Question

Which member in each of the following sets ranks higher?

$-{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}\mathrm{or}-\mathrm{CH}={\mathrm{CH}}_2$

Short answer

$-CH=C{H}_2$Ranks higher.

Step-by-step solution

E and Z designation

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

Rule for ranking substituents

• Consider each of the double-bond carbons separately and ranks the two substituents according to the atomic number of the first atom in each. An atom with a higher atomic number will rank higher.

• If the first atom same then, look at the second, third, or fourth atoms away from the double-bond until the first difference is found.

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

Ranking of given groups

As ranking is atomic number dependent, more the atomic number higher will be ranking. So,$-CH=C{H}_2$ ranks higher as its carbon has two carbon and one hydrogen attached while$-C{H}_{2}C{H}_{2}Br$ has carbon with two hydrogens and one carbon attached.