Solução dos exercícios Carey - parte A

Solução dos exercícios Carey - parte A

(Parte 1 de 10)

Solutions to Problems

Chapter 1

1.1. a. A dipolar resonance structure has aromatic character in both rings and would be expected to make a major contribution to the overall structure.

b. The “extra” polarity associated with the second resonance structure would contribute to the molecular structure but would not be accounted for by standard group dipoles.

c. There are three major factors contributing to the overall dipole moments: (1) the -bond dipole associated with the C−O and C−N bonds; (2) the -bond dipole associated with delocalization of electrons from the heteroatom to the ring; and (3) the dipole moment associated with the unshared electron pair (for O) or N−H bond (for N). All these factors have a greater moment toward rather than away from the heteroatom for furan than for pyrrole. For pyrrole, the C−N dipole should be larger and the N−H moment in the opposite direction from furan. These two factors account for the reversal in the direction of the overall dipole moment. The AIM charges have been calculated.

Hunshared pair bond electrons electronsO<N

N–H dipole bond O>N

AIM charges

2 Solutions to Problems

1.2. a. The nitrogen is the most basic atom.

PhCH=N+Ph H b. Protonation on oxygen preserves the resonance interaction with the nitrogen unshared electron pair.

c. Protonation on nitrogen limits conjugation to the diene system. Protonation on C(2) preserves a more polar and more stable conjugated iminium system. Protonation on C(3) gives a less favorable cross-conjugated system.

d. Protonation on the ring nitrogen preserves conjugation with the exocyclic nitrogen unshared electrons.

charge can be delocalizedcharge is localized on exocyclic nitrogen


1.3. a. The dipolar resonance structure containing cyclopentadienide and pyridinium rings would be a major resonance contributor. The dipole moments and bond lengths would be indicative. Also, the inter-ring “double bond” would have a reduced rotational barrier.

b. The dipolar oxycyclopropenium structure contributes to a longer C−O bond and an increased dipole moment. The C=O vibrational frequency should be shifted toward lower frequency by the partial single-bond character. The compound should have a larger pKa for the protonated form, reflecting increased electron density at oxygen and aromatic stabilization of the cation.

Solutions to Problems O–

Ph PhO Ph Ph c. There would be a shift in the UV spectrum, the IR C=O stretch, and NMR chemical shifts, reflecting the contribution from a dipolar resonance structure.

1.4. a. Amides prefer planar geometry because of the resonance stabilization. The barrier to rotation is associated with the disruption of this resonance. In MO terminology, the orbital with the C=O ∗ orbital provides a stabilized delocalized orbital. The nonplanar form leads to isolation of the nitrogen unshared pair from the C=O system.


CH3 b. The delocalized form is somewhat more polar and is preferentially stabilized in solution, which is consistent with the higher barrier that is observed. c. Amide resonance is reduced in the aziridine amide because of the strain associated with sp2 hybridization at nitrogen.

Ph C N+

The bicyclic compound cannot align the unshared nitrogen electron pair with the carbonyl group and therefore is less stable than a normal amide.

1.5. a. The site of protonation should be oxygen, since it has the highest negative charge density.

4 Solutions to Problems b. The site of reaction of a hard nucleophile should be C(1), the carbonyl carbon, as it has the most positive charge. c. A soft nucleophile should prefer the site with the highest LUMO coefficient.

The phenyl group decreases the LUMO coefficient, whereas an alkyl group increases it. Reaction would be anticipated at the alkyl-substituted carbon. 1.6. The gross differences between the benzo[b] and benzo[c] derivatives pertain to all three heteroatoms. The benzo[b] compounds are more stable, more aromatic, and less reactive than the benzo[c] isomers. This is reflected in both Hf and the HOMO-LUMO gap. Also the greater uniformity of the bond orders in the benzo[b] isomers indicates they are more aromatic. Furthermore benzenoid aromaticity is lost in the benzo[b] adducts, whereas it increases in the benzo[c] adducts, and this is reflected in the TS energy and H‡. The order of H‡ is in accord with the observed reactivity trend O > NH > S. Since these dienes act as electron donors toward the dienophile, the HOMO would be the frontier orbital. The HOMO energy order, which is NH > S > O, does not accord with the observed reactivity. 1.7. The assumption of the C−H bond energy of 104kcal/mol, which by coincidence is the same as the H−H bond energy, allows the calculation of the enthalpy associated with the center bond. Implicit in this analysis is the assumption that all of the energy difference resides in the central bond, rather than in strain adjustments between the propellanes and bicycloalkanes. Let BEc be the bond energy of the central bond:

This result indicates that while rupture of the center bond in [2.2.1]propellane is nearly energy neutral, the bond energy increases with the smaller rings. The underlying reason is that much more strain is released by the rupture of the [2.2.1]propellane bond than in the [1.1.1]propellane bond.

1.8. The various HH2 values allow assigning observed HH2 and Hisom as in the chart below. Using the standard value of 27.4kcal/mol for a cis-double bond allows the calculation of the heats of hydrogenation and gives a value for the

“strain” associated with each ring. For example, the HH2 of cis-cyclooctene is only 23.0kcal/mol, indicating an increase of 27 4−23 0 = 4 4kcal/mol of strain on going to cyclooctane. The relatively high HH2 for trans-cyclooctene reflects the release of strain on reduction to cyclooctane. The “strain” for each compound is a combination of total strain minus any stabilization for conjugation. The contribution of conjugation can be seen by comparing the conjugated 1,3- isomer with the unconjugated 1,4- and 1,5-isomers of cyclooctadiene and is about 4±1kcal/mol. Since the “strain” for cyclooctatetraene is similar to the other systems, there is no evidence of any major stabilization by conjugation.

5 Solutions to Problems

1.9. By subtracting the value for X=H from the other values, one finds the “additional” resonance stabilization associated with the substituent. There is some stabilization associated with the methyl and ethyl groups and somewhat more for ethenyl and ethynyl. This is consistent with the resonance concept that the unsaturated functional groups would “extend” the conjugation. The stabilization for amino is larger than for the hydrocarbons, suggesting additional stabilization associated with the amino group. The stabilizations calculated are somewhat lower than for the values for groups directly on a double bond. 1.10. The gas phase G gives the intrinsic difference in stabilization of the anion, relative to the corresponding acid. The reference compound, CH3CO2H, has the highest value and therefore the smallest intrinsic relative stabilization. The differ- ential solvation of the anion and acid can be obtained from p. 5 by subtracting the solvation of the acid from the anion. The numbers are shown below. The total stabilization favoring aqueous ionization, relative to acetic acid, is the sum of the intrinsic stabilization and the solvation stabilization. These tend to be in opposite directions, with the strongest acids having high intrinsic stabilization, but negative relative solvation.

X Net solvation Intrinsic total

6 Solutions to Problems

We see that the final stabilization relative to acetic acid gives the correct order of pKa. Interestingly, the solvation of the stronger acids is less than that of the weaker acids. This presumably reflects the effect of the stronger internal stabilization. These data suggest that intrinsic stabilization dominates the relative acidity for this series, with solvation differences being in the opposite direction. 1.1. These observations are the result of hyperconjugation between the nitrogen unshared electron pair and the axial C−H bonds. The chair conformation of the piperidine ring permits the optimal alignment. The weaker C−H bond reflects N → ∗ delocalizations. The greater shielding of the axial hydrogen is also the result of increased electron density in the C−H bond. The effect of the axial methyl groups is one of raising the energy of the unshared electrons on nitrogen and stabilizing the radical cation.

1.12. a–c. Each of these substitutions involves extending the conjugated system and results in an MO pattern analogous to allyl for fluoroethene and to butadiene for propenal and acrylonitile, respectively.

(a) (b) (c) d. The addition of the methyl group permits → ∗ and → ∗ interactions that can be depicted by the -type methyl orbitals. The orbitals can be depicted as the symmetry-adapted pairs shown. As a first approximation, one of each pair will be unperturbed by interaction of the adjacent orbital because of the requirement that interacting orbitals have the same symmetry.

e–f. The substituents add an additional p orbital converting the conjugated system to a benzyl-like system. In the benzyl cation, the 4 orbital is empty, resulting in a positive charge. In fluorobenzene, the pz orbital on fluorine

Solutions to Problems will be conjugated with the system and 4 will be filled. This results in delocalization of some -electron density from fluorine to the ring. The electronegative character of fluorine will place the orbitals with F participation at somewhat lower energy than the corresponding orbitals in the benzyl system. As a first approximation, the two benzene orbitals with nodes at C(1) will remain unchanged.

benzene benzyl fluorobenzene

Comment. With the availability of suitable programs, these orbitals could be calculated.

1.13. a. The resonance interactions involve → ∗ hyperconjugation in the case of methyl and n → ∗ conjugation in the case of NH2, OH, and F, as depicted below.

VB description of interaction with donor substituents

MO description of interaction with donor substituents b. There are two major stabilizing factors at work. One is the delocalization depicted for both the methyl group and heteroatoms. The order of this effect should be NH2 > OH > CH3, which is in accord with the observed order of the increase in stability. The other factor is the incremental polarity of the bonds, where an increment in stability owing to the electronegativity difference should occur. This should be in the order F > OH > NH2, but this order seems to be outweighed by the effect of the -electron delocalization.

8 Solutions to Problems c. The NPA charges are in qualitative agreement with the resonance/polar dichotomy. The electron density on the unsubstituted carbon C(2) increases, as predicted by the resonance structures indicating delocalization of the heteroatom unshared electron pair. The charge on the substituted carbon C(1) increases with the electronegativity of the substituent. As is characteristic (and based on a different definition of atomic charge), the AIM charges are dominated by electronegativity differences. There is some indication of the

-donor effect in that C(2) is less positive in the order NH2 < OH < F. 1.14. a. In the strict HMO approximation, there would be two independent and ∗ orbitals, having energies that are unperturbed from the isolated double bonds, which would be + in terms of the HMO parameters. b. There would now be four combinations. The geometry of the molecule tilts the orbitals and results in better overlap of the endo lobes. 1 should be stabilized, whereas 2 will be somewhat destabilized by the antibonding interactions between C(2) and C(6) and C(3) and C(5). 3 should be slightly stabilized by the cross-ring interaction. The pattern would be similar to that of 1,3-butadine, but with smaller splitting of 1 and 2 and 3 and 4. c. The first IP would occur from 2, since it is the HOMO and the second IP would be from 1. The effect of the donor substituent is to lower both IPs, but

IP1 is lowered more than IP2. The electron-withdrawing substituent increases both IPs by a similar amount. The HOMO in the case of methoxy will be dominated by the substituted double bond, which becomes more electron rich as a result of the methoxy substituent. The cyano group reduces the electron density at both double bonds by a polar effect and conjugation.

The HMO orbitals would each have energy α+β.

1.15. a. Since there are four electrons in the pentadienyl cation, 2 will be the HOMO.

b. From the coefficients given, the orbitals are identified as 2 and 3, shown below. 2 is a bonding orbital and is antisymmetric. 3 is a nonbonding orbital and is symmetric.

(Parte 1 de 10)