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Does trans 1,3-dichlorocyclobutane have zero dipole moment?
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Aaron Mishkin
Does trans 1,3-dichlorocyclobutane have zero dipole moment?
Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.
The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid. The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)
$\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
$\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
$\ce{C_{s}}$ (the molecule only has a plane of symmetry)
can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.
Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.
Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.
The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid. The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)
$\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
$\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
$\ce{C_{s}}$ (the molecule only has a plane of symmetry)
can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.
Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.
@JasonB What you write is correct, even when the molecule is inverting rapidly it does have an instantaneous dipole moment. I attempted to make that point when I wrote, "when flipping is rapid the dipole moment averages out to zero." Ive now changed the wording in the last paragraph to say things a bit better. If youd like to further modify the answer, please feel free to do so. In a situation like this I think it might be common usage to say that the molecule has zero or no dipole moment. It seems akin to the situation where a chiral conformation is in equilibrium with an achiralMore
Is it strictly correct to say that when flipping is rapid the molecule has zero dipole moment? Logically, it seems that it should have an oscillating dipole whose time average is zero (would it be stochastically oscillating?). So the ensemble average and the time average would both be zero but the instantaneous dipole would not.More
No, cyclobutane cannot be assumed to be planar, but it passes through a planar transition state when ring inverting. Heres an example of why you cant assume cyclobutane (or cyclohexane) is planar. As you noted, the two substituents are different in the puckered form. One substituent is axial and one equatorial, therefore they will have different reactivities. If you treated cyclobutane as planar you would not have axial and equatorial positions, both positions would be equivalent and you would expect equivalent reactivities, which is not correct.More
Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.
The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid. The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)
can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.
Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.
Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.
The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid. The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)
can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.
Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.
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