Question

Which of the following compounds posses the C II bond with the lowest bond dissociation energy? (1) Toluene (2) Benzenc (3) n-pentane (4) 2,2 -Dimethyl propane

Short answer

2,2-Dimethylpropane

Step-by-step solution

Identify the compounds

The compounds given are Toluene, Benzene, n-pentane, and 2,2-Dimethylpropane. Understand the structure of each compound.

Determine C-H bond dissociation energy in aromatic compounds

In aromatic compounds like Toluene and Benzene, the carbon-hydrogen bonds are part of the aromatic ring, which generally have higher bond dissociation energies due to the stability of the aromatic system.

Analyze C-C bond dissociation energy in alkanes

In alkanes such as n-pentane and 2,2-Dimethylpropane, the carbon-carbon bonds vary in stability. Branched alkanes (like 2,2-Dimethylpropane) have different bond dissociation energies compared to straight-chain alkanes (like n-pentane).

Consider steric hindrance effects

Steric hindrance in highly branched alkanes like 2,2-Dimethylpropane can lead to lower bond dissociation energy because the repulsion between bulky groups makes the bond easier to break.

Compare bond dissociation energies

Toluene and Benzene have higher bond dissociation energies due to aromatic stability. Between n-pentane and 2,2-Dimethylpropane, the latter has lower bond dissociation energy due to steric hindrance effects making the bond less stable.

Conclusion

Based on the considerations above, 2,2-Dimethylpropane has the C II bond with the lowest bond dissociation energy due to steric hindrance making the bond easier to break.

Key concepts

Aromatic Compounds

Aromatic compounds, like toluene and benzene, have unique stability due to their ring structure. This stability is often called 'aromaticity'. The electrons in these compounds are delocalized, providing extra stability that's not present in non-aromatic compounds.
For example:

• Toluene has a benzene ring with a methyl group attached.
• Benzene is just the ring with no additional groups.

This stability affects bond dissociation energy. Breaking a bond in aromatic compounds typically requires more energy, making them harder to break.

Alkanes

Alkanes are simple hydrocarbons with single bonds only. They come in two main forms:

• Straight-chain alkanes (e.g., n-pentane).
• Branched alkanes (e.g., 2,2-Dimethylpropane).

These compounds' bond dissociation energies can vary based on the structure. In straight-chain alkanes, bonds are usually more stable. In branched alkanes, the structure can lead to different levels of bond stability.

Steric Hindrance

Steric hindrance refers to the physical crowding of atoms within a molecule. This crowding can make bonds easier to break.
For example:

• 2,2-Dimethylpropane has a lot of steric hindrance due to the bulky groups around the central carbon.

This crowding can force bonds to break more easily, lowering the bond dissociation energy. In simpler terms, it's like trying to pull apart a jigsaw puzzle piece in a very crowded area, which is much easier than in a sparse area.

Chemical Stability

Chemical stability refers to how resistant a compound is to breaking down. Aromatic compounds are highly stable due to their ring structure and electron delocalization. On the other hand, alkanes' stability can vary:

• Straight-chain alkanes tend to be more stable than branched ones.
• Branched alkanes like 2,2-Dimethylpropane are less stable due to steric hindrance.

Understanding the stability can help predict which bonds will break first in a reaction.