SN1-vs-SN2 PDF

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Summary

Comparison between SN1 and SN2 - useful for Section F...

Description

SN1 Reaction

SN2 Reaction

SN 1 vs. SN2 Summary

Stereochemistry

Stereochemistry

Substitution occurs with a mixture of retention and inversion at a stereocenter

Substitution occurs with inversion of configuration at chiral centers

HO

Br 3

H 2O

1

2

Br

OH 3

1

2

3

2

1

4

+ H3O Br

2

3

Na

SN1 CN

C N 4

1

3

+ Na

2

Br

1

Bonds Formed

Bonds Broken

C 2–CN

C 2–Br

Rate Law

One stereoisomer

"Big Barrier"

Rate Law

Rate Law

Alkyl halide (electrophile)

The rate of the reaction is ONLY sensitive to the concentration of the substrate (and not the nucleophile)

The rate of the reaction is sensitive to the concentration of the substrate AND the nucleophile

retention

Br 4

inversion

Br

HO

H 2O 2

3

One stereoisomer inversion! This substitution reaction results in an inversion of configuration at C-2

4

1

3

Rate = k [R–Br]

+ H 3O Br

2 1

Rate

Rate

1

2

3

1

4

[

2

3

(doubling the concentration of water has no effect on the rate)

4

H 2O ]

Br

Fastest for tertiary, slowest for primary Br

2

3

50°C

1

Rate

HO

H 2O 2

3

3

CN

C N

Na

4

1

+ H3O Br

1.2 × 10 6

1

2

[

3

]

+ H3O Br

11.6

2

50°C

1

1

50°C

2

Mechanism

4

H

1

4

2

(R)

4

Br

1

3

< 0.001

2 1

Secondary

4

Na

CN

C N

2

3

4

1

1

Weak (generally neutral)

Strong (generally bearing a negative charge)

Solvent

Polar protic (e.g. alcohols)

Polar aprotic (e.g. DMSO, acetone)

Stereochemistry

Mix of retention and inversion

Inversion

4

2

3

Br

Methyl

Na

1

C N 4

1

H 3C

Br

Na

3

2

Once you've identified the leaving group, instpect the carbon it is attached to. How many carbons is that carbon connected to? That will tell you if the carbon is primary, secondary, or tertiary. If there are no attached carbons, that's the special case of "methyl" (SN2 for sure!)

CN

~20

CN

~1000

1

C N H 3C

1

The key skill to start with is identifying the leaving group Look for halogens (Cl, Br, I) or tosylates/mesylates (OTs, OMs) Alternatively, look for alcohols (OH) if acid is present

If the carbon is tertiary, it's likely SN1. You can rule out SN2 due to steric hindrance. If the carbon is primary, it's likely SN2. You can rule out SN1 due to the fact that primary carbocations are unstable [one exception: resonance stabilized carbocations].

2

3

Next, examine the nucleophile. A negatively charged nucleophile generally indicates an SN2 reaction. A neutral nucleophile (such as H 2O or ROH) generally indicates an SN1 reaction. Finally, check the solvent. A polar aprotic solvent (such as DMSO, acetone, acetonitrile, or DMF) generally indicates SN2, whereas a polar protic solvent such as H 2O or ROH generally indicates SN1 conditions.

One step (backside attack) In the "backside attack", the nucleophile attacks the substrate from the backside in a single step, resulting in inversion of configuration.

(S) alkyl halide

1

Path A

H

1

Br

3

2

4

OH2 (S) –H

Step 2: Attack of nucleophile on carbocation (fast) Can occur from either side of the flat carbocation ( Path A or Path B)

Step 3: deprotonation 1

H

3

3

2

4

4

(S)

Path A gives inversion (R)

H 2

Br

δ N C

–

H δ

+ δ– Br

H N C

1 2

3

OH2 Path B

OH2

N C

H

Carbocation

Step 3: deprotonation

4

1

Mechanism

Step 1: Loss of leaving group (slow)

1° > 2° >>3° (fastest)

3

3

–H

HO

2

C N

2

3

2

1

Stepwise - leaving group leaves (slow) forming a carbocation, which is then attacked by a nucleophile (fast)

H

(doubling the concentration of CN doubles the rate)

4

[ :CN ]

Na

Tertiary

Primary

3° > 2° >>1° (fastest)

Steric hindrance

Comparing SN1 vs. SN2 reactions

From "March's Advanced Organic Chemistry", 5th Ed. p. 431

H 2O 2 (R)

3

Slowest for tertiary, fastest for primary (methyl even faster) Rate Br NC

1

+ H3O Br

(likely occuring through SN2 mechanism)

H

2

1

OH

H 2O

1

2

3

Carbocation stability

1

1

4

Br

Bimolecular Unimolecular (substrate only) (substrate and nucleophile)

Nucleophile

CN]

1

OH

H 2O

Br Primary

3

Rate = k [R–Br] [

+ NaBr

Rate

Br Br Secondary 3

2

Substrate

Substrate

Tertiary

2

Rate

]

[

4

SN2

OH

Path B gives retention (S)

4

(S)

3 4

partial bonds!

Transition state

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(R)

• Explains bimolecular rate law (depends on conc. of nucleophile and substrate) • Explains inversion of stereochemistry • Explains sensitivity to steric hindrance (bulky groups slow down backside attack)

This is called the SN2 mechanism (Substitution, Nucleophilic, bimolecular)

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