SN1 SN2 E1 E2 Reactions - How To Know Which One Will Occur Based On Carbon Chain
SN1 SN2 E1 E2 Reactions are typically taught one at a time so that you recognize the starting molecule, reaction sequence and products. The difficult arises when you are asked to differentiate between the unimolecular and bimolecular substitution and elimination reactions given a random set of starting molecules, chemical reagents and solvents, where the specific reaction pathway is not specified
The trick to identifying which reaction will take place is to develop a logical scheme or analysis that will help you figure out which reaction is most likely to take place. There are 4 items to consider, when available view them all, however you may be able to identify the reaction after analysis of just 1 or 2 concepts. The 4 things to analyze are as follows 1- The alkyl chain that will be attacked 2- The strength of the leaving group 3- The nature and strength of the attacking nucleophile or base 4- The polarity and protic nature (or lack of) for the specific solvent
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 '2' type reactions, meaning SN2 or E2 differ slightly. An SN2 reaction occurs via backside attack and so you're looking for a molecule that has an easily accessible leaving group. There are 2 things to consider here. 1- The nucleophile prefers to attack a methyl or primary carbon. Secondary is still accessible but tertiary carbons cannot be attacked by a nucleophile 2- The nearby carbons should be minimized so as not to block the approach of the nucleophile to the carbon that will be attacked
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
The trick to identifying which reaction will take place is to develop a logical scheme or analysis that will help you figure out which reaction is most likely to take place. There are 4 items to consider, when available view them all, however you may be able to identify the reaction after analysis of just 1 or 2 concepts. The 4 things to analyze are as follows 1- The alkyl chain that will be attacked 2- The strength of the leaving group 3- The nature and strength of the attacking nucleophile or base 4- The polarity and protic nature (or lack of) for the specific solvent
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 '2' type reactions, meaning SN2 or E2 differ slightly. An SN2 reaction occurs via backside attack and so you're looking for a molecule that has an easily accessible leaving group. There are 2 things to consider here. 1- The nucleophile prefers to attack a methyl or primary carbon. Secondary is still accessible but tertiary carbons cannot be attacked by a nucleophile 2- The nearby carbons should be minimized so as not to block the approach of the nucleophile to the carbon that will be attacked
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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