ORGANOHALIDES + Nucleophilic Reactions (SN1/2, E1/E2/E1cB) CH21 PS CLASS

Preparation of Organohalides From ALKENES C=C [just review old lessons] FOR TERTIARY ALCOHOLS, we can simply use H-X (gas) X=Cl,Br in ether, 0°C

Preparation of Organohalides FOR TERTIARY ALCOHOLS, we can simply use H-X (gas) X=Cl,Br in ether, 0°C Follows SN1 so a carbocation is formed, be careful with rearrangements!

Preparation of Organohalides FOR PRIMARY/SECONDARY ALCOHOLS: SOCl2 / PBr3

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Alkyl Fluorides Also from ALCOHOLS + HF / Pryidine (CH3CH2)2NSF3

Grignard Reagents Reaction of R-X with Mg over ether/THF to form R-Mg-X organometallic compound.

Grignard Reagents: reduction of R-X

More samples:

Nucleophilic Reactions R-X, alkyl halides are ELECTROPHILES (positive or electron-poor) They react with NUCLEOPHILES/BASES (negative or electron-rich) Either substitution C-C-X becomes C-C-blah + X- or elimination reactions C-C-X becomes C=C + X-

SUBSTITUTION REACTIONS S – substitution: R-X + Nu  R-Nu + X- N – Nucleophilic 1 or 2  unimolecular or bimolecular rates INVERSION (change of stereochemistry) CAN HAPPEN!

Try this first…

SN2  BIMOLECULAR Bimolecular simply refers to the rate depending on BOTH reactants because of the nature of the mechanism Rate = k[RX][Nu] Rate depends on both because there is ONE SINGLE COLLISION BETWEEN RX and Nu to form a Nu-R-X transition state

SN2  BIMOLECULAR 100% INVERSION OF STEREOCHEMISTRY OCCURS! SUBSTRATE LEAVING GROUP

Factors that affect SN2 RXNS: STERIC EFFECTS TO INCOMING Nu: C=C-X (vinylic) and Ar-X (aryl) TOTALL UNREACTIVE

Factors that affect SN2 RXNS: THE NUCLEOPHILE

Factors that affect SN2 RXNS: THE LEAVING GROUP should be stable on its own as a free anion Comparing halides, we go down the column

Factors that affect SN2 RXNS: Alcohols and fluorides usually do not undergo SN2 because OH- and F- aren’t good leaving groups This is why we use SOCl2 and PBr3 … THEY CONVERT THE –OH INTO A BETTER LEAVING GROUP

Factors that affect SN2 RXNS: Reaction SOLVENT can also affect the reaction. We prefer POLAR APROTIC SOLVENTS POLAR but no –OH or –NH in the molecule (no H2O, NH3, etc…) Polar protic solvents form a CAGE around Nu

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SN1  UNIMOLECULAR Unimolecular: rate depends only on the substrate (mechanism), almost opposite of SN2 Rate = k[RX] Rate is only dependent on the slowest step which is the spontaneous dissociation of your leaving group. (molecules just don’t easily dissociate!)

SN1  UNIMOLECULAR

SN1  UNIMOLECULAR

SN1  UNIMOLECULAR STEREOCHEM IS LOST, A RACEMATE FORM IS MADE, but usually not 50:50

SN1  UNIMOLECULAR STEREOCHEM IS LOST, A RACEMATE FORM IS MADE, but usually not 50:50 An ION PAIR BLOCKS THE OTHER SIDE!

Factors that affect SN1 RXNS: SUBSTRATE:

Factors that affect SN1 RXNS: LEAVING GROUP: An –OH in acidic medium can become –OH2+ and leave as H2O which is very favorable

Factors that affect SN1 RXNS: NUCLEOPHILE: no effect, almost at all.

Factors that affect SN1 RXNS: SOLVENT: rates increase if you stabilize carbocation transition state. POLAR PROTIC!

Factors that affect SN1 RXNS: SOLVENT: rates increase if you stabilize carbocation transition state. POLAR PROTIC!

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Elimination Reactions More compliated (different mechanisms) The loss of H-X can lead to a MIXTURE of alkene products (C-C-X  C=C + HX) But we can predict the most stable/major poduct ZAITZEV’S RULE: base-induced eliminations will form more stable alkene

E2 elimination Again, bimolecular so a single collision between your Base B: and the alkyl halide.

E2 elimination Anti-periplanar is favored for transition state

E2 elimination Anti-periplanar is favored for transition state

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E1 reaction Unimolecular, ALSO spontaneously forms carbocation, but then followed by loss of H+ (taken by a base B: and not an attack by Nu:) COMPETES WITH SN1 reactions!

E1 reaction

E1 reactions No need for anti periplanar geometry

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E1cB Unimolecular, but this time CARBANION formed because a proton H+ is first removed by a base. cB stands for “conjugate base” because you deprotonate your carbon C-H into a C- and H+ Usually favored for poor leaving groups (e.g. –OH) Carbanion can be stabilized with C=O groups nearby

E1cB

E1cB PRESENCE OF C=O NEARBY CAN GIVE RESONANCE STABILIZATION!

PREDICTING WHAT PREDOMINATES:

Slight Clarifications: BASE vs. NUCLEOPHILE Affinity for a PROTON Strong base like R-O- or OH- Usually a LEWIS BASE In this context, how attracted to a CARBON

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