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Pyridine is less basic than triethylamine because (a) pyridine has aromatic character (b) nitrogen in pyridine is \(\mathrm{sp}^{2}\) hybridised (c) pyridine is a cyclic system (d) in pyridine, lone pair of nitrogen is delocalised.

Short Answer

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(b) nitrogen in pyridine is  hybridised; however, (d) impacts the basicity more.

Step by step solution

01

Understand Aromatic Character

Pyridine is an aromatic compound, which means it has a stable ring structure that involves delocalization of electrons. Being aromatic gives pyridine specific stability but does not directly affect its basicity significantly compared to non-aromatic amines.
02

Analyze Hybridization of Nitrogen

The nitrogen atom in pyridine is \(\) hybridized, meaning that the lone pair on the nitrogen is in an \(\) orbital. Hybridization can affect the availability of electrons for bonding. However, this situation typically makes the electrons more available because they are in a higher energy orbital compared to \(\) hybridization, thus not reducing the basicity of pyridine compared to aliphatic amines.
03

Consider the Cyclic System

While pyridine is cyclic, its cyclic nature itself does not inherently reduce the basicity. The basicity is more influenced by the nature and availability of the lone pair of electrons on the nitrogen, not merely the geometric shape of the molecule.
04

Delocalization of Nitrogen's Lone Pair

In pyridine, the nitrogen has a lone pair of electrons that is not involved in the aromatic \(\pi\) system, meaning these electrons are not delocalized. This is actually the reason pyridine is basic because it can donate this lone pair. However, compared to triethylamine, the lone pair is less available for bonding because it is part of an aromatic electron cloud, lowering its basicity relative to triethylamine.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aromatic Character
Pyridine is an example of an aromatic compound, which is a kind of molecule that has a special stability due to its ring structure. This stability comes from a property known as "aromaticity." Aromatic molecules follow Huckel's rule, which states that such compounds have a conjugated system of \(\pi\)-electrons distributed over a cyclic arrangement and typically contain \(4n+2\) \(\pi\)-electrons, where \(n\) is an integer.
This creates a stable electron cloud above and below the plane of the molecule, making it less reactive in terms of breaking the \(\pi\)-bonding. However, it is key to note that aromaticity itself does not have a significant direct effect on the basicity of pyridine compared to simple amines such as triethylamine, which are not aromatic. While aromaticity contributes to the stability of pyridine, it means that the electrons are engaged in maintaining this stable configuration, which indirectly influences the electron availability from the nitrogen atom for bonding purposes.
Hybridization of Nitrogen
In pyridine, the nitrogen atom is \(sp^2\) hybridized. This means its lone pair of electrons resides in an \(sp^2\) hybrid orbital, which is higher in energy compared to \(sp^3\) hybrid orbitals found in aliphatic amines like triethylamine. The \(sp^2\) hybridization involves one \(s\) orbital mixing with two \(p\) orbitals to form three equivalent \(sp^2\) orbitals, while the lone pair remains in one of these orbitals.
  • The hybridization affects how tightly the electrons are held by the nitrogen atom.
  • The energy of the \(sp^2\) hybridized orbitals makes electrons slightly less available for protonation compared to \(sp^3\).
Nonetheless, the hybridization mainly influences the shape and electron distribution in the molecule, so while it might not decrease the basicity in isolation, it works in conjunction with other factors to influence pyridine's overall basicity relative to non-cyclic and non-aromatic amines.
Delocalization of Electrons
Pyridine's nitrogen atom has a lone pair of electrons that is not part of the aromatic \(\pi\)-electron system. This contrasts with the \(\pi\)-delocalization that involves the carbon atoms around the pyridine ring. This separation means that the lone pair is relatively free to engage in bonding, contributing to pyridine's basicity.
However, the lone pair is also less available than in a non-aromatic amine because of the surrounding \(\pi\)-electron cloud influencing the electron dynamics around the nitrogen atom. Here are a few points on this:
  • The lone pair is part of the \(sp^2\) hybrid orbital but not spread out across the entire ring.
  • This positioning makes it potentially more reactive towards accepting a proton, theoretically increasing the basicity.
  • Despite this potential, due to its aromatic context, the electron cloud around the ring offers some competitive electron density that reduces the lone pair's full availability.
Ultimately, the nature of electron delocalization in pyridine contributes to its distinct behavior and accounts for its relatively lower basicity when compared to substances like triethylamine where electrons are fully available for bonding. This intricate balance in pyridine demonstrates how aromatic systems challenge straightforward conclusions about basicity and electron availability.

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Most popular questions from this chapter

During dehydration of alcohols to alkenes by heating with concentration \(\mathrm{H}_{2} \mathrm{SO}_{4}\) the initiation step is (a) protonation of alcohol molecule (b) formation of carbocation (c) elimination of water (d) formation of an ester

Tertiary alkyl halides are practically inert to substitution by \(\mathrm{SN}^{2}\) mechanism because of (a) insolubility (b) instability (c) inductive effect (d) steric hindrance

Match the following: List I List II (Reagents) (Nature) 1\. \((\mathrm{CN})\) (i) Neutral nucleophile 2\. (ONO) (ii) electrophile 3\. \(\mathrm{NH}_{3}\) (iii) Ambident nucleophile 4\. \(\mathrm{AlCl}_{3}\) (iv) both El and Nu The correct matching is: 1 \(2 \quad 3\) 4 (a) (i) (iii) (ii) (iv) (b) (iii) (iii) (i) (ii) (c) (iv) (iii) (iii) (i) (d) (ii) (iv) (i) (iii)

The increasing order of stability of the following free radicals is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}\) (b) \(\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (c) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\)

Which of the following alkyl halides would be the most reactive in an SN reaction? (a) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (b) CC(CBr)CBr (c) CC(C)(Br)c1ccccc1 (d) CC(Br)Cc1ccccc1

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