Instance of cyclopropane:

Let us now take the instance of cyclopropane. What is its IHD? If we look the molecular formula C₃H₆, we search that the numbers of hydrogen atoms are two less than that for the corresponding saturated aliphatic hydrocarbon, i.e. propane. Therefore, according to our definition its IHD should be 1. You know, in which this compound itself does not hold any multiple bond or unsaturation per se. Then how do we account for an IHD of 1? In that case, the IHD is because of the presence of a ring in the molecule. Therefore, a ring contributes to IHD in the similar way as a double bond. As same, we could consider the contributions of halogen atoms and nitrogen atoms on IHD. These considerations provide the following formula for the computation of the index of hydrogen deficiency for a molecule from its molecular formula:

IHD = Number of carbon atoms -[Number of hydrogen atoms/2]- [Number of halogen atoms/2]+ [Number of nitrogen atoms/2] + 1

You may remember here that the divalent atoms like as oxygen and sulphur are not taken within account while calculating the IHD. Therefore, for a compound with the molecular formula as C₄H₁₀O, the index of hydrogen deficiency would be zero as shown below:

IHD = 4 - 10/2 + 1 = 5 - 5 = 0

It implies that the molecule of C₄H₁₀O does not hold any unsaturation or a ring structure.