ok today we're going to look at this compound, we've got this alkyl bromide
and we're going to have this alkyl bromide undergo four different
transformations. these are four different reaction conditions
we're going to want to try to see what we can get, so as a hint we're going to
want to maybe go back and look at my lectures on SN1 SN2 E1 E2, especially
about choosing which mechanism is going to occur
ok so let's take a look at these four different reaction conditions and see if
we can choose which mechanism is going to occur. on the first one here remember
that usually the first thing that we want to look at especially considering
these are all secondary substrates, so it's not too much information we're
going to get there
we're going to look at the strength of the base or nucleophile. and so in this
case we have water, water is not especially strong. water is not going to
have the capacity to directly attack as a nucleophile or anything like that
so from a weak base we can rule out SN2 and E2
so if we ruled out SN2 and E2 that means it's going to be SN1 or E1
now in order to choose which one is going to occur, a hint can usually be
the temperature, and we know that because of the Gibbs free energy equation that a
colder temperature is going to favor substitution whereas a hotter
temperature is going to favor elimination, and so because this is weak
and cold
we're going to go SN1 with this one
and so SN1, we know that the first step is going to be that the leaving group
leaves, and so we get the carbocation there and then what's going to happen is
that we are probably going to get this hydride shift and so we will get this
guy here and there is our carbocation and then water is going to be able to
come in
and after it deprotonate we will get that tertiary alcohol there. so there is
our SN1 product
now let's go ahead and look at the next one here
tert-butoxide that is our dead giveaway. E2, we know that tert-butoxide is always
going to do E2
that is because it is a very strong base so it has the capacity to push this
molecule around, its going to do something, it's going to react, it doesn't
have to wait for a carbocation so it's either going to do E2 or SN2 but
because of how bulky that tert-butyl group is there's too much steric
hindrance associated with this molecule going ahead and coordinating or doing
SN2 so we know that tert-butoxide is always going to do
E2. now furthermore it has two options in terms of which proton it's going to
grab, there's three protons over here and there's one proton over there
so there's potentially two different elimination products and so we're
comparing the Zaitsev and the Hoffman elimination products but we know that
because of the steric hindrance of that base it is going to have a much easier
time, you have a much lower activation energy associated with approaching one
of these protons at the edge of the molecule and therefore forming the
Hoffman elimination product
ok so once again that is because of the strength of the base that we know that
either E2 or SN2, the steric hindrance that we know that it is elimination or
substitution, and then once again the steric hindrance that causes us to grab
this more sterically available proton than the one in the interior of the
molecule so that's number two. let's look at the third one, now once again we've
got water
now that's going to be, again that's weak so we're ruling out SN2 and E2, we're
going to be looking at SN1 and E1
now in this case we're looking at a hot temperature
once again because eliminations are entropically favorable that entropic
favorability is going to go up as the temperature increases and so hot
temperatures are going to favor elimination, we think that this one is
going to be an E1 reaction
so once again E1
it means we're going to first go ahead and have the leaving group leave
so there's our carbocation, once again probably this hydride shift is going to
occur
and so here we have that, and then we said that E1 was going to occur once
again because of the temperature and so water is going to go ahead and
choose a proton to extract. now in this case instead of Hoffman because of how
sterically hindered tert-butoxide is, water is a small molecule so it has
access to these protons on these methyls or it has access to a proton over here
and because thermodynamics favors a more substituted alkene we are in this case
going to probably predominantly get the Zaitsev elimination product so that is
the more substituted of the possible alkenes, we could have had the double
bond there or there but instead we have it there because that is a
trisubstituted alkene as opposed to a possible disubstituted alkene
so there is our E1 product. and now lastly as we can probably guess we
haven't done SN2 yet
and as it happens the CN minus nucleophile is an excellent
SN2 nucleophile and so that one is going to go ahead and just do
direct SN2 and we know that SN2 always occurs with inversion of stereochemistry
so that has to be on the dash bond there
so there we have our four reaction conditions, weak and cold is going
to give us
SN1, this is strong and sterically hindered so we've got E2, this is weak
and hot so we've got E1, and then we know that this is a strong nucleophile
and is an excellent SN2 nucleophile
so there are the four reactions that we've got on this substrate
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