the is renowned that dual bonds space "stiff" and make particles less likely come rotate approximately the bond. However why? What provides it so the the bond needs to be orientated a specific way?


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.