alkyl halides

1

Alkyl HalidesOrgano halogen

Alkyl halide

Aryl halide

Halide vynilik C

CCl3

H

ClCl

OH2CHO

I

I

I

I

HC

NH2

C

O

OH

C Cl

H

H

Cl

C Cl

Cl

Cl

Cl

C Cl

Cl

H

Cl

C C

H

H Cl

H

Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2

NUCLEOPHILIC SUBSTITUTION R X Y- R Y X-

1. Leaving groups

++ XR YYR Xstronger

baseweaker

base

K > 1

Br F+ NaF + NaBrSB WB

Br I+ NaI + NaBr (s)

WB SB

acetone

weaker base = better leaving group

reactivity: R-I > R-Br > R-Cl >> R-Fbest L.G.

most reactiveworst L.G.

least reactive

precipitatedrives rxn(Le Châtelier)

2. Mechanisms SN

general: Rate = k1[RX] + k2[RX][Y–]

RX = CH3X 1º 2º 3ºk1 increases

k2 increases

k1 ~ 0

Rate = k2[RX][Y–](bimolecular)

SN2

k2 ~ 0

Rate = k1[RX](unimolecular)

SN1

SN2 MechanismA. Kinetics

e.g., CH3I + OH– CH3OH + I–

find: Rate = k[CH3I][OH–], i.e., bimolecular

both CH3I and OH– involved in RLS

and recall, reactivity: R-I > R-Br > R-Cl >> R-F

C-X bond breaking involved in RLS

concerted, single-step mechanism:

CH3I + OH–

CH3OH + I–

[HO---CH3---I]–

B. Stereochemistry: inversion of configuration

H Br HO HNaOH

(R)-(–)-2-bromooctane (S)-(+)-2-octanol

Stereospecific reaction:Reaction proceeds withinversion of configuration.

Back-side attack:

HO C IH

HH

C

H

IHO

H H

CH

HOH

HI+

+ -

HO C I HO C I HO C I

inversion of configuration

C. Mechanism

D. Steric effects

e.g., R–Br + I– R–I + Br–

1. branching at the a carbon ( X–C–C–C.... )

a b g

Compound Rel. Rate

methyl CH3Br 150

1º RX CH3CH2Br 1

2º RX (CH3)2CHBr 0.008

3º RX (CH3)3CBr ~0

increasingsteric hindrance

C BrH

HH

I

H CH

C BrCH

H

H

H

H

HCH

I

minimal steric hindrance

maximum steric hindrance

Reactivity toward SN2: CH3X > 1º RX > 2º RX >> 3º RX

reactreadilyby SN2

(k2 large)

moredifficult

does notreact by

SN2(k2 ~ 0)

E. Nucleophiles and nucleophilicity

1. anions R X OH R OH X+ +

++ XR CNCNR X

R X R O H

H

+ +

++ R O R'

H

R X

H2O

R'OH

X

X

ROH + HX

ROR' + HX

hydrolysis

alcoholysis

2. neutral species

Summary:very good Nu: I–, HS–, RS–, H2N–

good Nu: Br–, HO–, RO–, CN–, N3–

fair Nu: NH3, Cl–, F–, RCO2–

poor Nu: H2O, ROHvery poor Nu: RCO2H

SN1 MechanismA. Kinetics

C

CH3

H3C

CH3

Br CH3OH CH3C

CH3

CH3

O CH3 HBr+ +e.g.,

3º, no SN2Find: Rate = k[(CH3)3CBr] unimolecular

RLS depends only on (CH3)3CBr

A. Kinetics

C

CH3

H3C

CH3

Br

CH3C

CH3

CH3

O CH3 HBr+

RLS: C

CH3

H3C

CH3

+ Br

C

CH3

H3C

CH3

HOCH3 C

CH3

H3C

CH3

OH

CH3

C

CH3

H3C

CH3

OH

CH3

-H+

A. Kinetics

Two-step mechanism:

RBr + CH3OH

R+

ROCH3 + HBr

B. Stereochemistry: stereorandom

CH3CH2

Br

H

CH3 CH3CH2

OH

H

CH3 CH3CH2

H

OH

CH3+H2O

racemic

CH3CH2 CH

CH3

+OH2

OH2

sp2, trigonal planar

C. Carbocation stability

R+ stability: 3º > 2º >> 1º > CH3+

R-X reactivity toward SN1: 3º > 2º >> 1º > CH3X

CH3+

1º R+

2º R+

3º R+

rearrangements possible

SN1 vs SN2A. Solvent effects

nonpolar: hexane, benzenemoderately polar: ether, acetone, ethyl acetate

polar protic: H2O, ROH, RCO2Hpolar aprotic: DMSO DMF acetonitrile

CH

O

N(CH3)2

CH3 C NCH3

S

O

CH3

SN1 mechanism promoted by polar protic solvents

stabilize R+, X– relative to RX

RX

R+X–

in less polar solventsin more polar solvents

A. Solvent effects

SN2 mechanism promoted by moderately polar & polar aprotic solvents

destabilize Nu–, makethem more nucleophilic

e.g., OH– in H2O: strong H-bonding to water makes OH– less reactive

OH– in DMSO: weaker solvation makes OH– more reactive (nucleophilic)

RX + OH–

ROH + X–

in DMSO

in H2O

B. Summary

RX = CH3X 1º 2º 3º

rate of SN1 increases (carbocation stability)

rate of SN2 increases (steric hindrance)

reactprimarilyby SN2

(k1 ~ 0, k2 large)

reactsprimarilyby SN1

(k2 ~ 0, k1 large)

may goby either

mechanism

SN2 promoted good nucleophile (Rate = k2[RX][Nu])-usually in polar aprotic solvent

SN1 occurs in absence of good nucleophile (Rate = k1[RX])-usually in polar protic solvent (solvolysis)

ELIMINATION REACTIONS

Dehydrohalogenation of alkyl halides

C

H

C

X

C C+ + BH +B X

strong base: KOH/ethanolCH3CH2ONa/CH3CH2OHtBuOK/tBuOH

Elimination

Follows Zaitsev orientation:

Br

+ +EtONaEtOH

61% 20% 19%

Br+EtONa

EtOH71% 29%

The E2 mechanism

• reaction is bimolecular, depends on concentrations of both RX and B–

Rate = k[RX][B–]

RLS must involve B–

• reactivity: RI > RBr > RCl > RF

RLS must also involve breaking the R—X bond

(and reaction doesn’t depend on whether RX is 1º, 2º, or 3º)

increasing R—X bond strength

elimination, bimolecular

1. Single step, concerted mechanism:

C C

X

H

C C

X

HB

C C

B H

X

B

Br

+ OH-

Zaitsev

2. stereoelectronic effects: anti elimination

spatial arrangement of electrons (orbitals)

In the E2 mechanism, H and X must be coplanar:(in order for orbitals to overlap in TS)

HC C

X

HC C

X

anti periplanar-most moleculescan adopt thisconformation more easilyE2 eliminations usually occur when H and X are anti

syn periplanar-but eclipsed!

2. stereoelectronic effects: anti elimination

CH3

Br

Br

CH3

+EtONaEtOH

""

major minor

major

but

2. stereoelectronic effects: anti elimination

CH3

Br

HH

CH3

Br

HH

but

Br must be axial to be anti to any b-H’s:

Br is anti to both H’s normal Zaitsev orientation

Br is anti only to H that givesnon-Zaitsev orientation

Recall:

Rate = k[RBr][B–] E2Reactivity: RI > RBr > RCl > RF (and no effect of 1º, 2º, 3º)

However:

Rate = k[RBr]E1 (no involvement from B–)Reactivity: RI > RBr > RCl > RF (RLS involves R–X breaking)

and: 3º > 2º > 1º (RLS invloves R+)

3. the E1 mechanism

minormajor

EtONaEtOH +

Br

minormajor

EtOH

+Br

3. the E1 mechanism

Step 1:(RLS)

Step 2:

Br

H

+ Br

EtOH

+ EtOH2

EtOH + HBr

- and R+ can rearrange eliminations usually carried out with strong base

Substitution vs Elimination

A. Unimolecular or bimolecular reaction?(SN2, E2)(SN1, E1)

Rate = k1[RX] + k2[RX][Nu or B]

• this term gets larger as [Nu or B] increases

bimolecular reaction (SN2, E2) favored by high concentration of good Nu or strong B

• this term is zero when [Nu or B] is zero

unimolecular reaction (SN1, E1) occurs in absence of good Nu or strong B

B. Bimolecular: SN2 or E2?

1. substrate structure: steric hindrance

decreases rate of SN2, has no effect on rate of E2 E2 predominates

Br

Br

Br

Br

NaOEt O

O

O

+

+

+tBuOK

91% 9%

13% 87%

100%

15% 85%

"

"

sterichindranceincreases

sterically hindered nucleophile

Rate = kSN2[RX][Nu] + kE2[RX][B]

B. Bimolecular: SN2 or E2?

2. base vs nucleophile

• stronger base favors E2• better nucleophile favors SN2

tBuOK

Br I

OCH3

OtBu

+

+

NaI

NaOCH3

100%

40% 60%

5% 95%

good Nuweak B

good Nustrong B

poor Nustrong B

C. Unimolecular: SN1 or E1?

Br H2O

(weak B,poor Nu)

H

OH2

OH2

OH

for both, Rate = k[R+][H2O]

no control over ratio of SN1 and E1

D. Summary

1. bimolecular: SN2 & E2

Favored by high concentration of good Nu or strong B

good Nu, weak B: I–, Br–, HS–, RS–, NH3, PH3 favor SN2

good Nu, strong B: HO–, RO–, H2N– SN2 & E2

poor Nu, strong B: tBuO– (sterically hindered) favors E2

Substrate:

1º RX mostly SN2 (except with tBuO–)2º RX both SN2 and E2 (but mostly E2)3º RX E2 only

b-branchinghinders SN2

2. unimolecular: SN1 & E1

Occurs in absence of good Nu or strong B

poor Nu, weak B: H2O, ROH, RCO2H

Substrate:

1º RX SN1 and E1 (only with rearrangement)2º RX3º RX

SN1 and E1 (may rearrange)

can’t controlratio of

SN1 to E1

1. Halogenation of Alkanes

R–H + X2 — R–X + HX a substitution reactionheat orlight

Reactivity: F2 > Cl2 > Br2 > I2

commontooreactive

toounreactive

(endothermic)

CH4 CH3Cl CH2Cl2 CHCl3 CCl4

+ HCl + HCl + HCl + HCl

Cl2

hnCl2

hnCl2

hnCl2

hn

Problem: mixture of productsSolution: use large excess of CH4 (and recycle it)

A. Free-radical chain mechanism

Step 1: Cl2 2Cl• (homolytic cleavage) Initiation

Step 2: Cl• + CH4 HCl + CH3•Step 3: CH3• + Cl2 CH3Cl + Cl•

net: CH4 + Cl2 CH3Cl + HCl

Sometimes: Cl• + Cl• Cl2 TerminationCH3• + CH3• CH3–CH3 (infrequent dueCH3• + Cl• CH3Cl to low [rad•])

Propagation-determines net reaction1000’s of cycles = “chain” reaction

B. Stability of free radicals: bond dissociation energies

BDECH3—H 104 kcalCH3CH2—H 98 kcalCH3CH2CH2—H 98 kcal (any 1º)(CH3)2CH—H 95 kcal (any 2º)(CH3)3C—H 91 kcal (any 3º)

easier to break bonds free radical more stable

R–H R• + H• H = BDE

CH3–CH2–CH3

CH3CH2CH2• CH3CHCH3• lower energy, more stable,

easier to form

98 kcal 95 kcalReactivity of C–H:

3º > 2º > 1º > CH3–H

C. Higher alkanes: regioselectivitySome alkanes give only one monohalo product:

CH3 CH3 CH3 CH2 ClCl2hn

Cl2hn

Cl2hn Cl

Cl

But:Cl2hn CH3CH2CH2Cl +CH3CH2CH3 CH3CHCH3

Cl

find: 43% 57%even though statistically: 75% 25%

(6 H) (2 H)

Syntheticallyuseful.

Not asuseful.

C. Higher alkanes: regioselectivity

Reactivity of C–H: 3º > 2º > 1º-for Cl2, relative reactivity is 5.2 : 3.9 : 1

Cl2hn CH3CH2CH2Cl +CH3CH2CH3 CH3CHCH3

ClPredicting relative amounts of monochloro product:

= x2º product1º product

reactivity of 2º Hreactivity of 1º H

number of 2º H’snumber of 1º H’s

= = =3.9 x 21 x 6

7.86

57%43%

C

CH3

CH3

H

C

H

H

C

CH3

CH3

H

Cl2

hvC

CH3

CH3

H

C

H

H

C

CH3

CH2

H

Cl C

CH3

CH3

H

C

H

H

C

CH3

CH3

Cl

C

CH3

CH3

H

C

H

Cl

C

CH3

CH3

H

1-chloro-2,4-dimethylpentane 2-chloro-2,4-dimethylpentane 3-chloro-2,4-dimethylpentane

# H 12 2 2reactivity factor x 1 x 5.2 x 3.9 12 10.4 7.8 sum = 12+10.4+7.8 = 30.2

percent 12/30.2 x 100 = 39.7% 10.4/30.2 = 34.4% 7.8/30.2 = 25.8

12 H, primary 2H, tertiary 2H, secondary

Cl2hn

Bromine is much more selective:

Cl2hn

Br2hn

Cl

Cl

+43% 57%

3% 97%

Relative reactivities for Br2: 3º 2º 1º1640 82 1

Syntheticallymore useful.

Br2hn