Question Bank of Module 02

Q-What are the requirement of compound to be aromatic?

The three general requirements for a compound to be aromatic are:

1. The compound must be cyclic
2. Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.
3. The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).

Q-Explain Aromaticity.

The name ‘aromatic’ was originated from the characteristic odor or ‘aroma’ of benzene-like compounds, chemists now have a completely different method of deciding whether a compound is aromatic or not. Based on the analysis of a number of compounds with unusual resonance stabilization energies, the following characteristics have been accepted as criteria for aromaticity.

1. The molecule must be cyclic, planar with uninterrupted cloud of π electrons above and below the plane of the ring.

2. It should have 4n+2 π electrons.

Here every atom in the ring must have a p orbital and the delocalization should result in an uninterrupted cyclic cloud of π electrons above and below the plain of the ring. The German Chemist Erich Hückel was the first one to recognize that an aromatic compound must have an odd number of pairs of electrons, which can mathematically be written as 4n+2 (n = 0,1,2,3 etc). Molecules which obey these rules are aromatic and those which follow these rules partially fall in the category of anti-aromatic and non aromatic compounds.

Q-Explain 4n + 2 Pi Rule.

Huckel 4n + 2 Pi Electron Rule:

A ring-shaped cyclic molecule is said to follow the Huckel rule when the total number of pi electrons belonging to the molecule can be equated to the formula ‘4n + 2’ where n can be any integer with a positive value (including zero).

Examples of molecules following Huckel’s rule have only been established for values of ‘n’ ranging from zero to six. The total number of pi electrons in the benzene molecule depicted below can be found to be 6, obeying the 4n+2 𝛑 electron rule where n=1.

This rule is also justified with the help of the Pariser-Parr-Pople method and the linear combination of atomic orbitals (LCAO) method.

Generally, aromatic compounds are quite stable due to the resonance energy or the delocalized electron cloud. For a molecule to exhibit aromatic qualities, the following conditions must be met by it:

• There must be 4n + 2 𝛑 electrons present in a system of connected p orbitals (where the electrons are delocalized) belonging to the molecule.
• In order to meet the first condition, the molecule must have an approximately planar structure wherein the p orbitals are more or less parallel and have the ability to interact with each other.
• The molecule must have a cyclic structure and must have a ring of p orbitals which doesn’t have any sp3 hybridized atoms.

Other examples of aromatic compounds that comply with Huckel’s Rule include pyrrole, pyridine, and furan. All three of these examples have 6 pi electrons each, so the value of n for them would be one.

Q- Enlist the criteria that must be attained in order the compound to be aromatic.

1.An aromatic molecule must be cyclic.

2. An aromatic molecule must be planar.

3.An aromatic ring must contain only sp2-hybridized atoms that can form a delocalized system of π molecular orbitals.

4.The number of π electrons in the delocalized π system must equal 4n + 2, where n is an integer.

Q-Explain the structure of Benzene.

The molecular formula of benzene is C6H6. Benzene is a clear, colorless, highly flammable and volatile, liquid aromatic hydrocarbon with a gasoline-like odor. Benzene is found in crude oils and as a by-product of oil-refining processes. In industry benzene is used as a solvent, as a chemical intermediate, and is used in the synthesis of numerous chemicals. Exposure to this substance causes neurological symptoms and affects the bone marrow causing aplastic anemia, excessive bleeding and damage to the immune system. Benzene is a known human carcinogen and is linked to an increased risk of developing lymphatic and hematopoietic cancers, acute myelogenous leukemia, as well as chronic lymphocytic leukemia.

Q-Explain the hybridization of Benzene.

Because each carbon is only joining to three other atoms, when the carbon atoms hybridise their outer orbitals before forming bonds, they only need to hybridise three of the orbitals rather than all four. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged. The new orbitals formed are called sp2 hybrids, because they are made by an s orbital and two p orbitals reorganising themselves. The three sp2 hybrid orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane. The remaining p orbital is at right angles to them. This is all exactly the same as happens in ethene. The difference in benzene is that each carbon atom is joined to two other similar carbon atoms instead of just one. Each carbon atom uses the sp2 hybrids to form sigma bonds with two other carbons and one hydrogen atom. The p electron on each carbon atom is overlapping with those on both sides of it. This extensive sideways overlap produces a system of pi bonds which are spread out over the whole carbon ring. Because the electrons are no longer held between just two carbon atoms, but are spread over the whole ring, the electrons are said to be delocalized. The six delocalized electrons go into three molecular orbitals - two in each. In common with the great majority of descriptions of the bonding in benzene, we are only going to show one of these delocalized molecular orbitals for simplicity.

Q-Explain Pyrrole.

Microwave spectroscopic measurements show that pyrrole is a planar molecule which is symmetrical about the y-axis, i.e. it has C2v symmetry. Recent determinations of the bond lengths and angles for pyrrole support the earlier values reported by Bak, although Cumper's values for the bond lengths of the C2—C3 and C3—C4 bonds warrant closer inspection.

Pyrrole can be considered to be a resonance hybrid of canonical structures (Ala)-(Alf). Unlike benzene, where two equivalent Kekule structures can be drawn, only one classical structure (Ala), which makes a major contribution to the resonance hybrid, can be written for pyrrole. Although the zwitterionic structures (Alc)-(Alf) make smaller contributions to the resonance hybrid, it is evident that the electron density at the a- and ß-carbon atoms should be significantly greater than predicted from the classical structure. The N—C2, N—C5, and C3—C4 bonds should be shorter, and the C2—C3 and C4~C5 bonds longer than normal single and double bonds. The lone pair on nitrogen is in the p orbital so it is involved in the 6 pi-electron aromatic system. Hence pyrrole is not very nucleophilic and is only weakly basic at nitrogen. Looking at the HOMO of pyrrole the lobes are much bigger at the 2- and 5- positions, this indicates that the reactions are most likely to take place at these positions.

Attack at C2 is also most favoured if you consider resonance forms as there are three resonance forms of the cation available, whereas there are only two resonance forms following attack at C3 so attack is less favoured at C3.

Q- What is the difference between hydrogen atom abstraction and oxygen-insertion pathway for oxidation of benzene to phenol?

A-

The hydrogen atom abstraction (HAA) is a radical (1-electron) process, while "the "oxygen insertion" is 2-electron process. Benzene is never oxidized through HAA pathway. Radicals, such as HO(.), form a complex with the aromatic ring, which then further oxidized.

Resonance energy measures the extra stability of conjugated systems compared to the same number of isolated C=C. 
Naphthalene

(i)                has a benzene ring plus an extra 2 conjugated C=C and so has a higher resonance energy than benzene

(ii)               itself (but not a great as two separated benzenes). Benzene is more aromatic and has a greater resonance energy than furan

(iii)            i > ii > iii