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? 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 carbocation too unstable has no β hydrogen. 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 carbons 2o = 2 carbons 1o = 1 carbon methyl Empty p orbital H hyperconjugation H C C C H H substituted carboncation methyl carboncation ¨ Primary carbocations 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 base = 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 alkene 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 Haloalkane produced by red algea substrate. HO Br The leaving group here would be Br-, a good leaving group. However, it is attached to an Sp2 carbon, no substitution or elimination reaction 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 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 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 Solvent 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 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 oxygen, 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 Solvent Effects: 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 hydrogens, but still Polar Aprotic SN2, E2 SN1, E1 contains a dipole Eg. CH2Cl2 Non-Polar SN1, SN2, E1, E2 Non-polar – has no dipoles Eg. hexanes, benzene ¨ S 1 and E1 reactions proceed more quickly in polar protic solvents 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 ¨ Polar aprotic solvents 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 water alcohols carboxylic acids 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 solvolysis 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 ether hexane 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 protic solvent SH SeH OH 2 3 1 q How would the rate of the reaction change if the solvent was changed from methanol 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. Temperature Effects 16 ¨ When all else is equal increased temperatures 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 entropy. 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 Acetone O Room Temp Stereochemistry 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: mixture of stubstituded alkene stereochemistry 1.