Q1(d). HCl on 1-phenylpropyne: E/Z, kinetic and thermodynamic products
The following reaction gives two products X and Y:
[structure] → (HCl, Al₂O₃) [X] + [Y]
(i) Assign E/Z configuration to the products X and Y.
(ii) Between X and Y, identify the kinetic and thermodynamic products.
(iii) Under the reaction conditions, the kinetic product slowly gets converted to a thermodynamic product, until the equilibrium is reached. Propose a probable mechanism for this conversion.
Answer in brief
X is (E) and Y is (Z). X, the syn adduct, is the kinetic product; Y, with phenyl trans to methyl, is the thermodynamic product. X turns into Y by reversible protonation to an α-chlorobenzylic cation, rotation and deprotonation.
Key points
- HCl protonates C-2 of X (Markovnikov again), giving the α-chlorobenzylic cation: the + on C-1 is stabilised by the phenyl ring and by a lone pair of Cl (C+-Cl ↔ C=Cl⁺).
- C1–C2 is now a single bond, so it rotates freely.
- Loss of H⁺ from C-2 re-forms the double bond in either geometry. Every cycle can return X or give Y; since Y is lower in energy, the mixture drifts towards it until the equilibrium ratio is reached.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. HCl adds to the alkyne in the Markovnikov sense: H goes to C-2 and Cl to C-1, the carbon that carries the phenyl group, because the positive charge that develops there is benzylic. Both products are 1-chloro-1-phenylprop-1-ene; they differ in geometry.
(i) E/Z
On C-1, Cl (atomic number 17) ranks above Ph (C); on C-2, CH₃ ranks above H. In X, Cl (top) and CH₃ (bottom) are on opposite sides: X = (E)-1-chloro-1-phenylprop-1-ene. In Y, Cl (bottom) and CH₃ (bottom) are on the same side: Y = (Z)-1-chloro-1-phenylprop-1-ene.
(ii) Kinetic and thermodynamic products
On the alumina surface HCl adds syn: H and Cl are delivered to the same face of the alkyne at nearly the same time. In X, H and Cl are cis, so X is the syn adduct and forms faster: X (E) is the kinetic product. In X the phenyl group is cis to CH₃; this steric clash twists the ring out of conjugation. In Y the phenyl is trans to CH₃ and can lie in the plane of the C=C. Y (Z) is the more stable, thermodynamic product.
(iii) Mechanism of E → Z isomerisation
- HCl protonates C-2 of X (Markovnikov again), giving the α-chlorobenzylic cation: the + on C-1 is stabilised by the phenyl ring and by a lone pair of Cl (C+-Cl ↔ C=Cl⁺).
- C1–C2 is now a single bond, so it rotates freely.
- Loss of H⁺ from C-2 re-forms the double bond in either geometry. Every cycle can return X or give Y; since Y is lower in energy, the mixture drifts towards it until the equilibrium ratio is reached.
Scheme: Reversible protonation turns the C=C into a single bond that can rotate.
Conclusion. X is (E) and Y is (Z). X, the syn adduct, is the kinetic product; Y, with phenyl trans to methyl, is the thermodynamic product. X turns into Y by reversible protonation to an α-chlorobenzylic cation, rotation and deprotonation.
Application. Surface-mediated (alumina, silica) additions of HX give clean syn adducts without strong acid, and the same kinetic-versus-thermodynamic reasoning sets the E/Z ratio of vinyl halides used in cross-coupling.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram
All parts of Question 1
- 1(a) [10]Annulene isomers; sydnone and tropolone10 marks
- 1(b) Solvolysis of anti- and syn-7-norbornenyl tosylates10 marks
- 1(c) Friedel–Crafts, Wittig, dihydroxylation and pinacol rearrangement10 marks
- 1(d) HCl on 1-phenylpropyne: E/Z, kinetic and thermodynamic products10 marks
- 1(e) Diels–Alder of o-xylylene; photochemistry of butadiene10 marks