a) Draw the structure of 2,2-dimethylbutane.
b) Draw the Newman projections to show the most stable conformation, sighting along the C1-C2 bond and C3-C4 bond.
c) Draw the Newman projections to show the most stable and least stable conformations, sighting along the C2-C3 bond. Calculate the rotation barrier sighting along the C2-C3 bond. (Use the kcal/mol as a dimension.)
The Correct Answer and Explanation is :
Let me walk you through the structure and conformations step by step:
Part a) Structure of 2,2-Dimethylbutane
The molecular formula for 2,2-dimethylbutane is C8H18. The structure is as follows:
- The parent chain is butane (C4H10), which is a straight chain of four carbon atoms.
- On the second carbon of the butane chain, there are two methyl groups (–CH₃) attached.
The structure of 2,2-dimethylbutane can be represented as:
CH₃
|
CH₃–C–CH₂–CH₃
|
CH₃
This structure has a central C2 carbon attached to two CH₃ groups, and the remaining C atoms are part of a straight four-carbon chain.
Part b) Newman Projections for C1-C2 and C3-C4 Bonds
C1-C2 Bond:
- The most stable conformation of a molecule is typically the staggered conformation where the atoms on adjacent carbons are as far apart as possible, minimizing steric repulsion.
- For the C1-C2 bond in 2,2-dimethylbutane, the most stable conformation would have the C1 and C2 carbon atoms positioned in a staggered manner.
Here is how to visualize it:
- Looking along the C1-C2 bond, the groups attached to the C2 carbon (the two methyl groups) will be staggered relative to the hydrogen atoms on C1.
H CH₃
\ /
H–C–C–CH₂
/ \
CH₃ H
C3-C4 Bond:
- Similarly, for the C3-C4 bond, the most stable conformation would again be the staggered one.
- The groups attached to C3 and C4 (which are hydrogens and methyl groups) would be positioned to minimize steric interference.
H CH₃
\ /
H–C–C–CH₃
/ \
CH₂ H
Part c) Newman Projections for C2-C3 Bond
Most Stable Conformation:
- Looking along the C2-C3 bond, the most stable conformation is the staggered arrangement where the two methyl groups (on C2) are positioned as far apart as possible from the hydrogen atoms on C3.
Here’s a visualization of the most stable conformation:
CH₃
|
CH₃–C–C–CH₂
|
CH₃
Least Stable Conformation:
- The least stable conformation is the eclipsed conformation, where the two methyl groups are aligned with hydrogens on C3. This leads to significant steric repulsion and torsional strain.
Here’s how the least stable conformation looks:
CH₃
|
CH₃–C–C–CH₂
|
CH₃
Rotation Barrier Calculation:
The rotation barrier along the C2-C3 bond is the energy required to overcome the steric and torsional strain between the different conformers.
In 2,2-dimethylbutane, the eclipsed conformation (where the two methyl groups on C2 align with the hydrogens on C3) will experience the highest energy strain due to steric repulsion. On the other hand, the staggered conformation (where the methyl groups are as far apart as possible) is the most stable and will have the lowest energy.
The energy barrier between the staggered and eclipsed conformations is typically in the range of 3-4 kcal/mol for molecules like 2,2-dimethylbutane.
Conclusion:
- The most stable conformation for 2,2-dimethylbutane along the C2-C3 bond is the staggered one with the methyl groups as far apart as possible.
- The least stable conformation is the eclipsed conformation where the methyl groups experience maximum steric repulsion with the hydrogen atoms on C3.
- The rotation barrier between these two conformations is typically 3-4 kcal/mol. This means it takes around this amount of energy to rotate the molecule from the most stable staggered conformation to the least stable eclipsed conformation.
This analysis illustrates the typical energy differences seen in conformational isomerism, where staggered conformations are generally more stable than eclipsed ones due to reduced steric hindrance and torsional strain.