S N 1 Reactions

SN1 Reactions

t-Butyl bromide undergoes solvolysis when boiled in methanol:

Solvolysis: “cleavage by solvent” nucleophilic substitution reaction in which the

solvent serves as the nucleophile

CH3 CCH3

CH3

Br + CH3OH CH3 CCH3

CH3

OCH3 + HBr

SN1 Reactions

The reaction between t-BuBr and methanol does NOT occur via an SN2 mechanism because: t-BuBr:

too hindered to be SN2 substrate CH3OH:

weak nucleophile

Solvolysis reactions occur via an SN1 mechanism:

SN1 Reactions

SN1 Reactions substitution nucleophilic unimolecular

Rate = k[R-X] 1st order overall

1st order in [R-X] zero order in [Nuc]

Only R-X is present in the transition state for the rate determining step

Nucleophile is NOT present in RDS

SN1 Reactions General Mechanism:

Step 1:

R X R+ + X-

Step 2:

R+ + Nuc - R Nuc

Rate determining step

SN1 Reactions

Reaction Energy Diagram for SN1 Reactions:Formation of carbonium ion is highly endothermic

According to Hammond’s Postulate, the transition state most closely resembles the carbonium ion.

SN1 Reactions

The reactivity of a substrate in an SN1 reaction depends on the stability of the carbonium ion formed:

3o > 2o > 1o > methyl

Allylic and benzylic halides undergo SN1 reactions because the resulting carbonium ions are resonance stabilized.

CH

CH

H CH2 Br

HC

HC

H

HC

HC

H

CH2 Br

CH

HH

CH

CH

C HH

CH

CH

H CH2 Br

HC

HC

H

HC

HC

H

CH2 Br

CH

HH

CH

CH

C HH

++

SN1 Reactions SN1 reactions involve:

weak nucleophile H2O not OH-

CH3OH not CH3O-

Substrates that form stable carbonium ion intermediates:

3o, benzylic, or allylic halide are most favored

2o (sometimes)

SN1 Reactions

Example: Draw the mechanism for the SN1 reaction of t-butyl bromide with methanol

Step 1: Slow ionization of R-X to form carbonium ion: rate determining step

SN1 Reactions

Step 2: Fast attack of nucleophile on the carbonium

ion

SN1 Reactions

Step 3: (Not needed for all SN1 reactions) Solvent molecule removes proton, leaving

the neutral product

SN1 Reactions-Stereochemistry The carbonium ion intermediate formed during

an SN1 reaction is sp2 hybridized and planar. The nucleophile can attack from either side

of the carbonium ion. A mixture of both possible enantiomers

forms.

Racemization: a process that gives both enantiomers of

the product not necessarily in equal amounts

SN1 Reactions-Stereochemistry

CH3 COCH2CH3

CH(CH3)2CH2CH3

CH3 C

CH(CH3)2

CH2CH3

OCH2CH3

CCH2CH3

CH(CH3)2CH3 CCH2CH3

CH(CH3)2CH3+

C

CH3CH2OH

CH2CH3

CH(CH3)2CH3

C

CH3CH2OH

CH2CH3

CH(CH3)2CH3

Attack from top

Attack from bottom

- H+

- H+


When the nucleophile attacks from the side where the leaving group was originally, retention of configuration occurs.

CH3

CH3

CH3 C

CH(CH3)2

CH2CH3

C

Br

CH(CH3)2

CH2CH3

C

CH2CH3

CH(CH3)2

OCH2CH3

OCH2CH3

CH3

CH3

CH3 C

CH(CH3)2

CH2CH3

C

Br

CH(CH3)2

CH2CH3

C

CH2CH3

CH(CH3)2

OCH2CH3

OCH2CH3

Attack from top

(R)(R)


When the nucleophile attacks from the back side (opposite to the original leaving group), inversion of configuration occurs.

CH3

CH3

CH3 C

CH(CH3)2

CH2CH3

C

Br

CH(CH3)2

CH2CH3

C

CH2CH3

CH(CH3)2

OCH2CH3

OCH2CH3

CH3

CH3

CH3 C

CH(CH3)2

CH2CH3

C

Br

CH(CH3)2

CH2CH3

C

CH2CH3

CH(CH3)2

OCH2CH3

OCH2CH3

(R) (S)

Attack from bottom


For most SN1 reactions, the leaving group partially blocks the front side of the carbonium ion more inversion of configuration less retention of configuration

SN1 Reactions-Rearrangements

Carbonium ions often undergo rearrangements, forming more stable cations. Structural changes resulting in a new

bonding sequence within the molecule

The driving force for a rearrangement is the formation of a more stable intermediate. 1o or 2o carbonium ion rearranges to a more

stable 3o carbonium ion or resonance-stabilized carbonium ion


A mixture of products often forms as a result of rearrangements during SN1 reactions. NOTE: Rearrangements cannot occur

during SN2 reactions since an intermediate is not formed.

CH3CHCHCH3

CH3CH2CCH3

CH3CHCHCH3

Br

CH3

CH3

CH3

OCH2CH3

OCH2CH3CH3CHCHCH3

CH3CH2CCH3

CH3CHCHCH3

Br

CH3

CH3

CH3

OCH2CH3

OCH2CH3

CH3CHCHCH3

CH3CH2CCH3

CH3CHCHCH3

Br

CH3

CH3

CH3

OCH2CH3

OCH2CH3+

CH3CH2OH

Rearranged productRearrangement occurs via

hydride shift.


Common rearrangements: Hydride shift (~H)

the movement of a hydrogen atom and its bonding pair of electrons

Methyl shift (~CH3) the movement of a methyl group and its

bonding pair of electrons


Hydride Shift Mechanism: Step 1: Formation of carbonium ion and

rearrangement:


Hydride Shift Mechanism: Step 2: Nucleophile attack and loss of

proton (if needed)

CH3

CH3

CH3

CH3

CCH3

CH3

CH2 Br

CCH3

CH3

CH2 Br

CCH3

CH2

CCH3

CH3

CH2

CH3

OCH2CH3

OCH2CH3CH3

CH3

CH3

CH3

CCH3

CH3

CH2 Br

CCH3

CH3

CH2 Br

CCH3

CH2

CCH3

CH3

CH2

CH3

EtOH


Example of a Methyl Shift (~CH3):

CH3

CH3

CH3

CH3

CCH3

CH3

CH2 Br

CCH3

CH3

CH2 Br

CCH3

CH2

CCH3

CH3

CH2

CH3

CH3

CH3

CH3

CH3

CCH3

CH3

CH2 Br

CCH3

CH3

CH2 Br

CCH3

CH2

CCH3

CH3

CH2

CH3

OCH2CH3

EtOH


Mechanism of ~CH3: Step 1: Simultaneous (often) shift of methyl

group and loss of leaving group:


Mechanism of ~CH3: Step 2: Attack of nucleophile and loss of

proton (if needed)


Example: Propose a mechanism for the following reaction.

CH3

Cl

CH3

OCH2CH3

CH3OCH2CH3

CH3

Cl

CH3

OCH2CH3

CH3OCH2CH3

CH3

Cl

CH3

OCH2CH3

CH3

OCH2CH3

CH2CH3OH

+

SN1 vs. SN2

SN2

Strong nucleophile

Primary or methyl halide

Polar aprotic solvents (acetone, CH3CN, DMF)

Inversion at chiral carbon No rearrangements

Weak nucleophile (may also be solvent)

Tertiary,allylic, benzylic halides

Polar protic solvent (alcohols, water)

Racemization of optically active compound

Rearranged products

SN1

E1 Reactions An elimination reaction involves the loss of two

atoms or groups from a substrate, usually forming a new bond.

Elimination reactions can occur via a first order (E1) or a second order (E2) process.

H

C

C CH

CH3 Br

CH2CH3

CH2CH3

CH3C

H CH2CH3

CH2CH3

H

C

C CH

CH3 Br

CH2CH3

CH2CH3

CH3C

H CH2CH3

CH2CH3

Na+ -OCH3

CH3OH

+ Br -

E1 Reactions E1 reactions:

Elimination, unimolecular

1st order kinetics Rate = k[R-X] RDS transition state involves a single

molecule

General conditions: 3o and 2o halides weak bases

E1 Reactions E1 Mechanism:

E1 Reactions E1 reactions almost always occur together

with SN1 reactions.

CH3 + CH3CH2OH

CH2 + CH3

C

CH3

Br

CH3

CCH3

CH3

C OCH2CH3

CH3

CH3

CH3 + CH3CH2OH

CH2 + CH3

C

CH3

Br

CH3

CCH3

CH3

C OCH2CH3

CH3

CH3

E1 SN1

E1 Reactions

CH3CH2-O-HH

+

E1 Reactions Once formed, a carbonium ion can:

recombine with the leaving group react with a nucleophile forming a

substitution product (SN1) lose a proton to form an alkene (E1) rearrange to form a more stable carbonium

ion and then: react with nucleophile lose a proton to form an alkene

E2 Reactions E2 reactions:

Elimination, bimolecular

2nd order kinetics Rate = k[R-X][B-] RDS transition state involves two molecules

General conditions: 3o and 2o halides strong bases

E2 Reactions In the presence of a strong base, elimination

generally occurs in a concerted reaction via an E2 mechanism

E2 Reactions SN2 reactions require an unhindered methyl or

1o halide steric hinderance prevents nucleophile from

attacking 3o halides and forming the substitution product

E2 reactions generally involve the reaction between a 3o and 2o alkyl halides and a strong base.

E2 Reactions The reaction of t-butyl bromide with methoxide ion gives

only the elimination product.

The base attacks the alkyl bromide much faster than the bromide can ionize.

E2 Reactions Many alkyl halides can eliminate in more than

one way. Mixture of alkenes produced

E2 Reactions Saytzeff Rule:

When two or more elimination products can be formed, the product with the most highly substituted double bond will usually predominate.

R2C=CR2 > R2C=CHR > RHC=CHR and R2C=CH2 > RHC=CH2

E2 Reactions

Example: Draw the structures for all possible products of the following reaction. Which one will predominate?

Br CH3 NaOCH2CH3

EtOH

E2 Reactions E2 reactions follow a concerted mechanism:

bonds breaking and forming simultaneously

specific geometry required to allow overlap of orbitals of bonds being broken and bonds being formed

E2 reactions commonly involve an anti-coplanar conformation.

E2 Reactions

E2 ReactionsExample: Predict the structure of the elimination product formed by the following reaction.

C C

HBr

H

CH3PhPh

NaOCH3

CH3OH

C

C

Ph =

CH

CH3PhBr

HPh

CPh

H

CH3

Ph

E1 vs E2E1

Weak base 30 > 2o Good ionizing solvent

polar, protic (water, alcohols)

Saytzeff product No required geometry

Rearranged products possible

Strong base required 3o > 2o

Solvent polarity not important

Saytzeff product Coplanar leaving groups

(usually anti) No rearrangements

E2

S N 1 Reactions

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