Assign stereochemistry (E or Z) to the double bond in the following compound, and convert the drawing into a skeletal structure (red = O):

The cis-trans naming system works only for disubstituted alkenes having two substituents other than hydrogen on the double bond. But when trisubstituted and tetrasubstituted of the double bond are present then the E and Z system is used for describing alkene stereochemistry.

• Multiple-bonded atoms are equivalent to the same number of single bonded atoms.

• Once the groups attached have been ranked look at entire molecule. If the higher ranked groups on each carbon are on the same side of the double bond then alkene is said to have Z geometry and if present on opposite sides then the alkene has E geometry.

As ranking is atomic number dependent, more the atomic number higher will be ranking.

So,

In${\text{-CH=CH}}_2$and$\text{-CH}{\left({\text{CH}}_3\right)}_2$:

• In${\text{-CH=CH}}_2$last carbon is attached to two hydrogen and two carbon.
• In$\text{-CH}{\left({\text{CH}}_3\right)}_2$last carbon is attached to three hydrogen.

So, ${\text{-CH=CH}}_2$rank 1.

In${\text{-CH}}_2\mathrm{OH}$and${\text{-CO}}_2{\text{CH}}_3$:

• In${\text{-CH}}_2\mathrm{OH}$, carbon is attached to two hydrogen and one oxygen.
• In${\text{-CO}}_2{\text{CH}}_3$carbon is attached to three oxygen and one carbon.

So,${\text{-CO}}_2{\text{CH}}_3$ rank 1.

From the structure of the given compound, it can be seen both high ranking groups are on the same side so the Z configuration.

Skeletal structure