Why does the reaction with #("CH"_3)_3"Br"# proceed with an #"S"_N1# and/or #"E"1# mechanism? What does it have to do with steric hindrance?
It can be considered as basically spatial crowding.
Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).
As an example, consider the reaction seen below, which one might hope is
Here, the nucleophile is cyanide (
We should predict that reaction does not work via an
(LEFT: steric hindrance; RIGHT: reduced steric hindrance)
That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).
Hence, this reaction proceeds more easily as a first-order mechanism (e.g.
#"S"_N1# or#E1# ), since the#"Br"^(-)# has the time (during the slow step) to come off on its own.The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:
#r(t) ~~ k_1["Br"^(-)]# (Depending on the choice of solvent, either
#"S"_N1# or#E1# could occur.)
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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