D. A. Evans Carboxylic Acids(& Esters): Anomeric Effects Again? Chem 206 I Conformations: There are 2 planar conformations Hyperconjugation: Let us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane (2 Confor (2 Conformer (E)Conformer R C-O lone pair is aligned to overlap Formic H^o△G°=+2 carmol The(E)conformation of both acids and esters is less stable by 2-3 kcal/mol. If :(E) Conformer this equilibrium were governed only by steric effects one would predict that the (E) conformation of formic acid would be more stable(H smaller than =O) Since this is not the case there are electronic effects which must also be co In the(E)cont considered. These effects will be introduced shortly lone pair is aligned to overlap o*C-R Rotational Barriers: There is hindered rotation about the =c-or bond These resonance structures suggest Since o*C-O is a better acceptor than GC-R hindered rotation about =c-or bond where R is a carbon substituent)it follows that This is indeed observed the(2)conformation is stabilized by this interaction R Esters versus Lactones: Questions to Ponder. o Rotational barriers are- 10 kcal/mol R G°-2-3 This is a measure of the strength of Esters strongly prefer to adopt the() conformation while CHaCH pi bond g lactones such as 2 are constrained to exist in the mation. From the preceding discussion explain the I Lone Pair Conjugation: The oxygen lone pairs conjugate 1)Lactone 2 is significantly more susceptible to nucleophilic attack at the carbonyl carbon than 1? Explain versus The filled oxygen p-orbital interacts with pi(and pi") 2) Lactone 2 is significantly more prone to enolization than 1? C=O to form a 3-centered 4-electron bonding system In fact the pKa of 2 is -25 while ester 1 is-30(DMSO). Explain 3)In 1985 Burgi, on carefully studying/Ow the X-ray structures of a number of SP2 Hybridization lactones, noted that the o-c-c (a)& o-C-o(B)bond angles were not equa a a Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridizati i Explain the indicated trend in bond is driven by system to optimize pi-bonding angle changes a-B=123°a-B=6.9°a-=45
3) In 1985 Burgi, on carefully studying the X-ray structures of a number of lactones, noted that the O-C-C (a) & O-C-O (b) bond angles were not equal. Explain the indicated trend in bond angle changes. a-b = 12.3 ° a-b = 6.9 ° a-b = 4.5 ° a b a b a b Lactone 2 is significantly more prone to enolization than 1? In fact the pKa of 2 is ~25 while ester 1 is ~30 (DMSO). Explain. 2) 1) Lactone 2 is significantly more susceptible to nucleophilic attack at the carbonyl carbon than 1? Explain. Esters strongly prefer to adopt the (Z) conformation while small-ring lactones such as 2 are constrained to exist in the (Z) conformation. From the preceding discussion explain the following: 2 1 versus Esters versus Lactones: Questions to Ponder. Since s* C–O is a better acceptor than s* C–R (where R is a carbon substituent) it follows that the (Z) conformation is stabilized by this interaction. (E) Conformer In the (E) conformation this lone pair is aligned to overlap with s* C–R. s* C–R s* C–O In the (Z) conformation this lone pair is aligned to overlap with s* C–O. (Z) Conformer ■ Hyperconjugation: Let us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane. ■ Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridization is driven by system to optimize pi-bonding. The filled oxygen p-orbital interacts with pi (and pi*) C=O to form a 3-centered 4-electron bonding system. SP2 Hybridization ■ Lone Pair Conjugation: The oxygen lone pairs conjugate with the C=O. Rotational barriers are ~ 10 kcal/mol This is a measure of the strength of the pi bond. barrier ~ 10 kcal/mol DG° ~ 2-3 kcal/mol Energy These resonance structures suggest hindered rotation about =C–OR bond. This is indeed observed: + ■ Rotational Barriers: There is hindered rotation about the =C–OR bond. The (E) conformation of both acids and esters is less stable by 2-3 kcal/mol. If this equilibrium were governed only by steric effects one would predict that the (E) conformation of formic acid would be more stable (H smaller than =O). Since this is not the case, there are electronic effects which must also be considered. These effects will be introduced shortly. DG° = +2 kcal/mol Specific Case: Formic Acid (Z) Conformer (E) Conformer ■ Conformations: There are 2 planar conformations. D. A. Evans Carboxylic Acids (& Esters): Anomeric Effects Again? Chem 206 O O R' R R O R' O O O H H O H H O R O R' O O – O R' R R O R O O O R R O C O R R C O O R R C R O R O R O R O R O C O R O R O O Et CH3CH2 O O O O O O O O O R •• •• •• ••
D. A. Evans Three-center Bonds Chem 206 Consider the linear combination of three atomic orbitals. The resulting molecular orbitals(MOs) usually consist of one bonding, one nonbonding Case 3: 2 p-Orbitals: 1 s-orbital and one antibonding MO antibonding Case 1: 3 p-Orbitals pI-onentation antibonding 2 nonbonding I Case 4: 2 s-Orbitals: 1 p-orbital Do this as bonding Note that the more nodes there are in the wave function, the higher its energy Examples of three-center bonds in organic chemistry H2C=CH-CH2 Allyl carbonium ion: both pi-electrons in bonding state A. H-bonds:(3-center, 4 electron) 0-H-0 The acetic acid dimer is H2c=CH—CH2Alyl 2 electrons in bonding obital plus 3 stabilized by ca 15 kcal/mol nonbonding MO B. H-B-H bonds:(3-center, 2 electron) Case 2: 3 p-Orbitals sigma-orientation antibonding i diborane stabilized by 35 kcal/mol nonbonding C. The SN2 Transition state: (3-center, 4 electron) The SN 2 transition state approximates a case 2 situation with a central carbon p-orbita The three orbitals in reactant molecules 1 nonbonding MO from Nucleophile (2 electr bonding 1 bonding Mo o C-Br(2 electrons) 1 antibonding MOσC-B
Consider the linear combination of three atomic orbitals. The resulting molecular orbitals (MOs) usually consist of one bonding, one nonbonding and one antibonding MO. Case 1: 3 p-Orbitals 3 Energy bonding nonbonding antibonding Note that the more nodes there are in the wave function, the higher its energy. + Allyl carbonium ion: both pi-electrons in bonding state ● Allyl Radical: 2 electrons in bonding obital plus one in nonbonding MO. – Allyl Carbanion: 2 electrons in bonding obital plus 2 in nonbonding MO. antibonding nonbonding bonding Energy 3 Case 2: 3 p-Orbitals pi-orientation sigma-orientation 2 + Case 3: 2 p-Orbitals; 1 s-orbital Examples of three-center bonds in organic chemistry A. H-bonds: (3-center, 4 electron) The acetic acid dimer is stabilized by ca 15 kcal/mol B. H-B-H bonds: (3-center, 2 electron) diborane stabilized by 35 kcal/mol C. The SN2 Transition state: (3-center, 4 electron) The SN 2 transition state approximates a case 2 situation with a central carbon p-orbital The three orbitals in reactant molecules used are: 1 nonbonding MO from Nucleophile (2 electrons) 1 bonding MO s C–Br (2 electrons) 1 antibonding MO s* C–Br D. A. Evans Three-center Bonds Chem 206 H2C CH CH2 H2C CH CH2 H2C CH CH2 O H H O O O CH3 CH3 B H B H H H H H B H B H H H H H C H H H Nu Br bonding nonbonding antibonding Case 4: 2 s-Orbitals; 1 p-orbital Do this as an exercise