Q. how to calculate the degree of unsaturation

Degree of unsaturation, also called double bond equivalents or rings plus pi bonds, counts how many pi bonds and/or rings are present in a molecule. For a molecular formula with counts of carbon \(C\), hydrogen \(H\), nitrogen \(N\), halogens \(X\) (F, Cl, Br, I), oxygen and sulfur ignored, and net charge \(z\), use the general formula:

\[ \text{Degree of unsaturation} \;=\; \frac{2C \;+\; 2 \;+\; N \;-\; H \;-\; X \;+\; z}{2} \]

Equivalent forms that are often convenient are

\[ \text{DoU} \;=\; C \;-\; \frac{H}{2} \;-\; \frac{X}{2} \;+\; \frac{N}{2} \;+\; 1 \;+\; \frac{z}{2} \]

How this is derived, step by step. Start from the formula for a saturated acyclic hydrocarbon (an alkane), which is \( \mathrm{C}_n\mathrm{H}_{2n+2} \). For a hydrocarbon only, the number of hydrogen atoms missing relative to the saturated case measures unsaturation. For hydrocarbons:

\[ \text{DoU} \;=\; \frac{2C \;+\; 2 \;-\; H}{2} \]

Include heteroatoms by treating halogens like hydrogens, since each halogen replaces one hydrogen, and treating nitrogens as contributing one extra valence (effectively reducing the required hydrogens by one per nitrogen). Oxygen and sulfur do not change the count. Including halogens and nitrogens gives the general numerator \(2C+2+N-H-X\). If the species is ionic, include the net charge \(z\) in the numerator as shown above.

Practical rules to apply when given a molecular formula.

Step 1. Identify and record counts: \(C,\ H,\ N,\ X\) and net charge \(z\). Ignore O and S.

Step 2. Plug into the formula.

Step 3. Compute the value. The result is the total number of rings plus pi bonds. Each double bond counts as 1, each triple bond counts as 2, and each ring counts as 1.

Worked examples.

Example 1. For \( \mathrm{C}_8\mathrm{H}_{10} \) : \(C=8,\ H=10,\ N=0,\ X=0,\ z=0\). Use the hydrocarbon formula.

\[ \text{DoU} \;=\; \frac{2(8) \;+\; 2 \;-\; 10}{2} \;=\; \frac{16 \;+\; 2 \;-\; 10}{2} \;=\; \frac{8}{2} \;=\; 4 \]

So there are four degrees of unsaturation.

Example 2. For \( \mathrm{C}_6\mathrm{H}_5\mathrm{Cl} \) : \(C=6,\ H=5,\ X=1,\ N=0,\ z=0\). Treat the halogen as an extra hydrogen for counting.

\[ \text{DoU} \;=\; \frac{2(6) \;+\; 2 \;-\; (5+1)}{2} \;=\; \frac{12 \;+\; 2 \;-\; 6}{2} \;=\; \frac{8}{2} \;=\; 4 \]

Example 3. For \( \mathrm{C}_2\mathrm{H}_5\mathrm{N} \) : \(C=2,\ H=5,\ N=1,\ X=0,\ z=0\). Use the full formula.

\[ \text{DoU} \;=\; \frac{2(2) \;+\; 2 \;+\; 1 \;-\; 5 \;-\; 0}{2} \;=\; \frac{4 \;+\; 2 \;+\; 1 \;-\; 5}{2} \;=\; \frac{2}{2} \;=\; 1 \]

Notes and common pitfalls. Oxygen and sulfur do not appear in the formula and are ignored. Halogens are counted as hydrogens. Nitrogen contributes half an extra hydrogen equivalent per nitrogen in the algebraic form, which is why the formula contains \(+N\) in the numerator. For charged species include the net charge \(z\) (positive or negative) in the numerator as shown. The final value should be a nonnegative integer or half-integer, but for typical neutral organic molecules it is an integer. Each integer DoU corresponds to one ring or one pi bond.