Organic Chemistry 202-SNU-RE • CEGEP, Montreal
Hydrocarbons are the entry chapter of organic chemistry, and it is where habits are formed: counting unsaturations before drawing, looking for the intermediate before predicting a product, checking a molecular formula before concluding.
This sheet leaves the course in your notes and gathers the moves that earn marks: calculating the degree of unsaturation, the three phases of a radical reaction, and the true rule of Markovnikov.
The thread of the chapter
All the reactivity of hydrocarbons can be read from the stability of the INTERMEDIATE: a radical for alkanes, a carbocation for alkenes, and the same scale serves both.
Calculate the degree of unsaturation BEFORE you start drawing: it tells you how many rings and multiple bonds you will have to place, and it eliminates half of the proposed structures.
Hydrocarbons are nonpolar: they are not miscible with water and float on it. This is the only solubility question that comes up in this chapter.
Chlorine is not very selective and attacks nearly everywhere; bromine is very selective and seeks the most substituted carbon. A statement that mentions bromine expects a single product.
Decolourization of bromine is the test for a multiple bond: an alkane does not decolourize it, an alkene and an alkyne do, and instantly.
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
What not to write
“ has degree , so it must be an alkene.”
What to write
“Degree can be a double bond OR a ring: cyclohexane has the same molecular formula as a hexene.”
Why: Closing a ring uses up two hydrogens, exactly like creating a bond. The calculation cannot tell the two apart, and it is the statement or a chemical test that settles it.
What not to write
“Neopentane is more compact, so denser, so it boils higher.”
What to write
“A branched molecule offers LESS contact area: London forces are weaker and neopentane boils at degrees against for pentane.”
Why: London forces depend on surface area, not on mass alone. Three isomers of the same molecular formula therefore rank by their shape, from the most linear to the most spherical.
What not to write
“: that is the second propagation step.”
What to write
“This step consumes TWO radicals and produces none: it is a termination.”
Why: A propagation sustains the chain: one radical in, one radical out. Simply counting the radicals on each side is enough to classify any step.
What not to write
“Chlorine will attack the tertiary carbon, like bromine.”
What to write
“Chlorine is not very selective and gives a mixture of all possible products; only bromine clearly favours the most substituted carbon.”
Why: The chlorine radical is very reactive, hence not very choosy: its transition state resembles the reactants. The bromine radical, less reactive, lets the stability of the radical formed decide.
What not to write
“We add following Markovnikov, so the first bromine goes to the carbon with more hydrogens.”
What to write
“The two atoms added are identical: there is nothing to orient, and the rule simply does not apply.”
Why: Markovnikov's rule concerns only UNSYMMETRICAL reagents, such as or . With , or , the question does not arise.
What not to write
“The is , so sodium hydroxide removes the terminal hydrogen.”
What to write
“A of is still far above that of water: a much stronger base is needed, sodium amide for example.”
Why: An acid-base reaction only goes toward the weaker acid. With , whose conjugate acid has a of , the equilibrium stays on the wrong side.
What not to write
“The bromine solution is decolourized, so there is a hydrocarbon in the tube.”
What to write
“Only multiple bonds decolourize bromine in the cold and without light: an alkane reacts only in light, and slowly.”
Why: The test distinguishes unsaturated from saturated compounds, which is its whole point. You must therefore state the conditions, no light and cold, otherwise radical halogenation muddles everything.
What not to write
“Ozonolysis cuts the carbon chain in two in the middle.”
What to write
“It cuts exactly the DOUBLE BOND, and each carbon that carried it becomes a carbonyl.”
Why: Ozonolysis is an identification tool: from two carbonyls you work back to the exact position of the double bond. The cut is made there and nowhere else.
What not to write
“The diene is conjugated, so the Diels-Alder reaction will take place.”
What to write
“The diene must also be able to adopt the s-cis conformation: a diene locked in s-trans does not react.”
Why: The six-membered ring formed requires the two ends of the diene to point to the same side. A steric lock then forbids the reaction, despite perfect conjugation.
Look at the starting functional group and the one you need to obtain.
If alkane to alkyl halide → under light, accepting a mixture with chlorine
Example: attacks mainly the tertiary carbon
If alkene to alcohol, with Markovnikov orientation → in acid medium, watching for rearrangements
Example: the goes on the most substituted carbon
If alkene to alkane → with a metal catalyst, SYN addition
Example: both hydrogens arrive from the same side
If alkene to vicinal dihaloalkane → , ANTI addition through the halonium bridge
Example: the two halogens end up on opposite faces
If you want to locate an unknown double bond → ozonolysis, then work back from the carbonyls obtained
Example: two identical ketones indicate a symmetrical alkene
All these additions consume one degree of unsaturation: the product must therefore have two more hydrogens, or two more substituents, than the reactant.
Start from the fragments obtained and join them back together.
If ozonolysis gives two different aldehydes → the double bond was between two carbons each carrying a hydrogen
Example: an unsymmetrical disubstituted alkene
If it gives a ketone and an aldehyde → one of the two double-bond carbons carried two substituents
Example: a trisubstituted alkene
If it gives a single product in double quantity → the alkene was symmetrical
Example: but-2-ene gives two molecules of ethanal
If combustion gives the amounts of and → count the carbons then the hydrogens, and conclude on the molecular formula
Example: mol of and mol of water for mol of compound give
The degree of unsaturation calculated from the molecular formula obtained must match the proposed structure. If the two do not agree, it is the structure that is wrong.
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
When to use it: The statement asks for the mechanism of a halogenation, or to justify the presence of a by-product.
Concluding sentence
“The second propagation step regenerates the halogen radical, which sustains the chain: a single initiation is enough to convert thousands of molecules.”
The trap: Using double-barbed arrows where the mechanism is radical: a radical moves with ONE electron, hence a single-barbed (fishhook) arrow.
Marking: 1 mark for the initiation, 1 mark per propagation step, 1 mark for the termination.
When to use it: The statement gives the masses of carbon dioxide and water formed from a known mass or amount of compound.
Concluding sentence
“ mol of carbon and mol of hydrogen for mol of compound give , with degree of unsaturation zero.”
The trap: Forgetting that each water molecule brings TWO hydrogens, which halves the count.
Marking: 1 mark for the amounts of substance, 1 mark for the atom count, 1 mark for the verified molecular formula.
Five minutes of checking recover more marks than one more problem started in a hurry.
The degree of unsaturation of the product
Recalculate the degree on the formula of the product: an addition uses up exactly one, an elimination creates one.
A product of addition that is still unsaturated signals a badly written reaction.
The atom count on both sides
Check that each element appears in the same number in the reactants and in the products, oxygen included.
A combustion balanced with a half coefficient is correct, provided you double the whole equation if the statement requires it.
One radical in, one radical out
Count the radicals on each side of each step written: and for a propagation, and for a termination.
Two radicals on the left and a stable product on the right: it is a termination, not a propagation.
The order of boiling points
Rank the isomers from the most linear to the most branched: the boiling point must fall in that order.
A neopentane announced as boiling higher than pentane betrays reasoning based on mass rather than on surface area.
The complete combustion of mol of a gaseous hydrocarbon produces g of carbon dioxide and g of water.
Determine its molecular formula, then write the balanced combustion equation.
Step 1
mol and mol.
Why
Convert first, without interpreting anything. Writing out the two molar masses earns the method mark, even if the calculation is done in your head.
Step 2
Carbon: mol, that is atoms per molecule.
Why
Each carbon dioxide molecule contains exactly one carbon, all of it coming from the fuel. The ratio gives the subscript directly.
Step 3
Hydrogen: mol, that is atoms per molecule.
Why
Each water molecule carries away TWO hydrogens: this is the factor that half the papers forget, and forgetting it halves the result.
Step 4
Molecular formula: , with degree of unsaturation .
Why
A degree of zero confirms an alkane, consistent with the general formula for . Two independent checks of the same result.
Step 5
Equation: .
Why
The half coefficient is perfectly legitimate; double the whole equation if the statement requires integers. The oxygen count, on each side, confirms the balancing.
Step 6
Check: mol of ethane does give mol of carbon dioxide and mol of water.
Why
Start again from the formula found and go back to the data of the statement: this is the inverse operation, and it catches any error in a subscript.
The conclusion, written out
“The hydrocarbon is ethane, of formula , and its complete combustion is written .”
The classic mistake on this problem: Counting mol of hydrogen instead of and concluding , a formula that corresponds to no stable compound.
Apply the formula that takes twice the number of carbons, plus two, plus the nitrogens, minus the hydrogens and the halogens, all divided by two. The result gives the total number of rings and multiple bonds. A triple bond counts for two degrees and an aromatic ring for four.
Because the dispersion forces between molecules depend on contact area. A linear chain lines up against its neighbours along its whole length, while a branched molecule is nearly spherical and touches them much less. Pentane thus boils at thirty-six degrees against nine and a half for neopentane, with an identical molecular formula.
By counting the radicals on each side. A propagation consumes one and produces another, which sustains the chain. A termination consumes two and produces none. An initiation consumes none and produces two, from a molecule broken by light or heat.
Only when the reagent added is unsymmetrical, such as hydrogen bromide or water. With hydrogen or bromine, the two atoms added are identical and there is nothing to orient. When it applies, it says that the proton attaches so as to form the most stable carbocation.
Much more than an alkane, but far less than water. Its acid dissociation constant corresponds to a pKa close to twenty-five, against fifty for an alkane and fifteen point seven for water. Sodium hydroxide is therefore not enough to remove it: a much stronger base is needed, such as sodium amide.
© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-hydrocarbons. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).
