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Organic chemistry reaction mechanisms

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  • 1. ORGANIC CHEMISTRY REACTION MECHANISMS (1) Substitution Nucleophilic Bimolecular (SN 2 ) Proposed by Edward Davis Hughes and Sir Christopher Ingold (1937). The reaction between a primary halide and a nucleophile follows second order Kinetics i.e., rate depends on the concentration of alkyl halide as well as nucleophile. e.g. rate µ [alkyl halide] [nucleophile] HO + Solid wedge represents the bond coming out of the paper dashed line represent bond going down the paper and a straight line represent bond in the plane of the paper. The incoming nucleophile interact with the alkyl halide causing the C—X bond to break while forming a new C—OH bond. These two processes takes place in a single step simultaneously and no intermediate is formed. Inversion of configuration takes place during the process. Carbon atom in the transition state is simultaneously bonded to five atom, therefore unstable and cannot be isolated. Tertiary halides donot undergo SN2 mechanism due to steric hinderance.
  • 2. Since the nucleophile attacks from opposite side of halide atom, the three alkyl groups do not permit the nucleophile to attack on carbon atom, the order of reactivity followed is Primary halide > Secondary halide > Tertiary halide H H H Nu: X H Nu: H C Nu: X H H Ethyl 1° (1) H C H X H H Methyl (30) H H H H Isopropyl 2° (0.02) C H Nu: H C H X H H C H C H H H tert-butyl 3° (0) (2) Substitution Nucleophilic Unimolecular (SN 1 ) Tertiary halide proceeds via SN1 mechanism. Rate of reaction depends only on the concentration of alkyl halide rate µ [alkyl halide] The reaction takes place in two steps Step-I. The polarised C—X bond undergo slow cleavage to produce a carbocation and a halide ion. Br Step-I is slowest and reversible. Step-II. The carbocation thus formed is attacked by nucleophile to complete the substitution reaction.
  • 3. Effect of Solvent : SN1 reactions are favoured in protic solvent (a) as step-I involves the C—Br breaking for which the energy is obtained through solvation of halide ion with proton of protic solvent. (b) Polar solvent promote ionization of halide ion. Since the reaction proceeds through formation of carbocation, so greater the stability of carbocation, faster will be the rate of reaction therefore ⊕ ⊕ b g b g ⊕ ⊕ ⊕ ⊕ ⊕ Ph3 C > Ph 2 CH > PhCH2 > allyl > CH3 3 C > CH3 2 CH > CH3 CH2 > CH 3 For this reason allylic and benzylic halides show high reactivity toward SN1 mechanism. The carbocation thus formed get stabilised through resonance. ⊕ ⊕ H2 C = CH—CH 2 ¬ →  ⊕ CH2 ⊕ H2 C—CH = CH 2 CH2 CH 2 ⊕ CH2 ⊕ For alkyl group, the reactivity of halides R—X follow the same order in both mechanisms. R—I > R—Br > R—Cl > R—F Vinyl halides neither undergo SN1 nor SN2 mechanism. SN 2 mechanism is hindered by the fact that carbon atom attains a negative charge and SN1 mechanism is hindered by resonance and no ionization possible.
  • 4. SN2 SN1 1. SN2 reaction follow 2nd order SN 1 reaction follow first order kinetics. kinetics. 2. Inversion of configuration takes Retention of configuration and also place. racemisation takes place. 3. No effect of solvent. More polar sovlent more is the rate or reaction. (3) Acid Catalysed Hydration of Alkene Alkene reacts with H O in presence of mineral acid as catalyst to 2 form alkenes. In unsymmetrical alkene the reaction proceed according to Markovnikov rule. > C = C <+ H2 O H+ C —C H OH OH H C + H O 3 CH = C —CH—CH 2 H CH 3 3 Step-1. Protonation of alkene to form carbocation by electrophilic attack of hydronium ion (H 3O+ ) + HO >C = C< + H—O—H H ⊕ —C—C— + H : : H 2
  • 5. Step-2. Nucelophilic attack of water on carbocation. Step-3. Deprotonation to form alcohol. (4) Addition of Grignard Reagent on Carbonyl Compounds Step-I. Nucleophilic addition of Grignard reagent to carbonyl group to form an adduct. δ+ δ− δ+ δ ++ δ > C = O+ R− Mg X − →  L M —OM g—X O P —C M P M P NR Q − 2+ − Bond in Grignard Reagent is highly polar carbon being non-metal and magnesium metal, So Mg reactes to oxygen to form adduct. Step-II. Hydrolysis of adduct yield alcohol. OH + – – ++ – —C—O Mg X —C—OH + Mg R R X (5) Acid Catalysed Dehydration of Alcohol Alcohols undergo dehydration by heating with concentrated H 2SO4. H 2 SO4 CH 3 CH 2OH  → H2C = CH2 + H 2O 443 K
  • 6. Secondary and tertiary alcohols are dehydrated under milder conditions. OH 85% H 3PO4 CH3 —CH—CH3  → CH 3CH = CH2 + H2 O  443 K CH CH2 3 20% H 3P O 4 H3 C—C —OH   → CH 3 —C —CH 3 + H 2O  350 K CH3 The reaction proceed in three steps— Step-1. Formation of protonated alcohol. H2 SO 4  → H⊕ + HSO4  Step-2. Formation of carbocation : — It is the slowest step and rate determining step. Step-3. Elimination of Proton.
  • 7. The acid used in step-1 is released in step-3. To drive the equilibrium to the right ethene is removed fast. (6) Reaction of Ether with HI Step-1. The reaction start with protonation of ether. Step-2. Iodide is a good nucleophile. It attack the least substituted carbon atom of the oxonium ion formed in step-1 and displaces an alcohol molecule by SN 2 mechanism. Thus in the cleavage of mixed ethers with two different alkyl group, the lower alkyl group forms alkyl iodide and larger forms alcohol. ⊕ M M M N ⊕ P P P Q I − + CH 3 —O—CH2 CH3 → I … CH 3 … O…CH 2CH3 → CH3 I + CH3 CH 2 OH H H When HI is in excess and reaction is carried at high temperature alcohol reacts with another molecule of HI to form another alkyl iodide.
  • 8. Step-3. When one of the alkyl group is tertiary, the halide formed is a tertiary halide. CH CH3 3  H3 C —C—O CH 3 + HI  → CH 3 OH + CH 3 —C —I CH3 CH3 Due to formation of tertiary carbocation (stable). CH CH 3 CH3 3 ⊕ Slow   H 3 C—C —O CH 3 + HI  → H3C —C—O—CH 3  → CH3OH + CH 3 —C⊕ CH3 CH3 H CH CH 3 CH3 —C ⊕ + I CH3 CH3 →  3 CH 3 —C—I SN1 CH3
  • 9. In anisole : + H — O— C H 3 O—CH 3 + HI + I The CH3 —O bond is weaker then C H5 —O bond because the carbon 6 atom of benzene ring is sp2 hybridised and there is a partial double bond character. There attack of I break the CH3—O bond from CH3I. OH ↓ H —O—CH 3 + I →  + CH3 I (7) Addition of HCN to >C = O The reaction proceeds by attack of nucleophile. OH > C = O + HCN  →  C CN Step-1. Generation of nucleophile. HO HCN + OH : CN + 2 Step-2. Nucleophilic attack of CN – on carbonyl group.
  • 10.  CH 3COOH + C2 H5OH H 2SO 4 → CH3COOC 2 H5 + H2 O (8) Esterification Step-1. Protonation of carbonyl oxygen activate the carbonyl group towards nucleophilic addition of alcohol. Proton transfer in the tetrahedral intermediate convert the OH– group into FOH I H K − + 2 Step-2. Transfer of Proton. ⊕ Step-3. OH 2 is a better leaving group and eliminated as H2 O. Protonated ester so formed finally loses a H+ (Proton) to give ester.
  • 11. (9) Mechanism Addition of NH3 , NH2OH, NH2 NH2 , C6H5 NHNH2 or NH 2 CONHNH2 to >C = O. Step-1. Addition of ammonia derivative to >C = O Step-2. Elimination of H 2 O to form product. Where X = – H, —R, —OH, —CONHNH2 or —NHC6H5 The pH of the reaction is controlled at 3.5, in strongly acidic medium proton is captured by amino grup to form salt .. ⊕  R NH 2 + H⊕  → R NH 3 In basic medium, OH – cannot attack to electro-negative oxygen atom. δ+ δ− − > C = O + OH  → No reaction  Hence no product is formed in strongly acidic or basic medium. (10) Some Important Reactions (A) Friedel Craft Reaction Addition of alkyl (R) or aryl group (COR) to benzene nucleus in presence of Anhydrous AlCl 3 (Lewis acid).
  • 12. (a) Alkylation — CH3 C H +l C Anhy. AlCl 3 + HCl (b) Acetylation or Acylation — COCH3 Anhy. AlCl C HC + OCl 3 + HCl COCH 3 H CO3 Anhy. AlCl + + HCl O H CO3 (c) Benzoylation — CH CO 65 COCl C + H6 + HCl 5 Mechanism Step-I. Generation of electrophile, AlCl 3 is Lewis acid and generate electrophile. ⊕ CH 3Cl + AlCl 3  → AlCl 4 + C H3 
  • 13. Step-II. Formation of intermediate. Step-III. H CH3 CH3 + + AlCl 4 HCl l ++AlC 3 Characteristics : (1) More stable carbocation will form the product. e.g.
  • 14. (2) Phenol and aniline form addition product with AlCl 3 donot form carbocation. (3) Chlorobenzene and vinyl cloride do not form carbocation. (4) Addition product will be determined by ortho, meta and para directing group e.g. CH3 CH3 C H +l C AlCl 3 CH3 CH3 + + HCl CH3
  • 15. (B) Aldol Condensation Aldehyde and ketones having one or more α –H-atom when warmed with dilute base undergo self addition reaction known as aldol condensation. OH NaOH CH 3 CH = O : + H CH 2 CHO  → CH 3 —CH—CH 2 CHO Whatever be the size of aldehyde, attack comes from α –H-atom and product is β . Ketones also undergo self addition to form ketol. (C) Cross Aldol Condensation Reaction When two different aldehydes are condensed together 4 products are formed.
  • 16. OH NaOH CH 3 CH2CH = O + H CH2 —CHO  → CH 3CH2 —CH—CH 2 —CHO Mechanism : Step-I.Formation of carbanion. − − CH3 CH = O + OH  CH 2 —CH = O ← → CH2 = CH —O + H2O α–H-atom is removed by base as HO. 2 Step-II. O– CH3 CH = O + CH 2CHO  → CH 3 —CH —CH2 —CHO  Step-III. – O OH H+ OH − − CH3 —CH—CH 2 CHO   → CH 3 CH—CH 2 CHO + OH  Cross Aldol Involving Aldehyde and Ketone O OH O − OH  CH 3CH = O + H CH 2 —C—CH3   → CH3 —CH—CH 2 —C —CH3 (D) Cannizzaro Reaction Disproportionation of an aldehyde lacking α –H-atom like HCHO, CH CHO, R C—CHO to salt of an acid and a primary alcohol is known as 6 5 3 Cannizzaro Reaction. 2 HCHO + NaOH  → CH 3OH + HCOONa 
  • 17. CHO 2 CHOH 2 + NaOH COONa + CH3 CH3 CH3 –  2 CH3 —C —CHO + NaOH  → CH 3 —C—CH2OH + CH3 —C—COONa+ CH3 CH3 CH3 Mechanism : Step-I.Reversible addition of OH– to >C = O. OH C6H5 —CH = O + OH−  → C 6H 5 —C —O  H Step-II.Transfer of hydride ion (H– ) to another aldehyde. OH O OH – O Slow C 6H5 —C—O + C —C6H5   → C 6H5 —C = O + H —C —C6 H5  H H Step-III H O C 6H5 —C —O + C 6H 5 —CH 2OH
  • 18. Step-III. The acid and alkoxide ion so obtained involve in proton exchange to yield more stable product of salt and alcohol. Similarly O O– O– OH − − H—C —H + OH  → H—C—H   → H —C—H + H2 O   – OH O– O H O H—C—H + C = O  → H—C  O– H + CH3OH O– Markownikov’s Rule (MKR) “In addition of HX molecule to an unsymmetrical alkene, H-atom goes to the C-atom which has already larger no of H-atoms attached to it.” 1. The molecule to be added is known as addendum e.g. + − + − + − + − + − + − + − − + H Cl , HB r , H I, HF , HCN , H NO3 , HH SO4 , H OC l 2. The positive part of adendum goes to the carbon atom which has already larger number of H-atoms attached to it. 3. The negative part of addendum goes to the C-atom which has lesser no of H-atoms attached to it. Br  CH 3 —CH = CH 2 + H Br  → CH3 —CH CH3
  • 19. 4. When the reaction is carried in presence of some peroxide, the product is reverse of MKR, called Kharasch or peroxide effect. Benzoyl  CH 3CH = CH2 + HBr   → CH 3CH2 CH 2 Br Peroxide Mechanism—MKR proceed via carbocation formation— 1. Larger alkyl group polarise theπ -bond Br  →  δ+ δ− ⊕ Br   CH 3 CH = CH + H⊕  → CH 3 CH —CH 3  → CH 3 —CHCH 3 ⊕ 2 . H attack first and generate secondary carbocation. 3. Br– attack to carbocation to form product. Another example is Br δδ + δδ − + Br− ⊕ CH3 CH2 CH = CHCH 3 + H → CH3 CH3 CHCH2 CH3  → CH3 —CH2CHCH2  CH3 Peroxide effect proceed via free radical mechanism O O O • hv • C 6H5 —C—O—O—C—C 6H5  → 2C6H5 —C —O  → C 6 H 5 + CO2   One free radical always generate another free radical • •  C 6 H 5 + HBr  → C 6H 5 + B r • • HBr •   CH 3 CH = CH 2 + Br  → CH 3 C HCH2Br  → CH 3 CH2 CH 2Br + B r l l l