Difficulty: Beginner to Intermediate | Prerequisites: Basic molecular formulae, bonding, valence
The index of hydrogen deficiency (IHD) tells you how far a molecular formula is from being fully saturated. It counts the total number of rings and pi bonds in a molecule before you ever look at a structure. In Organic Chemistry II, IHD is a first-pass tool: you see a formula, you calculate how "unsaturated" it is, and that constrains what functional groups and ring systems are possible. You should already be comfortable with Lewis structures, valence rules, and the idea that carbon forms four bonds.
IHD (degree of unsaturation) counts the number of rings plus pi bonds in a molecule. You calculate it from the molecular formula using a simple formula that adjusts for each type of atom. Each double bond or ring adds one unit; each triple bond adds two.
Index of hydrogen deficiency (IHD)
The number of pairs of hydrogen atoms "missing" from a molecular formula compared to the fully saturated, acyclic reference compound. Each unit corresponds to one ring or one pi bond.
In simple terms, it tells you how many rings and/or double bonds (or triple bonds) the molecule contains.
Degree of unsaturation (DoU)
Synonym for IHD. Some textbooks and professors prefer this term. Means the same thing.
Saturated compound
A molecule containing only single bonds and no rings. For a hydrocarbon, the saturated formula is CnH(2n+2). Think of it as the "maximum hydrogen" baseline.
Unsaturation
Any structural feature that reduces the hydrogen count below the saturated reference: a double bond, a triple bond, or a ring. Each of these costs one degree of unsaturation (except a triple bond, which costs two).
For a general molecular formula CaHbNcOdXe (where X = any halogen: F, Cl, Br, I):
IHD = (2a + 2 + c - b - e) / 2
Carbon (C): contributes 2a to the numerator. More carbons means a higher saturated hydrogen count, so more room for unsaturation.
Hydrogen (H): subtracted directly. Fewer hydrogens than the saturated reference means more unsaturation.
Nitrogen (N): adds +1 per nitrogen to the numerator (because each nitrogen increases the saturated hydrogen count by one).
Oxygen (O): ignored. Oxygen does not change the IHD calculation at all. It can insert into a C-H or C-C bond without changing the hydrogen count.
Halogens (F, Cl, Br, I): treated exactly like hydrogen. Each halogen subtracts 1 from the numerator, same as H does.
1 degree = one double bond OR one ring
2 degrees = one triple bond, OR two double bonds, OR two rings, OR one ring + one double bond
A benzene ring alone accounts for 4 degrees of unsaturation (3 double bonds + 1 ring)
a = 11 (carbons), b = 20 (hydrogens), c = 3 (nitrogens), d = 3 (oxygens, ignored), e = 7 (3 Br + 4 F = 7 halogens)
IHD = (2(11) + 2 + 3 - 20 - 7) / 2 = (22 + 2 + 3 - 20 - 7) / 2 = 0 / 2 = 0
Answer: A) 0. This molecule is fully saturated with no rings or pi bonds.
When you are given a structure rather than a formula, you have two options:
Count directly from the structure: count every ring (each ring = +1) and every pi bond (double bond = +1, triple bond = +2).
Derive the molecular formula from the structure, then apply the formula above.
The direct counting method is faster if the structure is simple. For complex structures with heteroatoms, deriving the formula and using the IHD equation is more reliable.
The drug Cetilistat has a structure containing a nitrogen-containing ring fused with other features. Counting from the structure: there are rings and pi bonds that total to IHD = 7. Answer: E) 7.
Students often forget that halogens are treated like hydrogen in the formula. Bromine, chlorine, fluorine, and iodine each subtract 1 from the numerator, just as H does. They do not add unsaturation.
Oxygen is sometimes mistakenly included in the calculation. Oxygen has zero effect on IHD. Whether you have zero oxygens or ten, the IHD stays the same.
A common error is forgetting that a triple bond counts as two degrees of unsaturation, not one. A nitrile (C=N triple bond) or an alkyne (C=C triple bond) each add 2 to the IHD.
Students sometimes confuse IHD with the number of pi bonds alone. Rings also contribute, and a ring has no pi bond at all. A cyclopropane ring contributes 1 IHD despite having only single bonds.
IHD questions appear on nearly every CHEM 225 exam and are typically quick marks if you have the formula memorised.
Expect both types: given a molecular formula (use the equation) and given a drawn structure (count rings + pi bonds).
Watch out for molecules with many heteroatoms (N, O, halogens). The formula handles them cleanly, but manual counting can trip you up.
An IHD of 4 or more should make you think about aromatic rings (a benzene ring = 4 IHD). This is a useful crosscheck when looking at drug structures.
True or false: Oxygen atoms affect the IHD calculation. False.
A molecule with IHD = 4 could contain a benzene ring. True (benzene = 3 double bonds + 1 ring = 4).
Fill in the blank: Each halogen atom is treated like ______ in the IHD formula. Hydrogen.
True or false: A triple bond contributes 1 degree of unsaturation. False (it contributes 2).
The IHD of cyclohexane (C6H12) is ______. 1 (one ring, no pi bonds).
Q: What is the IHD of a compound with the molecular formula C11H20Br3F4N3O3?
A: IHD = (2(11) + 2 + 3 - 20 - 7) / 2 = 0. The molecule is fully saturated. (Mock Exam Q1, Answer: A)
Q: A molecule drawn with one six-membered ring containing nitrogen, one ester group (C=O), and no other pi bonds or rings has what IHD?
A: 1 ring + 1 C=O pi bond = IHD of 2. (Relates to Mock Exam Q5, Answer: A)
Q: A benzene ring fused to a five-membered ring, with no additional pi bonds outside the aromatic system, has what IHD?
A: Benzene contributes 4 (3 double bonds + 1 ring). The fused five-membered ring adds 1 more ring. Total IHD = 5.
Q: You are given the drug Cetilistat's structure. Without knowing the molecular formula, how do you find the IHD?
A: Count every ring in the structure (+1 each) and every pi bond (+1 for each double bond, +2 for each triple bond). Sum them. For Cetilistat, the total is 7. (Mock Exam Q2, Answer: E)
IHD connects directly to functional group identification: once you know the degree of unsaturation, you can narrow down which functional groups are present. It also feeds into spectroscopy (IR, NMR, mass spec), where knowing the IHD before interpreting spectra saves significant time. In later chapters, the concept reappears when analysing reaction products to confirm that the right bonds formed.
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