1. The S N 2 and S N 1 Acid-Catalyzed Reactions of ROH  RX in (crude) animated form The crucial first step: Equilibrium formation of protonated alcohol ROH 2 + OH ROH H+H+ H + from H 2 SO 4.. 1) H + sees yummy lone pairs on O and tries to eat (bind to) them.. 2)… but can’t quite keep it down so endlessly releases and repeats attempts to consume OH electrons (…equilibrium) 3) To help H+ keep the electrons down we heat, driving equilbrium to right(reflux) OH 2 +

2. Br: - OH 2 + Initial stage S N 2 reaction of protonated 1 o alcohol starts by attack of Br(-) nucleophile  Br begins to approach protonated alcohol  Incipient bond forms from C----Br as lone pair of Br starts to donate  Three attached groups move away from Br  The C-OH 2 bond lengthens

3. 5-coordinate activated complex for S N 2 formed after attack of nucleophilic Br - Br: OH 2 - At this point C-O bond about to break away completely but is still attached. 1)two electrons in breaking bond goes to OH 2 (+) and neutralizes (+) 2a)Lone pair from Br contributes to C-Br bond forming on other side 2a) (-) from Br contributes to Br-C bond forming 2b) while Br continues to approach, OH 2 and other groups continue to move away to reach final inverted state Br-C nearly formed + Complex now deforms away from its unstable peak 2c) Br makes bond while OH 2 leaves Activated complex stage

4. Br - OH 2 + Start Un-inverted reactant (Left leaning) Br End inverted product (Right leaning) H2OH2O Leaving group has left Inversion for S N 2

5. Start of S N 1 reaction on 2 o or 3 o, protonated alcohol (step 2 after step 1 protonation) OH 2 2a) H 2 O leaves as H 2 O + …and leaves (+) on central carbon 1b) At same time all the other attached groups move towards central carbon to form a flattened, sp 2 like structure

6. Step 3 of S N 1 : attack of Bromine on intermediate 2 o or 3 o carbocation made in step 2 + + - Br 3a) Br - attacks carbocation from either side equally, causing deformation back to sp 3 shape Br - 3b) + and – charges neutralize 3c) C bonds to Br and molecules formed are opposite in inversion effect (racemic mix)

7. Example of rearrangement during S N 1 + H + reflux Initial (primary) carbocation formation Unprotonated alcohol protonated alcohol Initial =primary carbocation Rate limiting step Br - + Unrearranged product + H 2 O

8. Primary carbocation Rearrangements of primary carbocation 1,2 `hydride’ shift 3o3o 2o2o Secondary carbocation from 1,2 hydride shift rearrangement H moves to (+) (+) moves to site H vacates Br - Rearrangement product from 1,2 hydride shift

9. Rearrangements of primary carbocation 1,2 methyl shift CH 3 moves to (+) (+) moves to site CH 3 vacates Secondary carbocation from 1,2 methyl shift rearrangement 2o2o 3o3o Br - Rearrangement product from 1,2 methyl shift

10. Rearrangements only occur if degree of secondary carbocation increases vs primary 2o2o 3o3o 2o2o 1o1o YES NO

11. Both ROH and HX concentrations affect rate 0.)need reflux and extra acid (H 2 SO 4 ) to make ROH 2 + else no go 1)3 o ROH > 2 o ROH > 1 o ROH in rate Only ROH concentration affect rate No rearrangementsRearrangements occur Reaction rate follows the modest trend > Br > Cl > F No affect of halogen identity on rate Modestly polar aprotic solvents produce better yields Strongly Polar protic solvents produce better yields Product is inverted vs starting alcohol Product is racemized vs starting alcohol R-OH + HBr  R-Br +H 2 O Acid +base“salt” + 1 o ROH 2 o & 3 o ROH FACTS ABOUT …. water S N 2 MECHANISM S N 1 MECHANISM