Question

How will you convert-
i) Ethanoic acid into Methanamine
ii) Hexanenitrile into 1-aminopentane
iii) Methanol to ethanoic acid
iv) Ethanamine to Methanamine
v) Ethanoic acid into Propanoic acid
vi) Methanamine to ethanamine
vii) Nitromethane into dimethylamine
viii) Propanoic acid into ethanoic acid

Answer 1

Hint: Hoffmann bromamide reaction uses an alkali as a strong base to attack amide, leading to deprotonation and subsequent generation of anion. The reaction is used to convert primary amide to a primary amine with the reduction of one carbon from the chain. For the increase in one carbon in the chain can use the cyanide group. First introduce the cyanide group then convert that cyanide group to the desired group.

Complete step by step answer:
Ethanoic acid into Methanamine-
Ethanoic acid is first converted to acetamide, since it is not possible to direct conversion so with the help of acyl chloride, it is done. Then with the help of Hoffmann bromamide reaction, acetamide get converted to Methanamine
\[C{H_3}COOH\xrightarrow{{SOC{l_2}}}C{H_3}COCl\xrightarrow{{N{H_3}(excess)}}C{H_3}CON{H_2}\xrightarrow{{B{r_2}/NaOH}}C{H_3}N{H_2}\]
Hexanenitrile into 1- aminopropane -
First convert cyanide group into acid group, then convert acid to amide group through acyl chloride. The with the Hoffmann bromamide reaction we will convert this amide to amine
\[{C_5}{H_{11}}CN\xrightarrow{{{H^ + }/{H_2}O}}{C_5}{H_{11}}COOH\xrightarrow{{SOC{l_2}}}{C_5}{H_{11}}COCl\xrightarrow{{N{H_3}(excess)}}{C_5}{H_{11}}CON{H_2}\xrightarrow{{B{r_2}/KOH}}{C_5}{H_{11}}N{H_2}\]
Methanol to ethanoic acid –
First convert methanol to methyl chloride then to methyl cyanide. Then do acid hydrolysis of the cyanide, we will get ethanoic acid
\[C{H_3}OH\xrightarrow{{PC{l_5}}}C{H_3}Cl\xrightarrow{{NaCN/EtOH}}C{H_3}CN\xrightarrow{{{H^ + }/{H_2}0}}C{H_3}COOH\]
Ethanamine to Methanamine –
First convert ethylamine to ethanol with the help of its diazonium salt, then convert that alcohol group to acid by oxidation. Acid to amide with ammonia, then to reduce one carbon, we will do Hoffmann bromamide of amide. And we will get Methanamine
\[C{H_3}C{H_2}N{H_2}\xrightarrow{{NaN{O_2}/HCl}}C{H_3}C{H_2}N_2^ + Cl\xrightarrow{{{H_2}O}}C{H_3}C{H_2}OH\xrightarrow{{KMn{O_4}/{H^ + }}}C{H_3}COOH\xrightarrow{{N{H_3}(excess)}}C{H_3}CON{H_2}\xrightarrow{{B{r_2}NaOH}}C{H_3}N{H_2}\]
Ethanoic acid to Propanoic acid –
Here we have to increase one carbon. So we will convert it to cyanide by reduction of acid to alcohol and then convert to the cyanide. The by acid hydrolysis cyanide gets converted to the desired acid with increase in a carbon
\[C{H_3}COOH\xrightarrow[{{H_3}{O^ + }}]{{LiAl{H_4}/ether}}C{H_3}C{H_2}OH\xrightarrow{{PC{l_5}}}C{H_3}C{H_2}Cl\xrightarrow{{NaCN/EtOH}}C{H_3}C{H_2}CN\xrightarrow{{{H^ + }/{H_2}O}}C{H_3}C{H_2}COOH\]
Methanamine to ethanamine –
Here also we have to increase one carbon. We have to introduce cyanide to the group with the help of its diazonium salt. Cyanide group is then by reduction, we will get the desired group that is amine group.
$C{H_3}N{H_2}\xrightarrow{{NaN{O_2}/HCl}}C{H_3}N_2^ + Cl\xrightarrow{{{H_{2O}}}}C{H_3}OH\xrightarrow{{PC{l_5}}}C{H_3}Cl\xrightarrow{{NaCN/EtOH}}C{H_3}CN\xrightarrow[{Na(Hg)/{C_2}{H_5}OH}]{{{H_2}/Ni}}C{H_3}C{H_2}N{H_2}$
Nitromethane into dimethylamine –
First reduce the nitro group to an amine group. With the help of Carbylamine reaction convert the amine group to isocyanide group, then by the reduction we will get the desired product.
$C{H_3}N{O_2}\xrightarrow{{Sn/HCl}}C{H_3}N{H_2}\xrightarrow{{CHC{l_3}/KOH/\Delta }}C{H_3}NC\xrightarrow{{Na/EtOH}}C{H_3}NHC{H_3}$
Propanoic acid into ethanoic acid –
Here we have to reduce one carbon, so with the help of Hoffmann bromamide reaction we will reduce one carbon here. First we will convert acid to amide and then with the help of Hoffmann bromamide reaction we will get ethanamine. To convert ethanamine to acetic acid, we will do a diazonium reaction of the amine then treat it with water, we will get alcohol, then by oxidation we will get the desired acid group.
$C{H_3}C{H_2}COOH\xrightarrow{{N{H_3}(excess)}}C{H_3}C{H_2}CON{H_2}\xrightarrow{{B{r_2}/KOH}}C{H_3}C{H_2}N{H_2}\xrightarrow{{NaN{O_2}/HCl}}C{H_3}C{H_2}N_2^ + Cl\xrightarrow{{{H_2}O}}C{H_3}C{H_2}OH\xrightarrow{{KMn{O_4}/{H^ + }}}C{H_3}COOH$

Note: After going through all the above reactions that have been carried out we can note that the Carbylamine reaction is the synthesis of an isocyanides by the reaction of primary amine, base and chloroform. This reaction is also used to test the presence of primary amine, as it is only effective to primary amine.