The Polar Carbonyl Group
Every reaction in this note starts with one structural fact: the carbonyl group $\ce{C=O}$ is strongly polarised. Oxygen is far more electronegative than carbon, so the bonding electrons of the double bond — both the $\sigma$ and the more loosely held $\pi$ electrons — are drawn toward oxygen. The result is a permanent dipole in which the carbon carries a partial positive charge and oxygen a partial negative charge.
The NIOS general-principles unit describes this displacement explicitly. Under the electromeric effect, "in a carbonyl group it operates as follows: $\ce{C=O}$ → $\ce{C^+ - O^-}$", a complete shift of the $\pi$ pair toward oxygen that develops $+$ and $-$ charges within the molecule. Even in the ground state, before any reagent arrives, the carbon is electron-poor. That is exactly the "position of low electron density" a nucleophile is defined to attack.
The carbonyl carbon (δ+) is the electrophilic site; the nucleophile approaches roughly perpendicular to the planar sp2 carbon, on the side away from oxygen.
Geometrically, the carbonyl carbon is $sp^2$ hybridised and trigonal planar — the carbon, oxygen and the two attached groups all lie in one plane, with the empty trajectory above and below that plane open to attack. This flat, accessible, electron-poor carbon is the structural reason carbonyls are so reactive toward nucleophiles, in sharp contrast to the electron-rich $\ce{C=C}$ of an alkene, which instead invites electrophiles.
The General Addition Mechanism
Nucleophilic addition proceeds in two clean steps. In the first, slow step, the nucleophile donates its lone pair to the electrophilic carbonyl carbon. As that new bond forms, the $\pi$ electrons of the double bond are pushed entirely onto oxygen, which becomes a negatively charged alkoxide. The carbon rehybridises from planar $sp^2$ to tetrahedral $sp^3$. This tetrahedral alkoxide is the central intermediate of the whole chapter.
In the second, fast step, the negatively charged oxygen is protonated — by water, by an alcohol, or during acidic work-up — to give a neutral addition product in which both the nucleophile and a new $\ce{O-H}$ are bonded across the former double bond.
Written generally, with a negatively charged nucleophile:
$$\ce{R2C=O + Nu^- -> R2C(Nu)-O^- ->[H+] R2C(Nu)-OH}$$
The geometry change from planar to tetrahedral is the heart of the mechanism: a flat, exposed electrophilic carbon becomes a crowded sp3 centre bearing the new group and an alkoxide oxygen.
Notice the contrast with the addition seen in alkene chemistry. In electrophilic addition the first species to attack the multiple bond is the electrophile; here, because the multiple bond is already polarised toward oxygen, the carbon is electron-poor and the nucleophile attacks first. The name of the reaction always reflects the species that initiates it.
Acid and Base Catalysis
The same two-step skeleton runs under two different conditions, and NEET likes to test which one applies. A negatively charged or strongly basic nucleophile (such as $\ce{CN^-}$ or a Grignard carbanion) attacks the neutral carbonyl directly — this is base-style addition. A weak, neutral nucleophile (such as water or an alcohol) is too sluggish on its own, so the carbonyl is first activated by protonating the oxygen.
Protonation of oxygen makes the carbon even more electron-deficient (it now carries a full positive charge in the protonated resonance form), sharply raising its electrophilicity so the weak nucleophile can add. This acid-catalysed route is exactly why hydration and acetal formation are run with a trace of acid.
$$\ce{R2C=O + H+ <=> R2C=O^+H}$$
| Route | First event | Suits which nucleophile | Examples |
|---|---|---|---|
| Base / direct | Strong Nu attacks neutral C=O | Anionic / strongly basic | $\ce{CN^-}$, $\ce{RMgX}$, $\ce{H^-}$ from hydride |
| Acid-catalysed | Protonation of O first | Weak, neutral Nu | $\ce{H2O}$, $\ce{ROH}$, amine-derivatives |
Aldehydes vs Ketones: Reactivity Order
For nucleophilic addition the standard reactivity order is formaldehyde > other aldehydes > ketones:
$$\ce{HCHO > RCHO > RCOR'}$$
Two independent effects, both traceable to NIOS principles, push in the same direction. The electronic argument uses the inductive effect. NIOS lists alkyl groups among the $+I$ (electron-releasing) groups, in the order $\ce{(CH3)3C- > (CH3)2CH- > CH3CH2- > -CH3 > -H}$. A ketone has two such electron-donating alkyl groups pushing electron density onto the carbonyl carbon, partly neutralising its $\delta+$ charge; an aldehyde has only one alkyl group (and an H, which donates nothing). The aldehyde carbon therefore stays more electrophilic.
The steric argument concerns the tetrahedral intermediate. When the nucleophile adds, the carbon becomes crowded. Two bulky alkyl groups on a ketone crowd the approaching nucleophile and destabilise the tetrahedral product more than the single small group on an aldehyde. Less crowding plus more positive charge means aldehydes react faster.
Do not import the alkene reactivity logic.
In electrophilic addition to alkenes, more alkyl groups help by stabilising the carbocation, so substituted alkenes react faster. In nucleophilic addition to carbonyls the logic flips: alkyl groups hinder reaction, both electronically ($+I$ reduces the $\delta+$) and sterically. Aldehydes, with fewer alkyl groups, are the faster substrate.
More alkyl on a carbonyl → slower nucleophilic addition. The opposite trend to carbocation stability.
See how the electron-rich double bond reverses this story in Electrophilic Addition (Markovnikov & carbocations).
Why Acid Derivatives Substitute Instead
Aldehydes and ketones stop at the stable addition product because their carbonyl carbon carries only carbon and hydrogen substituents — there is no group that can leave as a stable anion. Carboxylic acid derivatives are different. In an ester, acid chloride, amide or anhydride, the carbonyl carbon also bears a potential leaving group: $\ce{-OR}$, $\ce{-Cl}$, $\ce{-NH2}$ or $\ce{-OCOR}$.
The nucleophile still adds first, giving the same tetrahedral intermediate. But now that intermediate can collapse: the lone pair on oxygen re-forms the $\ce{C=O}$ double bond and expels the leaving group. The net result is replacement of one group by another at the acyl carbon — nucleophilic acyl substitution, an addition–elimination sequence — rather than a permanent addition.
$$\ce{R-CO-L + Nu^- -> R-C(Nu)(L)-O^- -> R-CO-Nu + L^-}$$
| Substrate | Leaving group on carbonyl C | Outcome |
|---|---|---|
| Aldehyde / ketone | None (only C, H) | Stable addition product |
| Acid chloride | $\ce{-Cl}$ | Addition–elimination (substitution) |
| Ester | $\ce{-OR}$ | Addition–elimination (substitution) |
| Amide | $\ce{-NH2}$ | Addition–elimination (substitution) |
Key Addition Reactions
Each named reaction below is the same two-step mechanism with a different nucleophile. Recognise the nucleophile and you can predict the product.
Hydrogen cyanide → cyanohydrin
Cyanide ion is the nucleophile; it adds to give an alkoxide that is protonated to a $\alpha$-hydroxynitrile (cyanohydrin). The new $\ce{C-C}$ bond makes this a chain-lengthening reaction.
$$\ce{CH3CHO + HCN -> CH3CH(OH)CN}$$
Sodium bisulphite adduct
The sulphur of $\ce{HSO3^-}$ attacks the carbonyl carbon to give a crystalline bisulphite addition compound. Because it forms with aldehydes and methyl ketones and reverts to the carbonyl on treatment with dilute acid or base, it is used to purify and separate these carbonyls.
$$\ce{CH3CHO + NaHSO3 -> CH3CH(OH)SO3Na}$$
Grignard reagents → alcohols
A Grignard reagent $\ce{RMgX}$ delivers a carbanion-like alkyl group, a powerful nucleophile that forms a new $\ce{C-C}$ bond at the carbonyl carbon. Aqueous acidic work-up then protonates the magnesium alkoxide. The class of alcohol depends on the carbonyl: formaldehyde gives a primary alcohol, other aldehydes give secondary alcohols, and ketones give tertiary alcohols.
$$\ce{R'MgX + R2C=O -> R2C(R')-OMgX ->[H3O+] R2C(R')-OH}$$
Q. Which alcohol class results when ethylmagnesium bromide reacts with propanone (acetone), followed by acidic work-up?
A. Propanone is a ketone, $\ce{(CH3)2C=O}$. The ethyl carbanion adds to the carbonyl carbon, which already bears two methyl groups, so after work-up the carbon carries three carbon substituents and one OH — a tertiary alcohol, 2-methylbutan-2-ol. Any ketone + Grignard gives a tertiary alcohol.
Alcohols → hemiacetal and acetal
Under acid catalysis one molecule of alcohol adds to an aldehyde to give a hemiacetal (a carbon bearing both $\ce{-OH}$ and $\ce{-OR}$). A second alcohol then replaces the $\ce{-OH}$, via loss of water, to give a stable acetal (a carbon bearing two $\ce{-OR}$ groups). Acetal formation is reversible, which is why acetals serve as protecting groups for the carbonyl.
$$\ce{RCHO + R'OH <=>[H+] RCH(OH)OR' ->[R'OH][H+] RCH(OR')2 + H2O}$$
Water → hydrate (gem-diol)
Water adds reversibly to give a gem-diol (a 1,1-diol). For most carbonyls this equilibrium lies toward the carbonyl, but with strongly electron-withdrawing groups attached, the hydrate can dominate.
$$\ce{R2C=O + H2O <=> R2C(OH)2}$$
Addition–Elimination with Ammonia Derivatives
Ammonia and its substituted derivatives $\ce{H2N-Z}$ react with aldehydes and ketones by a special addition–elimination route. The nitrogen lone pair first adds to the carbonyl carbon (addition), giving a tetrahedral carbinolamine. Loss of water then forms a $\ce{C=N}$ double bond (elimination). The product depends entirely on what $\ce{Z}$ is.
Reagent H2N–Z | Z group | Product | NEET note |
|---|---|---|---|
| Ammonia / 1° amine | $\ce{-H}$ / $\ce{-R}$ | Imine (Schiff base) | $\ce{>C=NR}$ |
| Hydroxylamine | $\ce{-OH}$ | Oxime | $\ce{>C=N-OH}$ |
| Hydrazine | $\ce{-NH2}$ | Hydrazone | $\ce{>C=N-NH2}$ |
| Phenylhydrazine | $\ce{-NHC6H5}$ | Phenylhydrazone | colour test precursor |
| 2,4-DNP (Brady's reagent) | 2,4-dinitrophenyl | 2,4-dinitrophenylhydrazone | yellow–orange–red solid |
| Semicarbazide | $\ce{-NHCONH2}$ | Semicarbazone | crystalline derivative |
The generic transformation, with water lost as the leaving group, is:
$$\ce{R2C=O + H2N-Z -> R2C=N-Z + H2O}$$
2,4-DNP is a carbonyl test, not an alcohol test.
A coloured 2,4-dinitrophenylhydrazone precipitate confirms an aldehyde or ketone. It does not distinguish the two — for that you need Tollens' or Fehling's, which respond only to aldehydes. Do not confuse the orange 2,4-DNP solid with a positive alcohol or carboxylic-acid test.
2,4-DNP positive → carbonyl present (aldehyde OR ketone); use Tollens'/Fehling's to tell aldehydes apart.
NEET Strategy and Common Errors
Exam questions on this topic almost always reduce to one of three judgments: identify the nucleophile, predict the product geometry, or rank reactivity. Keep these anchors fixed.
| If asked… | Anchor to apply |
|---|---|
| Which attacks first? | Carbonyl C is $\delta+$ → the nucleophile attacks first |
| Aldehyde or ketone faster? | Fewer alkyl groups react faster: $\ce{HCHO > RCHO > RCOR'}$ |
| Why does an ester substitute? | It has a leaving group on the carbonyl C → addition–elimination |
| What does 2,4-DNP confirm? | Presence of a carbonyl group (aldehyde or ketone) |
| Grignard + ketone product? | Tertiary alcohol after acidic work-up |
Reductions you may see, such as $\ce{NaBH4}$ acting on a carbonyl, are mechanistically a hydride $\ce{(H^-)}$ nucleophilic addition: the hydride adds to the carbonyl carbon and the alkoxide is protonated on work-up to give an alcohol. Importantly, $\ce{NaBH4}$ reduces aldehydes and ketones but leaves esters and carboxylic acids untouched — a selectivity that has appeared directly in NEET.
Carbonyl addition in one screen
- The $\ce{C=O}$ bond is polar ($\ce{C^{\delta+}=O^{\delta-}}$); the carbon is the electrophilic, attack-prone site.
- Mechanism: nucleophile adds to the planar $sp^2$ carbon (slow) → tetrahedral $sp^3$ alkoxide → fast protonation to the addition product.
- Reactivity: $\ce{HCHO > RCHO > RCOR'}$, from combined steric crowding and the $+I$ effect of alkyl groups.
- Acid derivatives undergo addition–elimination (substitution) because the carbonyl carbon carries a leaving group.
- Key nucleophiles: $\ce{CN^-}$ (cyanohydrin), $\ce{HSO3^-}$, Grignard (alcohol), $\ce{ROH}$ (hemiacetal/acetal), $\ce{H2O}$ (hydrate).
- Ammonia derivatives $\ce{H2N-Z}$ give imines, oximes, hydrazones; 2,4-DNP gives a coloured carbonyl-confirming solid.