Question

What is the structure for $3,4 - \dim ethylpent - 4 - en - 1 - yne$ and $2,3 - \dim ethylpent - 1 - en - 4 - yne$ ?

Answer 1

The IUPAC name can be written by considering the root word from the longest carbon chain. The substituents must be written with positions also. If there were more than one functional group, then the alphabetical order should be followed to write the IUPAC nomenclature.

Complete answer:
Alkanes are the saturated hydrocarbons consisting of carbon-carbon single $\left( {C - C} \right)$ bonds. Alkenes are the unsaturated hydrocarbons consisting of one or more carbon-carbon double $\left( {C = C} \right)$ bonds. Alkenes are the unsaturated hydrocarbons consisting of one or more carbon-carbon triple $\left( {C \equiv C} \right)$ bonds.
Given IUPAC name is $3,4 - \dim ethylpent - 4 - en - 1 - yne$ which means there are two methyl groups at $3,4$ positions, double bond at ${4^{th}}$ position and triple bond at ${1^{st}}$ position. The base chain which is the longest carbon chain has five carbon atoms. Thus, the root word is pent.
The structure for the IUPAC name of $3,4 - \dim ethylpent - 4 - en - 1 - yne$ is

Given IUPAC name is $2,3 - \dim ethylpent - 1 - en - 4 - yne$ which means there are two methyl groups at $2,3,$ positions, double bond at ${1^{st}}$ position and triple bond at ${4^{th}}$ position. The base chain which is the longest carbon chain has five carbon atoms. Thus, the root word is pent.
The structure for the IUPAC name of $2,3 - \dim ethylpent - 1 - en - 4 - yne$ is

Both the above two structures were the same, but the numbering was done from alkene side in one structure and from alkyne side in another structure.

Note:
When any compound consists of two functional groups, the functional group that comes first in alphabetical order should be written first. Thus, in both the structures, the IUPAC name $2,3 - \dim ethylpent - 1 - en - 4 - yne$ is the correct one, as alkene comes first and has a lower numbering than alkynes.