Question

Calculate the degrees of unsaturation in a compound with the molecular formula C3H6 and draw two possible structures, including hydrogen atoms, for the formula.

Answer 1

Concepts and reason

This problem is based on the concept of isomers.

Isomers that have same molecular formula but different arrangement of atoms in space or different bonding between atoms. Isomers have same molecular formulae but different chemical and physical properties.

Fundamentals

Isomers can be determined easily if the molecular formula is known by using various methods like degree of unsaturation. Formula of degree of unsaturation is as follows:

${\rm{D}}{\rm{.B}}{\rm{.U}} = \frac{{{{\rm{N}}_{\rm{C}}}\left( {{{\rm{V}}_{\rm{C}}} - 2} \right) + {{\rm{N}}_{\rm{H}}}\left( {{{\rm{V}}_{\rm{H}}} - 2} \right)}}{2} + 1$

Here, ${{\rm{N}}_{\rm{C}}}$ is number carbon atoms, ${{\rm{V}}_{\rm{C}}}$ is valency of carbon atom, ${{\rm{N}}_{\rm{H}}}$ is number of hydrogen atoms and ${{\rm{V}}_{\rm{H}}}$ is valency of hydrogen atom.

The given molecular formula is ${{\rm{C}}_3}{{\rm{H}}_6}$ .

Degree of unsaturation $\left( {{\rm{D}}{\rm{.B}}{\rm{.U}}} \right) = \frac{{{{\rm{N}}_{\rm{C}}}\left( {{{\rm{V}}_{\rm{C}}} - 2} \right) + {{\rm{N}}_{\rm{H}}}\left( {{{\rm{V}}_{\rm{H}}} - 2} \right)}}{2} + 1$

Substitute, 3 for ${{\rm{N}}_{\rm{C}}}$ , 4 for ${{\rm{V}}_{\rm{C}}}$ , 6 for ${{\rm{N}}_{\rm{H}}}$ and 1 for ${{\rm{V}}_{\rm{H}}}$ .

$\begin{array}{c}\\\left( {{\rm{D}}{\rm{.B}}{\rm{.U}}} \right) = \frac{{3\left( {4 - 2} \right) + 6\left( {1 - 2} \right)}}{2} + 1\\\\ = \frac{{3\left( 2 \right) + 6\left( { - 1} \right)}}{2} + 1\\\\ = \frac{{6 - 6}}{2} + 1\\\\ = 1\\\end{array}$

Structures of the compound having the molecular formula of ${{\rm{C}}_3}{{\rm{H}}_6}$ are given below:

Ans:

Degree of unsaturation for the molecular formula ${{\rm{C}}_3}{{\rm{H}}_6}$ is 1.

The possible structures of the compound having the molecular formula of ${{\rm{C}}_3}{{\rm{H}}_6}$ are given below: