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Organic Chemistry 2 Unit 5: Aldehydes and Ketones: Nucleophilic Addition

The polar C=O of aldehydes and ketones is attacked by nucleophiles at carbon, and this single idea explains reductions, Grignard additions, hydrates, acetals, imines, the Wittig reaction and much of synthetic planning.

Covers: Structure, naming and reactivity of aldehydes and ketones, Preparation and oxidation of aldehydes and ketones, Hydride reduction and Grignard addition, Hydrates, acetals and protecting groups, Imines, enamines and reductive amination, Wittig reaction and cyanohydrins, Stereochemistry of carbonyl addition

Study guide

Aldehydes (RCHO, named with the suffix -al) and ketones (RCOR', suffix -one) contain a trigonal planar sp² carbonyl carbon. Oxygen is more electronegative, so the carbon is electrophilic (δ+) and the oxygen is weakly basic. Nucleophiles attack the carbon, breaking the π bond to give a tetrahedral alkoxide intermediate that is usually protonated. Aldehydes are more reactive than ketones for two reasons: they have less steric hindrance, and they have only one electron-donating alkyl group stabilizing the δ+ carbon. Formaldehyde is the most reactive of all. Electron-withdrawing groups near the carbonyl increase reactivity. Aldehydes and ketones are commonly made by PCC or Swern oxidation of alcohols, ozonolysis of alkenes, hydration of alkynes and Friedel-Crafts acylation.

Strong, irreversible nucleophiles add directly. NaBH₄ and LiAlH₄ deliver hydride (H⁻): aldehydes give primary alcohols and ketones give secondary alcohols. Grignard reagents (RMgX) and organolithiums (RLi) deliver carbanion-like R groups, forming new C–C bonds: formaldehyde gives primary alcohols, other aldehydes give secondary alcohols, and ketones give tertiary alcohols after an acid workup. Grignard reagents are also very strong bases, so they are destroyed by water, alcohols, carboxylic acids, amines and terminal alkynes; any O–H or N–H in the molecule must be absent or protected. To plan a Grignard synthesis of an alcohol, disconnect one bond to the carbinol carbon: that carbon becomes the C=O carbon, and the removed group becomes R in RMgX.

Weak nucleophiles add reversibly, often with acid or base catalysis. Water adds to give a gem-diol (hydrate); the equilibrium favors the hydrate only for very reactive carbonyls such as formaldehyde and chloral. Alcohols add to give a hemiacetal and then, with acid catalysis, an acetal: protonating the hemiacetal OH lets water leave, forming a resonance-stabilized oxocarbenium ion that a second alcohol attacks. Removing water drives acetal formation, and aqueous acid reverses it. Acetals are stable to bases, hydrides and Grignard reagents, so a cyclic acetal from ethylene glycol is the standard protecting group for an aldehyde or ketone. Cyanide adds to give a cyanohydrin, whose nitrile can be hydrolyzed to an α-hydroxy acid or reduced to an amine.

Primary amines (RNH₂) condense with aldehydes and ketones to give imines (C=N–R) and water; the reaction is fastest near pH 4 to 5, where there is enough acid to protonate the leaving OH but most amine is still unprotonated and nucleophilic. Secondary amines (R₂NH) cannot form a neutral C=N, so they lose a proton from the α-carbon instead to give an enamine (C=C–NR₂). Both reactions are reversed by aqueous acid. In reductive amination, an imine formed in situ is reduced with NaBH₃CN or NaBH(OAc)₃ to give an amine. Hydrazine forms a hydrazone, which in the Wolff-Kishner reduction (KOH, heat) loses N₂ to convert C=O into CH₂ under basic conditions.

The Wittig reaction converts C=O into C=C at a precisely defined position. A phosphonium salt, made by SN2 reaction of Ph₃P with an alkyl halide, is deprotonated by a strong base such as BuLi to form a phosphorus ylide, Ph₃P=CR₂. The ylide adds to the carbonyl to form an oxaphosphetane, which collapses to the alkene and triphenylphosphine oxide; the strong P=O bond drives the reaction. Unstabilized ylides usually give mainly Z-alkenes. Because the carbonyl carbon is planar, nucleophiles attack either face; an achiral ketone with two different groups gives a racemic product, while a substrate with an existing stereocenter gives unequal diastereomers. The Baeyer-Villiger oxidation with a peroxyacid converts ketones to esters, with the more substituted group (aryl or tertiary > secondary > primary > methyl) migrating to oxygen.

Key ideas

Formulas and rules

RCHO or R₂C=O + NaBH₄ or LiAlH₄, then H₃O⁺ → RCH₂OH (1°) or R₂CHOH (2°)
H₂C=O → 1° alcohol; RCHO → 2° alcohol; R₂C=O → 3° alcohol (with R'MgX, then H₃O⁺)
R₂C=O + 2 R'OH ⇌ R₂C(OR')₂ + H₂O (acid catalyst, remove water)
R₂C=O + R'NH₂ → R₂C=NR' (imine); R₂C=O + R'₂NH → enamine
R₂C=O + Ph₃P=CR'₂ → R₂C=CR'₂ + Ph₃P=O
K_hyd = [hydrate]/[carbonyl]; fraction hydrate = K/(1 + K)

Common mistakes

Before the test, make sure you can

Words to know

Carbonyl group
A C=O double bond with an electrophilic carbon and a nucleophilic, weakly basic oxygen.
Nucleophilic addition
Attack of a nucleophile on a carbonyl carbon that converts it from sp² to sp³.
Tetrahedral intermediate
The sp³ alkoxide or alcohol formed when a nucleophile adds to a carbonyl carbon.
Grignard reagent
An organomagnesium halide, RMgX, that acts as a carbon nucleophile and strong base.
Hydrate (gem-diol)
A compound with two OH groups on one carbon, formed by adding water to a carbonyl.
Hemiacetal
A carbon bonded to one OH and one OR, formed by adding one alcohol to an aldehyde or ketone.
Acetal
A carbon bonded to two OR groups, formed from a carbonyl and two alcohols with acid catalysis.
Protecting group
A temporary group, such as a cyclic acetal, that masks a functional group during another reaction.
Oxocarbenium ion
A resonance-stabilized cation R₂C=OR⁺ formed during acetal formation and hydrolysis.
Imine
A C=N compound formed from an aldehyde or ketone and a primary amine.
Enamine
A C=C–NR₂ compound formed from an aldehyde or ketone and a secondary amine.
Reductive amination
Formation of an imine followed by reduction, usually with NaBH₃CN, to make an amine.
Cyanohydrin
A compound with OH and CN on the same carbon, formed by adding HCN to a carbonyl.
Phosphorus ylide
Ph₃P=CR₂, a neutral species with adjacent opposite charges that reacts with carbonyls in the Wittig reaction.
Wittig reaction
Reaction of a phosphorus ylide with an aldehyde or ketone to form an alkene and Ph₃P=O.
Tollens' test
Oxidation of an aldehyde by Ag(NH₃)₂⁺, depositing a silver mirror; ketones give no reaction.
Baeyer-Villiger oxidation
Peroxyacid oxidation of a ketone to an ester, inserting O next to the group with greater migratory aptitude.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

What does NaBH₄ followed by aqueous workup do to a ketone?

  1. Oxidizes it to a carboxylic acid
  2. Reduces it completely to an alkane
  3. Reduces it to a primary alcohol
  4. Reduces it to a secondary alcohol
Show the answer and steps

Answer: D) Reduces it to a secondary alcohol

  1. NaBH₄ delivers a hydride to the carbonyl carbon, giving an alkoxide.
  2. Protonation gives R₂CHOH, a secondary alcohol, because the carbonyl carbon already had two R groups.
Question 2 (level 1)

What reagents and conditions convert an aldehyde into an acetal?

  1. Two equivalents of alcohol with a strong base
  2. One equivalent of water with a base catalyst
  3. One equivalent of alcohol with NaBH₄ and heat
  4. Two equivalents of alcohol, acid catalyst, water removed
Show the answer and steps

Answer: D) Two equivalents of alcohol, acid catalyst, water removed

  1. Acetal formation requires acid to protonate the hemiacetal OH so it can leave as water.
  2. Two alcohol molecules add, and removing water shifts the equilibrium toward the acetal; base stops at the hemiacetal stage.
Question 3 (level 3)

A student adds 1.2 equivalents of 3.0 M methylmagnesium bromide to 5.00 g of benzaldehyde (106.12 g/mol). What volume of Grignard solution is needed?

  1. 15.7 mL
  2. 13.1 mL
  3. 56.5 mL
  4. 18.8 mL
Show the answer and steps

Answer: D) 18.8 mL

  1. Moles of benzaldehyde = 5.00 / 106.12 = 0.0471 mol.
  2. Grignard needed = 1.2 × 0.0471 = 0.0565 mol.
  3. Volume = 0.0565 / 3.0 = 0.0188 L = 18.8 mL; 15.7 mL forgets the 1.2 equivalents, and 56.5 mL forgets to divide by the 3.0 M concentration.