Aldehydes, Ketones and Carboxylic Acids

Identify the reagents used for the following conversion [figure]

1. A = LiAlH4, B = NaOH(aq), C = NH2-NH2/KOH, ethylene glycol
2. A = DIBAL-H, B = NaOH(aq), C = NH2-NH2/KOH, ethylene glycol
3. A = LiAlH4, B = NaOH(alc), C = Zn/HCl
4. A = DIBAL-H, B = NaOH(alc), C = Zn/HCl

Answer

(4) A = DIBAL-H, B = NaOH(alc), C = Zn/HCl

Solution

Step A: ester -COOCH3 -CHO only ⇒ DIBAL-H

∵ LiAlH4 would reduce the ester and –CHO to -CH2OH

Step B: dialdehyde → five-membered α,β-unsaturated aldehyde ⇒ intramolecular aldol condensation (alc. NaOH)

∵ α-C of one -CH2CHO attacks the CHO of the other, then -H2O gives C=C

Step C: –CHO → -CH3 and the –OH is lost as a new C=C ⇒ Zn/HCl (Clemmensen, acidic: –OH dehydrates)

∵ Wolff–Kishner (NH2NH2/KOH) would keep the –OH

Answer: (4)