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HIGHLIGHTS OF NUCLEOPHILIC SUBSTITUTION REACTIONS
INVOLVING sp^3 CARBON
Sn2 REACTIONS
From a synthetic point of view, this is the most useful reaction. It provides a means to prepare many functional groups from alkyl halides, and therefore from alkanes through the free radical halogenation reaction.
Nu +^ C L C^ Nu^ +^ L
Nucleophile
Electrophile, or substrate L= leaving group
Nucleophilic substitution product
Alkane (^) Cl 2 or Br 2
R H X 2 R X
Alkyl halide (chloride or bromide)
light
R X + OH R OH
Alkyl halide Alcohol
R X + CN R CN
Alkyl halide Nitrile
etc.
+ X
+ X
The Sn2 mechanism:
a) is a single step process b) involves no intermediates c) involves only one transition state, which is of low polarity d) follows second order (bimolecular) kinetics. That is, rate= k [substrate][nucleophile]
In nucleophilic substitutions at sp^3 carbon, Sn2 mechanisms are favored by using:
a) sterically accessible substrates b) strong (negatively charged), small nucleophiles c) low to moderate polarity solvents
Stereochemically, if the electrophilic center in the substrate is chiral, the Sn2 reaction produces a product with inverted configuration.
Sn1 REACTIONS
From a synthetic point of view, the Sn1 reaction is less useful. It is prone to side reactions such as eliminations and carbocation rearrangements.
The Sn1 mechanism:
a) is a multistep process b) occurs with formation of carbocation intermediates in the rate determining step c) involves one transition state per step. The rate-determining step involves a high polarity transition state d) follows first order (unimolecular) kinetics. That is, rate= k [substrate]
In nucleophilic substitutions at sp^3 carbon, Sn1 mechanisms are favored by using:
a) sterically hindered substrates b) weak (neutral), small nucleophiles c) moderate to high polarity solvents that can stabilize the transition state and the carbocation intermediate
Stereochemically, if the electrophilic center in the substrate is chiral, the Sn1 reaction produces a racemic product. The relative proportions of the enantiomers depend on the specific reaction, but will typically be close to 50/50.
FACTORS THAT AFFECT THE COURSE OF NUCLEOPHILIC SUBSTITUTIONS AT sp^3 CARBON
1. STERIC NATURE OF THE SUBSTRATE. Steric accessibility of the electrophilic center in the substrate is probably the most important factor that determines if a nucleophilic substitution will follow an Sn1 or an Sn2 mechanism.
EXAMPLES OF Sn2 (sterically accessible) SUBSTRATES
CH 3 Br CH 3 CH 2 Cl
primary substrates
H 3 C
CH
CH 3
Cl
Br
unhindered secondary substrates
Br
primary allylic halides
EXAMPLES OF Sn1 (sterically hindered) SUBSTRATES
C
CH 3
CH 3
H 3 C Br
H 3 C Br
tertiary halides
Cl
hindered secondary halides
C
CH 3
CH 3
H 3 C CH 2 Cl
hindered primary halides
3. SOLVENT USED. It has already been mentioned that Sn2 mechanisms are favored by low to moderate polarity solvents such as acetone and N,N-dimethylformamide (DMF). Sn1 mechanisms are favored by moderate to high polarity solvents such as water and alcohols. It is frequently the case that in Sn1 reactions the solvent also doubles as the nucleophile. Water and alcohols are prime examples of this practice. 4. LEAVING GROUP. The nature of the leaving group has more of an effect on the reaction rate (faster or slower) than it does on whether the reaction will follow an Sn1 or an Sn2 mechanism. The most important thing to remember in this regard is that good leaving groups are weak bases.
a) All halogens, except for fluorine, are good leaving groups b) Groups that leave as resonance stabilized ions are also weak bases and therefore good leaving groups. c) Water is a good leaving group frequently used to prepare alkyl chlorides and bromides from alcohols.
The OH group in alcohols is not a good leaving group because it leaves as hydroxide ion, which is a strong base. However, if the hydroxyl group is protonated first with strong acid, it can leave as a water molecule, which is a good leaving group. Refer to the manuscript titled Introduction to Lewis Acid-Base Chemistry for a discussion and examples of this approach.
H 3 C
C
CH 3
H
C
N
O CH 3
O
CH 3
acetone (^) DMF
Sn2 solvents
H 2 O CH 3 OH^ CH 3 CH 2 OH
water methanol ethanol
Sn1 solvents
COMPETING (SIDE) REACTIONS IN NUCLEOPHILIC SUBSTITUTIONS
There are two major reactions that compete with nucleophilic substitutions. They are:
1. CARBOCATION REARRANGEMENTS (Sn1 only) 2. ELIMINATION REACTIONS (Sn1 and Sn2)
Carbocation rearrangements are examined first. Eliminations are examined in a separate paper.
CARBOCATION REARRANGEMENTS
Carbocations only form in Sn1 reactions. Carbocations are prone to skeletal rearrangements if this produces a more stable cation. Carbocation rearrangements occur mainly by two processes:
a) Hydride shift - migration of a hydrogen atom with electrons to an adjacent carbon b) Alky shift - migration of a carbon (usually as part of an alky group) with electrons to an adjacent carbon.
A quick way to tell whether a substrate will produce a carbocation prone to rearrangement is to look at the carbon that bears the leaving group. If this carbon is next to a higher order carbon (meaning secondary, tertiary, allylic, etc.) then the carbocation that results can rearrange to a more stable one, and will do so, probably yielding a product with different carbon connectivity. Examples follow.
The above example also shows the reason why, when the nucleophile is water or an alcohol, the group that replaces the leaving group in the product is the conjugate base of water (OH) or the alcohol (RO respectively).
Another example illustrates a similar point. Can you provide a step by step mechanism (it might be in the test, you never know)?
Br
The leaving group (Br) is on a secondary carbon, but this carbon is next to a tertiary carbon. The nucleophile is water, therefore the expected product is an alcohol. The product will consist of a mixture of the expected secondary alcohol (minor) and a tertiary alcohol (major) due to the rearrangement to a more stable cation shown below.
H 2 O
acetone
(polar solvent) OH
expected product (minor)
OH
rearranged product (major)
CH 3
Br CH 3 OH
(solvent and nucleophile)
CH 3
OCH 3
minor
H 3 C OCH 3
major
notice that the conjugate base of the nucleophile (in red) has replaced the leaving group
Br
Sn1 (^) secondary cation
hydride
r.d.s.
tertiary alcohol conjugat acid
H 2 O
+ H 3 O
H
H
shift
tertiary cation, more stable
H
H 2 O
O
H H
O
H H
OH
a proton transfer (acid-base reaction) yields the free alcohol
Sn
