Question

Given the structures of products A, B and C in the following reactions:
(i)${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br}}\xrightarrow{{{{KCN}}}}{\text{A}}\xrightarrow{{{\text{LiAl}}{{\text{H}}_{\text{4}}}}}{\text{B}}\xrightarrow[ {{{\text{0}}^{\text{o}}}{\text{C}}}]{{{\text{HN}}{{\text{O}}_{\text{2}}}}}{\text{C}}$

(ii) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\xrightarrow[{{\Delta }}]{{{\text{N}}{{\text{H}}_{\text{3}}}}}{\text{A}}\xrightarrow{{{\text{NaOH + B}}{{\text{r}}_{\text{2}}}}}{\text{B}}\xrightarrow{{{\text{CHC}}{{\text{l}}_{\text{3}}}{\text{ + Alc}}{\text{. KOH}}}}{\text{C}}$

Answer 1

The starting product of the first reaction is ethyl bromide. The starting product of the second reaction is acetic acid. Identify the role of each reagent and predict the products. The intermediates and the attckimg groups must be identified clearly which help in finding out the exact product.

Complete Step by step answer:
(i)Consider the reaction,
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br}}\xrightarrow{{{\text{KCN}}}}{\text{A}}\xrightarrow{{{\text{LiAl}}{{\text{H}}_{\text{4}}}}}{\text{B}}\xrightarrow[{{{\text{0}}^{\text{o}}}{\text{C}}}]{{{\text{HN}}{{\text{O}}_{\text{2}}}}}{\text{C}}$
Ethyl bromide reacts with potassium cyanide to form product A. Ethyl bromide on reaction with potassium cyanide to form ethyl cyanide. The reaction is a substitution reaction.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br}}\xrightarrow{{{\text{KCN}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}}$
Thus, the product A is ethyl cyanide $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}}} \right)$.
Ethyl cyanide reacts with lithium aluminium hydride to form product B. Ethyl cyanide reacts with lithium aluminium hydride to form propyl amine. Lithium aluminium hydride is a reducing agent and thus, it is a reduction reaction.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}}\xrightarrow{{{\text{LiAl}}{{\text{H}}_{\text{4}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}$
Thus, the product B is propyl amine $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}} \right)$.
Propyl amine reacts with nitrous acid at ${{\text{0}}^{\text{o}}}{\text{C}}$ to form product C. propyl amine on reaction with nitrous acid at ${{\text{0}}^{\text{o}}}{\text{C}}$ forms propyl alcohol. It is a substitution reaction.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}\xrightarrow[{{{\text{0}}^{\text{o}}}{\text{C}}}]{{{\text{HN}}{{\text{O}}_{\text{2}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}$
Thus, the product C is propyl alcohol $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}} \right)$.
Thus, the overall reaction sequence is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br}}\xrightarrow{{{\text{KCN}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CN}}\xrightarrow{{{\text{LiAl}}{{\text{H}}_{\text{4}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}\xrightarrow[{{{\text{0}}^{\text{o}}}{\text{C}}}]{{{\text{HN}}{{\text{O}}_{\text{2}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}$

(ii)Consider the reaction,
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\xrightarrow[{{\Delta }}]{{{\text{N}}{{\text{H}}_{\text{3}}}}}{\text{A}}\xrightarrow{{{\text{NaOH + B}}{{\text{r}}_{\text{2}}}}}{\text{B}}\xrightarrow{{{\text{CHC}}{{\text{l}}_{\text{3}}}{\text{ + Alc}}{\text{. KOH}}}}{\text{C}}$
Acetic acid reacts with ammonia in presence of heat to form product A. acetic acid on reaction with ammonia forms acetamide. It is a substitution reaction.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\xrightarrow[{{\Delta }}]{{{\text{N}}{{\text{H}}_{\text{3}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{CON}}{{\text{H}}_2}$
Thus, the product A is acetamide $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CON}}{{\text{H}}_{\text{2}}}} \right)$.
Acetamide reacts with sodium hydroxide in presence of bromine to form product B. acetamide on reaction with sodium hydroxide in presence of bromine forms methyl amine.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CON}}{{\text{H}}_2}\xrightarrow{{{\text{NaOH + B}}{{\text{r}}_{\text{2}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}$
Thus, the product B is methyl amine $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}} \right)$.
Methyl amine reacts with chloroform in presence of alcoholic potassium hydroxide to form product C. Methyl amine in reaction with chloroform in presence of alcoholic potassium hydroxide forms isocyanide.
The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}\xrightarrow{{{\text{CHC}}{{\text{l}}_{\text{3}}}{\text{ + Alc}}{\text{. KOH}}}}{\text{C}}{{\text{H}}_{\text{3}}}{{\text{N}}^ + } \equiv {{\text{C}}^ - }$
Thus, the product C is methyl isocyanide $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{{\text{N}}^ + } \equiv {{\text{C}}^ - }} \right)$.
Thus, the overall reaction sequence is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\xrightarrow[{t{\Delta }}]{{{\text{N}}{{\text{H}}_{\text{3}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{CON}}{{\text{H}}_{\text{2}}}\xrightarrow{{{\text{NaOH + B}}{{\text{r}}_{\text{2}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}\xrightarrow{{{\text{CHC}}{{\text{l}}_{\text{3}}}{\text{ + Alc}}{\text{. KOH}}}}{\text{C}}{{\text{H}}_{\text{3}}}{{\text{N}}^ + } \equiv {{\text{C}}^ - }$

Note: In the first reaction, ethyl bromide reacts with potassium cyanide to form ethyl cyanide which reacts with lithium aluminium hydride to form propyl amine which reacts with nitrous acid at ${{\text{0}}^{\text{o}}}{\text{C}}$ to form propyl alcohol.
In the second reaction, acetic acid reacts with ammonia in presence of heat to form acetamide which reacts with sodium hydroxide in presence of bromine to form methyl amine which reacts with chloroform in presence of alcoholic potassium hydroxide to form methyl isocyanide.