Match the conversion given in List-I with the reagents involved in List-II.
| LIST-I (Chemical Reactions) | LIST-II (Reagents) |
| A. But-2-ene to ethanal | I. DIBAL-H |
| B. Allyl alcohol to propenal | II. CrO3/ acetic anhydride/H3O+ |
| C. Ethanenitrile to ethanal | III. O3/H2O-Zn dust |
| D. p-Fluorotoluene to p-fluorobenzaldehyde | IV. PCC |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-I, D-II
Each conversion forms an aldehyde, but from a different starting group, so each needs a reagent that stops at the aldehyde stage:
A. But-2-ene to ethanal — This matches III. O3/H2O-Zn dust. Ozonolysis cleaves the C=C bond; zinc dust removes the \( \ce{H2O2} \) formed, so the aldehyde is not oxidised further. The symmetric alkene gives two molecules of ethanal.
\( \displaystyle \ce{CH3-CH=CH-CH3} \xrightarrow[\ce{H2O-Zn}]{\ce{O3}} \ce{2CH3CHO} \)
B. Allyl alcohol to propenal — This matches IV. PCC. PCC is a mild oxidising agent that turns a primary alcohol into an aldehyde and leaves the C=C bond untouched.
\( \displaystyle \ce{CH2=CH-CH2OH} \xrightarrow{\text{PCC}} \ce{CH2=CH-CHO} \)
C. Ethanenitrile to ethanal — This matches I. DIBAL-H. Diisobutylaluminium hydride reduces a nitrile to an imine, which hydrolyses to the aldehyde.
\( \displaystyle \ce{CH3-C#N} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) DIBAL-H}} \ce{CH3CHO} \)
D. p-Fluorotoluene to p-fluorobenzaldehyde — This matches II. CrO3/ acetic anhydride/H3O+. \( \ce{CrO3} \) in acetic anhydride converts the ring \( \ce{CH3} \) group into a benzylidene diacetate, which stops further oxidation; acid hydrolysis then gives the aldehyde.
\( \displaystyle \ce{F-C6H4-CH3} \xrightarrow[\ce{(CH3CO)2O},\ 273\text{-}283\ \mathrm{K}]{\ce{CrO3}} \ce{F-C6H4-CH(OCOCH3)2} \xrightarrow{\ce{H3O^+}} \ce{F-C6H4-CHO} \)
Shortcut: The pair C–I appears only in Options (1) and (2), and A–III then leaves only Option (1).