9OC1 Exercise 9 Study Guide: Aldol Reactions, Enolates, and Carbonyl Additions

Overview of Carbonyl Chemistry and Exercise Scope

• Course Information: Grundlagen der Organischen Chemie (OC1), Wintersemester 2026/2027, Übung 9 (Discussion Date: 04.01.2027).
• Instructors: Dr. Johanna Elter and Prof. Dr. Christmann.
• Central Themes:
  • Acid- and base-catalyzed Aldol reactions (additions and condensations), including crossed Aldol selectivity, thermodynamic vs. kinetic control, and stereochemical analysis.
  • Enolate halogenations ($\alpha$-halogenation and the haloform reaction).
  • Imine (Schiff base) formation and dehydration mechanisms.
  • Nucleophilic additions of carbon nucleophiles to carbonyls: cyanohydrin formation followed by hydride reduction, directed Aldol additions using lithium diisopropylamide ($\text{LDA}$), and terminal alkyne additions.
  • Protecting group chemistry (cyclic acetal/ketal formation with 1,3-diols) and functional group conversions (Williamson ether synthesis).
  • Retrosynthetic analysis and synthesis of polyconjugated bis-enones via double Aldol condensation.

Aufgabe 1: Acid- and Base-Catalyzed Aldol Reactions

Reaction 1.1 & 1.2: Benzaldehyde and Acetone (Claisen-Schmidt Condensation)

• Reactants:
  • Benzaldehyde ($\text{Ph}-\text{CHO}$): Non-enolisable partner due to the absence of $\alpha$-protons; can solely function as an electrophile.
  • Acetone ($\text{CH}_3\text{COCH}_3$): Symmetric ketone containing 6 equivalent enolisable $\alpha$-protons; functions as the nucleophilic partner.

Acid-Catalyzed Pathway ($H^+$)

• Mechanism:
  • Step 1: Enolization of acetone. Reversible protonation of the carbonyl oxygen of acetone yields the resonance-stabilized oxocarbenium ion ($\text{CH}_3-\text{C}(\text{OH}^+)-\text{CH}_3 \rightleftharpoons \text{CH}_3-\text{C}^+(\text{OH})-\text{CH}_3$). Subsequent deprotonation at the $\alpha$-carbon by a weak base ($\text{H}_2\text{O}$ or acetone) delivers the neutral enol tautomer, prop-1-en-2-ol ($\text{CH}_2=\text{C}(\text{OH})\text{CH}_3$).
  • Step 2: Electrophile activation. Reversible protonation of benzaldehyde generates the activated electrophilic species ($\text{Ph}-\text{CH}=\text{OH}^+ \rightleftharpoons \text{Ph}-\text{CH}^+-\text{OH}$).
  • Step 3: Carbon-carbon bond formation (Aldol addition). The enol of acetone attacks the activated carbonyl carbon of benzaldehyde. Deprotonation of the resulting oxonium species yields the $\beta$-hydroxyketone: 4-hydroxy-4-phenylbutan-2-one ($\text{Ph}-\text{CH}(\text{OH})-\text{CH}_2-\text{CO}-\text{CH}_3$).
  • Step 4: Acid-catalyzed dehydration ($E1$ elimination). Protonation of the $\beta$-hydroxy group converts it into a good leaving group ($-\text{OH}_2^+$). Spontaneous loss of water generates a resonance-stabilized, benzylically and allylically stabilized carbocation intermediate ($\text{Ph}-\text{CH}^+-\text{CH}_2-\text{CO}-\text{CH}_3 \rightleftharpoons \text{Ph}-\text{CH}=\text{CH}-\text{C}(\text{OH}^+)-\text{CH}_3$). Deprotonation at the $\alpha$-position yields the conjugated $\alpha,\beta$-unsaturated ketone: 4-phenylbut-3-en-2-one (benzylideneacetone).

Base-Catalyzed Pathway (Base)

• Mechanism:
  • Step 1: Enolate generation. Deprotonation of an $\alpha$-proton of acetone by a base (e.g., $\text{OH}^-$ or $\text{OR}^-$) produces the ambident enolate anion ($\text{CH}_2=\text{C}(\text{O}^-)\text{CH}_3 \rightleftharpoons ^-\text{CH}_2-\text{CO}-\text{CH}_3$).
  • Step 2: Nucleophilic addition. The enolate attacks the carbonyl carbon of benzaldehyde via nucleophilic addition, generating an alkoxide intermediate ($\text{Ph}-\text{CH}(\text{O}^-)-\text{CH}_2-\text{CO}-\text{CH}_3$).
  • Step 3: Protonation. Proton abstraction from solvent ($\text{H}_2\text{O}$ or alcohol) gives the $\beta$-hydroxyketone (4-hydroxy-4-phenylbutan-2-one) and regenerates the base catalyst.
  • Step 4: Base-catalyzed dehydration ($E1_{cB}$ elimination). The $\alpha$-protons of the $\beta$-hydroxyketone are significantly acidified by the adjacent carbonyl group ($pK_a \approx 19-20$). Base deprotonates the $\alpha$-carbon to form an enolate intermediate. Expulsion of hydroxide ($\text{OH}^-$) as a leaving group driven by the thermodynamic stability of the extended conjugated system delivers 4-phenylbut-3-en-2-one.

Stereochemistry and Selectivity Analysis

• Stereoisomers Formed:
  • Aldol addition stage: Generates 1 stereocenter ($\text{C}4$) -> 2 stereoisomers (enantiomeric pair: $(R)$- and $(S)$-4-hydroxy-4-phenylbutan-2-one).
  • Aldol condensation stage: Generates 1 disubstituted carbon-carbon double bond -> 2 geometric isomers: $(E)$- and $(Z)$-4-phenylbut-3-en-2-one.
  • Major isomer: The $(E)$-isomer is formed almost exclusively ($>99\%$) because the $(Z)$-isomer suffers severe steric repulsion between the phenyl ring and the acetyl group, which destabilizes planarity and disrupts optimal $\pi$-conjugation.
• Selectivity and Competing Reactions:
  • Chemoselectivity: Aldehydes are substantially more electrophilic than ketones due to less steric hindrance and less electron donation from alkyl substituents. Hence, nucleophilic attack of the acetone enol/enolate onto benzaldehyde is kinetically preferred over attack onto another acetone molecule (self-aldol to mesityl oxide).
  • Polycondensation: Benzylideneacetone still possesses an enolisable methyl group. It can undergo a second condensation with another molecule of benzaldehyde to form dibenzylideneacetone (1,5-diphenylpenta-1,4-dien-3-one). To selectively obtain mono-addition, acetone is typically used in excess.

Reaction 1.3 & 1.4: Cyclohexanone and Propanal

• Reactants:
  • Cyclohexanone: Cyclic ketone with 4 enolisable $\alpha$-protons across two equivalent positions.
  • Propanal: Linear aliphatic aldehyde with 2 enolisable $\alpha$-protons.

Acid-Catalyzed Pathway ($H^+$)

• Enolization Dynamics:
  • Cyclohexanone forms cyclohex-1-en-1-ol.
  • Propanal forms prop-1-en-1-ol.
• Preferred Crossed Reaction:
  • Propanal is substantially more electrophilic than cyclohexanone.
  • Cyclohex-1-en-1-ol nucleophilically attacks protonated propanal at the carbonyl carbon.
  • Subsequent proton transfer and acid-catalyzed elimination of water yields 2-propylidenecyclohexan-1-one.
• Mechanism Steps:
  1. Protonation of propanal carbonyl oxygen by $H^+$ activates the electrophilic center.
  2. Enol of cyclohexanone attacks the carbonyl carbon of propanal.
  3. Deprotonation delivers the addition product: 2-(1-hydroxypropyl)cyclohexan-1-one.
  4. Protonation of the hydroxyl group to form an oxonium ion ($-\text{OH}_2^+$), followed by loss of $\text{H}_2\text{O}$ and deprotonation at the ring $\alpha$-position, yields 2-propylidenecyclohexan-1-one.

Base-Catalyzed Pathway (Base)

• Selectivity Complexities:
  • Propanal has more acidic $\alpha$-protons ($pK_a \approx 17$) than cyclohexanone ($pK_a \approx 19-20$) and is sterically less encumbered.
  • Uncontrolled mixing in base results in four competing pathways:
  1. Propanal self-aldol: Deprotonation of propanal followed by addition to propanal gives 3-hydroxy-2-methylpentanal, which readily dehydrates via $E1_{cB}$ to (E)-2-methylpent-2-enal.
  2. Crossed aldol (cyclohexanone enolate + propanal): Forms 2-(1-hydroxypropyl)cyclohexan-1-one, which dehydrates to 2-propylidenecyclohexan-1-one.
  3. Crossed aldol (propanal enolate + cyclohexanone): Forms 2-(1-hydroxycyclohexyl)propanal, which can eliminate to 2-cyclohexylidenepropanal.
  4. Cyclohexanone self-aldol: Forms 2-(1-hydroxycyclohexyl)cyclohexan-1-one, eliminating to 2-(cyclohex-1-en-1-yl)cyclohexan-1-one or 2-cyclohexylidenecyclohexan-1-one.
  • Controlling factors: To favor the crossed product with cyclohexanone acting as the nucleophile, propanal must be added dropwise into an excess of basic cyclohexanone.

Stereochemistry and Product Evaluation

• Addition Stage (2-(1-hydroxypropyl)cyclohexan-1-one):
  • Contains 2 stereocenters: $\text{C}2$ of the ring and $\text{C}1'$ of the propyl side-chain.
  • Theoretical number of stereoisomers: $2^2 = 4$ stereoisomers (2 pairs of enantiomers: syn- and anti-diastereomeric pairs).
• Condensation Stage (2-propylidenecyclohexan-1-one):
  • The exocyclic double bond generates 2 geometric isomers: $(E)$- and $(Z)$-2-propylidenecyclohexan-1-one.
  • The $(E)$-isomer (ethyl group directed away from the ring carbonyl) is thermodynamically favored over the $(Z)$-isomer due to reduced steric clash with the carbonyl oxygen.

Reaction 1.5 & 1.6: Isobutyraldehyde (2-Methylpropanal) and Butanal

• Reactants:
  • Isobutyraldehyde ($(\text{CH}_3)_2\text{CH}-\text{CHO}$): Branched aldehyde with exactly one $\alpha$-proton.
  • Butanal ($\text{CH}_3\text{CH}_2\text{CH}_2-\text{CHO}$): Linear aldehyde with two $\alpha$-protons.

Pathway Comparison and Elimination Constraints

• Pathway A (Butanal as nucleophile, Isobutyraldehyde as electrophile):
  • Butanal enol/enolate attacks isobutyraldehyde at the carbonyl carbon.
  • Aldol addition product: 3-hydroxy-2-ethyl-4-methylpentanal.
  • Elimination capacity: The $\alpha$-carbon ($\text{C}2$) possesses a remaining $\alpha$-hydrogen. Under either acid or base conditions, elimination of water proceeds smoothly to yield the conjugated enal: (E)-2-ethyl-4-methylpent-2-enal.
  • Thermodynamic driving force: Dehydration provides an irreversible thermodynamic sink that pulls the equilibrium completely toward the condensation product.
• Pathway B (Isobutyraldehyde as nucleophile, Butanal as electrophile):
  • Isobutyraldehyde enol/enolate attacks butanal.
  • Aldol addition product: 3-hydroxy-2,2-dimethylhexanal.
  • Elimination capacity: The $\alpha$-carbon ($\text{C}2$) is fully substituted (quaternary center bearing two methyl groups). It has zero $\alpha$-protons.
  • Consequently, dehydration is chemically impossible; the reaction cannot eliminate to form an $\alpha,\beta$-unsaturated aldehyde and remains fully reversible (retro-aldol equilibrium).
• Pathway C (Butanal self-aldol):
  • Yields 2-ethylhex-2-enal upon dehydration. Highly competitive in base because butanal has less hindered $\alpha$-protons and high electrophilicity.
• Pathway D (Isobutyraldehyde self-aldol):
  • Yields 3-hydroxy-2,2,4-trimethylpentanal. Cannot eliminate and is severely sterically hindered; retro-aldol dominates.

Mechanistic Details for the Productive Crossed Pathway (Pathway A)

• Acid Catalysis ($H^+$):
  1. Enolization of butanal delivers (E/Z)-but-1-en-1-ol.
  2. Protonation of isobutyraldehyde carbonyl oxygen enhances its electrophilic reactivity.
  3. Nucleophilic attack of but-1-en-1-ol onto protonated isobutyraldehyde forms 3-hydroxy-2-ethyl-4-methylpentanal.
  4. Protonation of the hydroxyl group to $-\text{OH}_2^+$, followed by loss of water and deprotonation at $\text{C}2$, gives (E)-2-ethyl-4-methylpent-2-enal.
• Base Catalysis (Base):
  1. Base abstracts an $\alpha$-proton from butanal to generate the corresponding enolate.
  2. Enolate adds to the carbonyl of isobutyraldehyde to form an alkoxide.
  3. Protonation by water furnishes 3-hydroxy-2-ethyl-4-methylpentanal.
  4. Base deprotonates the sole remaining $\alpha$-proton at $\text{C}2$, yielding an enolate that expels $\text{OH}^-$ via an $E1_{cB}$ mechanism to form (E)-2-ethyl-4-methylpent-2-enal.

Stereochemistry

• Addition Product (3-hydroxy-2-ethyl-4-methylpentanal):
  • 2 asymmetric centers ($\text{C}2$ and $\text{C}3$) -> $2^2 = 4$ stereoisomers (2 diastereomeric pairs of enantiomers).
• Condensation Product (2-ethyl-4-methylpent-2-enal):
  • 2 geometric isomers: $(E)$- and $(Z)$-isomers.
  • Major product: $(E)$-isomer, which minimizes steric interactions between the bulky isopropyl group and the carbonyl oxygen / ethyl group.
• Non-eliminating Adduct (3-hydroxy-2,2-dimethylhexanal from Pathway B):
  • 1 asymmetric center ($\text{C}3$) -> 2 stereoisomers ($(R)$ and $(S)$).

Aufgabe 2: Halogenation, Imine Formation, and Carbonyl Reactions

2a: Haloform Reaction of Acetophenone ($I_2, NaOH$)

• Starting Material: Acetophenone ($\text{Ph}-\text{CO}-\text{CH}_3$).
• Reagents: Iodine ($\text{I}_2$) and aqueous Sodium Hydroxide ($\text{NaOH}$).
• Transformation: Classic Haloform Reaction (specifically the Iodoform test).
• Products: Sodium benzoate ($\text{Ph}-\text{COO}^- \text{Na}^+$) / Benzoic acid ($\text{Ph}-\text{COOH}$ after acidic workup) + Iodoform ($\text{CHI}_3$) as a canary yellow crystalline precipitate.
• Step-by-Step Mechanism:
  • Step 1: Base deprotonation. Hydroxide abstracts an $\alpha$-proton from the methyl group to form an enolate anion ($\text{Ph}-\text{CO}-\text{CH}_2^-$).
  • Step 2: Halogenation. The enolate attacks $\text{I}_2$, displacing iodide ($\text{I}^-$) to generate monoiodoacetophenone ($\text{Ph}-\text{CO}-\text{CH}_2\text{I}$).
  • Step 3: Accelerated subsequent halogenations. The presence of the electronegative iodine atom increases the inductive electron withdrawal, making the remaining $\alpha$-protons significantly more acidic than those of the unsubstituted starting ketone. Consequently, deprotonation and iodination occur sequentially faster for mono- and di-iodo intermediates, rapidly producing triiodoacetophenone ($\text{Ph}-\text{CO}-\text{CI}_3$).
  • Step 4: Nucleophilic acyl addition. Hydroxide ion attacks the carbonyl carbon of $\text{Ph}-\text{CO}-\text{CI}_3$, generating a tetrahedral intermediate ($\text{Ph}-\text{C}(\text{O}^-)(\text{OH})-\text{CI}_3$).
  • Step 5: Cleavage of the carbon-carbon bond. Collapse of the tetrahedral intermediate reforms the carbon-oxygen double bond and expels the triiodomethyl carbanion ($^-\text{CI}_3$). The $-CI_3$ group is an unusually stable carbon leaving group due to the strong inductive stabilization of three electronegative iodine atoms.
  • Step 6: Irreversible proton transfer. The strongly basic carbanion ($^-\text{CI}_3$) immediately and irreversibly deprotonates the newly formed benzoic acid ($\text{Ph}-\text{COOH}$), yielding benzoate ($\text{Ph}-\text{COO}^-$) and neutral iodoform ($\text{CHI}_3$).

2b: Reaction of Pivalophenone with $Br_2, H^+$ (Attempted $\alpha$-Bromination)

• Starting Material: Pivalophenone / 2,2-dimethyl-1-phenylpropan-1-one ($\text{Ph}-\text{CO}-\text{C}(\text{CH}_3)_3$).
• Reagents: Bromine ($\text{Br}_2$) in the presence of acid ($\text{H}^+$).
• Reaction Outcome: No reaction (Keine Reaktion).
• Exhaustive Justification:
  • Acid-catalyzed halogenation of carbonyl compounds strictly requires the formation of an enol intermediate.
  • Enol formation necessitates the removal of an $\alpha$-hydrogen atom from an $\text{sp}^3$-hybridized carbon adjacent to the carbonyl group.
  • Examination of pivalophenone reveals:
  • On the left: The phenyl group possesses only $\text{sp}^2$-hybridized ring carbons (no aliphatic $\alpha$-hydrogens).
  • On the right: The tert-butyl group ($-\text{C}(\text{CH}_3)_3$) features a quaternary $\alpha$-carbon bonded to three methyl groups, possessing zero $\alpha$-hydrogens.
  • Since pivalophenone lacks any enolisable $\alpha$-protons, enolization cannot take place, and $\alpha$-bromination is completely precluded.
  • Furthermore, electrophilic aromatic substitution (bromination of the phenyl ring) does not occur under these conditions because the carbonyl group is strongly electron-withdrawing (deactivating), and no powerful Lewis acid catalyst (such as $\text{FeBr}_3$ or $\text{AlBr}_3$) is present to activate molecular bromine.

2c: Condensation of Benzaldehyde with Neopentylamine (Imine Formation)

• Starting Materials:
  • Benzaldehyde ($\text{Ph}-\text{CHO}$
  • Neopentylamine / 2,2-dimethylpropan-1-amine ($\text{H}_2\text{N}-\text{CH}_2-\text{C}(\text{CH}_3)_3$
  • Catalyst: Catalytic acid (kat. $\text{H}^+$
• Reaction Class: Imine (Schiff base) condensation.
• Product: N-neopentyl-1-phenylmethanimine ($\text{Ph}-\text{CH}=\text{N}-\text{CH}_2-\text{C}(\text{CH}_3)_3$) + $\text{H}_2\text{O}$.
• Step-by-Step Mechanism:
  • Step 1: Carbonyl activation. Protonation of the carbonyl oxygen of benzaldehyde increases its electrophilicity.
  • Step 2: Nucleophilic addition. The primary amino group of neopentylamine attacks the activated carbonyl carbon, forming a tetrahedral intermediate with an ammonium center ($\text{Ph}-\text{CH}(\text{OH})-\text{NH}_2^+-\text{CH}_2\text{C}(\text{CH}_3)_3$).
  • Step 3: Proton transfers. Deprotonation of the nitrogen atom followed by protonation of the hydroxyl oxygen yields a neutral hemiaminal (carbinolamine) intermediate, which is then converted into a good leaving group ($-\text{OH}_2^+$).
  • Step 4: Dehydration and iminium formation. The lone pair on the nitrogen atom displaces the water molecule, establishing a carbon-nitrogen double bond and yielding an iminium ion ($\text{Ph}-\text{CH}=\text{NH}^+-\text{CH}_2\text{C}(\text{CH}_3)_3$).
  • Step 5: Deprotonation. Deprotonation of the iminium nitrogen delivers the neutral imine and regenerates the acid catalyst.
  • The reaction is an equilibrium driven forward by removing water (e.g., using molecular sieves or Dean-Stark distillation).

Aufgabe 3: Synthetic Transformations and Multi-Step Sequences

3a: Cyanohydrin Formation and Reduction of Butanal

• Starting Material: Butanal ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$
• Reagents:
  • Step i: Hydrogen cyanide ($\text{HCN}$
  • Step ii: Lithium aluminium hydride ($\text{LiAlH}_4$
• Chemical Transformations:
  • Step i (Cyanohydrin Synthesis): Nucleophilic addition of cyanide ($\text{CN}^-$) to the aldehyde carbonyl carbon generates an alkoxide intermediate, which upon protonation yields 2-hydroxypentanenitrile ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}(\text{OH})\text{CN}$
  • Step ii (Nitrile Reduction): $\text{LiAlH}_4$ is a strong reducing agent that cleanly reduces nitriles ($-C\equiv N$) into primary amines ($-\text{CH}_2\text{NH}_2$). The secondary alcohol is deprotonated by hydride during the reaction to form an aluminium alkoxide, which is hydrolyzed during aqueous workup.
• Final Product: 1-aminopentan-2-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}(\text{OH})\text{CH}_2\text{NH}_2$
• Classification: A $\beta$-amino alcohol (1,2-amino alcohol).

3b: Directed Aldol, Ketal Protection, and Williamson Ether Synthesis

• Multi-Step Synthetic Cascade:

Step 1: Directed Aldol Reaction

• Substrates: Acetone + $\text{LDA}$ (1.0 equiv, $-78^\circ\text{C}$), followed by Isobutyraldehyde ($(\text{CH}_3)_2\text{CH}-\text{CHO}$
• Chemistry:
  • Lithium diisopropylamide ($\text{LDA}$) is a strong, non-nucleophilic, sterically hindered base ($pK_a$ of diisopropylamine $\approx 36$). It deprotonates acetone quantitatively, irreversibly, and kinetically at $-78^\circ\text{C}$ to generate the lithium enolate of acetone ($\text{CH}_2=\text{C}(\text{OLi})\text{CH}_3$
  • Addition of isobutyraldehyde enables a directed crossed Aldol addition without self-condensation of the aldehyde.
  • The acetone enolate attacks the aldehyde carbonyl carbon. Aqueous workup gives the $\beta$-hydroxyketone:
  • Intermediate Product: 4-hydroxy-5-methylhexan-2-one ($\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)_2$

Step 2: Selective Ketal Protection

• Reagent: 2,2-Dimethylpropane-1,3-diol (Neopentyl glycol, $\text{HO}-\text{CH}_2-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{OH}$) + catalytic $\text{H}^+$ (acid)
• Chemistry:
  • Ketones react reversibly with 1,3-diols in the presence of an acid catalyst to produce highly stable six-membered cyclic ketals (1,3-dioxanes).
  • The secondary hydroxyl group ($-\text{OH}$ at $\text{C}4$) does not participate in cyclic ketal formation and remains intact.
  • Intermediate Product: 2-(2-hydroxy-3-methylbutyl)-2,5,5-trimethyl-1,3-dioxane.

Step 3: Williamson Ether Synthesis

• Reagents: i) Sodium hydride ($\text{NaH}$), ii) Benzyl bromide ($\text{BnBr}$, $\text{PhCH}_2\text{Br}$
• Chemistry:
  • Sodium hydride ($\text{NaH}$) quantitatively deprotonates the free secondary alcohol to form a nucleophilic sodium alkoxide, releasing hydrogen gas ($\text{H}_2\uparrow$
  • The alkoxide carries out an $S_N2$ nucleophilic substitution on benzyl bromide, displacing bromide ($\text{Br}^-$) to form a benzyl ether.
• Final Product: 2-(2-(benzyloxy)-3-methylbutyl)-2,5,5-trimethyl-1,3-dioxane.

3c: Alkyne Deprotonation, Carbonyl Addition, and Catalytic Hydrogenation

• Starting Material: Pent-1-yne ($\text{CH}_3\text{CH}_2\text{CH}_2-\text{C}\equiv\text{CH}$
• Reagents:
  • Step i: n-Butyllithium ($n\text{BuLi}$
  • Step ii: Acetone ($\text{CH}_3\text{COCH}_3$
  • Step iii: Hydrogen gas over Palladium on Carbon ($\text{H}_2, \text{Pd}/\text{C}$
• Chemical Steps and Mechanism:
  • Step i (Deprotonation): The terminal alkyne proton is acidic ($pK_a \approx 25$) due to the high s-character of the $\text{sp}$-hybridized carbon. $n\text{BuLi}$ ($pK_a$ of butane $\approx 50$) quantitatively deprotonates pent-1-yne to yield lithium pent-1-yn-1-ide ($\text{CH}_3\text{CH}_2\text{CH}_2-\text{C}\equiv\text{C}^-\text{Li}^+$
  • Step ii (Nucleophilic Carbonyl Addition): The alkynyl lithium nucleophile attacks the electrophilic carbonyl carbon of acetone. Aqueous workup furnishes a tertiary propargylic alcohol: 2-methyloct-3-yn-2-ol ($(\text{CH}_3)_2\text{C}(\text{OH})-\text{C}\equiv\text{C}-\text{CH}_2\text{CH}_2\text{CH}_3$
  • Step iii (Complete Hydrogenation): Palladium on activated charcoal ($\text{Pd}/\text{C}$) is an unpoisoned hydrogenation catalyst. Under an atmosphere of $\text{H}_2$, it completely hydrogenates the alkyne triple bond through the alkene stage down to a fully saturated carbon-carbon single bond. The tertiary alcohol is not reduced.
• Final Product: 2-methyloctan-2-ol ($(\text{CH}_3)_2\text{C}(\text{OH})-\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$

Aufgabe 4: Retrosynthesis and Mechanism of 2,6-Dibenzylidenecyclohexanone

Retrosynthetic Analysis

• Target Molecule: (2E,6E)-2,6-dibenzylidenecyclohexan-1-one.
• Structural Features: A central cyclohexanone core flanked by two exocyclic double bonds conjugated with phenyl rings on both $\alpha$-positions ($\text{C}2$ and $\text{C}6$).
• Disconnection Strategy:
  • Both exocyclic double bonds are disconnected across the Aldol condensation junction.
  • Disconnection yields:
  • One molecule of Cyclohexanone (symmetric ketone bearing four reactive $\alpha$-protons across positions 2 and 6).
  • Two molecules of Benzaldehyde ($\text{Ph}-\text{CHO}$, non-enolisable aromatic aldehyde).
• Required Stoichiometry:
  • Cyclohexanone: 1.0 equivalent.
  • Benzaldehyde: 2.0 equivalents (or a slight excess, e.g., $2.1 - 2.2\,\text{equivalents}$, to ensure complete double condensation).
  • Reaction Conditions: Basic catalysis (e.g., $\text{NaOH}$ or $\text{KOH}$ in $\text{EtOH}/\text{H}_2\text{O}$, classic Claisen-Schmidt conditions).

Detailed Reaction Mechanism

Phase 1: First Aldol Condensation (Mono-Addition)

• Step 1 (Enolate Generation): Hydroxide ion abstracts an $\alpha$-proton from position 2 of cyclohexanone, forming an enolate ($pK_a \approx 19-20$).
• Step 2 (Nucleophilic Addition): The cyclohexanone enolate attacks the carbonyl carbon of the first equivalent of benzaldehyde, producing an alkoxide intermediate.
• Step 3 (Proton Transfer): The alkoxide abstracts a proton from water or ethanol, delivering the mono-aldol adduct, 2-(hydroxy(phenyl)methyl)cyclohexan-1-one, and regenerating hydroxide.
• Step 4 (Dehydration via $E1_{cB}$): Hydroxide deprotonates the remaining $\alpha$-proton at $\text{C}2$. The resulting enolate expels hydroxide ($\text{OH}^-$) to establish the first exocyclic double bond, yielding (E)-2-benzylidenecyclohexan-1-one.

Phase 2: Second Aldol Condensation (Double-Addition)

• Step 5 (Second Enolate Generation): The mono-condensation product possesses two remaining enolisable $\alpha$-protons at the opposite $\alpha$-position ($\text{C}6$). Hydroxide deprotonates $\text{C}6$ to generate a second enolate.
• Step 6 (Second Nucleophilic Addition): This enolate attacks the carbonyl carbon of the second equivalent of benzaldehyde, producing a second alkoxide intermediate.
• Step 7 (Proton Transfer): Protonation by solvent gives 2-benzylidene-6-(hydroxy(phenyl)methyl)cyclohexan-1-one.
• Step 8 (Second Dehydration via $E1_{cB}$): Hydroxide deprotonates the remaining acidic proton at $\text{C}6$. Expulsion of $OH^-$ from the enolate completes the second elimination, furnishing (2E,6E)-2,6-dibenzylidenecyclohexan-1-one.

Stereochemical Configuration

• The double condensation generates two exocyclic double bonds, each possessing $(E)$/$(Z)$ geometric possibilities.
• The isolated product is exclusively the (2E,6E)-isomer:
  • In the $(2E,6E)$-configuration, both bulky phenyl rings point outwards, away from the carbonyl group and the cyclohexanone ring, which entirely avoids severe steric clashing with the carbonyl oxygen and ring methylenes.
  • This planar arrangement maximizes extended cross-conjugation ($\text{Ph}-\text{C}=\text{C}-\text{C}(=\text{O})-\text{C}=\text{C}-\text{Ph}$), providing substantial thermodynamic stabilization that causes the product to precipitate directly from the reaction mixture.