Organic ChemistryChapter 8

Substitution and Elimination• If an sp3 C is bonded to electronegative

atom Substitution reactions and Elimination reactions are possible

This chapter is all about substitution

SN2 and SN1 Reactions

SN2 - Reaction – bonds break and form at the same time

example

SN1 - CX bond breaks, forming a C+ then reacts with a nucleophile

XC+

C + + X-

NuCC + + Nu:

SN1

SN2

Nucleophilic Substitution Reactions

Either mechanism depends on the:• structure of the alkyl halide• reactivity of the nucleophile• concentration of the nucleophile• The solvent in which the Rx is carried out• The leaving group

SN2 Mechanism

• It’s a Substitution Reaction (S)• It’s Nucleophilic (N)• It’s rate is second order (2)

– Called bimolecular (rate is dependent on 2 reactants)

• (Substitution Nucleophilic Bimolecular)

CH3Br + HO - CH3OH + Br-

methyl bromide methyl alcohol

Rate = k [RX] [Nu:](Because rate is dependent of BOTH RX and Nu: it is 2nd. order.)

SN2 Mechanism• SN2 Mechanism involves a “backside attack”

SN2 MechanismThe “backside attack” causes an Inversion of Configuration

Steric Hindrance

• Groups that block the path from the nucleophile to the electrophilic atom produce steric hindrance

• This results in a rate differences or no reaction at all

methyl halide ethyl halide isopropyl halide t-butyl halide

Steric Hindrance

• Activation Energy is higher due to steric hindrance…..

Substitution Reactions Depend on a Good Leaving Group

• R-F alkyl fluorides• R-Cl alkyl chlorides• R-Br alkyl bromides• R-I alkyl iodides• Alkyl Halides make good “leaving groups”

– They are easily displaced by another atom– They allow the Conversion of alkyl halides to other functional

groups

SN2 Mechanism

• The Leaving Groups also affects rate• RI reacts fastest, RF slowest

– Iodide is the best “leaving group”– Fluoride is the worst “leaving group”

(…reacting with the same alkyl halide under the same conditions)

SN2 Reactions

SN2 Reactions

SN2 Reactions• Reaksi S N 2 mungkin reversibel• Bandingkan kebasaan (kekuatan nukleofil)

untuk melihat gugus pergi yang lebih baik.– Basa kuat akan menggantikan basa lemah

• Jika kebasaan hampir sama, Rx akan reversibel

CH2CH3 Br + I- CH2CH3 I + Br-

SN2 ReactionsCompare basicity to see which is a better nucleophile.

F Cl Br I-

Bertambah jari-jari ion(Atom yg lebih besar mampu menyebarkan muatan negative lebih baik drpd atom kecil, penyebaran muatan menyababkan penstabilan)

HF HCl HBr HI

pKa 3,45 -7 -9 -9,5

Naiknya kuat asam

C N O F

Meningkatnya keelektronegatifan unsur

R3C-H R2N-H RO-H F-H

Naiknya kuat asamAnion asam sangat kuat merupakan basa lebih lemah, sebaliknya anion asam sangat lemah merupakan basa sangat kuatR3C- R2N- RO- F-

Naiknya kuat basa

SN1 Reactions• Reaction of t-butyl bromide with water should be

slow– water is a poor nucleophile– t-butyl bromide is sterically hinderedHowever– Reaction is a million times faster than with CH3Br

t-butyl bromide

CCH3

CH3

Br

CH3

+ H2O

t-butyl alcohol

CCH3

CH3

OH

CH3

+ HBr

(Maybe not an SN2 reaction!)

SN1 Reactions•

SN1 Mechanism• Rate determining step does not involve

nucleophile

Step 2

Step 1

SN1 Mechanism

SN1 Reactivity• Relative Reactivities in an SN1 Reaction

1o RX < 2o RX < 3o RX

Increasing Reactivity

SN1 Stereochemistry

• Because a planer carbocation is formed, nucleophilic attack is possible on both sides, so both isomers are possible

SN1 Stereochemistry

SN1 should yield racemic mixture but it doesn’tThis is due to the steric hindrance of the leaving group

Stereochemistry• As the leaving group goes (Marvin K) it

blocks the path of any incoming nucleophiles

SN1 vs SN2

Inversion of configuration

racemization withpartial inversion

What Makes SN1 Reactions work the best

• Good Leaving Group– The weaker the base, the less tightly it is held

(I- and Br- are weak bases)

• Carbocation– How stable is the resulting carbocation?

• 3o > 2o > 1o > methyl

Increasing Stability

What Doesn’t Matter In anSN1 Reactions

• The Nucleophile• It has NO EFFECT on rate of Rx!!!

• Solvolysis Reactions • (the nucleophile is also the solvent)

Carbocation RearrangementsSince a carbocation is the intermediate, you may see

rearrangements in an SN1 Rx

No rearrangements in an SN2 Rx

Carbocation Rearrangement

• Methyl Shift

Benzylic, Allylic, Vinylic,and Aryl Halides

• Benzylic and allylic halides can readily undergo SN2 unless they are 3o – (steric hindrance)

Benzylic, Allylic, Vinylic,and Aryl Halides

• Benzylic and allylic halides can also undergo SN1 (they form stable carbocations)

• Meskipun RX primer : benzilik , alilik BISA bereaksi S N 1!

Vinylic,and Aryl Halides

• Vinylic halides and aryl halides– tidak mengalami reaksi S N 1 atau S N 2

e- mengusir masuk Nucleophile

BrBr

SN1 vs SN2 Review

SN1 vs SN2

Methyl, 1o RX …2o RX …3o RX …

Vinylic, aryl RX …1o, 2o benzylic, allylic RX …

3o benzylic, allylic RX …

SN2 onlySN1 and SN2SN1 onlyneither SN1 nor SN2SN1 and SN2SN1 only

Role of the Solvent• So….• In an SN1 reaction, the reactant is RX. The

intermediate is charged and is STABILIZED by a POLAR solvent

A POLAR solvent increases the rate of reaction for an SN1 reaction.

(However, this is true only if the reactant is uncharged.)

Kemampuan mensolvasi ion ditentukan oleh polaritas molekul pelarut itu atau tetapan dielektriknya.Pelarut yang sangat polar mempunyai tetapan dielektrik tinggi.

Umumnya pelarut sangat polar (air) membantu menstabilkan karbokation dengan jalan solvasi = mendorong reaksi SN 1.Pelarut yang kurang polar (spt: aseton) tidak membantu ionisasi shg memilih reaksi SN 2 dan E 2.

Pelarut yg dapat mensolvasi anion (menstabilkan anion) akan mengurangi nukleofitasnya dan sebaliknya. δ+ δ+ _CH3CH2OH----Cl-----HOCH2CH3 : etanol dpt mensolvasi ion negatif.

DMF (dimetil formamida) dan DMSO tidak memiliki H yg mampu mensolvasi ion negatif.

HCON(CH3)2 CH3SOCH3

*

Role of the Solvent In SN2• In an SN2 reaction, one of the reactants is the

nucleophile (usually charged). • The POLAR solvent will usually stabilize the

nucleophile.

A POLAR solvent decreases the rate of reaction for an SN2 reaction.

(However, this is true only if the nucleophile is charged.)

Pelarut protik

Pelarut protik adalah pelarut yang memiliki hidrogen terikat pada atom oksigen (seperti dalam hidroksil ) atau nitrogen (seperti pada amina ).

Molekul-molekul pelarut tersebut dapat menyumbangkan + H (proton). Sebaliknya, pelarut aprotik tidak dapat menyumbangkan hidrogen.

Contohnya adalah air , metanol , etanol , asam format , fluorida hidrogen dan amonia .

karakteristik Umum pelarut protik:

•Bisa membentuk ikatan hidrogen •pelarut memiliki hidrogen asam (meskipun mereka mungkin asam sangat lemah) •pelarut dapat menstabilkan ion

kation oleh berbagi pasangan elektron bebas anion oleh ikatan hidrogen

Pelarut aprotik kutub pelarut yang dapat melarutkan ion tetapi tidak merupakan hidrogen asam. Umum karakteristik pelarut aprotik: •pelarut tidak bisa membentuk ikatan hidrogen •pelarut tidak memiliki hidrogen asam•pelarut dapat menstabilkan ion Contohnya adalah dimetil sulfoxide, dimetilformamida , dioksan dan hexamethylphosphorotriamide , tetrahidrofuran .

Pelarut protik polar menguntungkan bagi reaksi S N 1, sedangkan pelarut aprotik polar menguntungkan bagi reaksi S N 2.

Pelarut Formula Kimia Titik didih Konstanta dielektrik

Pelarut Non-Polar

Heksana CH 3-CH 2-CH 2-CH 2-CH 2-CH 3 69 ° C 2.0

Benzena C 6 H 6 80 ° C 2.3

Toluena C 6 H 5-CH 3 111 ° C 2.4

1,4-dioksan /-CH 2-CH 2-O-CH 2-CH 2-O-\ 101 ° C 2.3

Khloroform CHCl 3 61 ° C 4.8

Dietil eter CH 3 CH 2-O-CH 2-CH 3 35 ° C 4.3

Pelarut aprotik polar

Diklorometana (DCM) CH 2 Cl 2 40 ° C 9.1

Tetrahidrofuran (THF) /-CH 2-CH 2-O-CH 2-CH 2 - \ 66 ° C 7.5

Etil asetat (EtOAc) CH 3-C (= O)-O-CH 2-CH 3 77 ° C 6.0

Aseton CH 3-C (= O)-CH 3 56 ° C 21

Dimetilformamida (DMF) HC (= O) N (CH 3) 2 153 ° C 38

Asetonitril (MeCN) CH 3-C ≡ N 82 ° C 37

Dimetil sulfoxide (DMSO)

CH 3-S (= O)-CH 3 189 ° C 47

Pelarut protik Polar

Asam formiat HC (= O) OH 101 ° C 58

n-Butanol CH 3-CH 2-CH 2-CH 2-OH 118 ° C 18

Isopropanol (IPA) CH 3-CH (-OH)-CH 3 82 ° C 18

n-Propanol CH 3-CH 2-CH 2-OH 97 ° C 20

Etanol (EtOH) CH 3-CH 2-OH 79 ° C 30

Methanol (MeOH) CH 3-OH 65 ° C 33

Asam asetat (AcOH) CH 3-C (= O) OH 118 ° C 6.2

Air Hoh 100 ° C 80

Polar Aprotic Solvents

• Polar Aprotic Solvents include:– DMF N,N-dimethylformamide– DMSO dimethylsulfoxide– HMPA hexamethylphosphoramide– THF Tetrahydrofuran– And even… acetone

Polar Aprotic Solvents

Polar Aprotic Solvents – do not H bond– solvate cations well– do NOT solvate anions (nucleophiles) well– good solvents for SN2 reactions

Polar Aprotic Solvents

• DMSO• DMF• Acetone• HMPA

Nucleophile Review

strong

weak

Br -, I-

HO-, CH3O-, RO-

CH3S-, RS -

CH3CO2-, RCO2

-

H2OCH3OH, ROHCH3CO2H, RCO2H

NH3, RNH2, R2NH, R3NCH3SH, RSH, R2S

EffectivenessNucleophile

moderate

CN-, N3-

SN1/SN2 Problems -1

• Predict the type of mechanism for this reaction, and the stereochemistry of each product

+

+ +OH

Cl

OCH3

CH3CHCH2CH3

CH3CHCH2CH3

CH3CHCH2CH3

CH3OH/H2O

HCl(R)-enantiomer

SN1/SN2 Problems -1

• Predict the type of mechanism for this reaction, and the stereochemistry of each product

+

+ +OH

Cl

OCH3

CH3CHCH2CH3

CH3CHCH2CH3

CH3CHCH2CH3

CH3OH/H2O

HCl(R)-enantiomer

SN1/SN2 Problems -2

• Predict the mechanism of this reaction

+

+

DMSOCH3

CH3

CH3CHCH2CN

CH3CHCH2Br Na+CN-

Na+Br-

SN1/SN2 Problems -2

• Predict the mechanism of this reaction

+

+

DMSOCH3

CH3

CH3CHCH2CN

CH3CHCH2Br Na+CN-

Na+Br-

SN1/SN2 Problems -3

• Predict the mechanism. If the starting material has the R configuration, predict the configuration of product

+

+

acetone

Br

SCH3

CH3CHCH2CH3 CH3S-Na+

CH3CHCH2CH3 Na+Br-

SN1/SN2 Problems -3

• Predict the mechanism. If the starting material has the R configuration, predict the configuration of product

+

+

acetone

Br

SCH3

CH3CHCH2CH3 CH3S-Na+

CH3CHCH2CH3 Na+Br-

SN1/SN2 Problems -4

• Predict the mechanism

+ acetic acidBr

OCCH 3

O

O

CH 3 COH

+ HBr

SN1/SN2 Problems -4

• Predict the mechanism

+ acetic acidBr

OCCH 3

O

O

CH 3 COH

+ HBr

SN1/SN2 Problems -5

• Predict the mechanism

+ toluene

Br-

(CH3)3PCH3(CH2)5CH2Br

CH3(CH2)5CH2-P(CH3)3+

SN1/SN2 Problems -5

• Predict the mechanism

+ toluene

Br-

(CH3)3PCH3(CH2)5CH2Br

CH3(CH2)5CH2-P(CH3)3+

64

• Negatively charged nucleophiles like HO¯ and HS¯ are used as salts with Li+, Na+, or K+ counterions to balance the charge. Since the identity of the counterion is usually inconsequential, it is often omitted from the chemical equation.

• When a neutral nucleophile is used, the substitution product bears a positive charge.

Alkyl Halides and Nucleophilic SubstitutionGeneral Features of Nucleophilic Substitution:

65

• Furthermore, when the substitution product bears a positive charge and also contains a proton bonded to O or N, the initially formed substitution product readily loses a proton in a BrØnsted-Lowry acid-base reaction, forming a neutral product.

• To draw any nucleophilic substitution product:Find the sp3 hybridized carbon with the leaving group.Identify the nucleophile, the species with a lone pair or bond.Substitute the nucleophile for the leaving group and assign charges (if necessary) to any atom that is involved in bond breaking or bond formation.

Alkyl Halides and Nucleophilic SubstitutionGeneral Features of Nucleophilic Substitution:

END