Source: CHEM 237 Organic Chemistry, University of Washington
Difficulty: Intermediate | Prerequisites: Lewis structures, acid-base fundamentals (Ch. 3), basic bonding and hybridisation
These chapters introduce the structural features of unsaturated molecules (double and triple bonds) and the reactions they undergo. You will learn how to determine the degree of unsaturation from a molecular formula, how to name and distinguish stereoisomers around double bonds, and how alkyne acidity connects back to hybridisation. The ozonolysis reactions here are your first real encounter with "cleavage" reactions that break carbon-carbon bonds, which will appear again in later chapters. If you are comfortable with Lewis structures and pKa, you are ready.
The index of hydrogen deficiency (IHD) tells you how many double bonds or rings a molecule has from its formula alone. Alkynes are more acidic than alkenes or alkanes because of their greater s-character. Alkynes can be selectively reduced to cis-alkenes (Lindlar catalyst) or trans-alkenes (Na in liquid NH₃). Ozonolysis cleaves double bonds to give carbonyls, with the workup conditions determining whether you get aldehydes or carboxylic acids.
Index of hydrogen deficiency (IHD)
Also called degree of unsaturation. The number of pairs of hydrogen atoms "missing" compared to the fully saturated formula. Each unit of IHD corresponds to one double bond or one ring. A triple bond counts as two.
In simple terms, IHD tells you how many double bonds and/or rings are hiding in a molecular formula.
E/Z isomers
Stereoisomers around a double bond, assigned using Cahn-Ingold-Prelog priority rules. E (entgegen) = higher-priority groups on opposite sides. Z (zusammen) = higher-priority groups on the same side.
Think of Z as "zame zide" (same side) to remember.
Cis/trans isomers
An older naming system for stereoisomers around a double bond. Cis = same side, trans = opposite side. Used when two identical substituents are on the double bond; E/Z is the more general system.
Alkyne
A hydrocarbon containing a carbon-carbon triple bond. General formula CₙH₂ₙ₋₂. The triple bond consists of one sigma bond and two pi bonds.
Enol
A functional group with a hydroxyl group (OH) attached to a carbon that is part of a C=C double bond. Written as OH-C=C. Think of it as a hybrid of "ene" (double bond) and "ol" (alcohol).
Lindlar's catalyst
A "poisoned" palladium catalyst (Pd on CaCO₃, treated with lead acetate and quinoline) that reduces alkynes to cis-alkenes selectively, without further reduction to alkanes.
In simple terms, it is a deliberately weakened catalyst that stops the reaction at the halfway point.
Ozonolysis
A reaction that cleaves carbon-carbon double bonds using ozone (O₃), producing carbonyl compounds. The products depend on the workup conditions.
Reductive workup
Ozonolysis followed by a mild reducing agent such as (CH₃)₂S (dimethyl sulfide). Produces aldehydes and/or ketones. Aldehydes are preserved (not oxidised further).
Oxidative workup
Ozonolysis followed by H₂O₂ in water. Oxidises any aldehydes to carboxylic acids. Ketones are unaffected.
Optically active
A compound that rotates plane-polarised light. Requires at least one chiral centre and no internal plane of symmetry.
Enantiomeric excess (ee)
A measure of optical purity. Calculated as: ee = (major − minor) / (major + minor) × 100%. An ee of 100% means a pure enantiomer; 0% means a racemic mixture.
Diastereomers
Stereoisomers that are not mirror images of each other. They have different physical properties (unlike enantiomers, which share most physical properties).
IHD = (2C + 2 + N − H − X) / 2
C = number of carbons
H = number of hydrogens
N = number of nitrogens
X = number of halogens (F, Cl, Br, I)
Each IHD unit = one ring or one double bond. A triple bond = 2 units of IHD.
Reference formulas to know:
Alkane (fully saturated, no rings): CₙH₂ₙ₊₂ → IHD = 0
Alkene (one double bond): CₙH₂ₙ → IHD = 1
Alkyne (one triple bond): CₙH₂ₙ₋₂ → IHD = 2
Benzene ring = 4 units of IHD (3 double bonds + 1 ring)
Oxygen and sulfur do not appear in the IHD formula; they do not change the hydrogen count
Double bonds prevent free rotation, so substituents can be locked on the same side or opposite sides
E/Z system (Cahn-Ingold-Prelog priorities):
Assign priority to the two substituents on each carbon of the double bond using atomic number (higher atomic number = higher priority)
Z: higher-priority groups on the same side
E: higher-priority groups on opposite sides
Example priority ranking from the notes: Br (#1) > O (#2), because Br has higher atomic number
Cis/trans is used when the double bond has identical groups on each carbon:
Cis = same side
Trans = opposite side
E/Z is always valid; cis/trans only works for simple cases. When in doubt, use E/Z.
The pKa of a terminal alkyne C-H is approximately 25
Compare: alkene C-H ≈ 44, alkane C-H ≈ 50
Acidity order: alkynes > alkenes > alkanes
This means alkynes are the most acidic hydrocarbons (though still weak acids overall)
The reason is s-character:
sp hybridised carbon (alkyne) has 50% s-character
sp² hybridised carbon (alkene) has 33% s-character
sp³ hybridised carbon (alkane) has 25% s-character
More s-character means the bonding electrons are held closer to the nucleus, stabilising the conjugate base (the carbanion) after deprotonation
The rule: more s-character → more acidic; more p-character → more basic
Lower pKa = higher acidity (same principle from Ch. 3)
Reduction to cis-alkene: Lindlar's catalyst
Reagent: H₂ with Lindlar's catalyst (Pd/CaCO₃, poisoned with lead acetate)
The catalyst is deliberately weakened ("poisoned") so it stops at the alkene stage
Produces cis (syn) addition of H₂ across the triple bond
The hydrogens add to the same face of the molecule
Reduction to trans-alkene: Na in liquid NH₃ (dissolving metal reduction)
Reagent: sodium metal in liquid ammonia
Results in anti addition of hydrogen atoms across the triple bond
Produces a trans-alkene
The mechanism involves radical intermediates, not catalytic hydrogenation
Key distinction: Lindlar → cis; Na/NH₃ → trans. This is a classic exam question.
Ozone (O₃) cleaves C=C double bonds, breaking the molecule into two carbonyl fragments
Step 1: Ozone reacts with the alkene to form an ozonide intermediate
Step 2: The workup determines the products:
Reductive workup (O₃, then (CH₃)₂S or Zn):
A disubstituted alkene carbon (R₂C=) → ketone
A monosubstituted alkene carbon (RHC=) → aldehyde
Preserves aldehydes as aldehydes
Oxidative workup (O₃, then H₂O₂/H₂O):
A disubstituted alkene carbon → ketone (unchanged)
A monosubstituted alkene carbon → carboxylic acid (aldehyde is oxidised further)
If the alkene carbon bears a hydrogen, the reductive workup gives an aldehyde; the oxidative workup gives a carboxylic acid
Practical tip: to predict ozonolysis products, mentally break the double bond, put an =O on each carbon where the double bond was, then check whether the workup would further oxidise any aldehydes
Ozonolysis also works in reverse for exam problems: given the carbonyl fragments, you can reconstruct the original alkene by joining the carbonyl carbons with a double bond
Index of hydrogen deficiency
IHD = (2C + 2 + N − H − X) / 2
Enantiomeric excess
ee = (major − minor) / (major + minor) × 100%
Alternatively: 100% = major + minor (where major and minor are the percentages of each enantiomer)
Ozonolysis product summary
Alkene carbon type | Reductive workup (O₃, then (CH₃)₂S) | Oxidative workup (O₃, then H₂O₂/H₂O) |
|---|---|---|
R₂C= (no H) | Ketone | Ketone |
RHC= (one H) | Aldehyde | Carboxylic acid |
H₂C= (two H's) | Formaldehyde | CO₂ + H₂O |
Ozonolysis is used industrially to determine the position of double bonds in complex natural products like fatty acids and terpenes. Lindlar's catalyst is essential in the synthesis of cis-alkenes found in pheromones and pharmaceutical intermediates, where the geometry of the double bond determines biological activity.
Students often forget that oxygen does not appear in the IHD formula. Adding an oxygen to a molecular formula does not change the IHD.
Students often confuse Lindlar's catalyst with Na/NH₃. Lindlar gives cis; Na/NH₃ gives trans. The mnemonic: Lindlar loves cis.
Students often assume ozonolysis always gives the same products regardless of workup. The workup is critical: reductive preserves aldehydes, oxidative converts them to carboxylic acids.
Students often mix up E/Z and cis/trans. E/Z uses CIP priority rules and works for all alkenes. Cis/trans only works when identical groups are compared.
⚠️ IHD calculation from a molecular formula is a quick-points exam question. Practise until it takes under 30 seconds.
⚠️ Know the difference between reductive and oxidative ozonolysis workup and be able to predict products for both.
⚠️ Be able to work ozonolysis backwards: given the products, reconstruct the starting alkene.
⚠️ Lindlar vs Na/NH₃ stereoselectivity is tested frequently. Know which gives cis and which gives trans.
⚠️ The notes flag: "practise identifying R and S like question 1 on quiz" and "practise with enantiomeric excess equations."
True or False: A molecule with IHD = 4 must contain a benzene ring. (False, it could be other combinations of rings and double bonds)
Fill in the blank: Lindlar's catalyst reduces an alkyne to a ______ alkene. (cis)
True or False: Ozonolysis with oxidative workup converts aldehydes to carboxylic acids. (True)
Fill in the blank: More s-character in a C-H bond means the bond is more ______. (acidic)
True or False: Oxygen atoms are included in the IHD formula. (False)
Q: Calculate the IHD for C₈H₈BrN.
A: IHD = (2(8) + 2 + 1 − 8 − 1) / 2 = (16 + 2 + 1 − 8 − 1) / 2 = 10/2 = 5.
Q: What products result from ozonolysis of 2-butene (CH₃CH=CHCH₃) with reductive workup?
A: Two equivalents of acetaldehyde (CH₃CHO). Each carbon of the double bond bears one H and one CH₃, so reductive workup gives an aldehyde from each.
Q: What products result from the same alkene with oxidative workup?
A: Two equivalents of acetic acid (CH₃COOH). The oxidative workup converts the aldehydes to carboxylic acids.
Q: How would you selectively convert 1-butyne to cis-2-butene?
A: First, deprotonate the terminal alkyne with a strong base (such as NaNH₂), then alkylate with CH₃I to form 2-butyne. Then reduce with H₂ over Lindlar's catalyst to give cis-2-butene.
Q: A mixture contains 85% (R)-enantiomer and 15% (S)-enantiomer. What is the enantiomeric excess?
A: ee = (85 − 15) / (85 + 15) × 100% = 70/100 × 100% = 70% ee.
This connects to acid-base chemistry (Ch. 3) because alkyne acidity is explained by pKa and the same s-character/hybridisation logic. The ability to deprotonate a terminal alkyne with a strong base is a direct application of pKa comparison.
This connects to stereochemistry (Ch. 6) because E/Z isomerism and CIP priorities introduced here are the foundation for R/S assignments at chiral centres. Enantiomeric excess and diastereomers become central topics in Ch. 6.
This connects to reaction mechanisms (later chapters) because ozonolysis introduces the concept of oxidation state changes at carbon, and the reductive/oxidative workup distinction foreshadows selectivity in reagent choice throughout organic synthesis.
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