Nucleophilic addition reactions

Every aldehyde and ketone carries the same reactive scar: a carbon-oxygen double bond in which oxygen hoards the electrons and leaves carbon starved of them. That starved carbon is an open invitation, and four very different reagents — hydrogen cyanide, sodium bisulphite, ammonia's nitrogen-rich cousins, and Grignard's reagent — all accept it the same way. Each one attacks the carbonyl carbon with a lone pair, and the π\pi electrons of C=O\text{C=O} retreat fully onto oxygen. This single move, nucleophilic addition, is the master reaction of the aldehydes-and-ketones chapter: almost every named product you must draw, from cyanohydrins to oximes to Grignard alcohols, comes out of the same two-step template. This article rebuilds that template from the electronics up, works the cyanohydrin and phenylhydrazone mechanisms in full curved-arrow detail, and pins down exactly why aldehydes beat ketones to the punch every time.

The core insight

A nucleophile attacks the electron-poor carbonyl carbon of an aldehyde or ketone, pushing the C=O\text{C=O} π\pi electrons onto oxygen to give a tetrahedral alkoxide that is then protonated (or, for nitrogen nucleophiles, dehydrated) into the final addition product, with aldehydes reacting faster than ketones because of lower steric bulk and higher electrophilicity.

The core idea

Why the carbonyl carbon is a target. Oxygen is more electronegative than carbon, so it pulls the shared π\pi electron cloud of C=O\text{C=O} toward itself, leaving carbon with a partial positive charge and oxygen with a partial negative one:

=\overset{\delta+}{\text{C}}=\overset{\delta-}{\text{O}}

where: δ+\delta+ and δ−\delta- mark the partial charges created by the electronegativity gap, not full unit charges. That partial positive carbon is the site every nucleophile in this chapter homes in on.

The universal two-step template. Whatever the nucleophile, the reaction runs through the same tetrahedral intermediate:

R2C=O+Nu−  ⟶  R2C(O−)(Nu)  → H+   R2C(OH)(Nu)\text{R}_2\text{C=O} + \text{Nu}^- \;\longrightarrow\; \text{R}_2\text{C(O}^-\text{)(Nu)} \;\xrightarrow{\ H^+\ }\; \text{R}_2\text{C(OH)(Nu)}

where: Nu−\text{Nu}^- is the attacking nucleophile (cyanide, bisulphite, an amine nitrogen, or a Grignard carbanion), the middle species is the tetrahedral alkoxide formed once the π\pi electrons land on oxygen, and the final step is protonation of that alkoxide (by acid or water) to give the neutral addition product.

Reactivity: aldehydes beat ketones. Two alkyl groups on a ketone's carbonyl carbon do double damage — they donate electron density that offsets the δ+\delta+ (lowering electrophilicity), and their bulk physically blocks the incoming nucleophile. An aldehyde has at most one alkyl group and a small hydrogen on the other side, so it is both more electrophilic and less hindered. The exam-standard order is: formaldehyde >> other aldehydes >> methyl ketones >> larger dialkyl ketones. This single ordering is why Tollens' and Fehling's tests reliably pick out aldehydes, and why bisulphite adducts and cyanohydrins form cleanly with aldehydes and methyl ketones but sluggishly or not at all with bulkier ketones.

The four named nucleophiles and their products:

  • HCN\text{HCN} → cyanohydrin, R2C(OH)(CN)\text{R}_2\text{C(OH)(CN)} — needs a trace of base/CN−\text{CN}^-, since HCN itself is a poor nucleophile source.
  • NaHSO3\text{NaHSO}_3 (saturated) → bisulphite addition compound, R2C(OH)(SO3Na)\text{R}_2\text{C(OH)(SO}_3\text{Na)} — a crystalline solid used to purify aldehydes and methyl ketones.
  • Ammonia derivatives (addition then loss of water): hydroxylamine NH2OH\text{NH}_2\text{OH} → oxime R2C=N–OH\text{R}_2\text{C=N–OH}; hydrazine NH2NH2\text{NH}_2\text{NH}_2 → hydrazone R2C=N–NH2\text{R}_2\text{C=N–NH}_2; phenylhydrazine → phenylhydrazone R2C=N–NHC6H5\text{R}_2\text{C=N–NHC}_6\text{H}_5.
  • Grignard's reagent R′MgX\text{R}'\text{MgX} → magnesium alkoxide, hydrolysed on workup to the alcohol R2C(OH)(R′)\text{R}_2\text{C(OH)(R}').
Figure
Energy profile comparing nucleophile attack on an aldehyde with a lower barrier against attack on a bulkier ketone with a higher barrier, both proceeding through a tetrahedral alkoxide intermediate before the final protonated product
Aldehyde versus ketone — same path, different barrier height reaction progress energy aldehyde: lower barrier ketone: higher barrier (steric + electronic) R2C=O + Nu- tetrahedral alkoxide R2C(OH)(Nu)
Derivation ABuilding the general nucleophilic addition mechanism

The goal is to derive, purely from the electronics of the carbonyl group and the curved-arrow rules already known for bond-breaking and bond-making, the exact two-step (or three-step, for nitrogen nucleophiles) sequence that governs all four named additions in this chapter, so that any new nucleophile can be handled by the same reasoning rather than memorised case by case.

Step 1. Locate the electrophile. In R2C=O\text{R}_2\text{C=O}, oxygen's higher electronegativity polarises the π\pi bond, leaving the carbonyl carbon δ+\delta+. Any species with a usable lone pair or negative charge — a nucleophile — is drawn to that carbon, not to oxygen, because carbon is where the electron deficiency actually sits.

Step 2. Attack and electron-pair reassignment. The nucleophile's lone pair forms a new σ\sigma bond to the carbonyl carbon. Because carbon cannot exceed four bonds, the π\pi electrons of C=O\text{C=O} must leave; they move entirely onto oxygen, converting the trigonal planar carbonyl carbon into a tetrahedral centre and leaving oxygen as a full alkoxide anion, O−\text{O}^-. This is the addition step proper, and it is identical in mechanism whether Nu−\text{Nu}^- is CN−\text{CN}^-, HSO3−\text{HSO}_3^-, a Grignard carbanion, or a neutral amine nitrogen.

Step 3. Quench the alkoxide. If the nucleophile was anionic (CN−\text{CN}^-, HSO3−\text{HSO}_3^-, the Grignard alkoxide), the negative oxygen simply picks up a proton from acid or water on workup, giving the neutral R2C(OH)(Nu)\text{R}_2\text{C(OH)(Nu)} product directly — this is where the sequence stops for HCN, NaHSO3\text{NaHSO}_3, and Grignard additions.

Step 4. Branch for nitrogen nucleophiles. If the attacking nucleophile was a neutral amine (hydroxylamine, hydrazine, phenylhydrazine), the nitrogen becomes positively charged the instant it bonds, so a fast internal proton transfer from that nitrogen to the alkoxide oxygen is needed first, giving a neutral carbinolamine, R2C(OH)(NHR′)\text{R}_2\text{C(OH)(NHR}'). This intermediate is not the final product: acid-catalysed loss of water then forms a new C=N\text{C=N} double bond, because the lone pair on nitrogen can push back into carbon exactly as the π\pi electrons once did, ejecting OH−\text{OH}^- (as water, after protonation). The result is the observed oxime, hydrazone, or phenylhydrazone, and it explains why these particular reactions are sometimes called addition-elimination even though the syllabus files them under nucleophilic addition. Verified: the same tetrahedral alkoxide intermediate, R2C(O−)(Nu)\text{R}_2\text{C(O}^-\text{)(Nu)}, appears in both worked examples below (Steps 2 of each), confirming Step 2 of this derivation against the two fully worked mechanisms.

Worked example

Problem: Show the full mechanism of acetaldehyde (CH3CHO\text{CH}_3\text{CHO}) reacting with hydrogen cyanide to form its cyanohydrin, then name the product.

Step 1. HCN is a weak acid and a poor nucleophile source on its own, so a trace of base (or the small equilibrium concentration of CN−\text{CN}^-) is required; the reaction is base-catalysed, and the true nucleophile is the cyanide ion generated by deprotonation of HCN.

Step 2. The cyanide carbon attacks the electrophilic carbonyl carbon of acetaldehyde, and the π\pi electrons of C=O\text{C=O} move fully onto oxygen:

CH3CHO+CN−→CH3–CH(O−)–CN\text{CH}_3\text{CHO} + \text{CN}^- \rightarrow \text{CH}_3\text{–CH(O}^-\text{)–CN}

Step 3. The alkoxide intermediate is protonated by HCN (or water), regenerating CN−\text{CN}^- to continue the catalytic cycle:

CH3–CH(O−)–CN+HCN→CH3–CH(OH)–CN+CN−\text{CH}_3\text{–CH(O}^-\text{)–CN} + HCN \rightarrow \text{CH}_3\text{–CH(OH)–CN} + CN^-

Step 4. State the answer. The product is 2-hydroxypropanenitrile (acetaldehyde cyanohydrin), CH3CH(OH)CN\textbf{CH}_3\textbf{CH(OH)CN} — a useful chain-extension intermediate, since the nitrile group can later be hydrolysed to a carboxylic acid or reduced to an amine, adding one carbon to the chain.

Competitive stretch

Acetone reacts with phenylhydrazine to give a phenylhydrazone; write the mechanism. The terminal, more nucleophilic nitrogen of C6H5NH–NH2\text{C}_6\text{H}_5\text{NH–NH}_2 attacks the carbonyl carbon of acetone, giving a zwitterionic tetrahedral intermediate, (CH3)2C(O−)(NH–NHC6H5)+(\text{CH}_3)_2\text{C(O}^-\text{)(NH–NHC}_6\text{H}_5)^+, in which oxygen is negative and the newly bonded nitrogen is positive. A fast internal proton transfer from that nitrogen to the oxygen gives the neutral carbinolamine, (CH3)2C(OH)(NHNHC6H5)(\text{CH}_3)_2\text{C(OH)(NHNHC}_6\text{H}_5). Acid-catalysed loss of water then forms the C=N\text{C=N} bond: (CH3)2C(OH)(NHNHC6H5)→−H2O(CH3)2C=N–NHC6H5(\text{CH}_3)_2\text{C(OH)(NHNHC}_6\text{H}_5) \xrightarrow{-H_2O} (\text{CH}_3)_2\text{C=N–NHC}_6\text{H}_5, the phenylhydrazone, exactly following Step 4 of the general derivation above.

Common traps

  • Forgetting that HCN addition needs a trace of base or cyanide-ion catalyst — writing HCN as directly attacking the carbonyl carbon skips the actual nucleophile.
  • Drawing amine-derivative reactions as a simple one-step addition without the water-loss step — the observed product (oxime, hydrazone) carries a C=N\text{C=N} double bond, not a C(OH)(NHR)\text{C(OH)(NHR)} single-bonded addition compound.
  • Assuming ketones and aldehydes react at similar rates — ketones are consistently slower and less complete due to steric bulk and reduced electrophilicity from two alkyl groups.
  • Forgetting to hydrolyse the magnesium alkoxide from Grignard addition with dilute acid or water to isolate the final alcohol product.

Ask the tutor: if any curved-arrow step — especially the base catalysis in HCN addition or the proton-transfer-then-dehydration branch for amine nucleophiles — isn't clicking, ask the AI tutor to draw it out arrow by arrow with you.

Why is HCN itself written as the reagent if the actual attacking species is the cyanide ion

That is the standard chemistry-syllabus shorthand: the overall reagent added to the flask is HCN, but the mechanism only works if a small amount ionises (or a trace of base is added) to generate the true nucleophile, CN−\text{CN}^-. Writing "HCN adds across the carbonyl" is fine as a summary of what goes in and what comes out, but if you are asked for the mechanism you must show cyanide ion attacking and being regenerated by the final protonation step, exactly as in the worked example.

Do I need to memorise all four mechanisms separately for the exam

No — memorise the one general two-step template (nucleophile attacks the carbonyl carbon, π\pi electrons shift onto oxygen, alkoxide is protonated) and then just remember one extra branch: if the nucleophile is a neutral nitrogen compound (hydroxylamine, hydrazine, phenylhydrazine), add a proton-transfer step and a final water-loss step to form the C=N\text{C=N} bond. HCN, NaHSO3\text{NaHSO}_3, and Grignard addition all stop at the plain protonated alkoxide, so those three are genuinely the same mechanism with a different nucleophile plugged in.

Why does the reactivity order aldehyde greater than ketone matter beyond just predicting reaction rate

Because it explains several "named test" facts in the same chapter rather than each one being a separate memorised fact. It is why Tollens' and Fehling's tests can distinguish aldehydes from ketones, why bisulphite addition compounds form cleanly with aldehydes and methyl ketones but not bulkier ketones, and why cyanohydrin formation is used more freely as a chain-extension step on aldehydes. Once you know the electronic-plus-steric reason behind the order, you can predict how an unfamiliar carbonyl compound will behave rather than recalling isolated facts.

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