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Propan-1-ol may be prepared by the reaction of propene with
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\(\mathrm{B}_2 \mathrm{H}_6, \mathrm{NaOH}-\mathrm{H}_2 \mathrm{O}_2\)
\(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{NaOH} / \mathrm{H}_2 \mathrm{O}_2]{\mathrm{B}_2 \mathrm{H}_6}
\mathrm{CH}_3-\underset{\text{Propanol}}{\mathrm{CH}_2-\mathrm{CH}_2}-\mathrm{OH}\)
\mathrm{CH}_3-\underset{\text{Propanol}}{\mathrm{CH}_2-\mathrm{CH}_2}-\mathrm{OH}\)
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