
Consider a carbon-carbon ($ceC=C$) bond.
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A dual bond consists of a $sigma$ bond and a $pi$ bond. A $pi$ bond is developed by party overlapping the unhybridized p-orbitals of two carbon atoms, above and below the plane of carbon atoms.
If now one of the carbon atoms of the twin bond is rotated v respect come the other, the p-orbitals will no longer overlap, and also the $pi$ bond must break. But the break of the $pi$ bond calls for 251 kJ/mol the energy, which is not noted by the collision that the molecule at room temperature. In turn the rotation around a carbon-carbon twin bond is not free, but it is strongly hindered or restricted.
In carbon-carbon single bond ($ceC-C$) only 12.55 kJ/mol of power is forced (this data is for ethane molecule). At room temperature, the collision of molecule supply adequate kinetic energy to get over this power barrier.
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Besides the thermodynamic obstacle with respect to double bond rotation, us could likewise look in ~ it indigenous a the opposite perspective.
Rotating a twin bond, without breaking the $pi$ shortcut first, calls for that you revolve the left or ideal orbitals that get involved in the $sigma$ and $pi$ bond (but not both atoms" orbitals, since that would certainly be rotating the molecule, not the bond).
Suppose the we room looking in ~ a $2p_x-2p_x$ $pi$ overlap because that a $ extC= extC$ bond. If we rotate the $2p_x$ orbital $90^o$ about the internuclear ($z$) axis, we deserve to transform the $2p_x$ orbital right into a $2p_y$ orbital. Ethene, because that example, is a molecule that $D_2h$ symmetry, and contains a $ extC= extC$ bond.
According come this D2h personality table, in ethene, the $2p_x$ orbit transforms under the $B_3u$ IRREP, however the $2p_y$ orbit transforms under the $B_2u$ IRREP.
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Two orbitals transforming under different IRREPs cannot overlap, so the $pi$ bond deserve to no much longer be made if among the orbitals is rotated.
Beyond that, if girlfriend rotate one of the $2p_x$ orbitals about the internuclear axis through some tiny angle (instead of particularly $90^o$), friend would change the orientation the the x-axis for just that orbital, and they would technically no be the very same orbital anymore due to the fact that each would certainly be complying with a various x- (and y-)axis convention.