Which of the following reactions does not produce ketone?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
A ketone has a C=O group bonded to two carbon groups (R–CO–R′), whereas an aldehyde has at least one hydrogen on the C=O carbon. Ketones can be made from nitriles with Grignard reagents, from acyl chlorides with dialkylcadmium, by hydration of alkynes other than ethyne, and by ozonolysis of alkenes whose double-bond carbon carries two carbon groups.
Option (1) 1 — Produces a ketone (acetophenone). The Grignard reagent adds to the C≡N group, and hydrolysis of the imine formed gives acetophenone:
\( \displaystyle \ce{CH3-C#N + C6H5MgBr} \xrightarrow{\text{dry ether}} \ce{CH3-C(=NMgBr)-C6H5} \)
\( \displaystyle \ce{CH3-C(=NMgBr)-C6H5 + 2H2O} \xrightarrow{\ce{H+}} \ce{CH3-CO-C6H5 + NH3 + Mg(OH)Br} \)
Option (2) 2 — Does not produce a ketone (gives two aldehydes). The double bond is at the end of the chain, \( \ce{CH2=CH} \)– on the ring; ozonolysis turns each carbon into C=O, and each carries a hydrogen:
\( \displaystyle \ce{C6H11-CH=CH2} \xrightarrow[\text{(ii) }\ce{Zn}/\ce{H2O}]{\text{(i) }\ce{O3}} \ce{C6H11-CHO + HCHO} \)
Option (3) 3 — Produces a ketone (1-phenylpropan-2-one). A dialkylcadmium converts an acyl chloride into a ketone:
\( \displaystyle \ce{(C6H5CH2)2Cd + 2CH3COCl -> 2C6H5CH2COCH3 + CdCl2} \)
Option (4) 4 — Produces a ketone (propanone). Water adds to propyne by Markovnikov’s rule; the enol formed changes to propanone:
\( \displaystyle \ce{CH3-C#CH + H2O} \xrightarrow{\ce{HgSO4}/\ce{H2SO4}} \ce{CH3COCH3} \)