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Nucleophilic addition to carbonyls

T-063Home CU-206Threads kinetics · structure
Statement

The reactivity of the carbon-oxygen double bond.

Why it matters

sn1-sn2 and e1-e2-elimination established nucleophilic and electrophilic reactivity at saturated sp3 carbon centres; the carbonyl group, C=O, is a distinct and extremely common electrophilic centre, made reactive by oxygen's electronegativity polarising the pi bond. Nucleophilic addition across that polarised bond is the single most important reactivity pattern of the carbonyl group, and it underlies much of the synthetic chemistry built on top of this unit, including protecting-groups' concern for masking exactly this reactivity when it would otherwise interfere with a different planned step.

Hypotheses
The carbonyl carbon carries a partial positive charge, arising from oxygen's greater electronegativity polarising the C=O pi bond.This polarisation is what makes nucleophilic, rather than electrophilic, attack the dominant mode of reactivity at this carbon, in direct contrast to electrophilic-addition-markovnikov's alkene chemistry, where the nonpolar C=C bond is instead attacked by electrophiles. The attacking nucleophile has an available lone pair (or pi electrons) able to form a new bond to the carbonyl carbon while pushing the C=O pi electrons fully onto oxygen.This electron-pushing step is what converts the sp2 carbonyl carbon into a tetrahedral sp3 alkoxide centre; without a genuine available lone pair the addition step cannot proceed. Whether the tetrahedral intermediate is the final, isolated product or instead collapses further, expelling a leaving group, depends on whether the group already attached to the carbonyl carbon can act as a leaving group.This fork — simple addition for aldehydes and ketones, versus addition-elimination for carboxylic acid derivatives that carry a genuine leaving group — is covered fully as its own separate lecture in the unit rather than folded entirely into this result.
Proof
1
\[ \text{R}_2\text{C}^{\delta+}{=}\text{O}^{\delta-} \]
The carbonyl carbon is sp2 hybridised, planar, and electrophilic on either face perpendicular to the C=O plane, owing to the bond's polarisation. A
2
\[ \text{Nu: attacks at} \approx107^\circ \text{ to the C=O axis (B\"urgi-Dunitz trajectory)} \]
The nucleophile's lone pair approaches along a specific trajectory reflecting optimal orbital overlap with the carbonyl's \(\pi^*\) antibonding orbital, rather than directly along the C=O axis — a well-established stereoelectronic detail with downstream consequences for facial selectivity on chiral substrates. B
3
\[ \text{R}_2\text{C=O} + \text{Nu:}^- \rightarrow \text{R}_2\text{C(Nu)-O}^- \]
As the new sigma bond forms, the C=O pi electrons are pushed fully onto oxygen, converting the sp2 carbonyl carbon into a tetrahedral sp3 centre and generating an alkoxide oxygen. A
4
\[ \text{R}_2\text{C(Nu)-O}^- \xrightarrow{\text{H}^+} \text{R}_2\text{C(Nu)-OH} \]
A fast proton-transfer step, from solvent or added acid, converts the alkoxide intermediate into the neutral tetrahedral addition product. A
5
\[ \text{Rate} \sim f(\text{electrophilicity of C=O},\ \text{nucleophile strength}) \]
reaction-energy-profiles' rate-determining-step logic applies directly: nucleophilic attack (Step 2) is usually the highest-energy, rate-limiting step, so substituents that raise or lower the carbonyl carbon's electrophilicity directly speed or slow the overall reaction. A
Result
\[ \text{Nu:}^- + \text{R}_2\text{C=O} \longrightarrow \text{R}_2\text{C(Nu)-O}^- \xrightarrow{\text{H}^+} \text{R}_2\text{C(Nu)-OH} \]

Reading. A nucleophile adds directly across the polarised carbonyl pi bond, converting a planar sp2 carbon into a tetrahedral sp3 alkoxide, which is then protonated to the neutral alcohol product.

Scope. Applies broadly to aldehydes and ketones, where the tetrahedral product is typically stable and isolable; carboxylic acid derivatives (esters, amides, acid chlorides) instead usually proceed onward through addition-elimination, re-forming a new carbonyl with the original substituent expelled as a leaving group.

Corollaries & converses
  • Aldehydes are generally more electrophilic, and hence more reactive to nucleophilic addition, than ketones: a ketone's second alkyl substituent is both more sterically bulky and more electron-donating than an aldehyde's hydrogen, and both effects reduce the carbonyl carbon's electrophilicity.
  • reaction-energy-profiles' rate-determining-step framework directly explains why more hindered or less electrophilic carbonyls react more slowly, since nucleophilic attack is generally the highest-energy transition state along the path.
  • Addition-elimination at the acyl carbon, the lecture immediately following this result, is this same mechanism's first half extended by a further collapse-and-expel step, whenever the carbon's original substituent can itself act as a leaving group.
Fails without
  • Attempt addition to a carbonyl carbon with no sterically accessible face along the Bürgi-Dunitz trajectory (Hypotheses, second assumption): steric blocking suppresses the reaction rate well below what the carbonyl's electronic electrophilicity alone would predict, sometimes preventing addition entirely.
  • Assume a simple ketone's tetrahedral alkoxide intermediate will collapse and expel an alkyl substituent as a leaving group: alkyl groups are extremely poor leaving groups, so no addition-elimination occurs; the alkoxide is instead simply protonated to the stable alcohol, exactly as the Hypotheses' third assumption distinguishes.
Common errors
  • Assuming nucleophiles attack directly along the C=O axis rather than along the genuine \(\approx107^\circ\) Bürgi-Dunitz trajectory reflecting real orbital overlap.
  • Confusing simple nucleophilic addition (aldehydes/ketones, product retains both the nucleophile and the original substituents) with addition-elimination (acid derivatives, product expels one original substituent as a leaving group) — these give structurally different products from different substrate classes.
  • Treating all carbonyl compounds as equally electrophilic, ignoring how electron-withdrawing or -donating substituents and steric bulk substantially modulate reactivity.
  • Drawing the alkoxide intermediate as the final, isolated product, forgetting the fast protonation step that occurs under typical aqueous or protic workup conditions.
Discussion

The tetrahedral intermediate central to this mechanism, and its role in distinguishing simple addition from addition-elimination at the acyl carbon, is one of the most thoroughly characterised transformations in organic chemistry, established through decades of kinetic and isotope-labelling studies through the twentieth century.

The Bürgi-Dunitz angle, named after Hans-Beat Bürgi and Jack Dunitz, who established the roughly \(107^\circ\) nucleophilic approach trajectory from a crystallographic survey in the 1970s, is a striking example of stereoelectronic control: the nucleophile's lone pair aligns for maximal overlap with the carbonyl's \(\pi^*\) orbital, and this trajectory has real downstream consequences for the stereochemical outcome (facial selectivity) of additions to chiral carbonyl substrates.

Common misconception: that a hydride reducing agent or Grignard reagent simply "replaces" the carbonyl oxygen. The mechanism is a genuine nucleophilic addition across the pi bond (Steps 1–4), with the oxygen retained throughout as an alkoxide and only protonated at the very end, never displaced or removed at any stage.

Worked examples
1
\[ \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN} \]
Cyanide adds to acetaldehyde's carbonyl carbon (Step 3), and the resulting alkoxide is protonated (Step 4) to give the neutral cyanohydrin product — a classic, simple illustration of the full mechanism. A
2
\[ \text{R}_2\text{C=O} + \text{NaBH}_4 \rightarrow \text{R}_2\text{CH-OH} \]
Hydride, delivered from borohydride, acts as the nucleophile in Step 3, adding to the ketone's carbonyl carbon; aqueous workup then protonates the resulting alkoxide (Step 4) to give the secondary alcohol — the same tetrahedral-intermediate mechanism, now with a hydride rather than a carbon or heteroatom nucleophile. A
\[ \text{Carbonyl} + \text{Nu:}^- \rightarrow \text{tetrahedral alkoxide} \xrightarrow{\text{H}^+} \text{alcohol product} \]

Reading. Whether the nucleophile is a small anion (cyanide) or a hydride reagent, the identical addition-then-protonation sequence accounts for the product.

Scope. The same two-step reading applies to any nucleophile adding to an aldehyde or ketone whose product is isolable as the simple tetrahedral alcohol, without a subsequent elimination step.

Problems
  1. Describe the mechanism for addition of water to formaldehyde, forming the gem-diol hydrate.
    SolutionWater's oxygen lone pair attacks formaldehyde's carbonyl carbon (Step 2–3), pushing the C=O pi electrons onto the original carbonyl oxygen and forming a new C–O bond, giving a protonated (oxonium) tetrahedral intermediate; loss of a proton from this new oxygen (rather than the addition of an external proton, as in Step 4) gives the neutral gem-diol, \(\text{H}_2\text{C(OH)}_2\).
  2. Rank an unhindered aldehyde, an unhindered ketone, and a sterically hindered (bulky) ketone by expected rate of nucleophilic addition.
    SolutionAldehyde fastest, unhindered ketone next, hindered ketone slowest. The aldehyde is both the most electrophilic (one substituent is only hydrogen, minimally electron-donating and non-bulky) and the least sterically hindered; the hindered ketone combines both a less electrophilic carbon (two electron-donating alkyl groups) and substantial steric blocking of the nucleophile's approach trajectory (Step 2), compounding to the slowest rate (Corollaries, first bullet).
  3. Explain why esters, unlike ketones, ultimately regenerate a carbonyl rather than retaining the tetrahedral alkoxide as the final product.
    SolutionAn ester's carbonyl carbon carries an alkoxy (\(\text{OR}'\)) substituent, which can act as a leaving group; once the tetrahedral alkoxide intermediate forms (Step 3), it can collapse by re-forming the C=O pi bond and expelling \(\text{OR}'^-\) (addition-elimination), regenerating a new carbonyl compound. A ketone's alkyl substituents are far poorer leaving groups, so its tetrahedral alkoxide instead simply gets protonated (Step 4) and is isolated as the stable alcohol product (Hypotheses, third assumption).
  4. Calculate the theoretical yield of 3-methylbutan-2-ol when 5.00 g of 3-methylbutan-2-one (methyl isopropyl ketone) is reacted with excess sodium borohydride (\(\text{NaBH}_4\)) followed by acidic workup. Assume 100% conversion.
    Solution First, determine the molar mass of 3-methylbutan-2-one (\(\text{C}_5\text{H}_{10}\text{O}\)). Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol. Molar mass = \(5(12.01) + 10(1.008) + 16.00 = 60.05 + 10.08 + 16.00 = 86.13 \text{ g/mol}\). Next, calculate the moles of ketone used: \(\text{Moles of ketone} = \frac{5.00 \text{ g}}{86.13 \text{ g/mol}} = 0.05805 \text{ mol}\). The stoichiometry of the reduction is 1:1 (one ketone molecule yields one alcohol molecule). Therefore, moles of 3-methylbutan-2-ol (\(\text{C}_5\text{H}_{12}\text{O}\)) produced = 0.05805 mol. Calculate the molar mass of 3-methylbutan-2-ol (\(\text{C}_5\text{H}_{12}\text{O}\)): Molar mass = \(5(12.01) + 12(1.008) + 16.00 = 60.05 + 12.096 + 16.00 = 88.15 \text{ g/mol}\). Finally, calculate the mass of the product: \(\text{Mass} = 0.05805 \text{ mol} \times 88.15 \text{ g/mol} = 5.117 \text{ g}\). Rounding to 3 significant figures (based on 5.00 g): \(\text{Yield} = 5.12 \text{ g}\).
  5. Explain why the addition of a Grignard reagent (\(\text{RMgX}\)) to a ketone followed by acidic workup yields a tertiary alcohol, whereas the addition of a Grignard reagent to an ester yields a tertiary alcohol with two identical alkyl groups from the Grignard reagent.
    Solution For a ketone (\(\text{R}_2\text{C=O}\)): 1. The Grignard reagent (\(\text{R'MgX}\)) adds to the carbonyl carbon, forming a tetrahedral magnesium alkoxide intermediate: \(\text{R}_2\text{C(OMgX)R'}\). 2. Acidic workup protonates the alkoxide to form the tertiary alcohol \(\text{R}_2\text{C(OH)R'}\). 3. Ketones do not have a leaving group attached to the carbonyl carbon (the substituents are alkyl or aryl groups, which are poor leaving groups). Therefore, the reaction stops at the alcohol stage. For an ester (\(\text{RCOOR''}\)): 1. The first equivalent of the Grignard reagent adds to the carbonyl carbon, forming a tetrahedral intermediate: \(\text{R-C(OMgX)(OR'')R'}\). 2. This intermediate is unstable because the alkoxy group (\(\text{OR''}\)) is a much better leaving group than an alkyl group. It collapses, expelling \(\text{OR''}^-\) and re-forming the carbonyl to produce a ketone: \(\text{RCOR'}\). 3. The newly formed ketone is more reactive than the original ester (less steric hindrance and no electron-donating alkoxy group). It immediately reacts with a second equivalent of the Grignard reagent (\(\text{R'MgX}\)). 4. This second addition forms a new tetrahedral alkoxide: \(\text{R-C(OMgX)(R')}_2\). 5. Acidic workup yields the tertiary alcohol \(\text{R-C(OH)(R')}_2\), where the two \(\text{R'}\) groups are identical because they both came from the Grignard reagent.