Q43P - Organic Chemistry (Mcmurry)

Question

How could you convert butanoic acid into the following compounds?

Write each step showing the reagents needed.

(a) 1-Butanol

(b) 1-Bromobutane

(c) Pentanoic acid

(d) 1-Butene 

(e) Octane

Step-by-Step Solution

Answer

The conversion of butanoic acid into the following compounds can be explained.

Step 1: Conversion of butanoic acid into 1-Butanol

The hydrolysis with acids can happen when treated with LiAlH₄, there is a reduction of butanoic acid to 1-Butanol. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H\underset{2.H_3{O}^{+}}{\overset{1.LiAl{H}_4}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH$

Step 2: Conversion of butanoic acid into 1-Bromobutane

The hydrolysis with acids can happen when treated with LiAlH₄, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr₃, there is a conversion of alcohol into 1-bromobutane. The steps involved are,

$C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H\underset{2.H_3{O}^{+}}{\overset{1.LiAl{H}_4}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\underset{}{\overset{PB{r}_3}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br$

Step 3: Conversion of butanoic acid into Pentanoic acid

The hydrolysis with acids can happen when treated with LiAlH₄, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr₃, there is a conversion of alcohol into 1-bromobutane. The butanenitrile hydrolysis with aqueous acids results in pentanoic acid formation. The steps involved are,

$\begin{array}{c}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H\underset{2.H_3{O}^{+}}{\overset{1.LiAl{H}_4}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\stackrel{PB{r}_3}{\to }C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br\\ \underset{2.H_3{O}^{+}}{\overset{1.NaCN}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}C{O}_{2}H\end{array}$

Step 4: Conversion of butanoic acid into 1-Butene

The hydrolysis with acids can happen when treated with LiAlH₄, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr₃, there is a conversion of alcohol into 1-bromobutane. Treatment of 1-bromobutane with KOH, eliminates HBr to get 1-butene. The steps involved are,

$\begin{array}{l}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H\underset{2.H_3{O}^{+}}{\overset{1.LiAl{H}_4}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\stackrel{PB{r}_3}{\to }C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br\\ \stackrel{alcoholic KOH}{\to }C{H}_{3}C{H}_{2}CH=C{H}_2\end{array}$

Step 5: Conversion of butanoic acid into Octane

The hydrolysis with acids can happen when treated with LiAlH₄, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr₃, there is a conversion of alcohol into 1-bromobutane. The conversion of 1-bromobutane into lithium-dibutyl copper in addition to octane. The steps involved are,

$\begin{array}{l}C{H}_{3}C{H}_{2}C{H}_{2}C{O}_{2}H\underset{2.H_3{O}^{+}}{\overset{1.LiAl{H}_4}{\to }}C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}OH\stackrel{PB{r}_3}{\to }C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br\stackrel{2Li}{\to }\\ 2C{H}_{3}C{H}_{2}C{H}_{2}C{H_2}^{-}L{i}^{+}\stackrel{Cul}{\to }(C{H}_{3}C{H}_{2}C{H}_{2}C{H}_2{)}_{2}C{u}^{-}L{i}^{+}+C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br\\ \to C{H}_3(C{H}_2{)}_{6}C{H}_3\end{array}$