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Does this compound have zero dipole moment?

enter image description here

I think it has a non-zero dipole moment. If I assume the ring to be planar, then the dipole moments of each $\ce{C-Cl}$ bond cancel out. (One is upward and the other is downward.)

But if I take the actual puckered structure of cyclobutane, one $\ce{C-Cl}$ bond is axial and the other is equatorial and hence they won't cancel out and the compound has a non-zero dipole moment.

But the answer given in my book says the compound has zero dipole moment.

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  • $\begingroup$ Is that possible when one Cl is at axial position and one Cl is in equatorial position? $\endgroup$
    – Aditya Dev
    Commented Oct 19, 2015 at 15:31
  • $\begingroup$ Should I consider ring flip? Because flipping switches chlorine from axial to equatorial and equatorial to axial. $\endgroup$
    – Aditya Dev
    Commented Oct 19, 2015 at 15:37
  • $\begingroup$ Is ring inversion in our IITJEE syllabus ? There are so many things at conflict here . $\endgroup$
    – Sujith Sizon
    Commented Oct 21, 2015 at 4:19
  • $\begingroup$ Yes. There or not there, it is indirectly present in questions $\endgroup$
    – Aditya Dev
    Commented Oct 21, 2015 at 7:11

1 Answer 1

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Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.

enter image description here

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.

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  • $\begingroup$ So cyclobutane can be assumed to be planar? Can I apply the same logic for cyclohexane? $\endgroup$
    – Aditya Dev
    Commented Oct 19, 2015 at 16:51
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    $\begingroup$ No, cyclobutane cannot be assumed to be planar, but it passes through a planar transition state when ring inverting. Here's an example of why you can't 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. $\endgroup$
    – ron
    Commented Oct 19, 2015 at 17:01
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    $\begingroup$ 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. $\endgroup$
    – Jason B.
    Commented Oct 20, 2015 at 7:20
  • $\begingroup$ @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." I've now changed the wording in the last paragraph to say things a bit better. If you'd 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 achiral $\endgroup$
    – ron
    Commented Oct 20, 2015 at 14:51
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    $\begingroup$ conformation. In this case we might say that the molecule is achiral, even though it may be in the chiral conformation at any given instant. $\endgroup$
    – ron
    Commented Oct 20, 2015 at 14:51

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