Explain why each compound is aromatic, antiaromatic, or nonaromatic

(a) Isoxazole

(b) 1,3-thiazole

(c) Pyran

(d) Pyrylium ion

(e) γ - pyrone

(f) 1,2-Dihydropyridine

(g)Cytosine

(h)

Short Answer

Expert verified

(a)

Isoxazole

(Aromatic)

(b)

1,3-thiazole

(Aromatic)

(c)

Pyran

(Nonaromatic)

(d)

Pyrylium ion

(Aromatic)

(e)

γ - pyrone

(Aromatic)

(f)

1,2-Dihydropyridine

(Nonaromatic)

(g)

Cytosine

(Aromatic)

(h)

Antiaromatic

Step by step solution

01

Aromatic, antiaromatic or nonaromatic:

Aromatic: A molecule must be cyclic, planar, completely conjugated, and must satisfy Huckel’s rule (4n+2 pi electrons) and contains a particular number of pi electrons.

Antiaromatic: A cyclic, planar, completely conjugated compound with 4n pi electrons.

Nonaromatic: A compound that lacks (one or more) of the four requirements to be aromatic.

02

Classification of given compounds:

(a) Isoxazole is an aromatic compound because it haselectrons and one of the lone pairs of oxygen participates in an aromatic sextet. So, it obeys Huckel’s rule.

Isoxazole

(Aromatic)

(b) 1,3-thiazole is an aromatic compound because it has electrons and one of the lone pairs of sulfur participates in the aromatic sextet. So, it obeys Huckel’s rule.

1,3-thiazole

(Aromatic)

(c) Pyran is a nonaromatic compound because it has no conjugated electron system. So, it does not obey Huckel’s rule.

Pyran

(Nonaromatic)

(d) Pyrylium ion is an aromatic compound because it has electrons and one of the lone pairs of oxygen participates in an aromatic sextet. Here, the carbocation also undergoes resonance. So, it obeys Huckel’s rule.

Pyrylium ion

(Aromatic)

(e) γ is an aromatic compound because it has 4π electrons and one of the lone pairs of oxygen participates in the formation of an aromatic sextet. So, it obeys Huckel’s rule.

γ - pyrone

(Aromatic)

(f) 1,2-Dihydropyridine is a nonaromatic compound because it has no conjugated π electron system and it contains one sp3 hybridized carbon atom. So, it does not obey Huckel’s rule.

1,2-Dihydropyridine

(Nonaromatic)

(g) Cytosine is aromatic because it has 4π electrons and one lone pair of nitrogen participates in the formation of an aromatic sextet. So, it obeys Huckel’s rule.

Cytosine

(Aromatic)

(h) The given compound is antiaromatic as it contains electrons through the 4n pi-electron rule.

Antiaromatic

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Most popular questions from this chapter

Repeat Problem 16-10 for the cyclopentadienyl ions.

(a) Draw one all-bonding MO, then a pair of degenerate MOs, and then a final pair of degenerate MOs.

(b)Draw the energy diagram, fill in the electrons, and confirm the electronic configurations of the cyclopentadienyl cation and anion.

(c) Add electrons to your energy diagram to show the configuration of the cyclopentadienyl cation and the cyclopentadienyl anion. Which is aromatic and which is antiaromatic?

One of the following compounds is much more stable than the other two. Classify each as aromatic, antiaromatic, or nonaromatic

(a)Methyl heptalene

(b) Methyl azulene


(c) Methyl pentalene


(a) Draw the molecular orbitals for the cyclopropenyl case. H H H (Because there are three p orbitals, there must be three MOs: one all-bonding MO and one degenerate pair of MOs.)

(b) Draw an energy diagram for the cyclopropenyl MOs. (The polygon rule is helpful.) Label each MO as bonding, nonbonding, or antibonding, and add the nonbonding line. Notice that it goes through the approximate average of the MOs.

(c) Add electrons to your energy diagram to show the configuration of the cyclopropenyl cation and the cyclopropenyl anion. Which is aromatic and which is antiaromatic?

Question: For each NMR spectrum, propose a structure consistent with the spectrum and the additional information provided.

a. Elemental analysis shows the molecular formula to be C8H7OCl . The IR spectrum shows a moderate absorption at 1602 cm-1 and a strong absorption at 1690 cm-1 .

b. The mass spectrum shows a double molecular ion of ratio 1:1 at m/z 184 and 186.

(a) Use the polygon rule to draw an energy diagram for the MOs of a planar cyclooctatetraenyl system.

(b) Fill in the eight pi electrons for cyclooctatetraene. Is this electronic configuration aromatic or antiaromatic? Could the cyclooctatetraene system be aromatic if it gained or lost electrons?

(c) Draw pictorial representations for the three bonding MOs and the two nonbonding MOs of cyclooctatetraene. The antibonding MOs are difficult to draw, except for the all-antibonding MO

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