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

Before we start, one shared picture you will refer back to constantly — the two sticky spots:

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

Level 1 — Recognition

L1.1

In the molecule below, label the α-carbon, the β-carbon, and mark with a "★" every carbon a nucleophile could attack. $C H_{2} = C H - C H O (acrolein / prop-2-enal)$

Recall Solution

Number from the carbonyl: the $C H O$ carbon is C1 (carbonyl C). The next carbon ( $= C H -$ ) is C2 = α. The end carbon ( $C H_{2} =$ ) is C3 = β. $β, C 3 C H_{2} = α, C 2 C H - C 1 C H O$ A nucleophile can attack C1 (carbonyl) ★ and C3 (β) ★. The α-carbon (C2) is never the electrophile — resonance parks the $δ^{+}$ on C1 and C3, not C2.

L1.2

Which of these are α,β-unsaturated carbonyls (i.e. $C = C$ conjugated to $C = O$ )? (a) $C H_{2} = C H - C H_{2} - C H O$ (b) $C H_{3} - C H = C H - C O - C H_{3}$ (c) cyclohex-2-enone (d) cyclohex-3-enone

Recall Solution

"Conjugated" means the $C = C$ and $C = O$ are separated by exactly one single bond so their p-orbitals overlap in a row.

(a) No — the $C = C$ and $C H O$ are separated by a $C H_{2}$ (that is a β,γ-unsaturated / homoconjugated aldehyde; the double bonds are not in a row).
(b) Yes — $C = C$ directly bonded to $C = O$ .
(c) Yes — the double bond at the 2-position sits right next to the ring carbonyl.
(d) No — the double bond at position 3 is one carbon too far; not conjugated.

Level 2 — Application

L2.1

Give the major product: $C H_{3} M g B r$ + cyclohex-2-enone, then $H_{3} O^{+}$ .

Recall Solution

$C H_{3} M g B r$ (Grignard) is a hard carbon nucleophile and adds irreversibly → 1,2-addition at the carbonyl carbon.

Step 1: $C H_{3}^{-}$ attacks C1 → magnesium alkoxide, $C = C$ untouched.
Step 2: $H_{3} O^{+}$ protonates the alkoxide oxygen.

Product: 1-methylcyclohex-2-en-1-ol — a tertiary allylic alcohol with the ring $C = C$ retained.

L2.2

Give the major product: $(C H_{3})_{2} C uL i$ + cyclohex-2-enone, then $H_{3} O^{+}$ .

Recall Solution

The cuprate is soft and polarisable → 1,4-addition at the β-carbon.

Step 1: $C H_{3}$ delivered to C3 (β) → a magnesium/lithium enolate.
Step 2: workup protonates the α-carbon; the enolate collapses back to $C = O$ .

Product: 3-methylcyclohexan-1-one — the ring $C = O$ retained, the $C = C$ gone.

Same substrate, opposite regiochemistry — decided entirely by the metal (hard Mg vs soft Cu). See the fork below:

L2.3

$L i A l H_{4}$ reduces cyclohex-2-enone. Predict the major addition mode and the product.

Recall Solution

$L i A l H_{4}$ delivers hydride $H^{-}$ , a small, hard, charge-dense nucleophile → 1,2-addition.

$H^{-}$ attacks the carbonyl carbon; after workup you get the alcohol with $C = C$ retained.
Product: cyclohex-2-en-1-ol (an allylic alcohol).

(To reduce only the $C = C$ conjugately you would use a soft hydride source or catalytic conditions — not the point here.)

Level 3 — Analysis

L3.1

Explain, with resonance, why the β-carbon is electrophilic while the α-carbon is not.

Recall Solution

Draw the two resonance structures (conjugation): $C β = C α - C = O ⟷^{+} C β - C α = C - O^{-}$ Oxygen is greedy for electrons, so it pulls the $π$ -electrons toward itself. Following the arrows, the positive charge lands on the β-carbon (C3) and the negative on oxygen. The α-carbon (C2) is caught in the middle of a double bond in both structures — it never carries the $δ^{+}$ . Mnemonic: "Oxygen hogs the minus, beta wears the plus." So the two electrophilic carbons are C1 (carbonyl) and C3 (β); the α-carbon is only ever a proton site.

L3.2

Using the HSAB principle, explain why a hard nucleophile prefers the carbonyl carbon and a soft one prefers the β-carbon.

Recall Solution

"Hard" = small, high charge density, poorly polarisable; "soft" = large, diffuse, easily polarisable.

The carbonyl carbon carries a large, localised $δ^{+}$ (a hard electrophilic site). Hard nucleophiles, with their concentrated charge, bind it well through strong electrostatic (charge–charge) attraction.
The β-carbon's $δ^{+}$ is diffuse and polarisable (a soft site). Soft, polarisable nucleophiles form a better covalent-type overlap there.
Rule: hard→hard (Nu to C1, 1,2), soft→soft (Nu to C3, 1,4).

L3.3

1,2-addition is faster but 1,4-addition gives the more stable product. Explain both halves and name the type of control each corresponds to.

Recall Solution

Faster (kinetic, 1,2): the carbonyl carbon is the more electrophilic and less hindered site, so the very first, lowest-barrier attack happens there. This is kinetic control. More stable (thermodynamic, 1,4): compare the bonds kept. $E (C = O) \approx 745 kJ/mol > E (C = C) \approx 615 kJ/mol$ The 1,4 product keeps the strong $C = O$ and sacrifices the weaker $C = C$ ; it also avoids a strained allylic alcohol. So $Δ (bond energy) \approx 745 - 615 = 130 kJ/mol$ in favour of keeping $C = O$ . When the addition is reversible (or at high temperature), the system equilibrates to this stabler 1,4 product — thermodynamic control.

Level 4 — Synthesis

L4.1

Design a Michael addition of diethyl malonate $C H_{2} (C O_{2} E t)_{2}$ onto methyl vinyl ketone (MVK, $C H_{2} = C H - C O C H_{3}$ ). Give the base, all arrow-pushing steps, and the final product.

Recall Solution

Step 1 — make the donor. Malonate's central $C H_{2}$ sits between two esters, so its C–H is acidic ( $p K_{a} \approx 13$ ). A mild base $N a O E t$ ( $p K_{a H} \approx 16$ for ethanol) removes it fully, giving the stabilised, soft, delocalised enolate (the Michael donor). Why $N a O E t$ and not $N a O H$ ? It matches the ester solvent/system and cleanly deprotonates without hydrolysing the esters. Step 2 — conjugate attack. The soft enolate carbon attacks the β-carbon of MVK (soft→soft, HSAB) → 1,4. This gives a new C–C bond and a ketone-enolate. Step 3 — protonation. The ketone-enolate grabs a proton (from EtOH), restoring the neutral $C = O$ . Product: $(E t O_{2} C)_{2} C H - C H_{2} - C H_{2} - C O - C H_{3}$ Count the two carbonyl carbons: the malonate carbonyls and the ketone carbonyl end up in a 1,5-dicarbonyl relationship — the Michael fingerprint.

L4.2

Propose a synthesis of 4-oxopentanenitrile $N C - C H_{2} - C H_{2} - C O - C H_{3}$ from but-3-en-2-one (MVK) using $H C N$ . State the conditions (kinetic vs thermodynamic) that give this product.

Recall Solution

$C N^{-}$ addition to a carbonyl is reversible, so temperature acts as a switch.

At low T you'd trap the fast 1,2 product (a cyanohydrin at the carbonyl carbon) — not what we want.
Warm / equilibrate: because the addition is reversible, the system drifts to the stabler 1,4 product, where the strong $C = O$ is retained. $C H_{2} = C H - C O - C H_{3} H C N high T, equilibrate N C - C H_{2} - C H_{2} - C O - C H_{3}$ Conditions: high temperature / equilibrating (thermodynamic control).

L4.3

You need 3-phenylcyclohexan-1-one from cyclohex-2-enone. Which organometallic and why?

Recall Solution

The product keeps the ring $C = O$ and has a phenyl at the β (C3) position → this is a 1,4-addition of a carbon nucleophile. A hard $P h M g B r$ would give mostly 1,2 (allylic alcohol) — wrong. Use the soft cuprate: $cyclohex-2-enone + P h_{2} C uL i then H_{3} O^{+} 3-phenylcyclohexan-1-one$ Soft Cu delivers $P h$ to the soft β-carbon (1,4), workup gives the ketone.

Level 5 — Mastery

L5.1 (Robinson-annulation flavour, two steps)

Diethyl malonate is treated with $N a O E t$ , then with MVK (Michael addition). The 1,5-diketo/ester product is then subjected to an intramolecular aldol step under base/heat. Predict the type of ring likely formed and justify using the 1,5-spacing.

Recall Solution

Step A (Michael, from L4.1): gives $(E t O_{2} C)_{2} C H - C H_{2} - C H_{2} - C O - C H_{3}$ , a 1,5-dicarbonyl relationship between the two carbonyl regions. Step B (intramolecular aldol): an enolate at one carbonyl's α-carbon attacks the other carbonyl carbon. Count the chain atoms that must close into the ring: a 1,5-dicarbonyl typically closes to a six-membered ring (a cyclohexenone after dehydration), which is favoured because six-membered rings have the lowest ring strain and form fastest. Answer: a six-membered (cyclohexenone-type) ring — this Michael-then-aldol sequence is the essence of a Robinson annulation.

L5.2 (counting / naming discipline)

For the conjugate unit $O (1) = C (2) - C (3) = C (4)$ , state exactly which atom the nucleophile bonds to and which atom the proton ends up on, in a 1,4-addition. Then say what goes wrong if a student instead writes "Nu on 4, H on 3."

Recall Solution

Number the four atoms of the conjugated unit: O(1) = C(2) – C(3) = C(4). In 1,4-addition the two things added land on atoms 1 and 4:

Nu bonds to atom 4 (that is the β-carbon),
H (proton) ends on atom 1 (the oxygen) — after tautomerisation this becomes the enol/carbonyl. "Nu on 4, H on 3" is wrong because atom 3 is the α-carbon; the proton in the net 1,4-addition formally caps the oxygen (atom 1). (In the final tautomer a proton does sit on the α-carbon, but the defining "1,4" count is Nu-4 / H-1.)

L5.3 (full prediction with justification)

Predict the major product and control regime for each, one line of reasoning each: (a) $L i A l H_{4}$ on but-3-en-2-one, low T. (b) $(n - B u)_{2} C uL i$ on but-3-en-2-one, then $H_{3} O^{+}$ . (c) $C H_{3} N H_{2}$ (a soft amine) on but-3-en-2-one.

Recall Solution

(a) $L i A l H_{4}$ = hard hydride, irreversible → 1,2 (kinetic). Product: but-3-en-2-ol (allylic alcohol, $C = C$ kept). (b) Cuprate = soft → 1,4. $n$ -Bu adds to the β-carbon; workup gives the ketone octan-2-one ( $C H_{3} C O - C H_{2} - C H_{2} - C_{4} H_{9}$ ; $C = O$ kept). (c) Amines are soft, and their addition is reversible → 1,4 (conjugate). Product: 4-(methylamino)butan-2-one $C H_{3} N H - C H_{2} - C H_{2} - C O - C H_{3}$ .

Recall Master checklist (self-test before moving on)

Two electrophilic carbons? ::: C1 (carbonyl) and C3 (β); α is only a proton site.
Hard Nu ( $R M g X$ , $L i A l H_{4}$ ) → ? ::: 1,2, kinetic, keeps $C = C$ (allylic alcohol).
Soft Nu (cuprate, amine, thiol, enolate) → ? ::: 1,4, thermodynamic, keeps $C = O$ .
Michael donor + acceptor give what spacing? ::: 1,5-dicarbonyl.
In $O (1) = C (2) - C (3) = C (4)$ , 1,4 puts Nu on __ and H on __. ::: Nu on 4 (β), H on 1 (O).
Cuprate vs Grignard on an enone? ::: cuprate 1,4, Grignard 1,2.