1.5 elimination reaction

TYBSc Chemistry

1. Power Point Presentation 0n Elimination Reaction T. Y. B. Sc. By Dr. G. D. Shirole Assistant Professor DEPARTMENT OF CHEMISTRY, A. S. C. COLLEGE, RAHATA

2. Elimination Reaction Defination – A reaction in which two atoms or group of atoms are removed from the reactant to form a product with high degree of unsaturation is known as elimination reaction. Generally the atoms or groups eliminated are from the neighboring atom i.e.1,2 position hence such reaction is known as 1,2 elimination reaction. Generally in 1,2 elimination the atom that is lost from C2 or C β is hydrogen (X=H) and some good leaving group from C1 or C α (Y=leaving Group). Such 1,2 elimination reaction is termed as β elimination reaction

3. Mechanism of 1,2 elimination The 1,2 elimination reactions which takes place in basic medium . Three things must be happened here- a.Breaking of C α –Y Bond : The substrate has a good leaving group Y. During the reaction C-Y Bond breaks heterolytically so that the leaving group loss with bonding pair of electrons. b.Breaking of C β –H Bond : The base used abstracts the β -hydrogen as a proton. During the reaction C β –H bond also breaks heterolytically so that the bonding pair of electron is retained by the β - carbon. c.Formation of the Л bond between C α -C β : The pair of electron retained by the β - carbon is used to form a Л bond bet C α -C β .

4. Three Possible Mechanisms 1) The E2 Mechanism: (Elimination Bimolecular) The breaking of C α –Y bond, formation of the Л bond between C α -C β and breaking of C β –H bond takes place simultaneously. Rate α [Substrate] [Base] This mechanism has some similarities with SN2 mechanism- It is bimolecular. All bond formation and bond breakings are concerted. Proceed through single T.S. It is one step process.

5. 2) The E1 Mechanism: (Elimination Unimolecular ) The breaking of C α –Y bond taks place in the slow step to form a carbonium ion, which is then followed by fast breaking of C β –H bond and formation of C α -C β Л bond . Rate α [Substrate] This mechanism has some similarities with SN1 mechanism- It is Unimolecular . Formation of carbonium ion occurs in the slow step. It is two step process.

6. 3) The E1cB Mechanism: (Elimination Proceeding through Conjugate base) The breaking of C β –H bond taks place first to form a carbanion or conjugate base of the substrate, followed by simultenious formation of C α -C β Л bond and breaking of C α –Y bond. Rate α [Substrate] [Base] E2 and E1 mechanisms are very commonly observed but E1cB mechanism is very rarely observed

7. The E2 Mechanism Consider the following 1,2 elimination – Kinetics :- Rate α [Substrate] [Base] Rate =K 2 [2-bromopropane] [ NaOMe ] Mechanism:- It is one step mechanism. The leaving group i.e. halogen atom lossed with bonding e pair and β –H lossed without bonding e pair, to form the double bond. All these operations occur simultaneously, in a single step via single T.S.

8. Transition State :- In the T.S. two bonds are broken (C-H and C-Br), the energy required for this bond breaking comes from bond making process i.e. formation of bond between proton and the base (B-H) and formation of a Л bond. As the base begins to pull β –H from the substrate, the β carbon with electron pair begins to form the Л bond. As the Л bond starts to form, the carbon halogen bond starts to break. The halogen atom (L.G.) lossed with bonding pair of electron Evidence for E2 Mechanism 1) KINETIC ISOTOPIC EFFECT:- If the rate of reaction is changed due to replacement of some atom in the molecule by its isotope, it is known as Kinetic Isotopic Effect or Primary isotopic effect.

9. Consider the rates of the following two reactions- The two reactions were carried out under identical conditions. When the rate constants were determined it was observed that K H /K D was close to 7. It means that comp. (A) reacts seven times faster than comp. (B) which contains D in place of H. This clearly indicates that in E2 elimination, breaking of C β –H bond must occur in the slow step . If breaking of C β –H bond were not in the slow step, both compound would react at same rate i.e. there will be absence of Kinetic Isotopic Effect

10. 2) THE ELEMENT EFFECT :- Nature of Leaving group As the leaving group also departs in the rate determining step of E2, changing the leaving group should change the E2 rate. With extremely good leaving groups, the E2 rates are expected to be very high. Consider the relative E2 rates in the following reactions:- R-CH 2 -CH 2 -X R-CH=CH 2 The relative rates are determined by changing the leaving group only Substrate Relative E2 rate R-CH 2 -CH 2 -F 1 R-CH 2 -CH 2 -Cl 70 R-CH 2 -CH 2 -Br 4 X 10 3 R-CH 2 -CH 2 -I 2.5 X 10 4 As leaving group ability is in the order I - > Br - > Cl - > F - . The relative rates of E2 elimination are also in the same order. This proves that the C-X bond breaks in the rate determining step. If this bond had broken in fast step then all above substrate reacted at same rate.

11. 3) ABSENCE OF HYDROGEN EXCHANGE :- The E2 & E1cB mechanism are both follow second order Kinetics . So Kinetically they are indistinguishable. The reaction follows E2 mech. and not E1cB, Hydrogen exchange expt. is performed. Ex. 2-phenyl ethyl bromide is subjected to E2 with NaOEt in EtOD as solvent. Reaction is allowed to proceed until half of the reactant is consumed and then stopped. Suppose that the reaction follow E1cB mech.-

12. If the reaction had really followed E1cB Mechanism, the unreacted substrate should contain deuterium. In actual experiment it was found that the unreacted substrate did not contain deuterium i.e . there was no hydrogen exchange. Therefore this reaction is not following E1cB. All the reactions which follow E2 mechanism, fail to show hydrogen exchange.

13. ORIENTATION & REACTIVITY IN E2 If the substrate contains more than one type of β - hydrogens , more than one product will be formed . In such cases one of the alkene is formed as major product and the other as minor.

14. THE SAYTZEFF RULE In order to decide which of the alkene will be obtained as major product, Alecxander Saytzeff praposed a rule in 1875 known as Saytzeff rule. It states that- “In an elimination reaction the more substituted alkene is obtained as the major product”.

15. JUSTIFICATION OF SAYTZEFF RULE Why the more substituted alkene obtained as major product ? Two major factors contribute to this- a) Stability of the Alkene :- As the No. of Alkyl Substituent Stability of the Alkene The stability of the alkene is measured In terms of the heat liberated when the double bond undergoes addition of hydrogen i.e heat of hydrogenation. Lower the heat of hydrogenation higher is the stability of alkene General stability order- Tetra-substituted > Tri-substituted > 1,1Di-substituted > 1,2Di-substituted > Mono-substituted > Ethylene With increase in the order of substitution – No. of ‘R’ group attached to the olefinic carbon +I effect No. of hyperconjugative structures also Stability of the Alkene

16. b) Lower energy of activation :- Consider the E2 transition state- As the Number of Alkyl Substituents attached to the olefinic carbon atom increase, the energy of activation required to form T.S. decreases & hence more substituted alkene is formed at faster rate. Thus the more substituted alkene obtained as major product because it is more stable and formed at faster rate due to lower energy of activation

17. MODIFIED SAYTZEFF RULE Modified Statement- “In an elimination reaction the more stable alkene is obtained as the major product”. The stability of alkene could also be due to Resonance stabilisation . e.g. In product (A) the double bond is in conjugation with the benzene ring and hence resonance will stabilize the product. Such stabilisation is not possible in (B) because of non-conjugation

18. Hofmann Elimination In case of some E2 elimination reactions instead of formation of more substituted alkene as major product, we get formation of less substituted alkene as the major product. Such a elimination reaction is k n as Hofmann elimination. In all the E2 elimination reactions which leads to the formation of more than one product- If more substituted alkene is obtained as major product it is termed as Saytzeff elimination and If less substituted alkene is obtained as major product it is termed as Hofmann elimination. Whether a given elimination reaction will follow the Saytzeff rule or Hofmann rule depands upon- 1) the nature of leaving group and 2) the nature of base used.

19. The Effect of Leaving Group(X) Consider the following elimination reaction- Effect of leaving group on elimination- Leaving Group (X) % of Saytzeff product % of Hofmann product i ) –Br 80 20 ii) -OTs 60 40 iii) –S + (CH 3 ) 3 25 75 iv) –N + (CH 3 ) 3 05 95 These results indicates that as the size of leaving group increases, % of Hofmann product increases. With smaller L.G. Saytzeff product always predominates and with highly bulky L.G. Hofmann product predominates.

20. The Effect of Attacking Base Effect of Size of Attacking Base on elimination- Base % of Saytzeff product % of Hofmann product i ) C 2 H 5 O - 70 30 ii) (CH 3 ) 3 C- O - 28 72 iii) (CH 3 ) 2 (C 2 H 5 )C- O - 22 78 iv) (C 2 H 5 ) 3 C-O - 20 80 As the size of Attacking Base increases , the T.S. for the Saytzeff elimination becomes more crowded than the Hoffmann Elimination and hence less substituted alkene is obtained as major product i.e. Hofmann product predominates.

21. The E1 Mechanism Consider the following elimination reaction- Kinetics :- Rate α [Substrate] Rate = K 1 [2-bromo-2-methyl-butane] Mechanism :- The First step involves hetrolysis of C α -Br bond to form carbocation which is well stabilized by +I effect of alkyl groups. This step is slow step and contains only one molecule and hence it is unimolecular reaction. The Second Step is a fast step and involves abstraction of β - hydrogen by base to form an alkene.

22. The carbocation formed in the first step may undergo following reactions in the second step- Combine with a nucleophile ( NaOEt ) to yield Sub. product by S N 1. Rearrange to a more stable carbocation (if possible). c. Eliminate a β -hydrogen to form an alkene by E1 reaction .

23. Evidence of E1Mechanism 1) No Kinetic Isotopic Effect :- The isotope effect is not expected in E1, as the C β -H is lost in the fast step and not in the slow step. When following expts . carried out, kinetic isotope effect was not observed . Experimentally it is shown that K H =K D This clearly indicates that , breaking of C-H/C-D bond does not takes place in the slow step or rate determining step.

24. 2) Structural effect :- The first step in E1 is same as SN1 i.e. formation of carbocation. Therefore the order of reactivity of alkyl halides must be the same as in SN1 i.e. Reactivity in E1 is – 3 o >2 o >1 o The rate of E1 as well as SN1 decreases as the stability of carbocation decreases in the order- 3 o >2 o >1 o

25. 3) Rearrangement :- If the structure of the substrate permits, the first order elimination (E1) are accompanied by rearrangement. The primary carbocation rearranges to the 2 o or 3 o whereas the 2 o carbocation will rearrange to 3 o carbocation Example-

26. Conclusion :- Formation of rearranged product (C) clearly indicates that the reaction did not follow E2 but proceed through formation of carbocation i.e. E1 route. In addition to above elimination products A & B we also get a rearranged elimination product by hydride shift as-

27. 4) Absence of β -H :- If the substrate does not contain β -H atom, E2 elimination is not possible at all. Formation of alkene takes place through carbocation i.e. by E1 route only . Example- Neopentyl bromide does not react under E2 conditions as it has no β -H. However, under E1 condition it does undergo elimination as shown Neopentyl bromide

28. ORIENTATION The elimination by E1 always show strong Saytzeff orientation, i.e. when more than one alkene can be formed, the highly branched or more stable alkene is the preferred product. The reactivity of the substrate in E1 depends upon slow step & orientation depends upon which β -H lost faster from the carbocation . Factors affecting E2 & E1 a)Structure of the substrate- As we proceed along the series 1 o ,2 o ,3 o for substrate, the reactivity by both E1 & E2 increases although for different reasons.- i ) Reactivity by E2 increases because of the greater stability of the highly branched alkene being formed. ii) Reactivity by E1 increases because of the greater stability of carbocation being formed in the rate determining step. b)Nature of Base- We have seen that rate of E2 reaction depends upon concentration as well as nature of base. In general E2 is favored by strong and higher conc. of base. The rate of E1 is independent of concentration as well as nature of base.

29. A) SN2 vs E2

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