Question

The third member of the family of alkenynes has the molecular formula?
A.${{\text{C}}_6}{{\text{H}}_6}$
B.${{\text{C}}_5}{{\text{H}}_6}$
C.${{\text{C}}_6}{{\text{H}}_8}$
D.${{\text{C}}_4}{{\text{H}}_4}$

Answer 1

The alkynes have both double bond and triple bond. The general formula of the alkynes is${{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}} - {\text{4}}}}$, where ${\text{n}}$ is a natural number.

Complete step by step solution:
Alkynes is an unsaturated class of organic compounds. It is acyclic in nature that means no ring like structure will be there. In these types of compounds we have both double and triple bonds. The general formula of this type of compound is:
${{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}} - {\text{4}}}}$ , where ${\text{n}}$ is a natural number. We cannot take ${\text{n}}$ equals to zero because in that case there will be no carbon left. In organic compounds carbon is the central element and if carbon is not here compound will not exist. Now let us calculate the number of this class of compounds by putting the value of ${\text{n}}$ starting from 1.
For ${\text{n }} = {\text{ }}1$ we will get
${{\text{C}}_1}{{\text{H}}_{{\text{2}} - {\text{4}}}}$ Solving we will see that value number of hydrogen comes negative ${{\text{C}}_1}{{\text{H}}_{ - 2}}$. This is not possible since we cannot have a negative number of atoms. Hence, ${\text{n }} = {\text{ }}1$ will not give us the first member of the series.
Putting ${\text{n }} = 2$ we will get ${{\text{C}}_2}{{\text{H}}_0}$ . This also does not make any compound. Moving on to ${\text{n }} = 3$, we will get
${{\text{C}}_2}{{\text{H}}_{2 \times 3 - 4}}$ Which gives us ${{\text{C}}_3}{{\text{H}}_2}$. This also is not valid because for one double bond and one triple bond we will get:
$C \equiv C = C{H_2}$
In the above we have valency of carbon as 5, which is not possible.
Putting ${\text{n }} = 4$ we will get ${{\text{C}}_4}{{\text{H}}_{2 \times 4 - 4}}$, on solving we will get ${{\text{C}}_4}{{\text{H}}_4}$, which makes the structure as:
$HC \equiv C - CH = C{H_2}$ This is a valid structure. Hence this gives the first member of this series at ${\text{n }} = 4$. So the third member will be at ${\text{n }} = 6$.
Putting the value we will get: ${{\text{C}}_6}{{\text{H}}_8}$.
The correct option is option C

Note: The word alkenynes consists of both “ene” and “yne”. “ene” represents double bond and “yne” represents triple bond. That is why this compound has both types of bonds. In the same way alkane have single bond and alkenes have double bond and alkynes have triple bond.