Loudon Chapter 8 & 9 Review: Substitutions and Eliminations: CL Nu CL H
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations: CL Nu CL H
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations: CL Nu CL H
Chapter 9 covers reactions you can do with alkyl halides. For the most part, these
break down into two categories: substitution and elimination. Substitution results in
replacing the halogen with some other group. Elimination results in removing the halogen,
along with a hydrogen on a neighboring atom (the H), to create a new double bond.
substitute eliminate
Cl Nu Cl
for Nu with base
H
the B H
The molecule thats having these reactions done to it is called the substrate, and the
halogen on it is called the leaving group. The other molecule, the one responsible for
changing the substrate, is called the base/nucleophile. Thats because it can play either of
these roles. Remember, a base goes after an H only, so if an elimination is happening
then its acting as a base. A nucleophile prefers to form an attachment to any other atom,
so this is the job the base/nucleophile is doing when is causes a substitution.
There are two different ways to perform each of these reactions. Substitutions are
called either SN1 or SN2, depending on the mechanism. Eliminations are called E1 or E2.
The number tells you how many molecules are involved in the rate-determining step; the
2-type reactions are bimolecular (a.k.a. concerted) while the 1-type reactions are
unimolecular (a.k.a. stepwise).
SN2: the attacking nucleophile is forming a bond to the substrate at the same time
the leaving group is detaching. The best way to picture this is as an umbrella
turning inside-out. The nucleophile must do a backside attack, by attacking the
carbon from the opposite face of where the leaving group is. This results in an
inversion of geometry- for instance, if the leaving group had a bold bond before,
then the new group will have a dashed bond. This also means that as long as the
leaving group and the incoming group have the same priority ranking in CIP rules,
the molecule will convert from R to S or vice-versa at the attacked carbon. Finally,
since the carbon is trying to hold on to five atoms at once, space is at a premium.
The fewer R groups and the more Hs there are on the central carbon, the faster
this reaction will go.
SN2 Nu Br Nu Br Nu
H
E2
Br
SN1: the leaving group falls off on its own, leaving a carbocation behind. Later,
the nucleophile comes along and sticks on. Everything we know about
carbocations applies here: more substituted is faster, stereochemical information
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations
CHEM 3311, Jacquie Richardson, Spring 2010 - Page 2
How can we predict which one of these four will actually happen?
1. Classify substrate as Me, 1o, 2o, 3o, or honorary 3o
2. Classify solvent as polar protic or polar aprotic
3. Classify base/nucleophile as strong or weak base
4. Classify base/nucleophile as good or poor nucleophile
5. Use this information to select which mechanism(s) will occur
6. Apply this mechanism to the substrate.
Classifying the substrate is relatively straightforward just group it based on how many
R groups are attached to the carbon bearing the leaving group. One exception is that even
if a molecule is primary or secondary, enough steric bulk attached to a carbon further
away can cause it to act like a tertiary carbon. This is called the neopentyl effect.
Br Br Br Br
Br
Me 1o 2o 3o "honorary 3o"
Solvents fall into one of two categories for these reactions: either polar protic or polar
aprotic. The difference is whether they can lose a proton readily under the conditions of
these reactions. Protic solvents are generally things like ROH, RCO2H, or H2O. Aprotic
solvents are generally the ones that go by acronyms: DMSO, DMF, THF, and acetone. In
general, polar protic solvents favor E1 and SN1 reactions, while polar aprotic solvents
favor E2 and SN2 reactions. Usually this preference is not strong enough to control the
choice of reaction, but be careful using a polar protic solvent with a very strong base
can often cause the base to react with the solvent first, which creates a different
base/nucleophile that will then go on to react with the molecule.
Strong bases are technically anything with a pKa greater than or close to 15.7, the pKa
of water. In general anything with a minus charge on C, N, or O is strong, unless theres
additional stabilization coming from somewhere (resonance, etc.). If its a strong base
then its assumed to be a good nucleophile, unless its very bulky.
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations
CHEM 3311, Jacquie Richardson, Spring 2010 - Page 3
Weak bases are anything with a pKa below 15.7. Weak base/good nucleophiles fall into
two categories: those at the stronger end of the weak base categories like N3- (pKa of 9.4)
and CN- (pKa of 4.7), and those with a minus charge on large atoms (I, Br or S). Weak
base/poor nucleophiles are anything outside of this category, including molecules with no
minus charge at all.
Good Nucleophiles HO-, RO- (if R isnt bulky), I-, Br-, HS-, RS-, CN-, N3-
RCC-
Poor Nucleophiles N Cl-, F-, RCO2-, H2O, ROH,
O RCO2H
(tBuO-) (LDA)
If you have a strong base: You must do SN2 or E2 only! Carbocations (from E1 or SN1)
cant exist in the presence of strong bases. E2 is the default, unless both the substrate and
the base/nucleophile are unhindered.
If you have a strong base/good nucleophile and
o substrate is Me or 1o, SN2 is favored
o substrate is 2o, SN2/E2 mix is favored
o substrate is 3o or very bulky, E2 is favored
If you have a strong base/poor nucleophile and
o substrate is Me, SN2 is favored
o substrate is 1o, 2o, or 3o, E2 is favored
If you have a weak base: E1/SN1 mix is the default, but SN2 can happen under the right
circumstances.
If you have a weak base/good nucleophile and
o substrate is Me, 1o, or 2o, SN2 is favored
o substrate is 3o or very bulky, SN1/E1 mix is favored
If you have a weak base/poor nucleophile and
o substrate is Me or 1o, SN2 is favored
o substrate is 2o, 3o, or very bulky, SN1/E1 mix is favored
A warning: none of this is set in stone. If you have a weak base thats getting close to
being a strong base, it might produce a little product from acting like a strong base, and
the rest of its product from acting like a weak base.
Here, youve got a 2o substrate, a protic solvent, and a strong base/good nucleophile. This
combination gives you a mixture of E2 and SN2 as the favored outcome. Using SN2 gives
you the first product shown, and E2 gives you the second product.
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations
CHEM 3311, Jacquie Richardson, Spring 2010 - Page 4
H2 O
Br OH
Here, the solvent is also working as the base/nucleophile. This is called solvolysis. You
have a 1o substrate, protic solvent, and weak base/poor nucleophile, giving you an SN2
outcome.
Br
EtOK Zaitsev product - usually major
Organometallics: If you take an alkyl halide and expose it to a metal, its possible to
replace the halogen with a metal atom. This doesnt go by SN1 or SN2, but a single-
electron-transfer mechanism. The organometallic thats formed has a big delta minus
charge on carbon, which makes it act like a very, very strong base. For this reason, you
have to do the reaction in an ether-type solvent like diethyl ether or THF.
Br Mg, THF MgBr Br Li, THF Li
Grignard reagent Organolithium
Loudon Chapter 8 & 9 Review: Substitutions and Eliminations
CHEM 3311, Jacquie Richardson, Spring 2010 - Page 5
These reagents can do lots of useful things which we wont get into until O Chem 2. For
now, well only look at one reaction: acid-base reactions with anything protic water,
alcohols, carboxylic acids, etc.
MgBr + H-OR H + -OR
Strong base Weaker base
Radical halogenations of alkanes: this is a way to create alkyl halides from alkanes
(unlike doing it from alkenes, which we saw in chapter 5). Here, instead of adding an H
and a Br to a double bond, were pulling an H off the molecule and replacing it with a
halogen. No radical initiator is needed the dihalogen will split into radicals on its own,
as long as you use UV light (written as h), or heat (written as ).
1) Initiation
Br Br 2 Br
2) Propagation
H Br
Br Br Br + Br
3) Termination: any step that involves 2 radicals getting together, for example:
This works best for chlorine and bromine. For iodine its too endothermic and takes
forever to go, and for fluorine its too exothermic and goes out-of-control until most Hs
are replaced. Bromine is significantly endothermic, so its slow and energy-poor. For this
reason, bromine will carefully select which H to pull off so that only the most stable
radicals are formed. The net result is that radical halogenations with Br2 only puts on a Br
at the most substituted carbons. Chlorine, on the other hand, is much less endothermic
and can afford to create radicals anywhere. Its a lot less predictable and basically useless
unless every H on the molecule is equivalent.
Br2, hv
Br Only major product
Cl
Cl2, hv Cl
Cl
Cl
H
tBuO- or
Cl C Cl Cl C Cl C C
Cl Cl Cl Cl
Cl Cl
dichlorocarbene
This creates a carbene. It looks like a carbon with a minus charge and a plus charge at the
same time, although you shouldnt draw it this way. Its a neutral carbon with both an
empty orbital (so its a great electrophile) and a lone pair (so its a great nucleophile).
The one reaction that well see it doing involves both behaviors at the same time.
Cl Cl
C Cl Cl
HCCl3
tBuONa