Download Chapter 11: Reactions of Alkyl Halides There are two basic types of

Chapter 11: Reactions of Alkyl Halides
There are two basic types of reactions that alkyl halides undergo:
1. Substitutions:
A-B + C-D → A-C + B-D
Ex.
CH3I + NaOH → CH3OH + NaI
Or:
OH
Br
+ PBr3
+ HO-PBr2
Or:
H
CH3
1. NaNH2
2. CH3I
+ NH3 + NaI
2. Eliminations: X → Y + Z
Ex.
Br H
B-
+ B-H + Br-
or:
OH
+ H2O
SUBSTITUTIONS:
Alkyl halides are AWESOME at doing substitution reactions – they have this attractive
carbon group (electron-poor, due to electronegative or polarizable halide attached, thus
attractive for nucleophiles) and they have the halide, which easily stabilizes an anion so
it can readily FALL OFF from the carbon group, allowing the nucleophile to bond to the
carbon:
δ+
R
electrophile attracts nucleophile
δ−
X
"leaving group" - group
that gets displaced
General Equation: Displacement of halide by Nucleophile
R-X + Nuc⊖ → Nuc-R + X⊖
Ex.
+ OH
Br
OH
+ Br
There are two general types of substitution mechanisms, that are dependent upon what
type of alkyl halide is undergoing reaction, what type of nucleophile is involved, what
reaction conditions are present.
1. SN2 (Substitution, Nucleophilic, Bimolecular)
Bimolecular Rate Equation: Rate of reaction is dependent upon TWO species in the ratedetermining step, including [RX] and [Nuc]. Double the concentration of one and the rate
doubles. Halve the concentration of one and the rate is halved.
Rate = k [RX][Nuc]
Concentration of both species affects the number of times collisions can occur. The more
times they can collide, the more times they can react.
For the reaction to occur, two things have to fall into place:
• Need enough energy during collision to overcome e- repulsions
• Need correct orientation for collision for bond formation/breakage to occur
correctly
Let’s look at the mechanism (that detailed description of how and where the e- are moving
during a reaction). There are two possible modes of attack by a nucleophile for the SN2
reaction. Let’s look at each…
a. Front-side attack:
H3CH2C
H
CH3
(R)
Br
OH
H3CH2C
H
CH3
(R)
OH
+ Br
"retention" of configuration
b. Back-side attack:
H3CH2C
H
CH3
(R)
Br
OH
HO
CH3
CH2CH3
(S)
H
+ Br
"inversion" of configuration
Which one is favored? OR more appropriately: which one ISN’T favored? Which one is
bad for both steric and electronic reasons?
Backside attack (inversion) is favored. Frontside attack (retention of configuration) is
never seen. Why?
a. Steric interference between incoming Nuc and out-going leaving group
H3CH2C
CH3
(R)
H
OH
Br
b. Electrostatic repulsions between negatively charged (e- rich) Nuc and departing leaving
group (X-).
H3CH2C
H
CH3
(R)
OH
δ−
Br
The SN2 type reaction is a one-step, concerted process (make new bonds, break old bonds
simultaneously). There aren’t any intermediates that form, although we can surmise what
the transition state of the reaction must look like:
H3CH2C
H
CH3
(R)
!"
Br
OH
H3CH2C CH3
HO
Br
HO
CH3
CH2CH3
(S)
H
H
+ Br
Note that the transition state involves both the alkyl halide and the nucleophile – THUS
– bimolecular reaction!
Remember: Transition states are not anything that can be isolated. They are a theoretical
summary of what’s probably occurring, based on all information presented before and
after the reaction.
Factors that affect the SN2 Reaction:
Remember – whatever stabilizes the transition state will lower the activation barrier and
cause the rate of the reaction to increase. (i.e. smaller mountain, faster to climb over!)
1. Steric Congestion of Alkyl Halide
- The more congested and hindered the alkyl halide, the slower the reaction.
Remember – ONLY alkyl halides with leaving groups on sp3 hybridized carbons can do this
reaction
Fastest
CH3-X > RCH2-X
Slower
Never
> R2CH-X >>>>>>>>>>>>>>> R3C-X
As you can probably surmised, tertiary halides DO NOT do the SN2 type reactions. The
Transition State is too crowded:
R
R'
Nuc
X
R"
Now, consider these two alkyl halides. Both are primary but they are not the same.
Determine which one will do an SN2 faster:
or
Br
Br
The second one is the slower of the two. With the extra branch attached at the β-carbon,
the second one has more sterics in the vicinity of the reacting carbon center.
2. Leaving Group – Almost always expelled with a full negative charge in an SN2 reaction.
The best leaving groups will be those that can best stabilize the anion (i.e. the weakest
bases).
Halides:
I⊖, Br⊖ > Cl⊖ >>> F⊖ (HO⊖, R2N⊖)
(best)
(worst)
Two other GREAT leaving groups:
Mesylates (methane sulfonate groups): (⊖OMs)
O
O
S
CH3
O
Tosylates (toluene sulfonate groups): (⊖OTs)
O
O
S
CH3
O
The anions for these sulfonates are resonance-stabilized:
O
S
O
O
O
O
R
S
O
O
R
O
S
R
O
and thus EXTREMELY stable anions and VERY weak bases – even better leaving groups
than I⊖.
So – the Updated List for Leaving Groups:
MsO⊖, TsO⊖ >
I⊖ , Br⊖
>
Cl⊖ >>> F⊖ (HO⊖, R2N⊖)
(best)
(worst)
The formation of these excellent leaving groups from alcohols can be discussed in more
detail in Ch. 17 in CHEM 332.
3. Solvent – affects rate by affecting reagent’s energy levels
2 Types of Solvents:
a. Protic (ROH, H2O) – Solvation occurs where the solvent molecules’ acidic proton will Hbond to the nucleophile, thus surrounding it (“solvating it”), shielding the nucleophile and
preventing it from being able to react.
R O
δ+
δ+
H
δ+
O H
R
O R
H
Nuc
δ+
H O
δ+H
O
R
R
b. Polar, Aprotic – Solvation also occurs, but this time the polar solvent molecules align
themselves so the negative end of the solvent molecule faces and therefore surrounds
(“solvates”) the corresponding metal cation initially associated with the nucleophile (i.e.
HO- typically comes as KOH or NaOH).
Once the counter ion is solvated, the nucleophile is “exposed” to react, no longer
associated with a cation, no longer paired in an ionic salt.
Examples of Polar, aprotic solvents: CH3CN (acetonitrile), DMSO (dimethylsulfoxide,
(CH3)2SO), DMF (N, N-dimethylformamide, (CH3)2NCHO).
H3C
H3C
C
N
δ-
C
N
δ
H3C
C
-
δ-
N
δ-
N
C
CH3
Nuc
δ-
N
M
δ
N
C
-
C
CH3
CH3
4. Nucleophile – “Nucleophilicity” is the measure of a compound’s ability to cause a
substitution reaction to occur (to be a good nucleophile)
Some things to keep in mind:
a. Nucleophilicity parallels basicity typically –
Ex.
HO
versus
HOH
CH3O
versus CH3OH
b. Nucleophilicity is reduced by the presence of EWG (electron-withdrawing groups) via
the inductive effect. If e- density is removed from the attacking atom, its nucleophilicity
is reduced.
Ex.
Cl
Cl
H
C
versus
H
Cl
C
H
Electron density pulled away from central, attacking atom, when electronegative chlorines
are attached! While still technically negatively charged (Formal Charge = -1), the electron
cloud has migrated away from the carbon atom to the chlorine atoms:
Cl
C
Cl
Cl
c. TRENDING: Nucleophilicity increases for atoms in the same family (column) on the
periodic table as they become larger (as you move DOWN a column on the periodic table).
Larger atoms will be more polarizable.
Ex.
F⊖ < Cl⊖ < Br⊖ < I⊖
d. Resonance stability of a nucleophile results in lower nucleophilicity. Negative charge
of nucleophile is delocalized, spread over two or more atoms, reducing its strength.
Ex.
O
OCH2CH3
versus
O
C
CH3
Anion averages over BOTH oxygen atoms, splitting up the strength of the anion:
O
O
C
O
CH3
O
C
CH3
1/2
1/2
O
O
C
Examples of SN2 reactions: Draw the products
CH3
H
H
OH
Cl
H
Br
I
I
Br
Cl
OCH3
Compare the following – Which reaction occurs faster via the SN2 reaction and Why?
a.
Br
b.
I
vs
Br
I
OTs
Cl
CH3
CH3
vs
c.
Br
OH
Br
vs
OCH2CH3
[Side Note, for your reading pleasure: The inversion of the stereochemistry in an SN2
reaction is rather like an umbrella flipping inside out in a bad storm. This inversion process
is called the Walden Inversion (Paul Walden, 1863-1957), who first noticed the inversion
of (+) malic acid to form (-) malic acid, and back again, using PCl5 and Silver (I) oxide in
water:
O
1. PCl5
OH
OH
HO
O
(+) Malic Acid
2 Ag2O, H2O
O
OH
OH
HO
O
(-) Malic Acid
He originally thought he’d made the same compound by using a leaving group and
then substitution but after checking the optical rotation, he realized he’d made the
enantiomer –inverting the chiral center in the process.
Emil Fischer (of the Fischer Projection Fischers) said it was “the most remarkable
observation made in the field of optical activity since the fundamental observations of
Pasteur.” Wondrous things, most assuredly…
A Walden cycle is considered to be any series of reactions that converts one
enantiomer into its mirror image. A series of studies took off in the 1920s and 1930s,
with one of the first being the conversion of (+)-1-phenyl-2-propanol to (-)-1-phenyl-2propanol. Originally it was thought that the same compound had been synthesized by using
a leaving group followed by the substitution reaction, but after checking the optical
rotation, it was discovered that the enantiomer had been synthesized. The chiral center
had been inverted the chiral center in the process:
An example of a Walden Cycle is shown below:
TsCl, pyridine
OTs
OH
(+)-1-phenyl-2-propanol
[α]D = +33.0
NaOAc
NaOH
OAc
OH
(-)-1-phenyl-2-propanol
[α]D = -33.2
The key to recognizing the inversion was noting that the optical activity was, although
equal in size, now OPPOSITE in direction! The first reaction added something to the
oxygen to make it leave. The third reaction chopped off something from the oxygen so
it would return to normal. So it must be the second reaction where the fun occurred…
End Side Note]
2. SN1 (Substitution, Nucleophilic, Unimolecular)
Unimolecular rate equation – Rate of this reaction is dependent on the concentration of
the alkyl halide, which is the only species involved in the rate-determining step.
Mech: the reaction can only go as fast as its slowest step, otherwise known as the “ratedeterming step”, shown below as “r.d.s.”:
C
X
r.d.s.
(slow)
C
+ X
Nuc
(fast)
C
Nuc
The rate of reaction is ONLY dependent on formation of carbocation intermediate
(slowest step of the reaction). It is the rate of the “spontaneous dissociation” that slows
down this reaction or speeds it up. Note that the Nucleophile is NOT involved in this step.
Rate = k [RX]
Question: How does the carbocation intermediate affect the stereochemistry of this
reaction?
Answer: Racemization (formation of a racemic mixture) occurs at the reaction center
when the carbocation forms:
Ph
H3C
OH
H
Br
H3CH2C
Ph
H3 C
Ph
H3 C
OH2
H3CH2C
CH2CH3
Br
Ph
H3C
Br
CH2CH3
+
Ph
H3 C
CH2CH3
Br
Question: How do you think someone figured out that the product was a racemic mixture?
Someone measured the optical activity of the product and found out that it didn’t have
any optical activity!
Factors that affect the rate of an SN1 Reaction:
1. Alkyl Halide: the more stable the carbocation, the faster the reaction will occur (more
stable intermediate = lower activation barrier). The rate-determining step is the
carbocation formation – the faster this occurs, the faster the overall reaction occurs.
Recall that alkyl groups stabilize carbocations via:
1. Inductive Effect
2. Hyperconjugation
More stable carbocations have lower activation barriers and thus faster reactions
Order of stability – 3º > 2º >>>>>> 1º or methyl
Methyl and primary carbocations are so unstable, they cannot form and thus these
generally will NOT do SN1 reactions.
Another stabilizing factor: Resonance Stabilization – the more the charge can be spread
out over multiple atoms, the more stable the charge will be (“delocalization”).
Allylic carbocations: Both (or all, if appropriate) resonance forms are reactive.
Ex.
Benzylic carbocations: Only the resonance form that maintains the aromatic ring reacts
(the other three are much higher energy without the stability of the aromatic system)
Ex.
Of course, systems that have multiple stabilizations of resonance (i.e. a carbocation that
might be both allylic AND benzylic) will be even MORE stable.
New Order of Stability:
3º or 2º with resonance
>
2º or 1º with resonance
>>>>>>>>>>> 1º and/or methyl
CH3
DID YOU NOTICE…? Those that are the most stable (3º) carbocations do SN1 but do not
do SN2 and those that do SN2 best (1º) will not do SN1. BUT WATCH those in the middle
that do BOTH!!!
2. Leaving Group: Same order of reactivity seen for SN2 reaction
TsO⊖, MsO⊖ > I⊖ > Br⊖ > Cl⊖ ≈ H2O > F⊖, ⊖OH, ⊖NH2
SN1 reactions are often run under acidic conditions, therefore H2O is often seen as a
Leaving Group for alcohols (refer to Ch 10, formation of tertiary halides from tertiary
alcohols):
(CH3)3C-OH H-X
(CH3)3C
(CH3)3C-OH2
X
(CH3)3C-X
3. Nucleophile: Does NOT affect the rate of the reaction because it is not part of the
rate-determining step
Generally, though, nucleophiles in SN1 reactions are WEAK bases, usually neutral molecules
like H2O or ROH or perhaps H2S and RSH (sulfides and thiols).
If you choose a stronger nucleophile, it is probably a strong base also… then you will see
either an SN2 reaction (from stronger nucleophile) or an elimination reaction (E2) (from
strong base) will occur instead!
4. Solvent: affects the rate of SN1 reaction by stabilizing the carbocation intermediate
as it forms (lowering the activation barrier, speeding the reaction), thus increasing the
rate of reaction.
Any POLAR solvent, protic or aprotic, can be used (ex. H2O, ROH, DMSO, etc). The
dipoles of the solvent molecules orient themselves so the electron-rich end of the dipole
of the solvent molecule faces the positively charged carbon. This surrounds the electrondeficient with more electron density and helps stabilize the positive charge.
H
H
H
δ
H
O
O δ-
H
δ-O
-
δ-
H
δ
O
-
O
H
H
H
H
Biggest drawback to the SN1: Use of carbocations that can rearrange!!
Summary: SN2 versus SN1
Factor
Primary
Secondary
Tertiary
Leaving Groups
Solvent
Nuc
SN 2
Yes
Yes
No
SN 1
No
Yes
Yes
same for both
Polar, aprotic
Strong
Polar
Weak
Strong Nucleophiles
• Usually anions with a full negative charge (easily recognizable by the presence of
sodium, lithium or potassium counterions)
• Participate in SN2-type substitutions
Examples:
NaOCH3 (any NaOR), LiCH3 (any RLi), NaOH or KOH, NaCN or KCN, NaC≡CR (acetylide
anion), NaNH2, NaNHR, NaNR2, NaI, KBr, NaCl, KI, NaN3
Weak Nucleophiles
• Typically neutral molecules
• Participate in SN1-type substitutions
Examples:
H2O, ROH, H2S, RSH
SN2 or SN1?
Ex.
NaI
OTs
Ex.
OTs
H2 O
Ex.
OTs
NaBr
Ex.
OTs
Ex.
NaOCH3
Br
Ex.
CH3OH
H2 O
Br