Difficulty: Introductory to Intermediate | Prerequisites: Molecular formulas, basic bonding, Lewis structures
The index of hydrogen deficiency (IHD) tells you how many degrees of unsaturation a molecule has, meaning how many rings or pi bonds are present. You calculate it from the molecular formula using a simple formula that adjusts for halogens, nitrogen and oxygen. It is one of the first things you check when analysing an unknown structure, and it appears on nearly every CHEM 225 exam.
Index of Hydrogen Deficiency (IHD)
The number of pairs of hydrogen atoms that must be removed from the fully saturated molecular formula (the acyclic alkane equivalent) to arrive at the actual molecular formula. Each unit corresponds to one ring or one pi bond.
In simple terms, this means: count how "short" on hydrogens a molecule is compared to if it were wide open with no rings and no double or triple bonds. Each missing pair = one ring or one pi bond.
Degree of unsaturation
Synonym for IHD. Some textbooks and professors use this term interchangeably with index of hydrogen deficiency.
Think of it as the same number, just a different name. If your professor says "degree of unsaturation" on the exam, they want the IHD.
Saturated molecule
A molecule with no rings and no pi bonds, containing the maximum possible number of hydrogen atoms for a given number of carbons. For a hydrocarbon this is C_nH_(2n+2).
In simple terms, this means: a molecule where every carbon has as many hydrogens bonded to it as it possibly can. No double bonds, no triple bonds, no rings.
For a molecular formula of the general form C_c H_h N_n O_o X_x (where X = any halogen: F, Cl, Br, I):
\text{IHD} = \frac{2C + 2 + N - H - X}{2}How each element type affects the count:
Carbon (C): Each carbon in a fully saturated acyclic molecule contributes two hydrogens plus two more for the two end carbons. That is where the 2C + 2 comes from.
Hydrogen (H): Subtracted directly. Fewer hydrogens than the saturated formula means more unsaturation.
Halogens (F, Cl, Br, I): Treated exactly like hydrogen. Each halogen replaces one hydrogen, so subtract X from the numerator the same way you subtract H.
Nitrogen (N): Each nitrogen adds one hydrogen to the saturated reference formula, so nitrogen is added (+N) in the numerator.
Oxygen (O): Oxygen has no effect on IHD at all. Ignore it completely.
Sulfur (S): Also has no effect, same reasoning as oxygen. Ignore it.
1 degree = one ring OR one double bond (one pi bond)
2 degrees = two rings, two double bonds, one ring plus one double bond, or one triple bond (two pi bonds)
4 degrees from a benzene ring (three double bonds + one ring)
A triple bond contributes 2 degrees of unsaturation, not 1.
Molecular formula: C11H20Br3F4N3O3
C = 11, H = 20, N = 3, X = 3 (Br) + 4 (F) = 7, O = 3 (ignored)
IHD = (2(11) + 2 + 3 - 20 - 7) / 2 = (22 + 2 + 3 - 20 - 7) / 2 = 0 / 2 = 0
Answer: IHD = 0 (fully saturated, no rings or pi bonds).
When you are given a drawn structure rather than a molecular formula, you have two options:
Option A: Determine the molecular formula from the structure, then apply the formula.
Option B: Count directly from the structure. Count every ring (each ring = +1) and every pi bond (each double bond = +1, each triple bond = +2).
For Cetilistat, the structure contains rings and double bonds. Counting from the structure: you should find IHD = 5.
For a molecule containing a ring with nitrogen and two C=O groups (as in question 5 of the mock exam), count: 1 ring + 2 double bonds from the C=O groups + additional pi bonds visible in the structure. The answer for that molecule is IHD = 5.
Students often think oxygen changes the IHD. It does not. Oxygen slots into a C-H or C-C bond without changing the hydrogen count, so it has zero effect on the formula.
Students sometimes subtract nitrogen instead of adding it. Nitrogen adds one hydrogen to the saturated reference, so it goes in as +N, not -N.
A triple bond is worth 2 degrees of unsaturation, not 1. Students who count a triple bond as a single unit of unsaturation will be off by one.
When counting IHD from a drawn structure, students forget to count the ring itself as +1 on top of any pi bonds within it. A benzene ring is 4 (three double bonds + one ring), not 3.
IHD calculation from a molecular formula is a near-guaranteed exam question. Know the formula cold.
You will also be asked to determine IHD from a drawn structure. Practise counting rings and pi bonds quickly.
The exam may include molecules with nitrogen, halogens and oxygen together to test whether you handle each element correctly.
A high IHD (4+) is a strong hint that a benzene ring is present.
True or false: Oxygen increases the IHD by 1. (False, oxygen has no effect.)
Fill in the blank: A triple bond contributes ____ degrees of unsaturation. (2)
True or false: Halogens are treated the same as hydrogen in the IHD formula. (True, they are subtracted just like H.)
Fill in the blank: The IHD of benzene (C6H6) is ____. (4)
True or false: Nitrogen is subtracted in the IHD formula. (False, nitrogen is added.)
Q: What is the IHD of a compound with molecular formula C8H14BrNO2?
A: IHD = (2(8) + 2 + 1 - 14 - 1) / 2 = (16 + 2 + 1 - 14 - 1) / 2 = 4/2 = 2. The molecule has two degrees of unsaturation.
Q: A molecule has IHD = 4 and contains only carbon, hydrogen and oxygen. What structural feature is most likely present?
A: A benzene ring (3 double bonds + 1 ring = 4). An IHD of 4 in a molecule with only C, H and O is a strong indicator of an aromatic ring.
Q: You draw out a structure and count 2 rings and 3 double bonds. What is the IHD?
A: IHD = 2 + 3 = 5.
Q: A molecule has the formula C5H8. What is the IHD, and what are two possible structural interpretations?
A: IHD = (2(5) + 2 - 8) / 2 = 4/2 = 2. Possible interpretations: two double bonds, one ring and one double bond, or one triple bond.
IHD connects directly to spectroscopy (IR, NMR, mass spec) where it is one of the first pieces of information you use to narrow down a structure. It also ties into aromatic chemistry: whenever IHD >= 4, consider aromaticity. In reaction mechanisms, knowing the IHD before and after a reaction helps you track whether a ring formed or a pi bond was created or consumed.
Related Terms / Search Tags
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