Chapter 6:Alkyl Halides
Nucleophilic Substitution and
Elimination
Nomenclature (6-2)
Alkyl Halides are also known as
Haloalkanes
Ú Molecules containing: F, Cl, Br, I
Ú Nomenclature
R
RCH2X
1o halide
RCHX
2o halide
R
RCX
R
3o halide
Dihalides:
X X
X
R
X
R
geminal
C C
vicinal
247
Structure of Alkyl Halides (6-4)
Ú Because X is more electronegative than C,
the C-X bond is strongly polarized
248
Synthesis of Alkyl Halides (6-6)
We have already seen that alkyl
halides are available via the
corresponding alkanes by radical
substitution. The reaction normally
favours the alkyl halides formed from
the most stable radical, however the
yields are usually low.
249
Ú However, free radical haloganation of alkenes is a
high yield process since it takes place via a very
stable allylic radical (assuming that allylic
hydrogen are present). The allylic position is the
carbon directly attached to the alkene.
250
Ú The mechanism of the reaction follows
the same 3 steps (initiation, propagation
and termination) that we have seen
before, but the high yield in these
reactions is due to the very stable allylic
radical produced in the process.
251
Ú Large quantities of Br2 must be avoided
as this reagent will also add to the double
bonds of alkene (chapter 8). A good way
to make large amounts of allylic bromide
is by using NBS (N-bromosuccinimide),
as this reagent keeps the concentration of
Br2 low.
252
Practice Question
Ú Complete the following reactions.
253
Ú There are many other ways to prepare
alkyl halides from other functional groups.
These will be covered later in other
chapters:
254
Reactions of Alkyl Halides (6-7)
Alkyl halides can give 2 major types of reactions
which often compete with each other
– Nucleophilic Substitutions (SN2 or SN1)
and Eliminations (E2 and E1)
255
Ú Because of this competition that is
always present, it is often difficult to
identify the major product in their
reactions.
Ú Analysis of reaction conditions is very
important to decide on the product
formed
256
Second Order Nucleophilic Substitutions (SN2) (6-8)
General Mechanism: SN2
257
Ú In SN2 the mechanism of this reaction is
“concerted” (i.e. bonds are formed and
broken at the same time). In other words,
no reactive intermediates are formed.
258
Ú This concerted mechanism is the reason for
the name, because the nucleophile and the
halide are both involved in the rate
determining step of the reaction. Another
name often used is: bimolecular nucleophilic
substitution (SN2).
259
– Nucleophile attacks the back side of the
halide
– Angle of attack of the nucleophile is 180o
from the C-X bond
– A transition state exists where partial
C-Nu and C-X bonds exist
– The halide departs (this is called the
leaving group).
– Reactivity of halides follow the trend:
CH3X > 1o > 2o (3o give no reaction)
260
Examples of SN2 Reactions (6-9)
261
Ú Factors Affecting SN2: Strength of the
Nucleophile (6-10)
– Increasing the negative charge on an
element, increases the nucleophilicity
– The more nucleophilic is the nucleophile,
the better the rate of SN2
262
Ú Factors affecting nucleophilicity
- A species with a negative charge is a stronger
nucleophile than one without a charge
HO- > H2O HS- > H2S H2N- > NH3
- Nucleophilicity decreases from left to right in
the periodic table (because the atoms are getting
smaller…therefore electrons are more tightly
held)
HO- > F:NH3 > H2O: (Me)3P: > (Me)3S:
- Nucleophilicity increases down the periodic
table (also based on the radius of atoms)
I- > Br- > Cl- > FHSe- > HS- > HOR3P: > R3N:
263
Ú Steric effect of the nucleophile
– The more sterically hindered is the
nucleophile, the poorer it is as a
reagent….slower rate. This effect is called:
steric hindrance.
264
Ú Solvent effect
solvent can also affect the nucleophilicity of
a nucleophile.
– Polar protic (with acidic proton in the structure,
OH, NH) H-bond with the nucleophile and
reduce its reactivity. Slower reactions.
265
Ú Polar aprotic solvents (no OH or NH)
cannot H-bond and keep the
nucleophile more reactive. These are
better solvents for SN2 reactions.
266
Ú Reactivity of the Substrate (6-11)
- Leaving group
The leaving group serves two purposes in a SN2
reaction:
(1) polarizes the C-X bond (makes C δ+)
(2) leaves the molecules taking 2 electrons
originally shared with the carbon electrophile.
A good leaving group has 3 characteristics:
(1) electron withdrawing (polarize C-X bond)
(2) stable (weak base) once it has left
(3) polarizable to stabilize the transition state.
The better the leaving group, the better will be
the rate of SN2 reactions
267
Ú The ability of a group to leave a molecule is a
measure of its ease of displacement.
Ú Since the leaving group takes along the electron
pair, its ability to leave can be correlated with its
capacity to accommodate a negative charge.
Ú Weak bases are good leaving groups
268
Ú Leaving group ability is inversely related to the base
strength.
Ú To recognize the weak bases, compare them to the
conjugate acids, since the weaker the base, the
stronger the conjugate acid…pka table in appendix
5 (pages 1261-1262)
269
Ú One way to make the previous reaction (in red)
work, is by changing the nature of the leaving
group chemically. Increasing the stability and
decreasing the basicity of the leaving group allows
methanol to be transformed into bromomethane.
Br-
H3C OH
H3C OH
H Br
H3C Br + OH-
H3C O+ H
H3C Br + H2O
H
Br
by protonating the hydroxyl group,
the leaving group is now stable and
a weak base (H2O) and the reaction
works.
270
Steric Factors on the Substrate
– Halides with less steric factors at the
reacting carbon will react faster in a
SN2 reaction. Primary halides (no
steric hindrance) will react this way
99% of the time.
CH3X > 1° > 2° >> 3°
– The same effect is observed at the
carbon adjacent to the C-X bond
271
Since the nucleophile must approach from a 180o angle
of the C-X bond, more substitution of the reacting
carbon slows down the reaction until no reaction is
observed for tertiary alkyl groups.
272
Table 6-5
273
Stereochemistry in SN2 Reaction (6-12)
Since the nucleophile attacks the alkyl
halide from the opposite side of the C-X
bond, for chiral molecule, there will be a
change in configuration
274
Practice Questions
Ú For each of the following pairs of SN2 reactions, indicate which
reaction occurs faster:
275
Ú What is the product of the reaction of ethyl
bromide with each of the following
nucleophiles?
276
Ú First-Order Nucleophilic Substitution: SN1
Reaction (6-13)
– Reaction is also known as Unimolecular
Nucleophilic Substitution (SN1) since the rate
determining step is the halide dissociation (this
first step is spontaneous). This results in the
formation of a carbocation (by solvolysis)
277
Ú The second step of the reaction is fast since
the electrophile (carbocation) is charged
and only a small amount of Ea is required.
278
Ú Tertiary alkyl halides normally react via
this mechanism, while primary halides
never do. The order is the opposite to the
SN2 reaction and follows the order of
stability of carbocations.
3° > 2° > 1° >> CH3X
279
Ú Similar to the SN2 reaction, many factors
influence the SN1 process.
Ú Leaving group
The easiest it is for the leaving group to depart,
the better is the rate of SN1 reaction. This is
because breaking the C-X bond is the ratedetermining step. Therefore, a better leaving
group will have more tendency to depart than a
poor one, increasing the rate of the reaction.
Trend is the same as for SN2: I- > Br- > Cl-
280
Strength of Nucleophile
Does not affect the rate of SN1 reaction,
since the nucleophile is not involved in the
rate-determining step below.
281
Solvent Effect
Polar protic (with HO group) solvents
accelerate the SN1 reaction (via the
solvation of the resulting intermediate
ions)
282
Stereochemistry of the SN1 Reaction (6-14)
Chiral carbon will be racemized (formation of
equal amounts of both enantiomers) in SN1
reactions. This is simply due to the geometry of the
carbocation. Since it is now hybridized sp2,
nucleophilic attack can take place from two
different directions resulting in retention or
inversion of configuration.
283
Rearrangement in SN1 Reactions (6-15)
Ú Since a carbocation is formed, you should be
careful to consider carbocation rearrangement.
– Whenever a carbocation is formed, it can
rearrange to a more stable species by hydrogen
or alkyl shifts. Smaller species shift more
easily.
Ease of hydride or
alkyl shift
H > CH3 > CH3CH2 > Phenyl
better
worst
– This process is always a possibility whenever a
carbocation is formed as a reactive
intermediate in any reactions.
284
Ú Driving force: more stable carbocation is
produced
285
Ú With primary halide, since carbocation
cannot be formed, a concerted
rearrangement may take place as seen
below in the methyl shift leading to a
tertiary carbocation.
286
Practice Question
Ú Arrange the following alkyl halides in order of decreasing
reactivity (most to least reactive) in SN1 reaction.
2-bromopentane, 2-chloropentane,
1-chloropentane, 3-bromo-3-methylpentane
287
Ú Which of the following alkyl halides form a substitution
product in a SN1 reaction that is different from the
substitution product formed in a SN2 reaction? Draw the
products.
288
Comparison of SN1 and SN2 Reactions (6-16)
SN2
SN1
CH3X > 1º > 2º
3º > 2º
Strong nucleophile
Weak nucleophile (may
also be solvent)
Polar aprotic solvent
Polar protic solvent
Rate = k[alkyl halide][Nuc] Rate = k[alkyl halide]
Inversion at chiral carbon
Racemization
No rearrangements
Rearranged products
289
Practice Questions
Ú Which reaction in each of the following pairs will take
place more rapidly?
H2O
CH3OH + HBr
CH3Br
+
-
CH3N H3 + Br
NH3
CH3Br
OH-
H2O
CH3OH + Br-
CH3OH + HBr
290
Ú Identify the most likely mechanism for the
following reactions, SN2 or SN1, and draw the
major organic product of each reaction.
I
EtOH
CH3COOH
Cl
Br
EtO-
291
First-Order Elimination E1 Reaction (6-17)
Ú Another type of reaction always compete with
the substitution reaction, the elimination. In
this case an alkene is produced. In the case of
the E1 reaction it is in competition with the SN1
process. In this case the nucleophile acts as a
base.
292
Ú Remember that the rate determining step is the dissociation
of the halide, so as long as a nucleophile is also basic, both E1
and SN1 will take place…a mixture of products is usually
obtained in those reactions. However, in the presence of
weak bases, SN1 usually predominates, while small amount of
a strong base will give more E1 product.
CH3
H3C
Cl
CH3
CH3
H3C C+
CH3
major
CH3
CH3
H3C
OH + H2C C
CH3
CH3
H2O
80%
20%
H2O (1% NaOH)
CH3
H3C
OH
CH3
40%
+ H2C C
CH3
CH3
60%
major
293
Ú Since carbocations are formed,
rearrangement is once again a possibility
leading to an isomeric product.
294
Zaitsev’s Rule (6-18)
Ú In elimination reactions, the most
substituted alkene usually predominates
(this is due to thermodynamics…more
substituted alkenes are more stable)
295
Ú Practice Question
Four alkenes are formed from the E1 reaction of 3bromo-2,3-dimethylpentane. Give the structures of
the alkenes, and rank them according to the
amounts that would be formed (most to least).
296
Second-Order Elimination: E2 Reaction (6-19)
Also known as bimolecular elimination,
the E2 reaction is always a competing
reaction when attempting to carry out
SN2 reactions. This is especially true
when the SN2 reaction site is sterically
hindered.
297
Ú Similar to the SN2 reaction, the
mechanism of the E2 elimination is
concerted (bonds break and form at
the same time).
H
Br
B-
a base deprotonate the carbon, the
electrons of the C-H bond are used to
form a new C-C
C C bond between the and
carbons while the leaving group departs
Ú Since the β carbon is the site of attack of
the base, E2 eliminations are often called a
β-elimination or 1,2-elimination.
298
Ú Since this reaction is in competition with SN2,
the reactivity of the alkyl halides towards the
E2 reaction follow the same trend as for the
SN2 reaction.
Ú The E2 reaction is also regioselective, i.e. when
more than one product can be formed, the
most substituted alkene is the major product
of the reaction, following Zaitsev’s rule
carbons
CH3OBr
20%
80%
299
Practice Question
Ú What would be the major product obtained from the
reaction of each of the following alkyl halides with
hydroxide ion in an E2 reaction?
300
Ú Which of the alkyl halides in each pair is more
reactive in an E2 reaction?
Br
Br
or
Br
Cl
or
Cl
or
Cl
301
Competition between E2 and E1 (9.4)
Ú Competition occurs only for 2o and 3o
halides since primary halides cannot react
via E1 because the carbocation that would
form is too unstable.
Ú For 2o and 3o halides, the same factors that
influenced SN1 or SN2 also affect E1/E2
competition.
– [strong base] + polar aprotic solvent = E2
– weak base + protic solvent = E1
302
Stereochemistry of E2 Reactions (9.5)
Ú The E2 reactions is very
specific as to the
reactive conformation.
An Anti-coplanar
arrangement of the β
proton and the leaving
group is preferred
rather than the syncoplanar due to
repulsion of between
the leaving group and
the base.
303
Ú For example:
304
Comparison of E1 and E2 Eliminations (6-21)
E1
* better with 3o halide
* base strenght in not
i
important
t t (usually
(
ll weak
k
base
* good ionizing solvent (protic
* Zaitsev product formed
* Rearrangement products
E2
* better with 3o halide
* strong base required
* polarity
l it off solvent
l
t nott
important
* Zaitsev products
*anticoplanar arrangement
necessary
* no rearrangements
305
Competition between Substitution and
Elimination
Since eliminations and substitutions always compete
with one another, it is sometimes difficult to predict
what product will be obtained. Here are a few
guidelines.
306
Practice Questions
Which reacts faster in an SN2 reaction? Why?
Which reacts faster in an SN1 reaction? Why?
Which reacts faster in an E1 reaction? Why?
Cl
Br
or
307
Study Problems
Ú 6-46
Show how each of the following compounds might be
synthesized by a SN2 displacement of an alkyl halide.
308
Ú 6-53
List the following carbocations in
decreasing order (most to least) of their
stability.
309
Ú 6-60
Predict the major organic products of the following
E1 eliminations.
Br
Br
310
Ú When 1-bromoethylcyclohexene undergoes solvolysis
in ethanol, three major products are formed. Give
mechanisms to account for these three products
311