Aldehydes, Ketones and Carboxylic Acids Part 3

Concept:Reduction of a nitrile with stannous chloride in acid followed by hydrolysis gives an aldehyde. Ozonolysis of an alkene cleaves the double bond, yielding carbonyl compounds.Chemical Equation / Formula:Propanenitrile ($\mathrm{CH_3CH_2CN}$) + $\mathrm{SnCl_2 / HCl}$ → imine intermediate → hydrolysis → propanal ($\mathrm{CH_3CH_2CHO}$).But-2-ene ($\mathrm{CH_3CH=CHCH_3}$) + $\mathrm{O_3}$ → ozonide → $\mathrm{Zn/H_2O}$ → two molecules of ethanal ($\mathrm{CH_3CHO}$).So $\mathrm{x = propanal}$ and $\mathrm{y = ethanal}$.Explanation:One mole each of propanal and ethanal undergo crossed aldol condensation in the presence of alkali, followed by heating to dehydrate.All possible aldol products (self and crossed) give unsaturated aldehydes after dehydration.Self‑aldol of propanal yields 2‑methylpent‑2‑enal (option A).Crossed aldol where ethanal enolate attacks propanal yields pent‑2‑enal (option D).Crossed aldol where propanal enolate attacks ethanal yields 2‑methylbut‑2‑enal (option B) after dehydration (elimination between the α and β carbons).3‑Methylbut‑2‑enal (option C) is not formed from any of these aldol reactions.Thus, the product that is not obtained is 3‑methylbut‑2‑enal.Shortcut:In crossed aldol, the only possible unsaturated products have the carbon‑carbon double bond at the position where the enolate attacked, and the resulting aldehyde always contains the original carbonyl carbon. 3‑Methylbut‑2‑enal would require a different substitution pattern not possible from these starting materials.<