Differentiating between substitution and elimination reactions may well be one of the most difficult topics that you will cover in your organic chemistry course. Part of the reason has to do with the way these reactions are taught. You learn the SN1 reaction, then SN2 then E1 and finally E2. Or perhaps you learn these out of order. And you gain a false confidence of knowing what to do for each one when presented by itself. Suddenly you are faced with a reaction that asks you to provide the products without specifying the reaction type, and this is where you get stuck.
There are 4 aspects that you want to consider when differentiating between substitution and elimination reactions, as well as between unimolecular and bimolecular reactions. These are as follows: Alkyl halide or carbon chain holding the leaving group; ability of the leaving group to break away from the molecule and remain stable in solution, strength of the attacking nuleophile or base, and finally the solvent where the reaction takes place. In this article I will help you understand the nature of the carbon holding the leaving group
The alkyl chain is analyzed for SN1 SN2 E1 E2 as follows. First determine if a stable carbocation can form allowing the unimolecular SN1 or E1 reaction to take place. The trend for carbocation stability is as follows: Tertiary carbons form very stable carbocations, secondary carbons form ok carbocations, primary and methyl carbons will not form a stable carbocation and therefore cannot undergo an SN1 or E1 reaction
The SN2 and E2 reactions cannot be lumped together the way we did with the unimolecular reactions. This is because the mode of attack is very different for substitution and elimination. An SN2 reaction occurs via a backside attack, meaning the carbon holding the leaving group must be accessible to such an attack. Thus an SN2 reaction will require an easy to access leaving group such as that bound to a methyl or primary carbon. Secondary can also take place but tertiary is too hindered
An E2 reaction is slightly different. Since the base attacks the nearby beta-hydrogen atom rather than the carbon holding the leaving group, substitution of this carbon is irrelevant. Instead we're looking for a beta-hydrogen that is easy to access, while at the same time will provide with the most substituted and thus stable pi bond. This means an E2 reaction can take place for tertiary, secondary and primary carbons, but it cannot take place on a methyl given that there are no beta carbons present
There are 4 aspects that you want to consider when differentiating between substitution and elimination reactions, as well as between unimolecular and bimolecular reactions. These are as follows: Alkyl halide or carbon chain holding the leaving group; ability of the leaving group to break away from the molecule and remain stable in solution, strength of the attacking nuleophile or base, and finally the solvent where the reaction takes place. In this article I will help you understand the nature of the carbon holding the leaving group
The alkyl chain is analyzed for SN1 SN2 E1 E2 as follows. First determine if a stable carbocation can form allowing the unimolecular SN1 or E1 reaction to take place. The trend for carbocation stability is as follows: Tertiary carbons form very stable carbocations, secondary carbons form ok carbocations, primary and methyl carbons will not form a stable carbocation and therefore cannot undergo an SN1 or E1 reaction
The SN2 and E2 reactions cannot be lumped together the way we did with the unimolecular reactions. This is because the mode of attack is very different for substitution and elimination. An SN2 reaction occurs via a backside attack, meaning the carbon holding the leaving group must be accessible to such an attack. Thus an SN2 reaction will require an easy to access leaving group such as that bound to a methyl or primary carbon. Secondary can also take place but tertiary is too hindered
An E2 reaction is slightly different. Since the base attacks the nearby beta-hydrogen atom rather than the carbon holding the leaving group, substitution of this carbon is irrelevant. Instead we're looking for a beta-hydrogen that is easy to access, while at the same time will provide with the most substituted and thus stable pi bond. This means an E2 reaction can take place for tertiary, secondary and primary carbons, but it cannot take place on a methyl given that there are no beta carbons present
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Now that we know how to analyze the alkyl chain, let's shift our focus to analysis of Leaving Groups and analysis of the different types of solvents including Polar Protic, Polar Aprotic and non-polar solvents
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