Organic Chemistry 202-SNU-RE • CEGEP, Montreal
The carbonyl is the crossroads of all functional organic chemistry: you reach it by oxidizing alcohols and you leave it toward alcohols, amines, acids and alkanes. A single polarity explains all these reactions.
This sheet leaves the course in your notes and gathers what decides an exam: the site attacked, the order of reactivity, the choice of reducing agent, and the reading of the identification tests.
The thread of the chapter
A single carbon decides everything: the carbonyl carbon, flat and electron-poor, which nucleophiles attack perpendicular to the plane.
The oxygen, rich in electrons, is attacked only by a PROTON in acid catalysis. Confusing the two sites is the error that makes the whole mechanism wrong from the very first arrow.
The two effects point the same way, which makes the order very reliable. A question that asks to compare two carbonyl compounds always expects both arguments, not just one.
The acetal is stable in BASIC medium and rehydrolyzes in aqueous acid: this is what makes it the protecting group of the carbonyl, and the question expects both steps.
The four tests are read together: the first says there is a carbonyl, the second and the third separate aldehyde from ketone, the fourth spots a methyl ketone.
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
“The nucleophile, negatively charged, binds to the oxygen of the carbonyl.”
What to write
“The oxygen is already rich in electrons: it is the CARBON, carrying , that the nucleophile attacks.”
Why: Two charges of the same sign repel each other. Only a proton, in acid catalysis, binds to the oxygen, and it does so precisely to make the carbon even more electrophilic.
What not to write
“The ketone carries two alkyl groups, so it is richer and more reactive.”
What to write
“These two groups hinder the approach AND donate electrons to the carbon: the ketone is LESS reactive than the aldehyde.”
Why: The reactivity of an electrophile increases when its positive charge increases and when its access is open. Alkyl groups work against both criteria at once.
What not to write
“ reduces the ester group to a primary alcohol.”
What to write
“ reduces ONLY aldehydes and ketones; for an ester, is needed, in anhydrous medium.”
Why: Borohydride is a mild hydride donor, reactive enough for an isolated carbonyl but not for a carbonyl already stabilized by a neighbouring oxygen. It is precisely this selectivity that makes it useful.
What not to write
“Propanone gives a silver mirror, like any carbonyl compound.”
What to write
“Tollens oxidizes the aldehyde to an acid: a ketone has no hydrogen on the carbonyl carbon and does not react.”
Why: Oxidation requires a hydrogen carried by the functional carbon, exactly as for alcohols. The ketone has none, which makes Tollens the test that separates the two families.
What not to write
“The yellow precipitate confirms that it is a ketone.”
What to write
“Iodoform responds only to METHYL KETONES and ethanol: a heavier symmetric ketone gives a negative test.”
Why: The test requires a group carried by the functional carbon, the only pattern able to give the three successive iodinations. It is a test of PATTERN, not of functional group.
What not to write
“Once the acetal is formed, the carbonyl is protected under all conditions.”
What to write
“The acetal is stable in BASIC medium and toward hydrides, but it rehydrolyzes in aqueous acid: this is how it is deprotected.”
Why: Formation is an equilibrium shifted by removing the water; adding it back reverses the equilibrium. A protection is always written together with its deprotection.
What not to write
“The secondary amine gives a substituted imine.”
What to write
“A secondary amine has only one hydrogen left on the nitrogen: elimination takes place toward the neighbouring carbon and gives an ENAMINE.”
Why: Imine formation consumes TWO hydrogens of the nitrogen, one for the addition and one for the dehydration. With only one available, the double bond forms elsewhere.
What not to write
“We acidify strongly to catalyze the formation of the imine.”
What to write
“The curve is BELL-SHAPED: near to . Too acidic, the amine is protonated and stops being nucleophilic.”
Why: Two requirements oppose each other: enough protons are needed to activate the carbonyl, and enough free amine to attack. The optimum is therefore a compromise, not an extreme.
What not to write
“Both turn the carbonyl into , so either will do.”
What to write
“Clemmensen works in strongly ACIDIC medium, Wolff-Kishner in hot BASIC medium: the choice depends on what the rest of the molecule can stand.”
Why: A molecule carrying an acetal does not survive the acidic medium; a molecule carrying an ester does not survive hot sodium hydroxide. The result is the same, the conditions are not.
Look at the functional group aimed at in the product.
If you want an alcohol without changing the chain → , mild and tolerant of water
Example: a ketone gives the corresponding secondary alcohol
If you want an alcohol with a LONGER chain → Grignard reagent, then hydrolysis
Example: the Grignard reagent brings its whole group
If you want an amine → reductive amination: imine formed, then reduced
Example: the carbonyl is not reduced directly to an amine
If you want to make the carbonyl disappear completely → Clemmensen in acidic medium, or Wolff-Kishner in basic medium
Example: the carbon becomes a simple
If you want to PROTECT the carbonyl while another reaction is carried out → acetal with a diol and acid catalysis, then acid hydrolysis at the end of the synthesis
Example: protect a ketone to reduce an ester with
Every protection is written with its deprotection: a synthesis that installs an acetal without removing it does not give the product asked for, and loses the conclusion mark.
Apply the tests in this order, from the most general to the most specific.
If 2,4-DNPH gives an orange precipitate → there is a carbonyl, aldehyde or ketone
Example: no distinction at this stage
If Tollens gives a silver mirror → it is an ALDEHYDE
Example: Fehling would give the same verdict, in brick red
If Tollens is negative but DNPH positive → it is a KETONE
Example: propanone
If iodoform gives a yellow precipitate → there is a methyl ketone pattern, or ethanol
Example: butanone is a methyl ketone
If all the tests are negative → there is no carbonyl: look for an alcohol, an ether or an acetal
Example: an acetal responds to none of these tests
The tests are combined with the degree of unsaturation calculated from the molecular formula: a degree of and a positive DNPH often suffice to conclude.
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
When to use it: The question asks for the mechanism of an addition to an aldehyde or a ketone.
Concluding sentence
“In basic medium, the ion attacks the carbonyl carbon; the alkoxide formed is then protonated during the final workup.”
The trap: Writing an attack of the nucleophile on the oxygen, or forgetting the final protonation step, which is worth a mark of its own.
Marking: 1 mark for the site attacked, 1 mark for the arrow of the addition, 1 mark for the final step, 1 mark for the neutral product.
When to use it: The question gives a molecular formula and a series of tests.
Concluding sentence
“The degree of unsaturation is , DNPH is positive and Tollens negative: it is a ketone; the positive iodoform requires a methyl group, hence butanone.”
The trap: Stopping at the first compatible test without checking that the structure also explains the NEGATIVE tests.
Marking: 1 mark for the degree of unsaturation, 1 mark per test interpreted, 1 mark for the verified structure.
Five minutes of checking recover more marks than one more problem started in a hurry.
Is the site attacked the carbon?
Reread the first arrow of the mechanism: it must end on the carbonyl carbon, unless it is a proton.
An arrow going from the nucleophile to the oxygen is wrong in all cases.
The carbon count after a Grignard reagent
Add the carbons of the carbonyl compound and those of the Grignard reagent: the product must carry the sum.
A three-carbon alcohol obtained from propanone and a methyl Grignard reagent is impossible.
The consistency of the tests
Check that the proposed structure also explains the negative tests, not only the positive ones.
An aldehyde proposed when Tollens is negative contradicts the statement.
Has the protection been removed?
Reread a multi-step synthesis: any acetal installed must be hydrolyzed before the end.
A final product still carrying the acetal is not the molecule asked for.
A compound of formula gives an orange precipitate with 2,4-DNPH, gives no silver mirror with the Tollens reagent, and forms a yellow precipitate with diiodine in basic medium.
Determine its structure and justify each observation.
Step 1
Degree of unsaturation: .
Why
The oxygen does not count in the formula. A single degree allows either a double bond or a ring: the first test will decide.
Step 2
DNPH positive: there is a carbonyl group, hence a bond and not a ring.
Why
This test responds to every aldehyde and every ketone, without telling them apart. It uses up the available degree of unsaturation and rules out alcohols, ethers and rings.
Step 3
Tollens negative: it is not an aldehyde, so it is a ketone.
Why
Oxidation by Tollens requires a hydrogen carried by the functional carbon. Its absence is here as useful a piece of information as a positive test.
Step 4
Iodoform positive: the carbonyl carbon carries a group.
Why
The test requires exactly this pattern, which alone allows the three successive iodinations followed by cleavage. It therefore points to a methyl ketone.
Step 5
Structure: butanone, .
Why
It is the only four-carbon ketone carrying a methyl on the carbonyl. Propanone has only three carbons, and butanal would be an aldehyde.
Step 6
Check: butanone does give DNPH positive, Tollens negative and iodoform positive.
Why
We go back from the structure to the three observations, the negative one included. It is this rereading that distinguishes a justified answer from a guessed one.
The conclusion, written out
“The compound is butanone: 2,4-DNPH establishes the carbonyl, the negative Tollens test rules out the aldehyde, and iodoform requires the methyl group carried by the functional carbon.”
The classic mistake on this problem: Concluding butanal as soon as DNPH is positive, without reading the negative Tollens result, which precisely rules out aldehydes.
Because the oxygen, much more electronegative, draws the electrons of the bond toward itself and ends up electron-rich, while the carbon becomes electron-poor. A nucleophile, which brings electrons, therefore goes to the depleted site. Only a proton binds to the oxygen, and it does so precisely to make the carbon even more attractive.
For two reasons that add up. First, it carries one alkyl group fewer, so access to the carbon is more open. Second, each alkyl group donates electrons to the carbon and lowers its positive charge, which makes it less attractive to a nucleophile. The ketone combines both handicaps.
Sodium borohydride is enough whenever there is only an aldehyde or a ketone to reduce, and it tolerates water and alcohols, which greatly simplifies handling. Lithium aluminium hydride is reserved for esters, acids, amides and nitriles, and it requires perfectly anhydrous ether.
The imine has a double bond between carbon and nitrogen, and comes from a primary amine, which has two hydrogens on the nitrogen. The enamine has a double bond between two carbons, with the nitrogen on one of them, and comes from a secondary amine, which has only one hydrogen available.
Both give an orange precipitate with dinitrophenylhydrazine, which establishes only the presence of a carbonyl. It is the Tollens reagent that separates them: it gives a silver mirror with aldehydes and has no effect on ketones, for lack of a hydrogen on the functional carbon. The Fehling test gives the same verdict.
© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-aldehydes-ketones. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).
