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Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik

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Presentation on theme: "Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik"— Presentation transcript:

1 Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik
Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2

1. Leaving groups weaker base = better leaving group reactivity: R-I > R-Br > R-Cl >> R-F best L.G. most reactive worst L.G. least reactive precipitate drives rxn (Le Châtelier)

3 2. Mechanisms SN general: Rate = k1[RX] + k2[RX][Y–] k1 increases RX = CH3X 1º 2º 3º k2 increases k1 ~ 0 Rate = k2[RX][Y–] (bimolecular) SN2 k2 ~ 0 Rate = k1[RX] (unimolecular) SN1

4 SN2 Mechanism A. 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: [HO---CH3---I]– CH3I + OH– CH3OH + I–

5 B. Stereochemistry: inversion of configuration
Stereospecific reaction: Reaction proceeds with inversion of configuration. (R)-(–)-2-bromooctane (S)-(+)-2-octanol C. Mechanism Back-side attack: inversion of configuration

6 D. Steric effects e.g., R–Br + I–  R–I + Br– 1. branching at the a carbon ( X–C–C–C.... ) a b g minimal steric hindrance Compound Rel. Rate methyl CH3Br 150 1º RX CH3CH2Br 1 2º RX (CH3)2CHBr 0.008 3º RX (CH3)3CBr ~0 increasing steric hindrance Reactivity toward SN2: CH3X > 1º RX > 2º RX >> 3º RX maximum steric hindrance react readily by SN2 (k2 large) more difficult does not react by SN2 (k2 ~ 0)

7 E. Nucleophiles and nucleophilicity
anions 2. neutral species hydrolysis alcoholysis Summary: very good Nu: I–, HS–, RS–, H2N– good Nu: Br–, HO–, RO–, CN–, N3– fair Nu: NH3, Cl–, F–, RCO2– poor Nu: H2O, ROH very poor Nu: RCO2H

8 SN1 Mechanism A. Kinetics e.g., 3º, no SN2
Find: Rate = k[(CH3)3CBr] unimolecular  RLS depends only on (CH3)3CBr

9 A. Kinetics

10 A. Kinetics Two-step mechanism: R+ RBr + CH3OH ROCH3 + HBr

11 B. Stereochemistry: stereorandom

12 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

13 SN1 vs SN2 A. Solvent effects nonpolar: hexane, benzene
moderately polar: ether, acetone, ethyl acetate polar protic: H2O, ROH, RCO2H polar aprotic: DMSO DMF acetonitrile SN1 mechanism promoted by polar protic solvents stabilize R+, X– relative to RX in less polar solvents in more polar solvents R+X– RX

14 them more nucleophilic
A. Solvent effects SN2 mechanism promoted by moderately polar & polar aprotic solvents destabilize Nu–, make them 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) in DMSO in H2O RX + OH– ROH + X–

15 B. Summary rate of SN1 increases (carbocation stability) RX = CH3X 1º 2º 3º rate of SN2 increases (steric hindrance) react primarily by SN2 (k1 ~ 0, k2 large) may go by either mechanism reacts primarily by SN1 (k2 ~ 0, k1 large) 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)

Dehydrohalogenation of alkyl halides Elimination strong base: KOH/ethanol CH3CH2ONa/CH3CH2OH tBuOK/tBuOH Follows Zaitsev orientation:

17 The E2 mechanism elimination, bimolecular 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

18 1. Single step, concerted mechanism:

19 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) anti periplanar -most molecules can adopt this conformation more easily E2 eliminations usually occur when H and X are anti syn periplanar -but eclipsed!

20 2. stereoelectronic effects: anti elimination

21 2. stereoelectronic effects: anti elimination
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 gives non-Zaitsev orientation

22 3. the E1 mechanism Recall: Rate = k[RBr][B–] E2 Reactivity: 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+)

23 3. the E1 mechanism - and R+ can rearrange  eliminations usually carried out with strong base

24 Substitution vs Elimination
A. Unimolecular or bimolecular reaction? (SN1, E1) (SN2, E2) 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

25 B. Bimolecular: SN2 or E2? Rate = kSN2[RX][Nu] + kE2[RX][B] 1. substrate structure: steric hindrance decreases rate of SN2, has no effect on rate of E2  E2 predominates steric hindrance increases sterically hindered nucleophile

26 B. Bimolecular: SN2 or E2? 2. base vs nucleophile stronger base favors E2 better nucleophile favors SN2 good Nu weak B good Nu strong B poor Nu strong B

27 C. Unimolecular: SN1 or E1? for both, Rate = k[R+][H2O]
 no control over ratio of SN1 and E1

28 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-branching hinders SN2

29 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 RX 3º RX can’t control ratio of SN1 to E1 SN1 and E1 (may rearrange)

30 1. Halogenation of Alkanes
heat or light R–H + X2 — R–X + HX a substitution reaction Reactivity: F2 > Cl2 > Br2 > I2 common too reactive too unreactive (endothermic) Cl2 hn Cl2 hn Cl2 hn Cl2 hn CH4  CH3Cl  CH2Cl2  CHCl3  CCl4 + HCl + HCl + HCl + HCl Problem: mixture of products Solution: use large excess of CH4 (and recycle it)

31 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 Termination CH3• + CH3•  CH3–CH3 (infrequent due CH3• + Cl•  CH3Cl to low [rad•]) Propagation -determines net reaction 1000’s of cycles = “chain” reaction

32 B. Stability of free radicals: bond dissociation energies
R–H  R• + H• DH = BDE BDE CH3—H 104 kcal CH3CH2—H 98 kcal CH3CH2CH2—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 CH3CH2CH2• CH3CHCH3 lower energy, more stable, easier to form 98 kcal 95 kcal Reactivity of C–H: 3º > 2º > 1º > CH3–H CH3–CH2–CH3

33 C. Higher alkanes: regioselectivity
Some alkanes give only one monohalo product: Synthetically useful. Not as useful. But: find: 43% 57% even though statistically: 75% 25% (6 H) (2 H)

34 C. Higher alkanes: regioselectivity
Reactivity of C–H: 3º > 2º > 1º -for Cl2, relative reactivity is 5.2 : 3.9 : 1 Predicting relative amounts of monochloro product: 2º product 1º product reactivity of 2º H reactivity of 1º H number of 2º H’s number of 1º H’s = x 3.9 x 2 1 x 6 7.8 6 57% 43% = = =

35 12 H, primary 2H, tertiary 2H, secondary
reactivity factor x x x 3.9 sum = = 30.2 percent /30.2 x 100 = 39.7% 10.4/30.2 = 34.4% /30.2 = 25.8

36 Bromine is much more selective:
Synthetically more useful. Relative reactivities for Br2: 3º 2º 1º

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