SN1 and SN2 Reactions

SN1 SN2

Rate =k[RX] =k[RX][Nuc:-]

Carbocation intermediate?

Y N

Stereochemistry mix Inversion of configuration

Rearrangement ~H, ~ CH3 possible

No rearrangements

SN1 and SN2 Reactions

SN1 SN2

Substrate 3°>2°>”1°” CH3X>1°>2°

Nucleophile Unimportant, but usually weak

Strong and unhindered

Leaving group Excellent Better than nucleophile

Solvent Polar and ionizing

Polar aprotic

SN1 and SN2 Reactions

Do not occur with vinyl halides or aryl halides. Consider the carbocation formed for SN1.

Consider the backside attack for SN2.

Elimination Reactions The substitution reaction mechanisms

you have learned are just part of the picture.

In the SN1 and SN2 reactions, the species that acts as a nucleophile may also act as a base and abstract a proton. This causes the elimination of HX and the formation of an alkene.

An elimination reaction can occur along with the substitution reaction.

Elimination

B:- is a species acting as a base.

Since HX is lost, this particular reaction is called a dehydrohalogenation.

E1 Reactions

E1 = elimination, unimolecular Rate = k[substrate]

(a first order process) The rate-limiting step is the

formation of the carbocation, the same as for SN1 (that’s why E1 competes).

E1 Reaction Mechanism Step 1: The substrate forms a carbocation

intermediate (rate-limiting step). Step 2: Methanol acts as a base and removes

H+(fast step).

E1 Reaction Profile

rate = k[(CH3)3Br]

k = Ae-EA(step 1)/RT

E1 Occurs with and Competes with SN1

When bromocyclohexane is heated with methanol, two products are possible.

Can you draw the mechanism that leads to each product?

E1 Reactions E1 reactions are exothermic. E1 reactions occur in at least two

steps and compete with SN1 reactions. The first step is slow. It is the

formation of the carbocation intermediate.

The second step is fast. It is the abstraction of H+ from the carbocation by the base.

Factors Affecting E1 Reactions

Structure of the substrate Can a stable carbocation be formed?

Strength of the base Nature of the leaving group The solvent in which the reaction

is run. Must be able to stabilize the

carbocation and the LG (which is usually an ion).

Factors Affecting E1 Reactions - Structure of the Substrate

The most important factor influencing the rate of E1 reactions is the stability of the carbocation formed.

Stability of carbocation: 3° > 2° >1°

Relative rates for E1: 3°>2°≈1°(resonance)

Factors Affecting E1 Reactions - Strength of the Base

The rate is not much affected by the strength of the base. Weak bases will work.

Factors Affecting E1 Reactions - the Leaving Group

The LG should be good.

Factors Affecting E1 Reactions - Solvent Effects

The solvent must be capable of dissolving both the carbocation and the leaving group.

E1 reactions require highly polar solvents that strongly solvate ions.

Typical solvents: water, an alcohol, acetone (to help the alkyl halide to dissolve), or a mix.

Rearrangements in E1 Reactions

The carbocation can undergo a structural rearrangement to produce a more stable species. hydride shift (~H) methyl shift (~CH3)

If ionization would lead to a 1° carbocation, look for a rearrangement to occur.

Ionization rates can be increased by the addition of reagents such as AgNO3 (how?); however, Ag is not cheap.

A Hydride Shift Can Occur in Either SN1 or E1 Reactions

~H CH3OH

-H+

What would the E1 products be?

SN1

-Br -

A Methyl Shift Can Occur in Either SN1 or E1 Reactions

~CH3CH3OH

-H+

What would the E1 product be?SN1

Zaitsev’s Rule

When two or more elimination products are possible, the product with the more substituted double bond will predominate.

Zaitsev’s Rule

When two or more elimination products are possible, the product with the more substituted double bond will predominate. Alkyl groups are electron-donating

and contribute electron density to the π bond.

Bulky groups can be more widely separated.

E1 Reactions - Summary The structure of the carbocation is the

most important factor: Relative rates for E1: 3°>2°.

The base is typically weak or moderate in strength.

The LG should be good. The solvent should be polar and protic to

stabilize the carbocation and LG. Products can exhibit rearrangements and

will follow Zaitsev’s Rule.