Elimination Reaction Organic Chemistry Surya.pptx

In pharmaceutical field , Elimination Reaction Organic Chemistry.
E1,E2& E1CB Mechanisms, pharmaceutical Application
Zaitsev's rule describe

1. Elimination Reaction PRESENTED BY Surya Kanta Maiti M . Pharmacy Department Of Pharmaceutical Chemistry Indira Gandhi National Tribal University

2. Elimination reaction is a type of organic reaction in which two substituents are removed from a molecule in either a one or two-step mechanism. - The one-step mechanism is known as the E2 reaction , and the two-step mechanism is known as the El reaction . Elimination reactions are exact reverse of addition reactions. Elimination reactions are endothermic reactions ie ,. These reactions takes place in the presence of heat . Elimination reactions Alpha elimination reactions Beta elimination reactions ELIMINATION REACTION

3. 1,1-Elimination or α -Elimination: It is an elimination reaction in which an organic compound loses two ligands from the same atom. Chloroform Carbene Beta elimination reactions E1 E2 E1cb β -Elimination:

4. There are three fundamental events in these elimination reactions: Removal of a proton (ii) Formation of the C=C ( π ) bond (iii) Breaking of the bond to the leaving group concerted (E2) elimination carbocation (E1) elimination

5. E1 MECHANISM: It is an unimolecular reaction. It is a two step process. In E1 reaction, the rate of alkene formation depends upon the concentration of alkyl halide only. It follows first order kinetics. Rate ∝ [Alkyl halide] , R = k [RX] 3ºalkyl halides undergo E1 Mechanism. E.g., 3º butyl bromide Tertiary butyl bromide E1 Reactions is Non-stereospecific- follows Zaitsev ( Saytseff ) Rule. Does NOT occur with primary alkyl halides (leaving groups)

6. Mechanism of E1 reaction: E1 mechanism is a two step process. Step I : In this step the molecule of alkyl halide undergoes ionization to give a carbocation (carbonium ion) and halide ion. Step II : In this step, the carbocation loses a proton from the adjacent carbon to yield the alkene. In a chemical reaction the slow step is rate determining step, so first step is rate determining step of E 1  reaction.

7. EXAMPLE

8. Step 1: The bromide dissociates and forms a tertiary (3°) carbocation. This is a slow bond-breaking step, and it is also the rate-determining step for the whole reaction. Since only the bromide substrate was involved in the rate-determining step, the reaction rate law is first order. This is the reaction rate only depends on the concentration of (CH 3 ) 3 Br and has nothing to do with the concentration of the base, ethanol. Step 2: The hydrogen on β-carbon (β-carbon is the one beside the positively charged carbon) is acidic because of the adjacent positive charge. The base, EtOH, reacts with the β-H by removing it, and the C-H bond electron pair moves in to form the C-C π bond. The base ethanol in this reaction is a neutral molecule and therefore a very weak base. Since a strong base favors E2, a weak base is a good choice for E1 by discouraging it from E2. Ethanol acts as the solvent as well, so the E1 reaction is also a  solvolysis  reaction. EXPLANATION

9. Effect of R- Reactivity order: (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3- In an E1 reaction, the rate determining step is the loss of the leaving group to form the intermediate carbocation. The more stable the carbocation is, the easier it is to form, and the faster the E1 reaction. -The rate of an E1 reaction increases as the number of R groups on the carbon with the leaving group increases.

10. ( ii) Leaving Group (LG ) The only event in the rate determining step of the E1 is breaking the C- LG bond. Therefore, there is a very strong dependence on the nature of the leaving group, the better the leaving group, the faster the E1 reaction will be. In the acid catalysed reactions of alcohols, the - OH is protonated first to give an oxonium ion, providing the much better leaving group, a water molecule. Base (B) Since the base is not involved in the rate determining step, the nature of the base is unimportant in an E1 reaction. Favored by weaker bases such as H2O and ROH. Type of Solvent Favored by polar protic solvents, which can stabilize the ionic intermediates.

11. E1 Mechanism for Alcohols Step 1 : An acid/base reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base. Step 2: Cleavage of the C- O bond allows the loss of the good leaving group, a neutral water molecule, to give a carbocation intermediate. This is the rate determining step (bond breaking is endothermic) Step 3 : An acid/base reaction. Deprotonation by a base (a water molecule) from a C atom adjacent to the carbocation center leads to the creation of the C=C

12. E1 Mechanism for Alkyl Halides Step 1: Cleavage of the polarised C-X bond allows the loss of the good leaving group, a halide ion, to give a carbocation intermediate. This is the rate determining step (bond breaking is endothermic) Step 2: An acid/base reaction. Deprotonation by a base from a C atom adjacent to the carbocation center leads to the creation of the C=C

13. In certain alkyl halides, the carbocation initially formed undergoes rearrangement to form more stable carbocation and thus highly branched alkene is formed as the major product. For example, dehydrohalogenation of 3-chloro-2,2-dimethylbutane yields 2,3-dimethylbut-2-ene as major product.

14. It is a bimolecular reaction. It is a single step process. It follows second order kinetics. Rate α [ Alkyl halide] [Base] , R = k[RX] [:B] 1º alkyl halides undergo E2 Mechanism. E2 REACTION: This reaction occurs when an alkyl halide is treated with a strong base such as hydroxide ion (OH-) and forms a carbon-carbon double bond. Example:

15. Mechanism of E2 reaction: E 2  mechanism is a one step process. Base attacks the hydrogen atom of β-carbon and begins to remove the H atom and at the same time as the carbon-carbon double bond starts to form, the –X group starts to leave as shown below in transition state. After the transition state, C-H bond and C-X bond are completely broken and carbon-carbon double bond is formed.

16. Identity of R group - More substituted halides react faster Rate: R3CX > R2CHX > RCH2X Effects of R- Reactivity order: (CH3)3C- > (CH3)2CH- > CH3CH2- > CH3- - In an E2 reaction, the reaction transforms 2 sp3 C atoms into sp2 C atoms. This moves the substituents further apart decreasing any steric interactions. As the number of R groups on the carbon with the leaving group increases, the rate of the E2 reaction increases.

17. Leaving Group (LG) The C- LG bond is broken during the rate determining step, so the rate does depend on the nature of the leaving group. However, if a leaving group is too good, then an E1 reaction may result. - Rate of reaction follows the order Rate of reaction follows the order, R−I > R−Br > R−Cl > R−F Base (B) Stronger bases favor the reaction. Since the base is involved in the rate determining step, the nature of the base is very important in an E2 reaction. More reactive bases will favor an E2 reaction.

18. It is a unimolecular reaction. It is a two step process, it forms carbanion as intermediate which is conjugate base of starting material. It follows second order kinetics . Rate ∝ [substrate] [Nucleophile]. It should have a poor leaving group Beta hydrogen is highly acidic in nature. It follows Hoffmann rule. E1CB REACTION:

19. Step 1 : Formation of carabanion by deprotonation Step 2: Removal of halide E1CB MECHANISM:

20. Type E1 E2 E1cb Reaction Unimolecular Reaction Biomolecular Reaction Rate Law 1st order (rate depends on substrate) Second order (depends on substrate & base conc.) Often 2nd order; kinetics vary by variant (anion, reversible, irreversible) Step Two Step reaction Single Step reaction Two Step reaction Intermediate/ Transition State Carbocation No Intermediate Carbanion Reactivity Order 3°>2°>1° 3°>2°>1° 3°>2°>1° Regioselectivity Saytzeff Rule Saytzeef Rule and Hoffman Rule Hoffman Rule Reagent Weak Base Strong Base Strong base, Solvent Polar Protic Solvent good because stabilized ionic intermediate. Polar Aprotic Solvent best Comparison of E1, E2, and E1CB elimination mechanisms in table form:

21. TYPE E1 E2 E1CB Stereochemistry Requirements No strict requirement; lacks anti-periplanar demand Strict anti-periplanar arrangement required for elimination No strict stereochemical requirement Competes with SN1 SN2 Rare, occurs when leaving group is poor but carbanion stabilized Example Dehydration of alcohols with H₂SO₄ Dehydrohalogenation of alkyl halides with strong base (e.g. NaOEt ) β- Elimination in carbonyl compounds (e.g. nitroalkanes, malonic esters)

22. Mechanism: Hofmann elimination involves the spontaneous breakdown of a quaternary ammonium compound, producing an alkene and tertiary amine without enzymatic intervention. Pharmaceutical Significance: Atracurium—used as a neuromuscular blocker—was purposefully designed to undergo this elimination reaction in vivo. It degrades predictably at physiological pH (~7.4) and temperature, offering reliable organ-independent clearance. Metabolic Pathways: Atracurium is cleared from the body via Hofmann elimination and ester hydrolysis. This dual-mode degradation occurs in both central and peripheral compartments, allowing reproducible drug inactivation regardless of liver or kidney function. Pharmaceutical Applications of Elimination Reactions Hofmann Elimination in Drug Metabolism Hofmann Elimination: Chemodegradation of Atracurium

23. ii. E2-Type Elimination in Drug Stability: Mometasone Furoate Reaction Example: Mometasone furoate, a topical corticosteroid, undergoes base- catalyzed E2-type elimination, leading to a 9,11-epoxide (denoted D1), followed by further cyclization (D2) under physiological-like conditions. Drug Stability Impact: The formation of D1 (epoxide) and D2 alters receptor binding affinity—D1 shows 4‑fold stronger affinity to glucocorticoid receptors compared to dexamethasone. Pharma Importance: Understanding such elimination pathways helps in formulating stable dosage forms and anticipating degradation profiles that may affect drug potency, safety, or shelf-life.

24. iii. Steroid Drugs Elimination reactions are widely used in the structural modification of steroids . Introduction of double bonds at specific positions (e.g., in corticosteroids, androgens, estrogens) helps improve biological activity, potency, and selectivity . iv. Pro-drug Activation Certain drugs are designed as pro-drugs and rely on in vivo elimination to release the active therapeutic form. Example: Some anticancer agents and antiviral pro-drugs undergo β-elimination in the body to generate the pharmacologically active compound. This strategy improves drug stability, solubility, or bioavailability until activation.

25. v. Alkene Functionalization Alkenes formed by elimination are versatile synthetic intermediates . These alkenes can undergo further transformations: Hydration → alcohols Epoxidation → epoxides (important in anticancer drugs) Halogenation → alkyl halides (used in antimicrobial synthesis) Thus, elimination provides a gateway to functional groups essential in medicinal chemistry . vi. Industrial Drug Manufacturing Elimination reactions are used on a large scale in the pharma industry. Key examples: Vitamin A synthesis – requires elimination to generate conjugated double bonds. Alkaloid synthesis – elimination helps in forming reactive unsaturated intermediates.

26. "The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the beta-carbon ( β ) having the fewest hydrogen substituents” Alexander M. Zaitsev, 1875 Zaitsev's Rule alpha ( α ) carbon: the carbon attached to the leaving group beta ( β ) carbon: the carbon adjacent to the carbon attached to the leaving group In this case the blue beta carbon has the fewest hydrogen, and therefore the alkene formed from this alkyl halide in an elimination reaction would be this