91Ó°ÊÓ

The primary alcohol is 1-butanol. There is an oxidation to butanoic acid when treated with aqueous acid of CrO₃. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}OH\underset{H_3{O}^{+}}{\overset{Cr{O}_3}{→}}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H$

There is a conversion of 1-bromobutane into 1-butanol. The primary alcohol is 1-butanol. There is an oxidation to butanoic acid when treated with aqueous acid of CrO₃. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br\stackrel{NaOH}{→}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\underset{H_3{O}^{+}}{\overset{Cr{O}_3}{→}}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H$

When there is a hydroboration of 1-Butene into BH₃, The following is the Anti-Markovnikov region chemistry. There is an oxidation to butanoic acid when treated with aqueous acid of CrO₃. The steps involved are,

$C{H}_{3}C{H}_{2}CH=C{H}_2\underset{2.H_2{O}_2,{}^{-}OH}{\overset{1.B{H}_3,THF}{→}}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\underset{H_3{O}^{+}}{\overset{Cr{O}_3}{→}}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H$

There is a yield of butanenitrile by bromide ion replacement using cyanide ion. The nitrile hydrolysis using aqueous acid will yield botanic acid. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}Br\underset{2.H_3{O}^{+}}{\overset{1.NaCN}{→}}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H$

The symmetrical alkene is 4-Octene. There is a breakage of a double bond between C₄ and C₅ for the yield of butanoic acid molecules when it is acidified with KMnO₄. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}CH=CHC{H}_{2}C{H}_{2}C{H}_3\underset{H_3{O}^{+}}{\overset{KMn{O}_4}{→}}2C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H$