The Electrophile. S N 2 and E2 least stable most stable least hindered most hindered. S N 1 and E1. > x > >

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1 The Electrophile 1 Recall that electrophile means electron- loving. When considering substitution and elimination reactions we must consider the carbon attached to the leaving group. Is it a primary, secondary, or tertiary carbon? S N 1 and E1 Electrophile Type Favors Disfavors Methyl S N 2 S N 1, E1, E2 Primary S N 2, E2 S N 1, E1 Secondary < < < methyl and primary carbocation too unstable to exist. No S N 1 or E1 S N 1, S N 2, E1, E2 Resonance Stabilized S N 1, S N 2 E1, E2 Primary Tertiary S N 1, E2, E1 S N 2 S N 2 and E2 least stable most stable least hindered most hindered LG methyl electrophile has no β hydrogen. No E2 > x > > LG LG LG Tertiary electrophile is too sterically crowded for direct nucleophilic attack. No S N 2

2 2 The Electrophile: S N 1 and E1 Reactions, Carbocation Stability The first step in a S N 1 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 C 3 C 3 C 3 What is happening when we say > > > inductive effects? 3 C C 3 3 C 3 o = 3 carbons 2 o = 2 carbons 1 o = 1 carbon methyl Primary carbocations may be formed if they have resonance stabilization hyperconjugation C C substituted carboncation Empty p orbital C methyl carboncation N N

3 Examining the Leaving Group and Electrophile 3 weak bases= GD leaving group strong base = BAD leaving group Potential Leaving Groups no S N 2 3 Favors S N 1, E1, E2 1 Favors S N 2, E2 2 Favors S N 1, S N 2, E1, E2 Electrophilic centers sterically hindered (no S N 2) not sterically hindered (allows for E2) Zaitsev s Rule: the most substituted alkene will be most stable and therefore the most favored product most substituted is the E2 product

4 The Leaving Group and the Electrophile: Examples 4 The leaving group here would be -, a bad leaving group No reaction is likely to take place here without further modifying the the substrate. 2 The leaving group here would be -, a good leaving group The electrophile is a secondary electrophile, SN1, SN2, E1, and E2 are all possible C C aloalkane produced by red algea The leaving group here would be -, a good leaving group. owever, it is attached to an Sp2 carbon, no substitution or elimination reaction will occur here.

5 Practice Question 5 Which substrate will undergo the fastest S N 2 reaction? A F F bad LG 1 o electrophile B I I good LG 3 o electrophile C bad LG 1 o electrophile D good LG 2 o electrophile

6 The Nucleophile 6 Recall that nucleophile means nucleus loving or positive-charge loving Nucleophiles 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 onsted base or simply a base. The proton is acting like an electrophile. Nucleophile Type Favors Disfavors Strong Nuc/Strong Base Strong Nuc/ Weak Base Weak Nuc/ Strong Base S N 2, E2 S N 2 E2 S N 1, E1 S N 1, E1, E2 S N 1, S N 2, E1 Weak Nuc/Weak Base S N 1, E1 S N 2, E2

7 7 The Nucleophile: Factors That Affect Nucleophilicity and Basicity ow do we know if a nucleophile is acting as a nucleophile or a base? Consider the following properties: Charge Steric indrance Strength of nucleophile/base Electronegativity Solvent

8 The Nucleophile: Charge and Electronegativity 8 Charge The greater the electron density on the nucleophile the greater its nucleophilicity Increased electron density results in an increased desire to donate those electrons; ie: increased nucleophilicity neutral = S N 1, E1 charged = S N 2, E2 The conjugate base is 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= Electronegativity= Nucleophilicity Basicity

9 9 Rank the Following Nucleophiles in rder of Increasing Nucleophilicity (1 being the weakest 5 being the strongest) N no charge, weakest less electronegative than oxygen, strongest

10 The Nucleophile: Steric indrance 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. 3 C! + LG bulky nucleophile = E2 N Non-Nucleophilic Bases N N N Li N If your nucleophile is bulky then its electrons may not be able to reach the antibonding orbitals at the carbon

11 Practice Question 11 t-butoxide has the following structure. Based on what you know about nucleophilicity what type of reaction or reactions are favored? A. Both S N 1, E1 B. Both S N 2, E2 C. Primarily E1 D. Primarily E2 E1 or S N 1 unlikely as this is a charged nucleophile (not A or C) bulky nucelophiles do not allow for S N 2 (not B)

12 Solvent Effects : The Electrophile 12 Solvent Type Favors Disfavors Polar Protic S N 1, E1 Polar Aprotic S N 2, E2 S N 1, E1 Non-Polar S N 1, S N 2, E1, E2 Polar protic has hydrogen bound to an electronegative atom Eg. 2 Polar aprotic no hydrogens, but still contains a dipole Eg. C 2 2 Non-polar has no dipoles Eg. hexanes, benzene S N 1 and E1 reactions proceed more quickly in polar protic solvents Intermediates are solvated through hydrogen bonding Leaving Group Stabilization δ+ δ- Carbocation Stabilization δ- δ- δ- δ- δ+ LG δ+ δ+ δ+ Polar protic solvents favour S N 1 and E1 reactions because the intermediates are solvated

13 Solvent effects: The Nucleophile 13 Nucleophilicity can be affected by solvent Polar protic solvents cause nucleophilicity to increase going down the periodic table Polar aprotic solvents cause nucleophilicity to decrease going down the periodic table

14 Solvent Effects: The Nucleophile 14 Polar protic 3 C water alcohols carboxylic acids Polar aprotic F < ydrogen bonding between lone pairs and polar protic solvents result in dispersion of negative charge. Lone pairs are tied up in -bond and less available to be donated I Na Favour S N 2, E2 Polar aprotic solvents have dipoles present which could interact with the nucleophile, however the partial positive charge created from the dipole is sterically hindered and therefore the nucleophile is unable to interact with these solvent. Polar aprotic solvents help in the solvolysis of cations which result in naked anions Non-polar solvent é interaction with solvent = ê nucelophilicity ether hexane Recall that like dissolves like. as many nucleophiles are charged or very polar they will not dissolve in non-polar solvents

15 15 Can you rank the following nucleophiles in order of increasing nucleophilicity? (1 being the weakest 5 being the strongest) 3 C Polar protic solvent S Se q ow would the rate of the reaction change if the solvent was changed from methanol to DMF NaS Me or DMF protic aprotic The protic solvent can interact with the nucelophile and keep it from reacting with the alkyl bromide. Therefore, DMF increases the rate of reaction.

16 Temperature Effects 16 When all else is equal increased temperatures will favor elimination reactions over substitution reactions Temperature Favors Disfavors igh E1, E2 S N 1, S N 2 Low S N 1, S N 2 E1, E2 ΔG 2 ΔG<0 reaction is spontaneous ΔG>0 reaction is non-spontaneous ΔG=0 reaction is in equilibrium ΔG=Δ-TΔS ΔG 1 ΔG LG LG B Nuc B Nuc LG LG 2 products 3 products Elimination reaction produces more products, this is favoured in the presence of heat as it increases the universe s entropy.

17 Mind Map 17

18 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. C 3 heat 1. LG? or is a better LG 2. Electrophile? 3 o S N 1, E1, E2 possible Which product is formed? C or D? 3. Nucleophile? Strong Recall Zaitsev s Rule! favours S N 2, E2 The most stable product is C. 4. Solvent? N.A. 5. Temperature? Yes eliminations preferred Therefore, E2 reaction occurs (not A or B) A B C D 3 C C 3

19 Substitution and Elimination 19 Reactions Acetone Room Temp Stereochemistry is inverted 1. LG: 2. Electrophile: 2 o, 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 N DMS eat 1. LG: 2. Electrophile: 3 o, Favours SN1, E1, E2 3. Nucleophile: Neutral but strong and Bulky, Favours E2 4: Solvent: Polar Aprotic, Favours SN2, E2 5. Temperature: eat, Favours E2 so E2 Reaction N

20 LG: 2. Electrophile: 3 o, Favours SN1, E1, E2 3. Nucleophile: Neutral, Favours SN1, E1 4: Solvent: N/A 5. Temperature: N/A so Mixture of SN1 and E1 SN1 products: mixture of stereochemistry E1 product: most stubstituded alkene favoured (Zaitsev s Rule) 3 C C 3 C 3 room temp 3 C C 3 C 3 3 C C 3 C 3 3 C C 3 C 3 1. LG: 2. Electrophile: 3 o, 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 21 heat 1. LG: 2. Electrophile: 2 o, Favours SN1, E1,SN2, E2 3. Nucleophile: Charged, favours SN2, E2 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction heat Zaitsev s Rule: most substitutted alkene is most stable and therefore most favoured product Less substituted alkene due to steric hinderance of t-butyl group on substrate and nucleophile 1. LG: 2. Electrophile: 2 o, Favours SN1, E1,SN2, E2 3. Nucleophile: Charged but bulky, favours,e2 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction too sterically hindered for base to deprotonate

22 22 What Reaction Is occuring? Which Isomer will undergo the fastest reaction? K FAST K K Favoured Chair conformation but no anti-coplanar hydrogen Less Favoured Chair conformation (axial t-butyl group) BUT anticoplanar hydrogen 1. LG: 2. Electrophile: 2 o, Favours SN1, E1,SN2, E2 3. Nucleophile: Charged but bulky, favours,e2 4: Solvent: N/A 5. Temperature: E2 so E2 Reaction 2 C C 2

23 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