Download RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS

RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS
SYNTHETIC DRAWBACKS OF Sn1 REACTIONS
In terms of synthetic value, any reactions whose mechanism involves carbocation formation suffer from
some drawbacks. Once formed, carbocations can undergo several process that may result in formation of
undesired side products. In the context of Sn1 reactions, some of the things that carbocations can do are:
1) They can go on to form the expected Sn1 products.
2) They can rearrange to form products whose connectivities are changed relative to that of the
original substrate.
3) They can undergo elimination (E!) reactions to form alkenes.
Examples:
CH3OH
+
∆
+
OCH3
Br
Sn1
product
CH3
OCH3
Rearramged
Sn1 product
CH3
Br
CH3OH
+
H3C
OCH3
∆
+
various alkenes
CH3
OCH3
+
CH3
+
etc.
This limits the synthetic usefulness of such reactions, for one has to deal with mixtures of products and the
separation of the desired ones.
In fact, Sn1 and E1 reactions typically go hand in hand and are difficult to disassociate, because they share
similar characteristics, and the conditions that favor one also favor the other. We’ve already learned the
characteristics of Sn1 reactions and the factors that favor them. We can extend that to E1 reactions as well:
Characteristics of the Sn1 and E1 mechanisms:
a) They are multistep processes
b) They occur with formation of carbocation intermediates in the rate determining step
c) They follow first order (unimolecular) kinetics. That is, rate=k[substrate]
Sn1 and E1 mechanisms are favored by using:
a) Sterically hindered substrates
b) Weak (neutral), small nucleophiles and heating
c) Moderate to high polarity solvents
The Sn1 mechanism leads to substitution products, and the E1 mechanism leads to formation of alkenes.
The same substrates that are prone to undergo Sn1 reactions also undergo E1 reactions. They are of two
major types:
a) Secondary and tertiary alky halides
b) Secondary and tertiary alcohols.
ALKYL HALIDES AS Sn1 and E1 SUBSTRATES
As mentioned before, conditions that favor Sn1 also favor E1 reactions. The first and rate-determining step
in the process is departure of the leaving group to form a carbocation. Let’s look at one of the examples from
the previous page, the reaction between 2-bromo-3-methylbutane and methanol. As has been mentioned
before, commonly used solvents in Sn1 reactions are water and alcohols. They frequently also double as
nucleophiles. In E1 reactions these same substances would act as bases to capture a proton in the elimination
step.
CH3OH
Br
+
∆
Br
Once formed, the carbocation can:
a) Go on to form the Sn1 product. Remember that when the nucleophile is a neutral protic solvent, its
conjugate base replaces the leaving group. This implies that a proton (H+) gets released in the process.
CH3OH
+
∆
H
OCH3
Sn1 product
The bromide ion released in the first step and the proton released in the second step can then get together
to form HBr, which is an inorganic product of the reaction. However, we will not focus on inorganic products.
As a matter of fact, inorganic products are frequently left out when writing organic reactions and mechanisms
to avoid clutter and keep the focus on the organic products.
b) Rearrange to a more stable cation. We already mentioned that this can happen by an alkyl or a hydride
shift, depending on which process yields a more stable cation. This cation can in turn form another Sn1
product. This Sn1 product is a structural isomer of the one formed in (a) because its connectivity has
changed.
H shift
CH3OH
Sn1 product
OCH3
c) Eliminate a proton to form an alkene. Remember that when there is a positive charge on carbon, the
neighboring protons become highly acidic. The solvent –acting as a weak base– can capture one of these
protons and cause an electron shift towards the positive charge that results in formation of a new pi bond,
or alkene product. When the protons surrounding the positive charge are nonequivalent, several alkenes
are possible, as illustrated below.
CH3
CH3
H3C
The protons shown in red are now acidic due to
the presence of the positive charge on carbon
H
Secondary cation before rearrangement
1
CH3
2
H
C H
H3C
H
H
H
O
CH3
CH3
The solvent can now
act as a base
H
C
H3C
H
H
alkene product
CH3
CH3
C H
H3C
H 2
H3C
H
H3C
H
CH3
alkene product
1
O
H
The rearranged cation can undergo similar elimination processes to yield two possible alkenes, but one of
them is the same as one of the alkenes formed above, with the pi bond located between carbons 2 and 3.
As an exercise, see if you can identify the protons which are acidic, how they are eliminated, and how
electron movement takes place to form the alkene structures shown below.
E1
(same as formed above)
E1
( same as
)
Thus we have the net reaction we introduced on page 1, showing the formation of Sn1 and E1 products:
CH3OH
+
∆
+
OCH3
Br
Sn1
product
etc.
+
OCH3
Rearramged
Sn1 product
various alkenes
Can you now explain (mechanistically) how the various products form in the second example given on p.1?
CH3
Br
CH3OH
CH3
OCH3 +
∆
H3C
CH3
CH3
OCH3
+
+
CH2
+
ALCOHOLS AS Sn1 and E1 SUBSTRATES
The classic textbook example of E1 elimination reactions is the acid-catalyzed alcohol dehydration. Strong
acid catalysis is needed to protonate the hydroxyl group of alcohols and turn it into a good leaving group.
For alcohols which are soluble in water, an aqueous solution of strong acid is usually used. Such solution
contains a high concentration of hydronium ion (the conjugate acid of water), which acts as the proton
donor. This first protonation step is an acid-base reaction, and as such it takes place very rapidly. The
protonation of 4-methylcyclohexanol illustrates this step.
OH2
OH
H3O+
CH3
+ H2O
CH3
Once protonated, the hydroxyl group can leave as water, leaving behind a positively charged carbon. This
is the rate determining step in the sequence.
OH2
+ H2O
CH3
CH3
The elimination step takes place after formation of the carbocation, with water acting as a base.
H
H3O+
+
+ H2O
CH3
CH3
Notice that each step in this mechanism is reversible. This means that the acid-catalyzed alcohol dehydration
is an equilibrium process. As such, equilibrium must be manipulated to shift the outcome towards the desired
product. This reaction enables us to make alkenes from alcohols, or alcohols from alkenes. Typically, formation
of alkenes is favored by use of concentrated acid, whereas formation of alcohols is favored by use of dilute
acid. The following sequence shows the steps in reverse, starting with an alcohol, and arriving at an alkene.
1
2
+
H
H
H
+ H2O
O
CH3
CH3
H
alkene protonation
OH2
+ H2O
CH3
H
O
H
CH3
1
OH
2
+ H3O+
+ H2O
CH3
CH3
OH
H3O+
NET REACTION:
CH3
CH3
ELIMINATIONS INVOLVING ASYMMETRICAL SUBSTRATES
In some eliminations the products include several possible alkenes. The next question is then, which ones
form preferentially? Is there a rule for predicting which alkene will predominate? Let’s look at the following
examples given before, but this time let’s focus only on the alkene products.
CH3OH
+
∆
+
Br
Saytzeff’s rule enables us to make a prediction. According to this rule, in elimination reactions, the most
highly substituted alkene usually predominates. For details, refer to the Wade textbook, section 6-19 (5th
ed.) or 6-18 (6th ed.).
The most highly substituted alkene is the one with the most alkyl groups directly attached to the carboncarbon double bond. This of course does not include hydrogens attached to the carbon-carbon double bond.
In the above example, we have mono, di, and trisubstituted products. According to Saytzeff’s rule, the
trisubstituted product will predominate because it is the most stable.
H
H
monosubstituted
alkene
H
disubstituted
alkene
H
H
trisubstituted
alkene
H
Why is the trisubstituted alkene more stable than the other two? The relative stabilities of alkenes is measured
by their heat of hydrogenation (much like the relative stabilities of cycloalkanes is measured by their heats
of combustion, see ch. 3). The relationship is inverse: the higher the heat of hydrogenation, the lower the
stability, because the higher the heat of hydrogenation, the higher the potential energy of the alkene.
For a full discussion of this trend with tables, refer to the Wade textbook, section 7-7 (A-D) in both the 5th
and the 6th eds. Please note that in the case of cis and trans isomers, the higher stability of the trans isomer
is due to a steric effect. The alkyl substituents in the cis isomer are closer to each other than in the trans
isomer, leading to increased steric crowding.
H3C
CH3
CH3
cis
H3C
trans
Let’s look at the last example. In this case we have di and trisubstituted products. Once again, the trisubstituted
product will be predominant because it is the most stable.
CH3
Br
CH3OH
∆
H
CH3
H
H
C
CH3
CH3
H
CH2
+
+
CH3
disubstituted
H
trisubstituted
(most stable)