CN(-) N3(-)

Bad Nucleophile

(CH3)3CO(-) → the broomstick LDA

F(-) H2O ROH

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Methyl → SN2, not stable enough to form carbocation Primary - SN2 (except E2 with (CH3)3CO(-) or another hindered base - Either of the -2 options - Not stable enough to form carbocation → SN2 - BUT SN2 affected by steric hindrance → E2 if bulky base - If beta branching is fully substituted → no rxn for SN2 or E2 - Ex: (CH3)3CBr --N3(-)--> Even though Br is primary, nothing happens Secondary - lot of options - Strong base → E2 (prefers to pick up H) - Weak base, good Nu → SN2 (does not like H, but strong enough to attack directly) - Weak base, bad Nu → SN1 (does not like H, and not strong enough to attack directly) Tertiary - Strong base - E2 - Weak base - SN1 (very stable → can form carbocation) Leaving groups: OTF > OTS > OMS > I > Br > Cl > F Nucleophiles in aprotic solvent: F > Cl > Br > I - Protic solvents: H2O, ROH, RCOO(-) ionized carboxylic acid Nucleophiles in protic solvent: I > Br > Cl > F - Aprotic solvents: acetone, DMF, DMSO, HMPA - Probably has a weird name...