GOC-III(Introductory Reaction Mechanism) Introductory Mechanisms in Organic Reactions A reaction mechanism is a step by step narration of the bond-breaking and bond-making processes that occur when reagents react to form the products. The proof of the mechanism, that the reaction is really going via this route, not via the other route, has been established by many experimental corroborations. Let us study the mechanisms of several organic reactions. Substitution Reactions: Substitution reactions are of three types (a) Free Radical Substitution (SF) (b) Nucleophilic Substituion(SN) (c) Electrophilic Substituion(SE) Free Radical Substitution(SF) Usually alkanes or alkenes undergo free radical substitution at the sp3 hybrid carbon, where a stable carbon free radical is formed. For introductory purpose we take halogenation of alkane in presence of heat and light for explaining SF mechanism. light/heat R H + X2 R X + HX A H atom in alkane is substituted by a halogen atom (X) when alkane reacts with halogen in presence of uv/visible light or heat or both. This reaction proceeds with the participation of free radicals(not any ionic species) and hence its name SF. A free radical mechanism proceeds via the following three steps (i) Initiation (ii) Propagation (iii) Termination Alkane reacts with halogen like Cl2 and Br2(written as X2) in the vapour phase in presence of light or heat to produce haloalkane(alkyl halide) This reaction goes via the above three steps. Let us discuss one by one. Initiation: light/heat XX 2 X The reagent X2 undergoes homolytic cleavage of the bond in presence of light or heat or both to produce two halogen free radicals(better call them atoms). These are primary free radicals which will take part in the propagation step. Note that X–X(halogen-halogen) bond is weaker than C–C and C–H bonds and hence light/heat preferentially breaks X–X bond. Propagation: R H ++ X R HX (i) alkyl radical + R X X R X + X (ii) alkyl halide The halogen atom abstracts a H atom from alkane(R–H) to produce alkyl free readical(its a carbon free radical) and HX. If there are more than one non-eqivalent –H atoms in a molecule, more than one alkyl free radicals will be formed with the order 30 > 20 > 10 (according to stability of alkyl free radical) This radical again reacts with a diatomic halogen molecule(X2) to abstract a X atoms and produce the product R–X(alkyl halide) and a halogen atom. This halogen atom attacks another neutral alkyl halide and repeat the cyclic process again and again till the reactions stops. In each propagation step, a radical(or Dr. S. S. Tripathy 1 GOC-III(Introductory Reaction Mechanism) atom) is consumed but another radical(or atom) is produced and thus the chain propagation continues reapeating the above two sequential cyclic process. These two steps together belong to the propagation. Note that in step (i) of propagation, the halogen atom does not abstract a alkyl radical to produce the product R–X and a hydrogen atom. This is an unfavourable process and its ΔG0 is hugely positve. In stead the step that happens shown above invoves lowering of free energy. R H + X R X + H (does not take place) Other minor termination processes also occurs, which explain the formation of dihalo and polyhaloalkanes alongwith monohaloalkane(major prouduct) when one mole of halogen reacts with one mole of alkane. A specific case of chloriantion of methane is given below. H CH2 Cl + Cl CH2 Cl + HCl CH2 Cl + Cl Cl CH2Cl2 + Cl We can write two more chain propagating pairs to show the formation of CHCl3 and CCl4. Termination: When propagation is complete, then the chain reaction is broken or terminated. Free radicals and atoms are not stable. There will be some residual radicals and atoms of every kind at the end of propagation steps. All these radicals present in the will undergo mutual combination with each other and several byeproducts are formed during. R + R RR R + X RX X + X XX The combination of two alkyl radicals give a higher alkane(R–R) having double the number of carbon atoms. Sometimes disproportionation reaction occurs if the carbon free radical has at least two carbon atoms to form an alkene and alkane. See this example. H CH CH + CH CH CH2 CH2 + CH3 CH3 2 2 3 2 ethane ethane So ethane, on halogenation gives ethane and ethene, alongwith butane(C2H5–C2H5), ethyl halide(Et–X) and X2 as the small byeproducts formed in the termination steps. Is it not interesting ??!! Unless we know the mechanism of this reaction, can we rationalise the formation of these unwanted byeproducts while we halogenate an alkane ? Exampe Br Br light/heat 2 CH3 CH2 CH3 ++ Br2 CH3 CH CH3 CH3 CH2 CH2 + 2HBr major minor Propane reacts with bromine in presence of heat and light to produce 2-bromopropane as the major product(>97%) and 1-bromopropane as the minor product(<3%). Let us see how it happens,? 2 Dr. S. S. Tripathy GOC-III(Introductory Reaction Mechanism) h Initiation: Br Br 2 Br Propagation: H CH3 CH CH3 + Br CH3 CH CH3 + HBr 20 stable free radical Br CH CH CH Br Br 3 3 + CH3 CH CH3 + Br In this case, there are two types of non-equivalent H atoms : one in the primary carbon atom(two terminal H atoms) and in the other seconndary carbon atom(middle carbon). The bromine atom formed in the initiation step will preferntially abstract a H atom from sec- position to produce a more stable 20 free radical. You already know from GOC-I, that the stability order of alkyl free radicals follows the order 30 > 20 > 10 which is explained by both hyperconjugation and +I effects. Hence isopropyl free radical is formed as major carbon free radical(>97%) while n-propyl radical is formed as minor(less than 3%). H CH3 CH2 CH2 ++ Br CH3 CH2 CH2 HBr n-propyl radical(minor) (unstable) Termination: CH3 CH + CH CH3 CH3 CH CH CH3 CH CH3 CH3 3 CH3 CH3 CH Br CH3 CH Br CH3 CH3 Br + Br Br2 H CH2 CH + CH CH3 CH2 CH CH3 + CH3 CH2 CH3 propene propane CH3 CH3 (disproportionation) So four byeproducts are formed in termination, out of which the two namely isopropyl bromide is already formed as the major product and Br2 was taken as reactant. But most interesting byeproducts are 2,3- dimethylbutane formed by mutual combination of isopropyl radicals and a mixture of propane and propene formed by the disproportionation reaction. N.B: Here we have ignored the primary free radical i.e n-propyl free radical that would form in less than 3% extent. That willl form, besides n-propyl bromide as the minor haloalkane, insignificant amount of the byeproducts from termination steps. We have not included those equations here. I suggest you to write those in your rough copy for the sake of practice. The byeproducts will be n-hexane, propane, propene, the latter two are already there. There can be also combination between isopropyl and n-propyl radicals to form 2-methylpentane. Got it? N.B: More on halogenation of alkanes, their reactivity vs. selectivities will be disucssed in the chapter ‘alkanes’. This is introduction to SF reaction mechanism. NO RADICAL REARRAGEMENT:: Unlike carbocations, which are prone to rearrangement from a less stable to more one, radicals do not undergo rearrangement of such type. This has been experimentally Dr.proved. S. S. Tripathy 3 GOC-III(Introductory Reaction Mechanism) N:B: Free radical substitution of alkane is an example of CHAIN REACTION, a reaction which involves a series of steps(initiation and propagation), each of which generates a reactive species(in this case free radicals) that brings about the next step. In chain reaction, the quantum yield(QY) is very high. One single photon of light(uv/visible) brings about the formation of several thousand product molecules. If QY for a reaction is 10,000, then 10,000 product molecules are formed from merely one photon of light. One photon dissociates one halogen molecule(X2) and each halogen atom(X) starts a chain consisting of 5000 repetitions of propagating cycles before it stops. Graph: The above graph shows variation of potential energy in the two propagation steps in chlorination of methane. The net energy evolved is 104.6 kJ/mole(–104.6 kJ/mole). The first step of converting methane and chlorine atom to methyl radical and HCl is endothermic absorbing 8.4 kJ/mole but second step is highly exothermic evolving 112.9 kJ/mole(–112.9 kJ/mole). Inititation step is endothermic(BDE) of 241.3 kJ/mole. Since the QY is very high, for the endothermic dissociation of one mole mole of Cl2, several thousands of methane and other chlorine molecules react to form several thousands of product molecules. Hence its a hugely exothermic process in totallity. Nucleophilic Substititution(SN) Reaction: Since nucloephile is involved in bringing about the substitution, it is called SN reaction. Usually alkyl halides( R–X) and other compound bearing good leaving groups like X–(halide ions) undergo substitution by this mechanim. Principle: A strong base acting as a nucleophile can displace a weak base acting as leaving group. See the forllowing examples. CH3 Cl + OH CH3 OH + Cl stronger base weaker base (nucleophile) (leaving group – – Since OH is stronger base than Cl (as HCl is a stronger acid than H2O: see BL theory given elsewhere), the above SN reaction is feasible. A stronger base OH– acting as nucelophile can displace a weaker base Cl– actiing as leaving group.