Substitution vs Elimination Reactions:
Which one and why?

What favors an $S_{N}2$ reaction vs. an $E2$ reaction?

Table 41.2: Similarities between $S_{N}2$ and $E2$ reactions.

	$S_{N}2$ and $E2$
# of steps	1
Reaction rate order	2
Limiting step	Transition step formation in an unhindered substrate to allow for attack by a strong nucleophile.

Table 41.3: Differences between $S_{N}2$ and $E2$ reactions.

	$S_{N}2$	$E2$
Site of nucleophilic attack	On the carbon bearing the leaving group	On the $\beta$ hydrogen
Substrate	1$^{o}$ halide	2$^{o}$ or 3$^{o}$ halide
Solvent	Weakly basic, highly polarizable	High conc. of a strong, polarizable base that is sterically hindered
Temperature	Lower temperatures	Higher temperatures

What favors an $S_{N}1$ reaction vs. an $E1$ reaction?

The distinctions here are less obvious and less easily controlled. For this reason, if an elimination product is desired, force the $E2$ elimination using a strong base.

Table 41.4: Similarities between $S_{N}1$ and $E1$ reactions.

	$S_{N}1$ and $E1$
# of steps	2
Reaction rate order	1
Limiting step	Carbocation formation in the substrate
Substrates	Ones that can form stable carbocations, i.e. 3$^{o}$ halide
Nucleophile	Weak bases
Solvent	Polar, aprotic solvents

Table 41.5: Differences between $S_{N}1$ and $E1$ reactions.

	$S_{N}1$	$E1$
Temperature*	Lower temperatures	Higher temperatures

* Elimination reactions have higher free energies of activation because more bonds are broken, therefore, at higher temps, elimination ($E1$ and $E2$) will be favored over ($S_{N}1$ and $S_{N}2$).

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