Question

Which of the following compounds possesses the \(\mathrm{C}-\mathrm{H}\) bond with the lowest bond dissociation energy? (a) N-pentane (b) Toluene (c) Benzene (d) 2,2 -dimethylpropane

Short answer

2,2-dimethylpropane has the lowest C-H bond dissociation energy.

Step-by-step solution

Understanding Bond Dissociation Energy

Bond dissociation energy is the energy required to break a specific bond in a molecule. A bond with lower dissociation energy requires less energy to break and is generally weaker.

Analyzing N-pentane

N-pentane is a straight-chain alkane with only single C-H bonds. Typically, alkanes have relatively strong C-H bonds because the bonds involve sp3 hybridized carbon.

Analyzing Toluene

Toluene is an aromatic compound with a methyl group attached to a benzene ring. The C-H bonds in the methyl group are like those in alkanes but the sp2 hybridized bonds in the aromatic ring are typically stronger due to resonance and conjugation effects.

Analyzing Benzene

Benzene is an aromatic compound with a fully conjugated ring structure. The C-H bonds in benzene are bonded to sp2 hybridized carbons in the aromatic ring, contributing to the bond strength through resonance stabilization.

Analyzing 2,2-dimethylpropane

2,2-dimethylpropane, also known as neopentane, is an alkane with tertiary carbons, which means it has many substituents around the carbon atom. Tertiary carbons form weaker C-H bonds because the surrounding groups can stabilize radicals formed when the bond is broken.

Determining the Weakest Bond

Considering all the compounds, 2,2-dimethylpropane with its tertiary C-H bonds likely has the lowest bond dissociation energy because these bonds are weaker than the primary C-H bonds in pentane and weaker than benzene and toluene’s aromatic sp2 C-H bonds.

Key concepts

Alkanes

Alkanes are a group of hydrocarbons which are composed entirely of single bonds between carbon and hydrogen atoms. In these molecules, each carbon is typically sp3 hybridized, which means that each carbon forms four sigma bonds. This hybridization results in a tetrahedral geometry around each carbon atom, allowing for a zig-zag structure along the chain.

One key characteristic of alkanes is their relative lack of reactivity when compared to other hydrocarbons like alkenes or alkynes. The single C-H and C-C bonds in alkanes are quite stable due to their sigma bond nature, making the molecules less reactive.

In terms of bond dissociation energy, alkanes generally have relatively strong C-H bonds. This is due to the electron density being distributed amongst sp3 hybridized orbitals that form these single bonds. However, the strength can vary slightly, especially in cases involving different types of carbon atoms such as primary, secondary, or tertiary carbons. The presence of tertiary C-H bonds, found in branched alkanes like 2,2-dimethylpropane, can lead to weaker bonds due to the stabilization of resulting radicals.

Aromatic Compounds

Aromatic compounds, like benzene and toluene, have distinct structural characteristics that distinguish them from other types of hydrocarbons. These compounds contain a conjugated ring system which allows for a stabilization process known as aromaticity.

One of the most famous aromatic compounds is benzene, with a ring made of six carbon atoms. Each carbon is part of a planar hexagonal structure where these carbons are sp2 hybridized. This sp2 hybridization involves each carbon forming three sigma bonds and one p-orbital contributing to a delocalized pi-electron cloud, leading to great stability.

• Benzene's C-H bonds are typically stronger than those in alkanes due to the added stability from resonance.
• Toluene, which is benzene with a methyl group attached, also shares this aromatic stability.

The presence of the aromatic system increases the bond dissociation energy of the C-H bonds directly involved with the sp2 hybridized carbons, making them resistant to breaking.

sp3 Hybridization

The concept of sp3 hybridization is integral to understanding the geometry and characteristics of alkanes. In sp3 hybridization, one s orbital and three p orbitals from a carbon atom hybridize to form four equivalent sp3 hybrid orbitals. These orbitals arrange themselves in a tetrahedral geometry for optimal spatial distribution.

This tetrahedral arrangement is responsible for the typical structure of alkanes, where each carbon forms four single bonds, maximizing bond strength and stability due to their sigma bonding nature.

Alkanes like n-pentane and 2,2-dimethylpropane show sp3 hybridization. Interestingly, in branched alkanes like 2,2-dimethylpropane, the tertiary C-H bonds can exhibit a lower bond dissociation energy than typical primary or secondary C-H bonds, due to radical stabilization by the surrounding alkyl groups upon bond breaking.

sp2 Hybridization

In aromatic compounds, sp2 hybridization plays a crucial role. Here, like in benzene and toluene, each carbon atom contributes to the conjugated ring system through two major hybridization processes: forming sigma bonds using sp2 hybrid orbitals and allowing one p orbital to engage in pi-bonding.

• sp2 orbitals form three sigma bonds, contributing to a planar structure optimal for resonance.
• The unhybridized p orbitals overlap to form a delocalized electron cloud above and below the plane of the atoms, a hallmark of aromatic compounds.

This delocalized electron cloud contributes significantly to the stability of the molecules, enhancing bond dissociation energy compared to non-aromatic hydrocarbons.

In compounds like benzene, the C-H bonds associated with sp2 hybridized carbons tend to be more robust due to this resonance stabilization. This property is further pronounced when compared to the C-H bonds found in alkanes such as n-pentane, as seen in the relative comparison of their bond dissociation energies.