The SN Bond breaking is Bond for f ft Iii

Products

The SN2 Mechanism

Summary:

Regiochemistry:

Stereochemistry:

Example:

NaN3

H O + BrC

H

H

H

Transition state

Na

I

Bond that

Bondthatis is breaking

for

storeH f

u ft IiiNucleophile must react at

he back of the C Br bond

this angle and direction ofattack helps break the C Br bond

itH C HTheconfiguration

at C isinverted

BAt

The nucleophile attacks by makinga new bond to C from the back of theC X bond just as the X leaves

N AINVERSION at siteofreactors

4 Inde.phN3ueaEfsP

Products

The E2 Mechanism

Summary:

Regiochemistry:

Stereochemistry:

Example:

NaOH

H3C O + C

H

H

H

Transition state

Na

H3C

CH3

C

Br

CH3

CH3

ClCH3

BondsbeingBonds being

broke

EoutEBr

GHK

43kH H RCH3

C F CCH

The H that is leaving and

the Br must be HK I H B.inanti periplanar

Base removes an H atom as a piband forms and the Br atom leaves

The H and Dr must be

anti peri planarZaitsev's Rule most stable alkene productDetermined by anti periplanar transition state

1734 5 3

µH kHz

Products

The SN1 and E1 Mechanisms

C

Cl

C

H

H

H

C

C

H HH

H

H

C

H

H

H

H3C O H

Products

H3C O HE1 H3C O H

SN1

Summary:

Regiochemistry:

Stereochemistry:

Example:

CH3CH2OH

(S)

Br

H heat

yHaloalkae with CHz 17

Breaka bond

t H tcH3

it.it oE'it fifth city1 H f lH H K H H L Ha t t Make a 11 H

bond it

Take a Take a protonproton awayaway

KE't it H H a inH H d d c C HH C c I l l H

y H H H C HH it if I

t t Hy F HHFor sterically hindered haloalkanes the C X

bond breaks to give a carbocation intermediate thateither reacts as an electrophile Sv 1 or

has a proton taken away EI

EI Zaitsev's RuleSv1 Scrambled not quite I exactly

fr E't sit3 I A4 v Bocage atCHS

To

Very

Overy

GHz

Hy c K KotBy1 orCH

TBF

Substitution/Elimination Decision Map

Primary Haloalkane

Methyl Halide

Tertiary Haloalkane

E2

SN1/E1Very Weak Base ?

SN2

tBuOK ? Yes E2

No SN2

Yes

No

For SN2 Remember Chiral Center InVERSiONFor E2 Remember anti-periplanar and ZaitsevFor SN1 Remember Chiral Center ScramblingFor E1 Remember Zaitsev

Note: With Very Weak Bases, SN2 can compete here, but for the purposes of this class, assume SN1 / E1 predominate

Note: If tBuOK is the very strong base, an appreciable amount of a non-Zaitsev product can be formed because the bulky tBuOK will tend to react with the most accessible H atom.

Secondary Haloalkane

or

Allylic/Benzylic Halides

SN1/E1Very Weak Base ? Yes

Very Strong Base ? Yes E2

No SN2

*

**

*

**

1 Methyl halo alkane always SNZ

2 to halo alkanes Sw 2 with all but

tBu0 k

Kota

try zNEE bond

Sir

nBr I

3 20 halo alkanes Sir 2 with allbut strong bases or veryweak bases

EZ with strong bases

sv1 E1 with very weak bases

strong base

CH CX t CH CE H

Br E220 haloalkane 2,523daLt

Rj 9 INVERSION

very weak base

Nap A Bratz

SNHEI Et NZeitz

SNI

4 30 halo alkanes EZ withall but veryweak bases

Na OH 41700

I Na

EL

Zaitsev

very weak base 1I HOTISNI EI SNIEI

T

EPIC NEW REACTIOND Alkylation of terminal attkyane

anim with a primary halotaldkane

T P CHz CEC CHICHICHCH C t CH CHIK B.ir

Pr kae 3

New C C bond

Making new GC bonds allows us to

construct larger molecules fromsmaller fragments

Tine capsuleThis is an SNLreaction The halo alkanemust be primary toavoid E3

E Conversion o

vicinal dihalo a Kane

to an alkyne in base

ratio ofmolecules

toBr Br 2 equivalents

NaN H2Hsc CHz H C CEC CH

H H Very straybase Alkyne

Vicinal dihaloalkane

isa E'EIII

When creating a terminal alkyneyou need to all 3 equivalents of

NaN Hz then in a second step addmild acid Halite

For highly substituted halo alkanes

you must first look for t Zaitsev

most substituted alkene productthen possibly rotate

a C C bond to find theanti periplanar geometry that

predicts Evs Z

424 H3 CHIH s

Hye Ia Br D Br CH

E Ciotat H EI'tperiplanor

Not a reactiveconformation

Rotak central C C bondBr

HSE 1B't H C featsHzcHfac C CH HGH tH3

CHIH HH

Br and H are

anti periplanarThis is the reactiveconformation

HE Mr NaOCH Itsy 4 3

HZCHIAE.ec C C c

if bi HEH EZ HE KICKS

Anti perilanarZ product

H IHy H it

E teaHtt

These twoaxons

are These two H atomsa

ratare anti periplanarto the Br atomso they could react

No E2This reactsbecause there

are NO because thereare

anti peri planar anti periplananH atoms H atoms

Thg Ei7z

BrH IHy H it

Fsr t.IEf

It

nucleophile canNucleophile prevented attack fromfrom attacking from behind the

behind the C Br c Br bond

bond steric blockade like any20 haloalkane

i