MCQ
$HO - (CH_2)_6 - OH,$ this conversion can be achieved by


- A$O_3 , Zn,$ then $LiAlH_4$
- B$O_3 /H_2O_2,$ then $LiAlH_4$
- Ccold dil. $KMnO_4 , HIO_4$ then $LiAlH_4$
- ✓all of these


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$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{SOCl}_{2} \longrightarrow \mathrm{A} \xrightarrow[AlCl_3]{Benzene} \mathrm{B} \xrightarrow[-OH]{KCN} \mathrm{C}$
Correct statement regarding $B$ product?