# In infrared spectroscopy, why is there no C=C reading for E alkenes?

For the (E)-2-pentene, there was no C=C reading, about 1600. For the (Z)-2-pentene, there is a C=C. Why is there a reading on the Z but not on the E?

## 1 Answer

The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment. If we examine the $\ce{C=C}$ stretch in cis- and trans-2-butene we find that

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• cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we would expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule
• trans-2-butene does not have a dipole moment (0 D) and upon stretching the double bond the dipole moment will remain zero, there is no change; therefore we would not expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule

A general rule based on these dipole moment arguments is that, for symmetrically di-substituted olefins $\ce{C=C}$ stretches in trans olefins are non-existent or very weak, the absorption is much stronger in the cis isomers.

We can extend this reasoning to molecules with dipole moments, if the dipole moment is very small. Trans-2-pentene does have a dipole moment but it is very small and it is less than the dipole moment in the cis-isomer and the dipole moment in the trans-isomer (being smaller to begin with) will change less upon stretching the double bond than it will for the cis isomer. Therefore we would expect a very weak absorption in trans-2-pentene, while the cis isomer should show a stronger absorption.