α,β-Unsaturated carbonyls — Michael addition, 1,2 vs 1,4 addition: Worked examples

This page is the "throw everything at it" drill for the parent topic. We build a scenario matrix of every kind of case an exam can hand you, then work one example for each cell — forecasting first, then reasoning, then verifying.

Everything here rests on one fact from the parent note: an α,β-unsaturated carbonyl has two electrophilic carbons — the carbonyl carbon (C1) and the β-carbon (C3) — because resonance leaks the positive charge from oxygen all the way to the β-carbon.

The scenario matrix

Think of every problem as one cell in this grid. The "axis" of a chemistry problem is which lever the question is pulling: the nucleophile's hardness, the temperature, the reversibility, the sterics, or a real-world/exam twist.

#	Case class	The lever being pulled	Expected outcome	Worked in
A	Hard C-nucleophile (Grignard/RLi)	nucleophile hardness	1,2 (allylic alcohol)	Ex 1
B	Soft C-nucleophile (cuprate)	nucleophile hardness	1,4 (saturated ketone)	Ex 2
C	Same substrate, both metals	side-by-side contrast	opposite regiochemistry	Ex 3 (figure)
D	Reversible Nu + temperature ( $C N^{-}$ , low vs high T)	[[Kinetic vs thermodynamic control	kinetic vs thermodynamic]]	1,2 cold, 1,4 hot
E	Stabilised carbanion (malonate enolate) — true Michael	soft delocalised donor	1,4 → 1,5-dicarbonyl	Ex 5
F	Heteroatom soft Nu (amine/thiol)	soft N/S nucleophile	1,4 (aza-/thia-Michael)	Ex 6
G	Degenerate input: no conjugation ( $C = C$ not next to $C = O$ )	limiting/zero case	β-carbon is not electrophilic → no conjugate path	Ex 7
H	Hydride reagents ( $L i A l H_{4}$ vs Stryker/L-Selectride)	hard vs soft hydride	1,2 vs 1,4 reduction	Ex 8
I	Real-world word problem (drug synthesis, 1,5-dicarbonyl for a ring)	applied planning	Michael then aldol	Ex 9
J	Exam twist: β,β-disubstituted enone + bulky Nu	sterics override	forced back to 1,2	Ex 10

Every cell A–J is covered below. Guess before you read each solution.

Ex 1 — Case A: Hard carbon nucleophile

Worked example

$C H_{3} M g B r$ + cyclohex-2-en-1-one, then $H_{3} O^{+}$ Forecast: Grignard is a carbon nucleophile — will it add 1,2 or 1,4? Write your guess.

Step 1 — Classify the nucleophile. $C H_{3} M g B r$ is a Grignard reagent: a small, charge-localised, hard carbanion-like species. Why this step? Hardness is the lever; classify before you predict.

Step 2 — Match to a centre (HSAB). Hard Nu → hard, localised δ⁺ centre = the carbonyl carbon (C1). Why this step? The parent rule: hard→hard. Also RMgX adds irreversibly, so it locks in the kinetic product, and the carbonyl carbon is the faster (kinetic) site.

Step 3 — Draw the addition. $C H_{3}^{-}$ adds to C1; electrons go onto O, giving an alkoxide. $C = C$ is untouched. Why this step? This is ordinary nucleophilic addition to a carbonyl.

Step 4 — Workup. $H_{3} O^{+}$ protonates the alkoxide → 1-methylcyclohex-2-en-1-ol, a tertiary allylic alcohol (C=C retained). Why this step? Neutralise the charged intermediate.

Verify: Product keeps the C=C (allylic alcohol) — the fingerprint of 1,2-addition. Carbon count: $C_{6}$ ring $+ C H_{3} = C_{7} H_{12} O$ . Degrees of unsaturation of $C_{7} H_{12} O = (2 \times 7 + 2 - 12) /2 = 2$ (one ring + one C=C) ✓. No C=O left, so not 1,4.

Ex 2 — Case B: Soft carbon nucleophile

Worked example

$(C H_{3})_{2} C uL i$ + cyclohex-2-en-1-one, then $H_{3} O^{+}$ Forecast: Same $C H_{3}$ delivered — but a different metal. 1,2 or 1,4?

Step 1 — Classify the nucleophile. A lithium dimethylcuprate delivers $C H_{3}$ through a large, polarisable copper — a soft nucleophile. Why this step? The polarisable Cu makes the transferred carbon "soft", unlike the naked $C H_{3}^{-}$ of a Grignard.

Step 2 — Match to a centre. Soft Nu → soft, diffuse δ⁺ = the β-carbon (C3). Why this step? HSAB: soft→soft → conjugate (1,4) addition.

Step 3 — Addition then tautomerise. $C H_{3}$ lands on C3 → an enolate intermediate → tautomerises to the C=O form. The C=C is destroyed. Why this step? The initial 1,4-adduct is always an enolate; tautomerisation regenerates the strong carbonyl.

Step 4 — Workup: protonate → 3-methylcyclohexan-1-one (C=O kept, C=C gone).

Verify: Product formula $C_{7} H_{12} O$ , degrees of unsaturation $= 2$ = one ring + one C=O ✓. Same molecular formula as Ex 1 but a different structure — regiochemistry chosen purely by the metal.

Ex 3 — Case C: Same substrate, both metals side by side (figure)

Figure — α,β-Unsaturated carbonyls — Michael addition, 1,2 vs 1,4 addition

Step 1 — Locate the two δ⁺ sites (yellow dots). C1 (carbonyl carbon) and C3 (β-carbon). Why? Resonance spreads the plus — see the two arrows curving from O.

Step 2 — Hard path (blue arrow). Grignard → C1 → 1-methylcyclohex-2-en-1-ol.

Step 3 — Soft path (red arrow). Cuprate → C3 → 3-methylcyclohexan-1-one.

Verify: Both add one $C H_{3}$ (mass identical, $C_{7} H_{12} O$ ), but the double bond that survives differs — C=C for the hard reagent, C=O for the soft. This is the cleanest demonstration that regiochemistry is set by nucleophile hardness, not by the substrate.

Ex 4 — Case D: Reversible nucleophile + temperature switch

Worked example

$H C N$ (i.e. $C N^{-} / H^{+}$ ) on but-3-en-2-one $C H_{2} = C H - C O - C H_{3}$ Forecast: What changes between an ice bath and a warm flask?

Step 1 — Note that cyanide addition is reversible. $C N^{-}$ adds and can leave. Why this matters? Reversibility is the switch that lets kinetic vs thermodynamic control operate.

Step 2 — Low temperature (kinetic). The faster attack is at the carbonyl carbon → 1,2 product = cyanohydrin $C H_{2} = C H - C (O H) (C N) - C H_{3}$ (C=C kept). Why? At low T, molecules cannot cross back over the barrier; the first-formed (faster) product accumulates.

Step 3 — High temperature / long time (thermodynamic). The system equilibrates; the more stable product wins → 1,4 product = 4-oxopentanenitrile $N C - C H_{2} - C H_{2} - C O - C H_{3}$ (C=O kept). Why? Keeping the strong C=O (≈745 kJ/mol) instead of the weaker C=C (≈615 kJ/mol) lowers the energy.

Verify: Both products have formula $C_{5} H_{7} N O$ (add H–CN, i.e. $+ C H N$ , to $C_{4} H_{6} O$ : $C_{4} H_{6} O + C H N = C_{5} H_{7} N O$ ) ✓. Bond-energy check: swapping a C=C for a C=O in the 1,4 product gains $745 - 615 = 130$ kJ/mol, confirming 1,4 is thermodynamically favoured ✓.

Ex 5 — Case E: The classic Michael (stabilised carbanion)

Worked example Diethyl malonate + methyl vinyl ketone (MVK),

$N a O E t$ catalyst Forecast: Where does the malonate carbon attack, and what dicarbonyl spacing results?

Step 1 — Make the Michael donor. $N a O E t$ removes the acidic central C–H of $C H_{2} (C O_{2} E t)_{2}$ ( $p K_{a} \approx 13$ ) → a resonance-stabilised enolate. Why this step? Two flanking esters delocalise the negative charge, giving a soft, stable carbanion — the ideal 1,4-adder.

Step 2 — Conjugate (1,4) attack. The soft enolate carbon attacks the β-carbon of MVK $C H_{2} = C H - C O - C H_{3}$ . Why this step? Soft donor + soft β-carbon → HSAB says 1,4.

Step 3 — Protonate the acceptor enolate. The new ketone-enolate grabs a proton → neutral 1,5-dicarbonyl. Why? Restores the stable C=O.

Product: $(E t O_{2} C)_{2} C H - C H_{2} - C H_{2} - C O - C H_{3}$ — a 1,5-dicarbonyl (Michael fingerprint; this sets up a later aldol ring closure).

Verify: Count the chain between the two carbonyls. Donor carbonyl C = position 1; then C, C, C; acceptor carbonyl C = position 5 → carbonyls are 1,5 ✓. Atom balance: malonate $C_{7} H_{12} O_{4}$ + MVK $C_{4} H_{6} O$ = adduct $C_{11} H_{18} O_{5}$ ✓ (simple addition, no atoms lost).

Ex 6 — Case F: Heteroatom soft nucleophile (aza-Michael)

Worked example Dimethylamine

$(C H_{3})_{2} N H$ + acrolein $C H_{2} = C H - C H O$ Forecast: Amines are nucleophilic at nitrogen — 1,2 (to give a hemiaminal/imine) or 1,4?

Step 1 — Classify the nucleophile. A secondary amine nitrogen is large and polarisable → soft. Why this step? Softness predicts the site.

Step 2 — 1,4 (conjugate) addition. $(C H_{3})_{2} N$ adds to the β-carbon → enolate → tautomerise to aldehyde. Why this step? Soft N → soft β-carbon (aza-Michael); the reversibility of amine addition also lets it settle on the stable 1,4 product.

Product: $(C H_{3})_{2} N - C H_{2} - C H_{2} - C H O$ (3-(dimethylamino)propanal). Aldehyde C=O retained.

Verify: Atom balance: acrolein $C_{3} H_{4} O$ + $(C H_{3})_{2} N H$ $C_{2} H_{7} N$ = adduct $C_{5} H_{11} N O$ ✓. Degrees of unsaturation of $C_{5} H_{11} N O = (2 \times 5 + 2 + 1 - 11) /2 = 1$ = the single C=O only (C=C gone) ✓, confirming 1,4.

Ex 7 — Case G: Degenerate input (no conjugation)

Worked example

$(C H_{3})_{2} C uL i$ + hex-5-en-2-one $C H_{2} = C H - C H_{2} - C H_{2} - C O - C H_{3}$ Forecast: This has a C=C and a C=O — but does the soft cuprate do a conjugate addition?

Step 1 — Check for conjugation. The C=C and the C=O are separated by two $C H_{2}$ groups; they are not conjugated. Why this step? Conjugate (1,4) addition requires the C=C and C=O to share a resonance path. Here there is none.

Step 2 — Consequence: no δ⁺ on any β-carbon. Without conjugation, resonance cannot push positive charge onto the alkene carbon. The alkene is an ordinary, unreactive-toward-cuprate double bond. Why this step? This is the degenerate/zero case — the "β-carbon electrophile" simply does not exist.

Step 3 — What actually happens. The cuprate is mild and does not attack the isolated carbonyl efficiently either (cuprates are poor at simple 1,2 additions). So no productive reaction at the alkene; the substrate is essentially inert to conjugate addition.

Verify: Test the conjugation requirement directly — number the atoms between C=O and C=C: carbonyl C, $C H_{2}$ , $C H_{2}$ , then $C H = C H_{2}$ . That is a 1,5 relationship of the two π systems (an insulating $\geq 1$ saturated carbon breaks conjugation). Conjugation needs them adjacent (α,β). Not satisfied → no 1,4 pathway ✓.

Ex 8 — Case H: Hard vs soft hydride

Worked example Cyclohex-2-en-1-one reduced by (a)

$L i A l H_{4}$ vs (b) L-Selectride/Stryker's reagent Forecast: Hydride is $H^{-}$ . Which reagent gives the allylic alcohol and which the saturated ketone?

Step 1 — Classify each hydride. $L i A l H_{4}$ delivers a small, hard $H^{-}$ → hard. Bulky/soft hydride sources (L-Selectride, Stryker's Cu–H) are soft. Why this step? Same hardness lever as before, now for hydride.

Step 2 — Predict each site. (a) Hard $H^{-}$ → carbonyl carbon → 1,2 reduction → 2-cyclohexen-1-ol (allylic alcohol, C=C kept). (b) Soft H → β-carbon → 1,4 reduction → enolate → tautomerise → cyclohexanone (C=O kept, C=C gone). Why this step? Hard→carbonyl, soft→β, exactly the parent rule extended to $H^{-}$ .

Verify: (a) $C_{6} H_{8} O + H_{2} = C_{6} H_{10} O$ with DoU $= (2 \times 6 + 2 - 10) /2 = 2$ = ring + C=C ✓ (allylic alcohol). (b) $C_{6} H_{8} O + H_{2} = C_{6} H_{10} O$ , DoU $2$ = ring + C=O ✓ (saturated ketone). Same formula, opposite surviving double bond — regiochemistry set by hydride hardness.

Ex 9 — Case I: Real-world word problem (build a ring)

Worked example A chemist needs a

cyclohexenone (a six-membered ring) from cheap open-chain pieces. She has cyclohexanone and MVK, plus base. Design it. Forecast: Guess the two-reaction sequence before reading.

Step 1 — Michael addition (make the 1,5-dicarbonyl). Base makes the enolate of cyclohexanone; it is a soft, stabilised carbanion → 1,4-adds to MVK at the β-carbon. Why this step? A Michael reaction reliably stitches two carbonyl fragments into a 1,5-dicarbonyl, the required precursor for a ring closure.

Step 2 — Intramolecular aldol condensation. The 1,5-dicarbonyl undergoes an intramolecular aldol condensation (this whole two-step sequence is the Robinson annulation), forming a new six-membered ring and losing water → a fused bicyclic enone (an octalone). Why this step? Aldol closes the ring; dehydration gives the conjugated enone she wanted.

Verify (atom/count): cyclohexanone $C_{6} H_{10} O$ + MVK $C_{4} H_{6} O$ → Michael adduct $C_{10} H_{16} O_{2}$ ✓. Aldol condensation loses one $H_{2} O$ : $C_{10} H_{16} O_{2} - H_{2} O = C_{10} H_{14} O$ (the bicyclic enone) ✓. DoU of $C_{10} H_{14} O = (2 \times 10 + 2 - 14) /2 = 4$ = 2 rings + C=C + C=O ✓ — exactly a fused-ring enone.

Ex 10 — Case J: Exam twist — sterics override HSAB

Worked example

$(C H_{3})_{2} C uL i$ + 4,4-dimethylcyclohex-2-en-1-one vs a β,β-disubstituted acceptor, $β, β$ -dimethyl-substituted enone $(C H_{3})_{2} C = C H - C O - C H_{3}$ (mesityl oxide) Forecast: Cuprate is soft → you'd say 1,4. But what if the β-carbon is too crowded?

Step 1 — Apply HSAB first. Soft cuprate → normally 1,4 at the β-carbon. Why this step? Always start with the electronic (HSAB) prediction.

Step 2 — Check sterics at the β-carbon. In mesityl oxide the β-carbon carries two methyl groups — it is highly hindered. A bulky reagent cannot reach it. Why this step? Sterics is the override lever: when the electronically-preferred site is physically blocked, the reaction is forced elsewhere.

Step 3 — Outcome. With a very hindered β-carbon and/or bulkier organometallics, conjugate addition is suppressed and attack is diverted toward the less hindered carbonyl carbon (1,2), or the reaction slows/stalls. The clean lesson: electronics propose, sterics can veto.

Verify (self-consistency): Count substituents on the β-carbon of mesityl oxide $(C H_{3})_{2} C =$ : two methyls → a fully substituted (neopentyl-like) β-centre, the classic case where soft reagents are pushed off 1,4. This does not violate HSAB; it shows HSAB is the default that sterics can override — the exam's favourite trap. Formula of mesityl oxide: $C_{6} H_{10} O$ , DoU $= 2$ (C=C + C=O) ✓ confirming it is a genuine α,β-unsaturated ketone.

Recall Feynman recap of the whole matrix

Every one of these ten problems is the same question: "Is the incoming partner small-and-picky (goes to the near carbonyl, 1,2) or big-and-easygoing (goes to the far β-carbon, 1,4)?" Then two footnotes: (1) if the reagent is reversible, heat lets the stabler far-end product win; (2) if the far end is physically blocked, even a big easygoing partner is forced to the near end. Learn the default (HSAB), then the two override switches (temperature, sterics), and you can place any new reagent into the matrix.