1. Physical Organic Chemistry CH-4 Nucleophilic aromatic substitution & Elimination reactions Prepared By Dr. Khalid Ahmad Shadid Islamic University in Madinah Department of Chemistry

2. Nucleophilic aromatic substitution reactions  Electrophilic substitution reaction generally occur in an aromatic compounds. Aryl halides are less reactive in Nucleophilic substitution reaction due to:  high electron density in benzene ring.  bond in C-X stronger and shorter  Aryl cation unstable therefore no SN1  There is no transition state with same plane of the ring C-Br hence no SN2

3.  Nucleophilic aromatic substitutions reaction occur in Addition-Elimination reaction.  The electron withdrawing group EWG in ortho and para position to hydrogen will stabilize carbanion ion.  No reaction without EWGs.  Chlorobenzen will never react with sodiumethoxide, but it will react with EWG like notro. Nucleophilic aromatic substitution reactions

4.  Another example; the substitution reaction of chlorine by hydroxyl. The reaction temperature decrease when number of EWG increase  If EWG in meta position, the reaction will not give a product

6. Benzyne mechanism  The aromatic halides like chlorobenzene and bromobenzene are not react with nucleophiles in normal condition, but will react while benzyne intermediate form.

7.  Benzyne intermediate occur in: 1. Dieles-Alder reaction 2. When there is no alpha hydrogen in reactant. 3. In isotops labeling

8. Elimination Reactions  Elimination reaction: to eliminate two atoms, two groups, or one atom and one group without substituted with another atom or group.  The elimination of HX molecule from alkyl derivatives. While X is a halogen or ester… etc. the hydrogen atom on adjacent carbon with X  Elimination reactions and nucleophilic substitution are similar in cases of affecting factors.  Hence it’s a competitive reaction which produce alkenes ( β- elimination)

9. Elimination Reactions  α- elimination: elimination of groups from one carbon and produce carbene

10. First step: formation of carbocation فعالية الكاربوكاتيونات (CH 3 ) 3 C- > (CH 3 ) 2 CH- > CH 3 CH 2 - > CH 3 - UNIMOLECULAR ELIMINATION REACTIONS E 1  In this reaction the substrate will determine the rate of reaction Substrate K = rate of reaction Mechanism: Second step : Lose of proton to form double bond

11.  Double bond form When a proton near a positive charge (by elimination of proton)  Due of carbocation formation in this type of reaction SN1 reaction will form also. UNIMOLECULAR ELIMINATION REACTIONS E 1

12.  Reaction of tertiary butyl bromide with alkoxide ion to form prppene: 1 st step: (rate determining step) C-X cleavage due to a good leaving halide group to form carbocation 2 nd step: (fast step) A proton elimination with a strong base to firm alkene UNIMOLECULAR ELIMINATION REACTIONS E 1

13.  E1 and SN1 are similar in reaction: happen in an ionized solvent and good leaving group.  2-chloro-2-metheylbutane to give different alkenes

14. UNIMOLECULAR ELIMINATION REACTIONS E 1 Formation more stable carbocation  Intermediate carbocation of E1 and SN1 can rearrange to more stable intermediate Example: solvolysis of neopentyl iodide to form 2-methyl-2-butane. Happen when methyl group migrate, hence carbocation intermediate converted from primary to tertiary more stable

15.  Carbocation intermediate rearranged by migration of Hydrogen UNIMOLECULAR ELIMINATION REACTIONS E 1

16. BIMOLECULAR ELIMINATION REACTIONS E 2  In this reaction the substrate and nucleophile will determine the rate of reaction  Elimination of bimolecule in the same time in one step  Happen when adjacent proton to leaving group. Base will eliminate proton and C-X cleavage by leaving group then formation of a double bond

17.  E2 and SN2 are competitive. When base is Nucleophile.  To reduce competitive and to increase E2 we use non nucleophilic base  The condition of SN2 always will form elimination. The reaction of 2-bromopropane with sodium ethoxide in ethanol. The elimination rate depend on both substrate and nucleophile. Then its second order reaction. BIMOLECULAR ELIMINATION REACTIONS E 2 Kinetic and mechanism

18. Structural effects  E2 depend on a good leaving group like halides, ammonium ions, sulphoniume. Like SN2.  E2 prefers Strong base.  SN2 prefers weak base I -, (except for nonpolar and aprotic) BIMOLECULAR ELIMINATION REACTIONS E 2

19.  For Alkyl groups: 1.C-H single bond in beta position of Leaving group. 2.E2 easily happen primary R< secondary R< tertiary R 3.E2 can react fast with tertiary alkyl not like SN2 due to steric hindrance. 4.E2 reaction is fast because there is no steric hindrance unless base molecule is big. BIMOLECULAR ELIMINATION REACTIONS E 2

20. Structural effects  H-X elimination from alkene halides or Arene halides (both strong bonds) are less reactive than alkyl halides. This can happen in a few conditions like alkene preparation.  E2 can be favored over SN2 by: 1.strong base nucleophile 2.Big nucleophile 3.Increase alkyl substitution on alpha carbon 4.Increase temperature High temperature without solvent

21. Stereochemistry of E2 Reactions  E2 is stereoselective

22. The Competition between Elimination and Substitution S N 2 and E2 favored over S N 1 and E1 by a strong base/Nu S N 2 is slowed by steric hindrance, but E2 is not strong base, strong Nu strong base means E2, not S N 1

23. S N 2 and E2 Stronger bases favor E2 over S N 2 stronger base weaker base

24. S N 2 and E2 higher temperatures favor elimination  G =  H - T  S SN2SN2 weaker bases less steric hindrance lower temperature E2 stronger bases more steric hindrance higher temperature

25. S N 1 and E1 favored over S N 2/E2 by absence of strong base/Nu often neutral or acidic conditions tertiary or secondary substrates in polar solvents S N 1 is usually major, but some E1 always occurs also

26. Methyl Substrates: CH 3 L S N 2 only Primary Substrates: RCH 2 L good for S N 2 with almost any nucleophile no S N 1/E1 can cause E2 with a sterically hindered strong base potassium tert-butoxide (KOt-Bu)

27. Secondary Substrates: R 2 CHL S N 2 favored with good Nu that is not too basic (especially in aprotic solvents) CH 3 CO 2 –, RCO 2 –, CN –, RS – E2 favored with strong bases HO –, RO – (NaOH, NaOEt) S N 1 favored by absence of good Nu in polar solvent often neutral or acidic conditions some E1 product is usually formed a solvolysis reaction

28. Tertiary Substrates no S N 2 (too hindered) E2 favored with strong bases HO –, RO – (NaOH, NaOEt) S N 1 favored by absence of good Nu in polar solvent often neutral or acidic conditions some E1 occurs

29. Hofmann’s rule: The major alkene product has fewer alkyl groups bonded to the carbons of the double bond (the less highly substituted product). The effect of directing in Elimination reactions Hofmann’s and Zaitsev’s Rule  unsimilar alkyls on alkyl halides like 2-chloro-2-methylbutane, can form one alkene or more. Depending on the relativity rate of beta elimination  The use of HO - or NH 2 - will form more stable alkene which contain less number of Hydrogen and more number of alkyl groups bonded to double bond carbon alkene is Zaitsev’s product.  The other product which contain more number of hydrogen is Hofmann’s product  Zaitsev’s Rule: The major alkene product is the one with more alkyl groups on the carbons of the double bond (the more highly substituted product).

30.  Change of Base in this reaction will change yields

31. Base + (CH 3 ) 2 CBrCH 2 CH 3..........> (CH 3 ) 2 C=CHCH 3 + H 2 C=C(CH 3 )CH 2 CH 3 I II EtO- 70% and 30%; Me 3 CO- 28% and 72%, Et 3 CO- 12% and 88% The effect of directing in Elimination reactions Hofmann’s and Zaitsev’s Rule  Hofmann’s and Zaitsev’s products will vary and depends on: 1.How easily of proton elimination from two adjacent beta carbons near leaving group 2.the stability of olefins produces 3.Effect of strain on replacing Leaving group 4.How base is big in elimination rxn base

32. CH 3 CH 2 CH(S+Me2)CH 3..........> CH3CH=CHCH3 (26%) + CH 3 CH 2 CH=CH 2 CH 3 CH(N+Me 3 )CH 2 CH 2 CH 3......> CH 2 =CHCH 2 CH 2 CH 3 (Major) + CH 3 CH=CHCH 2 CH 3 minor)  When substrate is a an ammonium salts, sulfur or quaternary phosphonium, will produce less substituted alkene (Hofmann)  Steric hindrance on alkyl halihes will prevent proton elimination hence the products is Zaitsev’s products which contain many substituted groups. The product with less substituted is more preferred (Hofmann)

33.  Elimination and substitution reaction with Alcohols and ethers occur only in a strong acids.  Alkenes preparation from alcohols by E1and E2 reactions will depend on alcohol, acid, solvent and temperature. Elimination Reactions with acidic catalyst

34.  Reaction Tertiary butyl alcohol in E1: 1 st step: reversible and fast addition of proton to hydroxyl to make it a good leaving group 2 nd step: C-O cleavage and H2O as a good leaving group to form carbocation. Rate determining step. 3 rd convert carbocation to alkene by eliminate proton using water

35. GOOD LUCK