Question

How will you obtain:
$\left( 1 \right)$ Formic acid from Acetic acid
$\left( 2 \right)$ Acetaldehyde from Acetylene

Answer 1

For in this conversion we use too many methods and can be converted into a desired product by various steps. For example, to obtain the formic acid from acetic acid. There are three steps for this conversion. Which we will discuss in complete solution:

Complete step by step answer:
In the given question there are two conversions. Let's solve one by one.
$\left( 1 \right)$ For the conversion of acetic acid to formic acid. This conversion can be done by many methods.
Method $1$ :
Step $1$ : Reaction of acetic acid with sodium azide gives out methylamine. This reaction is known as Schmidt reaction
$C{H_3}COOH + Na{N_3} \to N{H_2}C{H_3}$
Step $2$ : Reaction of methylamine with $HN{O_2}$ to give out methanol
$N{H_2}C{H_3} + HN{O_2} \to C{H_3}OH$
Step $3$ : Oxidation of methanol via potassium dichromate in the presence of acid to give formic acid.
$C{H_3}COOH\xrightarrow[{{H_ + }}]{{{K_2}Cr{O_7}}}HCOOH$
$\left( 2 \right)$ For conversion of acetaldehyde from acetylene. Acetylene can be converted into acetaldehyde in a single step process or multi steps.
Method $1$ : (single step process)
Reaction of acetylene with water in presence of sulphuric acid and mercury sulphate to give acetaldehyde.
${C_2}{H_4} + {H_2}O\xrightarrow[{HgS{O_4}}]{{{H_2}S{O_4}}}C{H_3}CHO$
This reaction is known as the Kucherov reaction.
Method $2$ : (Multi steps process)
Step $1$ : Acetylene reacts with lithium aluminium hydride to give Ethane.
${C_2}{H_4}\xrightarrow{{LiAl{H_4}}}C{H_3}C{H_3}$
Step $2$ : Ethane reacts with chlorine in presence of sunlight to give out chloroethane
$C{H_3}C{H_3} + C{l_2}\xrightarrow{{Sunlight}}C{H_3}C{H_2}Cl$
Step $3$ : Reaction of chloroethane with aqueous potassium hydroxide to give out ethanol and then ethanol react with $PCC$ to produce acetaldehyde.
$C{H_3}C{H_2}Cl + Aq.KOH \to {C_2}{H_5}OH$
${C_2}{H_5}OH + PCC \to C{H_3}CHO$ .

Note:
For the conversion of acetaldehyde from acetylene if we take methyl acetylene as a reactant in Kucherov reaction then the resultant product is propanal. If the alkynes are highly substituted then we will get ketone as the product.