Revision sheet: organic molecules, functional groups and isomers (202-SN1-RE)
In 202-SN1-RE (formerly 202-NYA), this revision sheet covers organic molecules in general chemistry: recognizing the seven common functional groups, writing a molecule in three ways, counting isomers, telling constitutional isomerism from geometric isomerism, and deducing a molecule's physical properties from its functional group.
The thread: a molecular formula does not designate a molecule, it is the connectivity that determines the properties. Two substances that count the same atoms can boil a hundred degrees apart, and it is by recognizing the functional group, and nothing else, that you predict which of the two does.
Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.
The thread of the chapter
A molecular formula does not designate a molecule: it is the connectivity that determines the properties. Two substances with the same atoms can boil a hundred degrees apart.
The seven functional groups and their characteristic group
•ALCOHOL R−OH: the oxygen at the end of the chain, carrying a hydrogen. ETHER R−O−R′: the same oxygen, but INSERTED between two carbons.
•ALDEHYDE R−CHO: carbonyl C=O at the end of the chain, the carbon keeping a hydrogen. KETONE R−CO−R′: the same carbonyl, flanked by two carbons, with no hydrogen.
•CARBOXYLIC ACID R−COOH: carbonyl and hydroxyl on the same carbon. ESTER R−COO−R′: the same pattern, the hydrogen replaced by a chain.
•AMINE R−NH2: nitrogen in place of oxygen.
•In all three pairs, the functional group changes only through the POSITION of the same atom. It is always this displacement that makes the difference in properties.
Three pairs, and in each the same atom changes place: the oxygen at the end or in the middle, the carbonyl at the end or in the middle, the hydrogen of the acid replaced by a chain.
Writing and counting
•Carbon ALWAYS forms four bonds. Bonded to one, two, three, four carbons, it carries 3, 2, 1, 0 hydrogens: primary, secondary, tertiary, quaternary.
•Saturated acyclic alkane: CnH2n+2. Cycloalkane: CnH2n. An unsaturation or a ring costs exactly TWO hydrogens.
•Saturated acyclic alcohol or ether: CnH2n+2O, the oxygen being inserted without costing any hydrogens. Checking this count BEFORE looking for isomers avoids looking for ones that do not exist.
•Three ways of writing: expanded, every bond visible; condensed, the bonds to hydrogen implied; skeletal, each vertex and each end standing for a carbon, the heteroatoms staying written.
•C4H10O has SEVEN isomers, four alcohols and three ethers. A heteroatom multiplies the isomers much faster than an extra carbon, because it adds FUNCTIONAL isomerism to skeletal and positional isomerism.
The same molecule three times: every bond, then the hydrogens grouped, then the skeleton alone, where each vertex and each end is a carbon.
The class of an alcohol is read from the number of carbons attached to the carbon that carries the −OH, never from the branching of the chain. 2-Methylpropan-1-ol is primary despite its branching.
From functional group to properties
•At similar molar mass, the order of boiling points follows the order of the intermolecular forces: London forces only, then dipole-dipole, then hydrogen bonding.
•Butane 58.14 g/mol boils at −0.5 °C, propanone 58.09 g/mol at 56 °C, propan-1-ol 60.11 g/mol at 97 °C. Less than 4% apart in mass, nearly a hundred degrees apart in boiling point.
•An ALCOHOL always boils markedly higher than the ETHER that is its isomer: only the alcohol carries a hydrogen on the oxygen, so only the alcohol forms hydrogen bonds between molecules. Ethanol 78 °C against dimethyl ether −24 °C.
•Like dissolves like. A ketone cannot DONATE a hydrogen bond, having no hydrogen on its oxygen, but it can ACCEPT them from water, which is enough to make it miscible.
•The solubility of alcohols collapses as the chain gets longer: infinite for methanol, 73 g/L for butan-1-ol, 0.5 g/L for octan-1-ol, a factor of 146 for four carbons.
The carbon chain is nonpolar, the functional group is polar. Everything depends on which of the two wins, and the same reasoning explains the structure of a soap.
The mistakes that cost marks
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.
1.Concluding from a molecular formula to a molecule
the whole identification question
What not to write
“My elemental analysis gives C2H6O: so it is ethanol.”
What to write
“C2H6O has TWO isomers, ethanol, which boils at 78 °C, and dimethyl ether, which boils at −24 °C. A chemical test or a physical measurement is needed to decide.”
Two molecules, same atoms: the oxygen carries an H in ethanol, it sits between two carbons in the ether, and the boiling points are a hundred degrees apart.
Why: A molecular formula only gives the count of the atoms. The properties depend on the CONNECTIVITY, and a hundred degrees of boiling gap for the same atoms is the clearest proof. No elemental analysis settles it alone.
2.Confusing an aldehyde and a ketone
2 to 3 marks, and a laboratory test misinterpreted
What not to write
“CH3−CO−CH3 carries a carbonyl, so it is an aldehyde, and it will react with Fehling's solution.”
What to write
“The carbonyl is flanked by TWO carbons: it is a ketone. The carbonyl carbon has no hydrogen, so it is not oxidized and reacts with neither Fehling's nor Tollens' reagent.”
The same C=O: at the end of the chain, its carbon keeps a hydrogen, it is an aldehyde that is oxidized; in the middle, it has none, it is a ketone.
Why: The two functional groups share the same C=O group: what separates them is its POSITION, at the end of the chain or in the middle, and so the presence or absence of a hydrogen on this carbon. It is this hydrogen that oxidation removes, and without it nothing happens.
3.Looking for cis-trans isomerism without checking both conditions
2 marks, and two structures drawn for nothing
What not to write
“But-1-ene has a C=C double bond, so it exists as cis and trans.”
What to write
“It does have the rigidity, but its first carbon carries TWO identical hydrogens: the two drawings are the same molecule flipped over. It has no geometric isomer.”
But-1-ene does have its double bond, but its first carbon carries two H: the two drawings are the same molecule flipped over, there is neither cis nor trans.
Why: Two conditions are necessary: a rigidity, double bond or ring, AND two DIFFERENT substituents on each of the two atoms involved. But-1-ene fails the second, butane the first, and the right cause must be named in each case.
4.Classifying an alcohol by the branching of its chain
1 mark, and the announced reactivity is wrong
What not to write
“2-Methylpropan-1-ol is branched, it carries a methyl group, so it is a secondary alcohol.”
What to write
“Look at the carbon that carries the −OH: it is bonded to only ONE other carbon, so it is a PRIMARY alcohol, despite the branching.”
The class is read from the carbon that carries −OH: it is bonded to only one carbon, the alcohol is primary, whatever the branching further along.
Why: The class of an alcohol has nothing to do with the general shape of the molecule: it is read only from the neighbourhood of the functional carbon. Of the four alcohols of C4H10O, two are primary, one secondary, one tertiary, and it is this class that decides whether oxidation gives an aldehyde, a ketone or nothing at all.
5.Comparing two boiling points by molar mass
2 marks, and the whole argument has to be redone
What not to write
“Propan-1-ol weighs 60.11 g/mol and propanone 58.09: it boils higher because it is heavier.”
What to write
“Two grams per mole of difference, that is 3%, do not explain 41 degrees. It is the HYDROGEN BONDING of the alcohol, which the ketone cannot form, that makes the difference.”
Three molar masses within four percent, three boiling points spread over a hundred degrees.
Why: Molar mass acts only through London forces, and its effect is weak within a few grams. The trio butane, propanone, propan-1-ol is built for exactly this: similar masses, different intermolecular forces, and nearly a hundred degrees of gap.
6.Generalizing the miscibility of alcohols
2 marks, and a liquid-liquid extraction badly explained
What not to write
“Alcohols carry an −OH like water, so they are all miscible with water.”
What to write
“This is true for the first members of the series, but the solubility collapses with chain length: 73 g/L for butan-1-ol and 0.5 g/L for octan-1-ol.”
The −OH group does not change, the nonpolar tail gets longer: the solubility in water goes from infinite to 73 g/L then to 0.5 g/L.
Why: The −OH group stays the same as the chain gets longer, but the nonpolar part grows and ends up winning. Octan-1-ol behaves practically like a hydrocarbon. It is this same balance between polar head and nonpolar tail that makes a soap work.
7.Counting the same isomer twice
2 marks, and the same wrong count recurs with the alcohols
What not to write
“C5H12: pentane, 2-methylbutane, 3-methylbutane, 2,2-dimethylpropane, that makes four isomers.”
What to write
“3-Methylbutane does not exist: it is 2-methylbutane seen from the other end. C5H12 has only THREE isomers.”
Flipped over, 2-methylbutane seems to carry its methyl on carbon 3: we number from the end that gives the lowest number, and the two drawings are a single molecule.
Why: A chain is numbered from the end that gives the lowest number, and two numberings of the same molecule do not make two molecules. The remedy is to draw, then to check that no two drawings can be superimposed by rotation or flipping before counting them separately.
8.Forgetting that a ring costs two hydrogens
2 marks, and half of the possible structures are forgotten
What not to write
“C6H12: it must be an alkene, since it is not C6H14.”
What to write
“C6H12 can be an alkene OR a cycloalkane, cyclohexane for example. A ring and a double bond cost exactly the same number of hydrogens, namely two.”
A double bond and a ring each cost two hydrogens: C6H12 can be either.
Why: Closing a chain ties up one valence at each of its two ends, just as a double bond ties up one on each of the two carbons. The molecular formula therefore cannot tell them apart, and a chemical test, such as the decolourization of bromine water, is needed to decide.
Which method to choose
Which functional group does this molecule carry
The heteroatom present in the formula, and its position
If a nitrogen carrying hydrogens → AMINE
Example: CH3−CH2−NH2
If a singly bonded oxygen, at the end of the chain with a hydrogen → ALCOHOL
Example: CH3−CH2−OH
If a singly bonded oxygen, inserted between two carbons → ETHER
Example: CH3−O−CH3
If a carbonyl at the end of the chain, the carbon keeping a hydrogen → ALDEHYDE
Example: CH3−CHO
If a carbonyl between two carbons → KETONE
Example: CH3−CO−CH3
If carbonyl and hydroxyl on the same carbon → CARBOXYLIC ACID, or ESTER if the hydrogen is replaced by a chain
Example: CH3−COOH and CH3−COO−CH3
The order of the questions matters: first look for the heteroatom, then see whether there is a carbonyl, and only then its position. Asking the questions in this order avoids every confusion of this chapter.
Does this molecule have geometric isomers
The presence of a rigidity, then the nature of the substituents
Butane fails on rigidity, but-1-ene on asymmetry; only but-2-ene meets both conditions and exists as cis and trans.
If no double bond or ring → NO: single bonds rotate freely
Example: butane CH3−CH2−CH2−CH3
If a double bond, but one carbon carrying two identical substituents → NO: the two drawings are the same molecule flipped over
Example: but-1-ene CH2=CH−CH2−CH3
If a double bond, and two different substituents on EACH side → YES: there is a cis isomer and a trans isomer
Example: but-2-ene CH3−CH=CH−CH3
If a ring carrying two substituents on different carbons → YES: the ring plays the same rigidifying role as the double bond
Example: 1,2-dimethylcyclopropane
The two negative branches fail for DIFFERENT reasons, and a question that asks you to justify expects you to name the right one: lack of rigidity for one, lack of asymmetry for the other.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
Counting the isomers of a molecular formula
When to use it: The question gives a molecular formula and asks for all possible structures.
1First check whether the molecule is saturated and acyclic, by comparing with the count CnH2n+2 or CnH2n+2O.
2Place the heteroatom if there is one, and distinguish its possible positions: at the end of the chain or inserted.
3Draw the possible carbon skeletons for each position, from the most linear to the most branched.
4Eliminate duplicates by checking that no drawing can be superimposed on another by rotation or flipping.
5Count, name, and group by functional group to present the result.
Concluding sentence
“C4H10O satisfies 2(4)+2=10: the molecule is saturated and acyclic. Four alcohols, two on the straight chain and two on the branched chain, and three ethers according to the split 2+2 or 1+3. Seven isomers in all.”
The trap: Counting the same molecule twice, numbered from the other end. The so-called 3-methylbutane is 2-methylbutane flipped over, and C5H12 has only three isomers, not four.
Marking: 1 mark for the saturation check, 2 marks for the alcohols, 2 for the ethers, 1 for the total, 1 for the classification as primary, secondary and tertiary if the question asks for it.
Comparing two boiling points
When to use it: The question gives two or three molecules and asks you to rank or explain.
1Calculate the molar masses and say whether they are close or not.
2Identify the functional group of each molecule.
3Name the dominant intermolecular force of each, London forces only, dipole-dipole or hydrogen bonding.
4Rank by increasing strength, and check that the ranking obtained is indeed that of the given temperatures.
Concluding sentence
“The three molar masses fall within 2 g/mol, so they explain nothing. Butane has only London forces, propanone a permanent dipole, propan-1-ol hydrogen bonds: the order of the temperatures, −0.5, 56 and 97 °C, follows exactly this order.”
The trap: Invoking molar mass when the masses are close. This trio is proposed precisely to neutralize that argument, and the marker expects you to say so.
Marking: 1 mark per molar mass, 1 mark per force correctly named, 2 marks for the justified ranking. Saying that the masses are close is almost always worth a mark of its own.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The valence of four on every carbon
Count the bonds of each carbon in the drawing: it must always have exactly four, a double bond counting for two.
In CH3−CH=CH−CH3, each middle carbon has one double bond, one single bond and one hydrogen: four.
The hydrogen count against the general formula
Compare with the alkane count CnH2n+2. Each missing pair of hydrogens signals a ring or an unsaturation.
C6H12 has two hydrogens fewer than C6H14: a ring or a double bond, not both.
The uniqueness of each isomer drawn
Mentally flip each structure and compare it with the previous ones. Two superimposable drawings count as only one molecule.
3-Methylbutane is 2-methylbutane flipped over: it does not count.
Consistency between the functional group and the announced property
If you conclude that a molecule boils high, check that it does carry a hydrogen on an oxygen or a nitrogen. Otherwise it does not form hydrogen bonds.
Propanone has an oxygen but no hydrogen on it: it does not DONATE hydrogen bonds.
The chemical test consistent with the functional group found
Each functional group has its test. If the test described in the question does not match the functional group announced, the identification is wrong.
Fizzing with hydrogen carbonate signals a carboxylic acid, never an alcohol.
The typical problem, taken apart
A compound identified by its molecular formula and three tests
A colourless liquid has the molecular formula C3H6O2.
Its aqueous solution turns indicator paper red; poured onto sodium hydrogen carbonate, it gives off a gas that turns limewater cloudy; heated with methanol in an acidic medium, it gives a compound with a fruity smell.
Atomic masses: C 12.01; H 1.01; O 16.00. Molar volume of a gas at 25 °C and 101.3 kPa: 24.5 L/mol.
The molar mass will be used in the final calculation, and it already allows candidates to be ruled out if the question proposes some.
Step 2
Indicator paper turning red signals an acid, which rules out the ester, the alcohol, the ether and the ketone.
Why
A single test is rarely enough, but this one eliminates all the neutral functional groups on the list at once.
Step 3
The gas that turns limewater cloudy is carbon dioxide: the acid is strong enough to displace carbonic acid from its salt.
Why
This is the test that distinguishes a carboxylic acid from an alcohol, which does nothing with hydrogen carbonate.
Step 4
The fruity smell obtained with methanol in an acidic medium is that of an ester: it is an esterification.
Why
It requires a carboxylic acid and an alcohol. The third test therefore confirms the first two.
Step 5
Once the −COOH group is placed, only two carbons remain to be put in a chain: there is only one possible acid, propanoic acid CH3−CH2−COOH.
Why
The count is what turns a bundle of clues into a certain identification. The two other C3H6O2 isomers among the families of this chapter are esters, and they would not turn the paper red.
Step 6
CH3CH2COOH+NaHCO3→CH3CH2COONa+H2O+CO2, in a 1 to 1 ratio.
Why
The carbonic acid formed immediately decomposes into water and carbon dioxide, and it is this decomposition that produces the fizzing.
Step 7
n=74.093.70=4.994×10−2 mol of acid, so as much carbon dioxide.
Why
The hydrogen carbonate is in excess: the acid is the limiting reactant, and the one-to-one ratio avoids any further calculation.
Step 8
V=4.994×10−2×24.5=1.22 L.
Why
Sanity check: fifty millimoles of gas give a little over a litre, which fits the fizzing observed for a few grams.
The conclusion, written out
“The compound is propanoic acid CH3−CH2−COOH, of molar mass 74.09 g/mol. It is the only one of the three isomers of C3H6O2 in the families of this chapter to carry a carboxylic acid group, the other two being esters. With an excess of hydrogen carbonate, 3.70 g of this acid releases 1.22 L of carbon dioxide.”
The classic mistake on this problem: Concluding after the first test that the compound is a carboxylic acid. A dissolved mineral acid would also turn the paper red, and only the fizzing with hydrogen carbonate, confirmed by the esterification, points to the organic functional group.
Learn by heart
•Alcohol at the end, ether inserted: the same oxygen, two functional groups.
•Aldehyde at the end with a hydrogen, ketone in the middle without: this hydrogen decides oxidation.
•Carbon always forms four bonds, without exception.
•CnH2n+2 for the alkane, CnH2n for the cycloalkane: a ring costs two hydrogens, a double bond too.
•A molecular formula does not designate a molecule: C4H10O has seven isomers.
•The class of an alcohol is read from the carbon that carries the −OH, not from the branching.
•Geometric isomerism: rigidity AND two different substituents on each side.
•A single bond rotates, a double bond does not.
•London, then dipole-dipole, then hydrogen bonding: the order of the forces is the order of the boiling points.
•An alcohol always boils higher than the ether that is its isomer.
•The solubility of an alcohol collapses as the chain gets longer.
Frequently asked questions
How do you tell an aldehyde from a ketone?
Both carry the same carbonyl group, a carbon doubly bonded to an oxygen. What separates them is its position: at the end of the chain for the aldehyde, which then keeps a hydrogen on this carbon, and between two carbons for the ketone, which keeps none. This single hydrogen allows the aldehyde to be oxidized to a carboxylic acid, which the ketone cannot do. In the laboratory, Fehling's solution or Tollens' reagent settle it immediately.
Why does an alcohol boil so much higher than the ether that is its isomer?
Because the alcohol carries a hydrogen directly bonded to its oxygen, very strongly polarized, which latches onto the oxygen of the neighbouring molecule: this is hydrogen bonding. This whole network has to be broken to vaporize the liquid, and that takes a lot of energy. The ether has no hydrogen on its oxygen, and its molecules are held together only by much weaker dipole-dipole forces. For the same molecular formula, the gap reaches a hundred degrees.
What are the conditions for a molecule to have cis and trans isomers?
There are two, and both are necessary. First a rigidity, that is a carbon-carbon double bond or membership of a ring, which prevents rotation. Then an asymmetry: each of the two atoms involved must carry two substituents that differ from each other. Butane fails the first, its single bonds rotating freely; but-1-ene fails the second, one of its carbons carrying two identical hydrogens.
How do you decide whether an alcohol is primary, secondary or tertiary?
Look only at the carbon that carries the hydroxyl group, and count the other carbons it is bonded to. One, the alcohol is primary; two, it is secondary; three, it is tertiary. The branching of the rest of the chain plays no part: 2-methylpropan-1-ol is primary despite its methyl group, because the carbon bearing the group has only one carbon neighbour. This class then decides the oxidation product.
Why aren't all alcohols soluble in water?
Because an alcohol molecule has two parts of opposite nature. The hydroxyl group is polar and forms hydrogen bonds with water, while the carbon chain is nonpolar and has no affinity for it. As long as the chain is short, the group wins and the alcohol is miscible in all proportions. As soon as it gets longer, the nonpolar part dominates and the solubility collapses: it goes from infinite for methanol to half a gram per litre for octan-1-ol.
Is a molecular formula enough to identify an organic molecule?
No, never. The molecular formula only gives the count of the atoms, whereas the properties depend on how they are joined. Ethanol and dimethyl ether share the formula of two carbons, six hydrogens and one oxygen, and yet one is a liquid that boils at seventy-eight degrees and the other a gas. A chemical test, a physical measurement or a spectroscopy is therefore always needed to conclude, and this is the reason for the expanded and condensed formulas.
Practise it
Corrected exercises: Organic molecules, functional groups and isomers, 202-SN1-RE
A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.
Something missing from this sheet?A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.
Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space.
Rather not create an account? Write to me directly.
Get in touch for a first session. We work through general chemistry at the level of real CEGEP tests, with the complete solution in front of you and the trap named each time it costs marks.
