Question

Each cyclic structure shown below contains a heteroatom that has lone-pair electrons.  For each structure, answer the following questions: Are the lone-pair electrons delocalized? Is the structure aromatic?

Answer 1

General guidance

Concepts and reason

The problem is based on the concept of aromaticity in organic compounds. An organic compound is said to be aromatic if it follows certain rules.

Fundamentals

The rules that must be followed by a compound to be aromatic are as follows:

(a) The molecule must be cyclic.

(b)The molecule must have planar structure.

(c) The molecule must have conjugated $\\pi$ electrons system.

(d) The molecule must satisfy $\\left( {{\\rm{4}}n{\\rm{ + 2}}} \\right)\\pi$ electron system here, n is integer.

Those compounds which are cyclic, planar and have conjugated $\\pi$ electrons system but they follow $\\left( {4n} \\right)\\pi {\\rm{ }}$electron system are known as antiaromatic compounds.

Compounds which do not follow any one of the above rules and $\\left( {4n} \\right)\\pi {\\rm{ }}$electron rule are non-aromatic compounds.

Step-by-step

Step 1 of 8

The first structure is of pyridine which has nitrogen as a hetero atom. It is cyclic and planar. Nitrogen has lone pair of electron out of the plane. So, the lone pair of nitrogen in pyridine is localized and cannot take part in the conjugation.

The resonance structure of pyridine is as follows:

Explanation | Hint for next step

Nitrogen have five electrons in which the outermost valence electrons form bond with adjacent two carbon atoms by $s{p^2}$hybridized orbital triangular planar in structure and have its lone pair out of the plane, while the unhybridized p-orbital which contains an electron will form the $\\pi$ bond with the adjacent unhybridized p-orbital of carbon. So, the lone pair does not participate in the conjugation. Hence the lone pair is localized.

Step 2 of 8

There are 3 $\\pi$ bonds that mean 6 electrons participating in conjugation which satisfy the $\\left( {{\\rm{4}}n{\\rm{ + 2}}} \\right)\\pi$ rule. Also, the compound is cyclic and planar so, it satisfies all the rules of aromaticity. Hence the structure of pyridine is aromatic.

In structure, lone pair electrons are localized and it is aromatic.

For the first structure, total number of $\\pi$ electron is 6 which follow the $\\left( {4n{\\rm{ + 2}}} \\right)\\pi$electrons. Overall the given structure constitutes 6 membered rings, and hence they remain in planar form. The compound is aromatic.

Step 3 of 8

The second structure is of pyrrole which has nitrogen as a hetero atom. Nitrogen have the lone pair in the plane. So, the lone pair of nitrogen in pyrrole is delocalized and can take part in the conjugation.

The resonance structure of pyrrole is as follows:

Explanation | Hint for next step

Nitrogen have five electrons in which the outermost valence electrons form bond with adjacent two carbon atoms by $s{p^2}$hybridized orbital triangular planar in structure and have its lone pair in the plane. So, the lone pair of electron participates in the conjugation. Hence the lone pair is delocalized.

Step 4 of 8

The structure of pyrrole is cyclic, planar have conjugated $\\pi$ electrons of the double single lone pair conjugation system and the total number of $\\pi$ electrons which take part in conjugation is 6 which also satisfy the $\\left( {{\\rm{4}}n{\\rm{ + 2}}} \\right)\\pi$ rule. So, it satisfies all the rules of aromaticity. Hence the given structure of pyridine is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.

In compound, total number of $\\pi$ electron is 6 which follow the $\\left( {4{\\rm{n + 2}}} \\right)\\pi$electrons. Overall the structure has 5 membered rings, and hence they remain in planar form.

Step 5 of 8

The third structure is of pyryliumcation, it has oxygen as a hetero atom. Oxygen has the lone pair out of the plane. So, the lone pair of oxygen in pyrylium cation is localized and cannot take part in the conjugation.

The resonance structure of pyrylium cation is as follows:

Explanation | Hint for next step

Oxygen have six electrons in which the outermost valence electrons form sigma bond with adjacent two carbon atoms by $s{p^2}$hybridized orbital triangular planar in structure and have its lone pair out of the plane, while the unhybridized p-orbital which contains an electron will form the $\\pi$ bond with the adjacent unhybridized p-orbital of carbon. So, the lone pair do not participate in the conjugation. Hence the lone pair is localized.

Step 6 of 8

The structure of pyrylium cation is cyclic, planar, have conjugated $\\pi$ electrons of the double single double conjugation system and the total number of $\\pi$ electrons which take part in conjugation is 6 which also satisfy the $\\left( {{\\rm{4}}n{\\rm{ + 2}}} \\right)\\pi$ rule. So, it satisfies all the rules of aromaticity. Hence the given structure of pyridine is aromatic.

In structure, a lone pair electron is localized and is aromatic.

For the structure, total number of $\\pi$ electron is 6 which follow the $\\left( {4n{\\rm{ + 2}}} \\right)\\pi$electrons. Overall the given structure constitutes 6 membered ring, and hence they remain in planar form.

Step 7 of 8

The fourth structure is of furan, it has oxygen as a hetero atom. Oxygen have two lone pairs out of which one is in the plane and other is out of plane. So, one of the lone pair of oxygen in furan is delocalized and can take part in the conjugation.

The resonance structure of furan is given below:

Explanation | Hint for next step

Oxygen have six electrons in which the outermost valence electrons form sigma bond with adjacent two carbon atoms by $s{p^2}$hybridized orbital triangular planar in structure and have its lone pair in the plane, So, the lone pair participate in the conjugation. Hence the lone pair is delocalized.

Step 8 of 8

The structure of furan is cyclic, planar have conjugated $\\pi$ electrons of the double single double conjugation system and the total number of $\\pi$ electrons which take part in conjugation is 6 which also satisfy the $\\left( {{\\rm{4}}n{\\rm{ + 2}}} \\right)\\pi$ rule. So, it satisfies all the rules of aromaticity. Hence the given structure of pyridine is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.

For structure, total number of $\\pi$ electron is 6 which follow the $\\left( {4n{\\rm{ + 2}}} \\right)\\pi$electrons. Overall the given structure constitutes 5 membered ring, and hence they remain in planar form.

Answer

In structure, lone pair electrons are localized and it is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.

In structure, a lone pair electron is localized and is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.

Answer only

In structure, lone pair electrons are localized and it is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.

In structure, a lone pair electron is localized and is aromatic.

In structure, lone pair electrons are delocalized and it is aromatic.