Problems on Functional Groups and Chemical Bonding

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Chapter: Organic Chemistry : Functional Groups and Chemical Bonding

Organic Chemistry : Functional Groups and Chemical Bonding - Problems, Questions and answers on Functional Groups and Chemical Bonding


PROBLEMS

 

1 Excluding alkyl groups, name and point out the functional groups in the following molecules:


Answer:


 

2. Give the bonding scheme (orbitals, etc.) and geometry for the following functional groups:

(a) alkyl nitrile (use R for alkyl group)

(b) alkyl azide (use R for alkyl group)

(c) nitro alkane (use R for alkyl group)

(d) N -methyl pyrrole (it is aromatic)

Answer:


 

3. For the following compounds, give the approximate bond angles around the atoms indicated by an arrow:


Answer:


 

4. For the following compounds add all lone pairs of electrons to the struc-tures and then specify the type of orbital in which they are located:


CH3OCH3     

CH3CHO

CH3O

CH3C ≡O+

CH3CH=NCH3

CH3CH2NH2

CH2=N

CH3CN

CH3F

CH3Cl

CH3Br 

CH3CNO

CH3NC

CH3SCN

CH3NCS

Answer:


 

5. On the basis of electronic structure and orbital energies, supply predictions for the following and explain your answer:

a. Which will be more nucleophilic towards methyl iodide?


b. Which will be more basic?


c. Which anion will be more stable?


Answer:


Because the electrons are in an sp2 orbital which has less s character than the e in the sp orbital in the other compound.

(b)


Because of being in an orbital of 25% s character compared to 33% s character in the ketone, the e are less tightly bound and more easily donated to a proton.

(c)


Because the e are in an sp orbital with greater s character than the lone pair in the sp2 orbital of the other compound.

 

6. Which of the following compounds or ions are aromatic? Draw orbital diagrams to demonstrate why.


Answer:


 

7. Consider the tropanyl anion T and the cyclopentadienyl anion C. Which one is more stable and why? Predict the structure of each based on your analysis.


Answer:


 

8. Compound K is found almost entirely in its enol form E. Why?


Answer:

Tautomer K is a resonance-stabilized structure as shown; however, tautomer E has a principal resonance contributor which is an aromatic species. The need to separate charges decreases the stability somewhat, but the aromatic system is a large stabilizing feature. Thus E is of much lower energy than K and the equilibrium between K and E is shifted to E.


 

9. Explain why compound B can be considered a doubly aromatic molecule.


Answer:


By using the lone pair on nitrogen, each ring can have six π electrons and thus satisfy Huckel’s rule. Thus each ring is an aromatic ring and the molecule can be considered doubly aromatic. The need to separate charges to accomplish this decreases the aromatic stabilization somewhat. Nevertheless the aromatic stabi-lization makes B’ a significant resonance contributor. As a consequence B has a great deal of charge separation and thus a large molecular dipole moment of 9.6D.

 

10. It is found that bromodiazirine A undergoes loss of bromide to produce a cation much more easily than bromocyclopropane C. Can you think of a reason why?


Answer:


Both bromides give benzylic cations which are resonance stabilized by the benzene rings. The nitrogens of A are more electronegative than the carbons of C which should destabilize the ion somewhat. The greatest stabilization comes from the fact that the three-membered cationic ring of A is a 2π aromatic system. This aromatic stabilization of the cation makes its formation much more rapid than the cation from C.

 

11. Compounds L1 and L2 are both lactones. Can you think of a reason why it is much more difficult to remove the α proton of L1 than to remove the α proton of L2?


Answer:


Removal of a proton from L1 gives an anion whose lone pair is orthogonal to the π bond of the carbonyl group by virtue of the rigid geometry of the bicyclic system. Consequently the lone pair cannot overlap with the carbonyl π bond and delocalization via resonance is not possible — it is effectively a localized anion. Removal of a proton from L2 gives rise to a lone pair in a p orbital which can overlap with the carbonyl π bond and thus resonance delocalization is possible. Thus the anion from L2 is resonance stabilized and is thus formed more easily.

 

12. Explain why squaric acid ionizes completely (dissociates two protons) in water and is nearly as strong as sulfuric acid.


Answer:


The bis anion of squaric acid is highly resonance stabilized by four equivalent resonance contributors as well as an aromatic resonance contributor. This results in both protons being easily removed.

 

13. Explain why guanidine is one of the strongest noncharged organic bases known.


Answer:


Protonation of guanidine gives an ion that is stabilized by four resonance forms, three of which are equivalent. This gives a large amount of resonance stabilization to the protonated form, making guanidine easily protonated and thus a good base.

 

14. Using resonance arguments, explain why cyclopentadiene is more acidic than indene.


Answer:


The cyclopentadiene anion is stabilized by five equivalent resonance structures. The anion is an aromatic anion by virtue of it being a six-π -electron system. The indenyl anion is stabilized by a total of seven resonance contributors. However, they are nonequivalent and all but one require that the aromatic cloud of the benzene ring is disrupted. Thus, while the negative charge is well delocalized, the resonance stabilization is less than that of the cyclopentadiene system. Thus the proton is not as easily removed, making indene a weaker acid.

 

15. It has been found by NMR measurements that the α-methylene groups () of N -acetylpyrrolidine 1 are not equivalent whereas the α-methylene groups in N -(2-propenyl)-pyrrolidine 2 are equivalent. Provide an expla-nation based on resonance.


Answer:

If the CH2 groups of 1 are not equivalent, then the rate of rotation of the acetyl group is slow so that one methylene group is in the vicinity of the C=O group and the other is in the vicinity of the CH3 group. Such is not the case for 2. The rotation of the isopropenyl group is rapid and the methylene groups experience an averaged environment. Because of the electronegativity of oxygen, contributions of 1a to 1 are more important than the contributions of 2a to 2. Thus 1 has greater C–N double-bond character and it is difficult to rotate about that bond. The C–N double-bond character is small in 2, so rotation is facile, leading to an averaged environment.



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