The option(s) with correct sequence of reagents for the conversion of P to Q is(are) [figure]
- i) Lindlar’s catalyst, H2; ii) SnCl2/HCl; iii) NaBH4; iv) H3O+
- i) Lindlar’s catalyst, H2; ii) H3O+; iii) SnCl2/HCl; iv) NaBH4
- i) NaBH4; ii) SnCl2/HCl; iii) H3O+; iv) Lindlar’s catalyst, H2
- i) Lindlar’s catalyst, H2; ii) NaBH4; iii) SnCl2/HCl; iv) H3O+
Answer
(1) i) Lindlar’s catalyst, H2; ii) SnCl2/HCl; iii) NaBH4; iv) H3O+, (3) i) NaBH4; ii) SnCl2/HCl; iii) H3O+; iv) Lindlar’s catalyst, H2, (4) i) Lindlar’s catalyst, H2; ii) NaBH4; iii) SnCl2/HCl; iv) H3O+
Solution
Changes needed from P to Q: both C≡C → cis C=C; ring C=O → CH–OH; –CN → –CHO; –CO₂Et → –CO₂H; –OCOCH₃ → –OH.
Route of option (D), step by step:
[figure] (P) Lindlar's catalyst, H2
[figure]
Lindlar catalyst reduces each C≡C only to a cis C=C.
NaBH4 [figure]
NaBH4 reduces the ketone to a secondary alcohol; it does not touch the ester or the nitrile.
SnCl2/HCl [figure]
Stephen reduction: –C≡N → aldimine salt –CH=NH₂⁺Cl⁻ (the –CHO is still masked).
H3O+ [figure] (Q)
H3O+ hydrolyses the aldimine salt to –CHO, the ester to –CO₂H and the acetate to –OH.
Key rule: free –CHO must not meet NaBH4, so NaBH4 must act before the H3O+ step that releases –CHO.
(A) Lindlar, then SnCl2/HCl (the aldimine separates as its tin chloride double salt, so the masked CHO is not attacked), NaBH4 reduces the ring C=O, and H3O+ finally frees –CHO and hydrolyses the ester and acetate → Q obtained.
(B) The last reagent is NaBH4 and no H3O+ step follows the Stephen reduction: the aldimine salt is never hydrolysed, and NaBH4 would reduce the C=N (or any –CHO set free) to –CH₂NH₂ / –CH₂OH, so the –CHO of Q is not obtained → wrong.
(C) NaBH4 first, then Stephen reduction + hydrolysis; Lindlar at the end reduces only the C≡C (not –CHO) → Q obtained.
(D) Shown above → Q obtained.
Answer: (A), (C), (D)