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How will you convert:
A.Ethanoic acid into methanamine
B.Hexane nitrile into 1-aminopentane
C.Methanol to ethanoic acid
D.Ethanamine to methanamine
E.Ethanoic acid into propanoic acid
F.Methanamine to ethanamine
G.Nitromethane into dimethyl amine
H.Propanoic acid into ethanoic acid

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Last updated date: 13th Jun 2024
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Answer
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Hint: To reduce the number of carbon atoms we use hoffmann bromamide degradation. This will help us to reach the lower hydrocarbon chain from the higher hydrocarbon chain.

Complete step by step solution:
Let us to the convergence one by one:
Option A
Ethanoic acid reacts with thionyl chloride to form acid chloride. On reaction with excess ammonia the acid chloride converts into acid amide. To reduce the number of carbon we do bromide degradation and hence methane amine forms the reaction occurs as follow:
\[{\text{C}}{{\text{H}}_3}{\text{COOH}}\xrightarrow{{{\text{SOC}}{{\text{l}}_2}}}{\text{C}}{{\text{H}}_3}{\text{COCl}}\xrightarrow{{{\text{N}}{{\text{H}}_3}({\text{excess}})}}{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}\xrightarrow{{{\text{B}}{{\text{r}}_2}/{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}\]
Option B
First hydrolysis occurs to convert cyanide into amide. Then again using bromide degradation we will reduce number of carbon and forms 1-aminopentane:
\[{\text{C}}{{\text{H}}_3}{\left( {{\text{C}}{{\text{H}}_2}} \right)_4}{\text{CN}}\xrightarrow{{{{\text{H}}_3}{{\text{O}}^ + }}}{\text{C}}{{\text{H}}_3}{\left( {{\text{C}}{{\text{H}}_2}} \right)_4}{\text{CON}}{{\text{H}}_2}\xrightarrow{{{\text{B}}{{\text{r}}_2}/{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\left( {{\text{C}}{{\text{H}}_2}} \right)_4}{\text{N}}{{\text{H}}_2}\]
Option C
To convert methanol into ethanoic acid, first of all methanol is treated with Phosphorus pentachloride. Then methyl chloride is formed by substituting hydroxide group from chloride then reaction with alcoholic potassium cyanide forms methyl nitrile. The hydrolysis of methyl cyanide will give us the carboxylic acid that is ethanoic acid:
\[{\text{C}}{{\text{H}}_3}{\text{OH}}\xrightarrow{{{\text{PC}}{{\text{l}}_5}}}{\text{C}}{{\text{H}}_3}{\text{Cl}}\xrightarrow{{{\text{KCN(alc}}{\text{.)}}}}{\text{C}}{{\text{H}}_3}{\text{CN}}\xrightarrow{{{{\text{H}}_3}{{\text{O}}^ + }}}{\text{C}}{{\text{H}}_3}{\text{COOH}}\]
Option D
Ethylamine is treated with sodium nitrite with hydrochloric acid. Amide is converted to alcohol. The ethyl alcohol formed is oxidised to acetic acid using potassium permanganate as an oxidizing agent. Using excess of ammonia acid is converted to amide then again using bromide degradation acid amide is converted to methane amide.
\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}\xrightarrow[{{{\text{H}}_2}{\text{O}}}]{{{\text{NaN}}{{\text{O}}_2}/{\text{HCl}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}\xrightarrow{{{\text{KMN}}{{\text{O}}_4}/{{\text{H}}^ + }}}{\text{C}}{{\text{H}}_3}{\text{COOH}}\xrightarrow{{{\text{N}}{{\text{H}}_3}({\text{excess}})}}{\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}\xrightarrow{{{\text{B}}{{\text{r}}_2}/{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}\]
Option E
Ethanoic acid is first reduced to ethanol using lithium aluminium hydride. Then reaction of alcohol with Phosphorus pentachloride will yield ethyl chloride, then again reacting it with alcoholic solution of sodium cyanide will give us cyanide. Hydrolysis of cyanide will give us carboxylic acid that is propanoic acid:
\[{\text{C}}{{\text{H}}_3}{\text{COOH}}\xrightarrow[{{{\text{H}}_3}{{\text{O}}^ + }}]{{{\text{LiAl}}{{\text{H}}_4}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}\xrightarrow{{{\text{PC}}{{\text{l}}_5}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{Cl}}\xrightarrow{{{\text{NaCN(alc}}{\text{.)}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{CN}}\xrightarrow{{{{\text{H}}^ + }/{{\text{H}}_2}{\text{O}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}}\]
Option F
Methyl amine is first converted into methanol. Then methanol is reacted with Phosphorus pentachloride to form methyl chloride and then reaction with sodium cyanide forms methyl cyanide reduction of methyl cyanide will give us ethane amine:
\[{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}\xrightarrow[{{{\text{H}}_2}{\text{O}}}]{{{\text{NaN}}{{\text{O}}_2}/{\text{HCl}}}}{\text{C}}{{\text{H}}_3}{\text{OH}}\xrightarrow{{{\text{PC}}{{\text{l}}_5}}}{\text{C}}{{\text{H}}_3}{\text{Cl}}\xrightarrow{{{\text{NaCN(alc}}{\text{.)}}}}{\text{C}}{{\text{H}}_3}{\text{CN}}\xrightarrow[{{\text{Na(Hg)}}/{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}}]{{{{\text{H}}_2}/{\text{Ni}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}\]
Option G
Nitromethane is first converted using tin hydrochloride to methyl amine. Then methyl amine is reacted with chloroform to form methyl isocyanide, which is a again reduced to form dimethylamine
\[{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{O}}_2}\xrightarrow[{\left[ {\text{H}} \right]}]{{{\text{Sn}}/{\text{HCl}}}}{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}\xrightarrow{{{\text{CHC}}{{\text{l}}_3}/{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\text{NC}}\xrightarrow[{\left[ {\text{H}} \right]}]{{{\text{Na}}/{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}}}{\text{C}}{{\text{H}}_3}{\text{NHC}}{{\text{H}}_3}\]
Option H
To convert propanoic acid into ethanoic acid, first of all ethanoic acid is converted to acid amide by using excess of ammonia. Then bromide degradation occurs to remove one carbon. after bromide degradation it is reacted with sodium nitrite in the presence of hydrogen chloride and then oxidizes to form the methanoic acid:
\[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}}\xrightarrow{{{\text{N}}{{\text{H}}_3}({\text{excess}})}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{CON}}{{\text{H}}_2}\xrightarrow{{{\text{B}}{{\text{r}}_2}/{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{N}}{{\text{H}}_2}\xrightarrow[{{{\text{H}}_2}{\text{O}}}]{{{\text{NaN}}{{\text{O}}_2}/{\text{HCl}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}\xrightarrow{{{\text{KMN}}{{\text{O}}_4}/{{\text{H}}^ + }}}{\text{C}}{{\text{H}}_3}{\text{COOH}}\]

Note: We have used the bromide degradation a lot. Hoffmann bromamide reaction mechanism generally includes the use of a strong base to attack the amide, yields to the deprotonation and the subsequent generation of an anion. This reaction is used for the conversion of a primary amide to a primary amine with one less carbon atom. This is accomplished by heating the primary amide with a mixture of a halogen, a strong base, and water.