Degree of UnsaturationJEE Main

Rings + π bonds from a formula — interactive Chemistry simulation for IIT-JEE.

Concept

The degree of unsaturation (index of hydrogen deficiency) tells you how many rings plus π bonds a molecule has, straight from its molecular formula — before you ever draw it. Each ring or double bond removes one pair of hydrogens from the saturated maximum.

Key formula

\text{DoU} = \frac{2C + 2 + N - H - X}{2}

Derivation

A saturated acyclic molecule $C_nH_{2n+2}$ has the most hydrogens possible. Every ring or π bond costs two hydrogens, so counting how far $H$ falls below $2C+2$ (adjusted by $+N$ for nitrogen and $-X$ for halogens, oxygen being neutral) gives twice the number of rings + π bonds.

Dividing by 2 yields the DoU. A benzene ring, for example, scores 4 — one ring plus three double bonds.

Scenarios to explore

Degree of Unsaturation — Rings + π bonds straight from a molecular formula.

Real-world applications

Narrowing down structures from a molecular formula.
Interpreting mass-spectrometry and combustion data.
Quick sanity checks when proposing organic structures.

JEE exam tips

Benzene (C₆H₆) → DoU 4: a ring + three π bonds.
A triple bond counts as 2 degrees; a ring or double bond counts as 1.
A non-integer or negative DoU means the formula is impossible.

Common mistakes

Subtracting oxygen — it does not change the count.
Adding halogens instead of subtracting (they behave like H).
Forgetting to add nitrogen in the numerator.

Exam traps to avoid

Each C=O still counts as one degree of unsaturation.
Oxygen and sulfur (divalent) are simply ignored in the formula.