Question

How butanoic acid is prepared from the following starting chemicals;
A. An Alcohol
B. An Alkyl Halide
C. An alkene

Answer 1

Hint:: From the name of the butanoic acid we can guess that it’s a 4 carbon carboxylic acid, as it has a suffix ‘oic acid’, which is usually used for compounds having a carboxylic acid as the functional group (${\text{COOH}}$). Now to prepare it from all the given organic compounds recall the chemical property of these compounds.

Complete step-by-step answer:i.In the first case, we have alcohol, and we have studied that when we use strong oxidizing agent alcohol directly gets oxidized to the corresponding carboxylic acid, the reaction to get butanoic acid from n-butanol can be written as;

${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{OH}}\xrightarrow{{{\text{KMn}}{{\text{O}}_4}}}{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{COOH}}$

ii. Now, we have an alkyl halide, which can also be called haloalkane, to get butanoic acid from haloalkanes we will react n-chloropropane with KCN and then follow hydrolysis, the reaction can be written as:

${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{Cl}}\xrightarrow{{{\text{KCN}}}}{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CN + KCl}}$

${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CN}}\xrightarrow{{{{\text{H}}_3}{{\text{O}}^ + }}}{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{COOH}}$

iii. In the last, we have an alkene, and we are gonna make it into a butanoic acid. Now will use a simple step, we will react propene with carbon monoxide (CO) and steam (${{\text{H}}_2}{\text{O}}$), in presence of ${{\text{H}}_3}{\text{P}}{{\text{O}}_3}$ and a good amount of pressure with a temperature of ${350^ \circ }{\text{C}}$ this is also known as Koch reaction, the reaction can be written as:

${\text{C}}{{\text{H}}_2}{\text{ = CH}} - {\text{C}}{{\text{H}}_3} + {\text{CO + }}{{\text{H}}_2}{\text{O}}\xrightarrow[{{{350}^ \circ }{\text{C, Pressure}}}]{{{{\text{H}}_2}{\text{P}}{{\text{O}}_3}}}{\text{ C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{COOH}}$

Note:Look carefully in the last two cases i.e. in the case of an alkyl carbon and an alkene, we have used a reagent having a 3-carbon reactant, but the corresponding product formed has 4-carbon, this is because of the carbon present in another reactant present i.e. KCN in case of n-chloropropane and CO is the case of n-propene. Remember that these are some of the many ways we can synthesis, use the reaction you’re comfortable with but try to make the product in 1-2 steps.