sign in sign up
<<
Home , Electrophile, Polar aprotic solvents, Protic solvent, Solvent, Solvent effects

The Electrophile

1

¨ Recall that electrophile means “electron- loving.”

¨ When considering substitution and elimination reactions we must consider the attached to the . Is it a primary, secondary, or tertiary carbon?

Electrophile Type Favors Disfavors

Methyl SN2 SN1, E1, E2

Primary SN2, E2 SN1, E1

Secondary SN1, SN2, E1, E2

Resonance Stabilized SN1, SN2 E1, E2 Primary

Tertiary SN1, E2, E1 SN2

SN1 and E1 SN2 and E2 least stable most stable least hindered most hindered LG H H LG > > LG < < < H > H H H LG x methyl and primary methyl electrophile Tertiary electrophile is too too unstable has no β . sterically crowded for direct nucleophilic attack. No S 2 to exist. No SN1 or E1 No E2 N The Electrophile: SN1 and E1 Reactions, Carbocation Stability 2

¨ The first step in a SN1 and E1 reaction is to form a carbocation. If the carbocation is too unstable, then regardless of how good the leaving group is the carbocation will not form.

¨ To determine the stability of a carbocation we must consider inductive and resonance effects What is happening when we say CH3 CH3 CH3 H > > > “inductive” effects? H3C CH3 H3C H H H H H 3o = 3 2o = 2 carbons 1o = 1 carbon methyl Empty p orbital

H

hyperconjugation H C C C H H

substituted carboncation methyl carboncation ¨ Primary may be formed if they have resonance stabilization

N N O OH H Examining the Leaving Group and Electrophile

3

weak bases= GOOD leaving group

Br Cl strong = BAD leaving group O OH

Potential Leaving Groups no SN2

sterically hindered (no SN2) 3 ° Favors S 1, E1, E2 N Br Zaitsev’s Rule: the most Br Cl substituted will be Cl H O OH most stable and therefore 1° Favors S 2, E2 O OH N not sterically hindered the most favored product (allows for E2) 2 ° Favors SN1, SN2, E1, E2 Electrophilic centers Cl H O OH

Cl H Br H Br H H H O OH Cl Cl O H Br H H H H H Cl Cl O OH O OH HO most substituted is H H H the E2 product O OH The Leaving Group and the Electrophile: Examples

4

The leaving group here would be Cl-, a good leaving group The electrophile is a secondary electrophile, SN1, SN2, E1, and E2 are all possible Cl The leaving group here would be OH-, a bad 2° C leaving group No reaction is likely to take CH O place here without further modifying the the produced by red algea . HO Br The leaving group here would be Br-, a good leaving group. However, it is attached to an Sp2 carbon, no substitution or will occur here. Practice Question

5

Which substrate will undergo the fastest SN2 reaction?

F bad LG ✗ A F 1o electrophile ✓

I good LG ✓ B I 3o electrophile ✗

C OH bad LG ✗ OH 1o electrophile ✓

D Cl good LG ✓ 2o electrophile ✓ Cl The

6

¨ Recall that nucleophile means “nucleus loving” or “positive-charge loving”

¨ donate a pair of electrons to an electrophile. (Lewis base!)

¨ When the nucleophile donates a pair of electrons to a proton it is called a Bronsted base or simply a base. The proton is acting like an electrophile.

Nucleophile Type Favors Disfavors

Strong Nuc/Strong SN2, E2 SN1, E1 Base

Strong Nuc/ Weak SN2 SN1, E1, E2 Base

Weak Nuc/ Strong E2 SN1, SN2, E1 Base

Weak Nuc/Weak Base SN1, E1 SN2, E2 The Nucleophile: Factors That Affect Nucleophilicity and Basicity 7

¨ How do we know if a nucleophile is acting as a nucleophile or a base?

¤ Consider the following properties:

Charge Steric Hindrance

Strength of nucleophile/base

Electronegativity The Nucleophile: Charge and

8

¨ Charge ¤ The greater the electron on the nucleophile the greater its nucleophilicity ¤ Increased electron density results in an increased desire to donate those electrons; ie: increased nucleophilicity

The conjugate base is neutral = SN1, E1 charged = SN2, E2 always a stronger nucleophile

¨ Electronegativity ¤ Electronegativity relates how “tightly” held a pair of electrons are. The more tightly they are held the less the ability for them to be donated is.

Electronegativity= Nucleophilicity

Electronegativity= Basicity Rank the Following Nucleophiles in Order of Increasing Nucleophilicity (1 being the weakest 5 being the strongest) 9

O OH NH

2 1 3

no charge, less electronegative than weakest , strongest

The Nucleophile: Steric Hindrance

10

¨ Nucleophilicity decreases with increasing steric bulk ¤ Nucleophiles generally are reacting at electrophilic carbon. The orbitals that participate in the reactions are less accessible than those orbitals on hydrogen.

H3C LG !+ bulky nucleophile = E2 H H

H H Non-Nucleophilic Bases Li N N N N N

¤ If your nucleophile is bulky then its electrons may not be able to “reach” the antibonding orbitals at the carbon Practice Question

11

¨ t-butoxide has the following structure. O

Based on what you know about nucleophilicity what type of reaction or reactions are favored?

E1 or S 1 unlikely as this is a charged A. Both S 1, E1 N N nucleophile (not A or C) B. Both SN2, E2 C. Primarily E1 bulky nucelophiles do not allow for S 2 (not B) D. Primarily E2 N : The Electrophile

12

Solvent Type Favors Disfavors Polar protic – has hydrogen bound to an electronegative atom Polar Protic SN1, E1 Eg. H2O Polar aprotic – no , but still Polar Aprotic SN2, E2 SN1, E1 contains a Eg. CH2Cl2 Non-Polar SN1, SN2, E1, E2 Non-polar – has no Eg. ,

¨ S 1 and E1 reactions proceed more quickly in polar protic N δ+ ¤ Intermediates are solvated through hydrogen bonding H

Oδ- Leaving Group Stabilization H H Carbocation Stabilization O H H Hδ+ H Polar protic solvents favour δ- O O H δ+ SN1 and E1 reactions H δ- O H LG H O δ+ because the intermediates H are solvated H δ- δ+H δO- O H O H H H Solvent effects: The Nucleophile

13

¨ Nucleophilicity can be affected by solvent

¨ Polar protic solvents cause nucleophilicity to increase going down the periodic table

¨ cause nucleophilicity to decrease going down the periodic table Solvent Effects: The Nucleophile

14 Polar protic Polar aprotic O O OH OH H H H3C HO alcohols carboxylic

H O H H Polar aprotic solvents have dipoles present O H which could interact with the nucleophile, O O O H O H O however the partial positive charge created F H I H H H < O O from the dipole is sterically hindered and O O Na H therefore the nucleophile is unable to interact H H O O O O with these solvent. H H O H Hydrogen bonding between lone pairs and polar protic solvents result Polar aprotic solvents help in the of cations which result in “naked” anions in dispersion of negative charge. Lone pairs are tied up in H-bond and Favour SN2, E2 less available to be donated Non-polar solvent

O Recall that “like dissolves like.” as many nucleophiles are charged or very polar they will not dissolve in non-polar solvents é interaction with solvent = ê nucelophilicity Can you rank the following nucleophiles in order of increasing nucleophilicity? (1 being the weakest 5 being the strongest) 15

O OH H2O O OH H3C O 3 1 5 4 2

Polar

SH SeH OH 2 3 1 q How would the rate of the reaction change if the solvent was changed from to DMF

NaSH protic The protic solvent can interact with the MeOH nucelophile and keep it from reacting Br or with the alkyl bromide. Therefore, DMF DMF aprotic increases the rate of reaction. Effects

16

¨ When all else is equal increased will favor elimination reactions over substitution reactions Temperature Favors Disfavors High E1, E2 SN1, SN2

Low SN1, SN2 E1, E2

ΔG<0 reaction is spontaneous ΔG>0 reaction is non-spontaneous ΔG=0 reaction is in equilibrium ΔG2 ΔG=ΔH-TΔS ΔG1 H Nuc H LG 2 products LG Nuc Δ H G B LG H HB 3 products LG

Elimination reaction produces more products, this is favoured in the presence of heat as it increases the universe’s . Mind Map

17 Practice Question

18

¨ What is the most likely reaction to occur under the following conditions?

¨ What is the most likely product of the following reaction. H C A 3 O Br CH3O HO Br heat 1. LG? OH or Br B Br is a better LG HO OCH3 2. Electrophile? 3o Which product is formed? C or D? SN1, E1, E2 possible 3. Nucleophile? Strong Recall Zaitsev’s Rule! C The most stable product is C. favours SN2, E2 HO 4. Solvent? N.A. 5. Temperature? Yes eliminations preferred D Therefore, E2 reaction occurs (not A or B) HO Substitution and Elimination Reactions 19

O

H O Cl H O O Room Temp is inverted

1. LG: Cl 2. Electrophile: 2o, Favours SN1, SN2, E1, E2 3. Nucleophile: Charged, Favours SN2, E2 4: Solvent: Polar Aprotic, Favours SN2, E2 5. Temperature: Room Temp, Favours SN2 so SN2 Reaction

H H H Br Br H N DMSO Heat H H

1. LG: Br N 2. Electrophile: 3o, Favours SN1, E1, E2 heat 3. Nucleophile: Neutral but strong and Bulky, Favours E2 4: Solvent: Polar Aprotic, Favours SN2, E2 5. Temperature: Heat, Favours E2 heat so E2 Reaction 20 H H Br O O H H H2O

E1 product: most SN1 products: of stubstituded alkene stereochemistry 1. LG: Br favoured (Zaitsev’s 2. Electrophile: 3o, Favours SN1, E1, E2 Rule) 3. Nucleophile: Neutral, Favours SN1, E1 4: Solvent: N/A 5. Temperature: N/A so Mixture of SN1 and E1 OH HO

CH3 CH3 CH3 H CH 3 H3C O O H3C O H3C Cl H C 3 H O CH H CH3 CH 3 3 room temp CH3

1. LG: Cl 2. Electrophile: 3o, Favours SN1, E1, 3. Nucleophile: neutral, favours SN1, E1 4: Solvent: Polar protic, Favours SN1, E1 5. Temperature: Room Temp, Favours SN1 so SN1 Reaction 21

Br H Br OH heat H Zaitsev’s Rule: most O H 1. LG: Br substitutted alkene is 2. Electrophile: 2o, Favours SN1, E1,SN2, E2 most stable and therefore most favoured product 3. Nucleophile: Charged, favours SN2, E2 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction Less substituted alkene due to steric hinderance of t- Br on substrate and nucleophile Br H heat H

O H H 1. LG: Br O o 2. Electrophile: 2 , Favours SN1, E1,SN2, E2 too sterically hindered for 3. Nucleophile: Charged but bulky, favours,E2 base to deprotonate 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction 22 What Reaction Is occuring? Which Isomer will undergo the fastest reaction?

Br

Br K O FAST

H

K O

O H H Br K O Br H H H H Br Favoured Chair conformation Less Favoured Chair conformation but no anti-coplanar hydrogen (axial t-butyl group) BUT anticoplanar hydrogen

H 1. LG: Br 2. Electrophile: 2o, Favours SN1, E1,SN2, E2 H2 3. Nucleophile: Charged but bulky, favours,E2 C H 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction

Br CH2 H Additional Practice Questions

23

¨ Textbook Problems Solomons+Fryhle+Snyder (12 edition )

¨ Leaving Group and Electrophile ¤ Q. 6.20, 6.22

¨ Nucleophile and Solvent Effects ¤ Q.6.21, Q6.29

¨ Deciding between SN1 and SN2 ¤ Q. 6.28, 6.35, 6.40

¨ Challenge Problems ¤ 6.41, 6.42, 6.46, 6.47

¨ Group Discussion Problems