Since the discovery of benzene, a lot of compounds have been identified that behave similarly. They — Organic Chemistry Chemistry Question
Non-Benzoid Aromatic Systems
Since the discovery of benzene, a lot of compounds have been identified that behave similarly. They all have some common features. According to Hückel's rule, an aromatic system must have the following properties:
- cyclic
- fully conjugated
- planar
- 4n + 2 π electrons
Write down the number of π-electrons in each of the compounds shown below.
[VISUAL]
Model Answer
Each double bond and each heteroatom (O, N) with lone pairs donates 2 π-electrons as well as a negative charge. Boron or a positive charge does not donate any electrons to the π-system but provide an empty p-orbital for delocalization.
[VISUAL]
From left to right, the number of π-electrons in the structures shown are:
1. Boron compound: 6
2. Phenol/oxygen-heterocycle derivative: 6
3. Nitrogen heterocycle: 8
4. Oxygen heterocycle: 6
5. Cyclopentadienyl-type anion: 6
6. Five-membered ring with boron: 6
7. Five-membered ring with heteroatoms (or positive charge): 4
Which compounds are aromatic?
Model Answer
According to Hückel's rule, compounds are classified as (a) aromatic or (na) non-aromatic based on the number of π-electrons (4n+2 for aromatic):
1. 6 π-electrons: aromatic (a)
2. 6 π-electrons: aromatic (a)
3. 8 π-electrons: non-aromatic (na)
4. 6 π-electrons: aromatic (a)
5. 6 π-electrons: aromatic (a)
6. 6 π-electrons: non-aromatic (na)
7. 4 π-electrons: non-aromatic (na)
Which of the following two compounds would you expect to have a greater dipole moment? Support your answer by writing the corresponding (plausible) resonance structures.
[VISUAL]
Model Answer
Charge separation is more favourable in compound b), because there is one mesomeric resonance structure in which both rings are formally aromatic according to Hückel’s rule. In all other resonance structures at least one of the rings is formally anti-aromatic (4n π-electrons). Hence, compound b) resembles electronically a cycloheptatrienyl cation fused to a cyclopentadiene anion and therefore possesses a large dipole moment.
[VISUAL]
Which of the following three compounds can be protonated more easily? Assign the three pKb values (8.8, 13.5, 3.1) to these three compounds:
[VISUAL]
Model Answer
1. Pyrrole (pKb = 13.5):
The lone pair of nitrogen in pyrrole is involved in the aromatic π-system. Protonation destroys the aromatic sextet (only 4 π-electrons left, π-system not fully conjugated any more, because the protonated nitrogen is sp3-hybridized). Pyrrole is hence only a very weak base.
2. Pyridine (pKb = 8.8):
The lone pair of nitrogen in pyridine is not involved in the aromatic π-system; protonation is easier than in pyrrole. Nitrogen, however, is sp2–hybridized and therefore less electronegative and more difficult to protonate than in a normal amine in which nitrogen is sp3–hybridized.
3. Triethylamine (pKb = 3.1):
Triethylamine is the most basic compound in this series. The higher the p–character of the lone pair, the easier is protonation.
[VISUAL]
Cyclopentadiene (C5H6) is not an aromatic compound because it is not completely conjugated. However, in contrast to acyclic dienes, it can quite easily react with a strong base such as sodium ethoxide to form a crystalline salt.
[VISUAL]
Write down a structure for compound A.
Model Answer
As a hydrocarbon, cyclopentadiene is unusually acidic (pKa = 16). The increased acidity is due to the stability of the cyclopentadienide anion containing 6 π-electrons and in which the delocalization is extended over all 5 carbon atoms in a complete cyclically conjugated system. Hence, the anion is aromatic. Just as in benzene, the anion is symmetric (D5h–symmetry), all C–C and all C–H bonds are the same. Therefore, the 1H NMR spectrum only shows one signal.
[VISUAL]
Compound A is sodium cyclopentadienide (C5H5- Na+).
Is A aromatic according to Hückels–rule?
Model Answer
Yes, A (the cyclopentadienide anion) is aromatic because it contains 6 π-electrons (4n + 2 where n = 1) in a cyclic, fully conjugated, planar system.
How many signals in the 1 H NMR do you expect for A?
Model Answer
Since the anion is highly symmetric (D5h-symmetry) and all C-C/C-H bonds are identical, all five protons are chemically and magnetically equivalent. Therefore, the 1H NMR spectrum only shows one signal.
If A reacts in the following sequence, a stable, deep red compound X will form:
[VISUAL]
Hint: C has the following elemental composition: C 85.69 %, H 5.53 %.
Write down structures for the compounds B, C and X.
Model Answer
The reactions in the sequence are:
1. Addition of the Grignard reagent (phenylmagnesium bromide) to benzaldehyde yields the secondary alcohol benzhydrol (B): HO-CH(Ph)2.
2. Oxidation of B (using an oxidizing agent like KMnO4 or K2Cr2O7) yields the ketone benzophenone (C): O=C(Ph)2, which matches the elemental composition of C 85.69 % and H 5.53 %.
3. The cyclopentadienide anion (A) acts as a strong nucleophile and attacks the carbonyl carbon of C. Subsequent elimination of water (via an E1cB mechanism) gives the fulvene derivative X, which is 6,6-diphenylfulvene (C18H14).
[VISUAL]
Suggest a plausible reagent Z.
Model Answer
Reagent Z must be an oxidizing agent capable of converting a secondary alcohol (B, benzhydrol) to a ketone (C, benzophenone). Plausible reagents include KMnO4 or K2Cr2O7.
Cyclopentadiene has to be freshly distilled before use in the above synthesis, because it dimerizes upon prolonged standing. Suggest a structure for this dimer.
Model Answer
Cyclopentadiene is a 1,3–diene that easily reacts in a Diels–Alder [4+2]–cycloadditon. In this reaction, it is so reactive that one molecule of 1,3-cyclopentadiene (reacting as a diene) combines with another molecule (reacting as an olefin/dienophile) to form dicyclopentadiene. This bicyclic dimer is the endo adduct.
[VISUAL]