Question

Rank substituents in order of their priority when assigning the E or Z label to an alkene.

Rank substituents in order of their priority when assigning the E or Z label to an alkene. highest priority lowest priority

Answer 1

Concepts and reason

The problem is based on the concept of E-Z configuration.

In organic chemistry, E-Z configuration describes the absolute stereochemistry of double bonds.

Fundamentals

E-Z configuration is extension of the cis-trans configuration of double bonds. The priorities are assigned on the basis of atomic number. The atom which has highest atomic number and is directly linked to the double bonded carbon atom is assigned the first priority. If the highest priority groups are on the same side of double bond it is assigned Z configuration and if the highest priority groups are on different side, the configuration is E.

Atomic number of ${\rm{I}}$ is 53, atomic number of ${\rm{F}}$ is 9 and atomic number of ${\rm{C}}$ is 6. Therefore, ${\rm{I}}$ has the highest priority in E-Z configuration.

In ${\rm{C}}{{\rm{H}}_2}{\rm{OH}}$ and ${\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}$ , both are attached to the ${\rm{C}}$ atom. Therefore, priority is given as per the atom which is next to ${\rm{C}}$ . ${\rm{C}}{{\rm{H}}_2}{\rm{OH}}$ has higher priority than ${\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}$ . ${\rm{H}}$ has atomic number 1. Therefore, it has least priority.

The order of priority from highest priority to lowest priority is as follows:

${\rm{I > F > C}}{{\rm{H}}_2}{\rm{OH > C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}{\rm{ > H}}$

Ans:

The order of priority from highest priority to lowest priority is as follows:

${\rm{I > F > C}}{{\rm{H}}_2}{\rm{OH > C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}{\rm{ > H}}$