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What is Stephen's reaction?
Answer
419.8k+ views
Hint: Stephen reaction is also known as Stephen aldehyde synthesis or Stephen Reduction Reaction. It was discovered by an English chemist Henry Stephen. It is an organic redox reaction which gives aldehyde as the end product.
Complete step by step solution:
Alkyl nitriles undergo reduction with the help of tin (II) chloride (\[S{{n}_{2}}Cl\]) and hydrochloric acid and form an intermediate imine salt which undergoes hydrolysis with water, giving the final result as aldehydes. Tin(ll) chloride reacts with hydrochloric acid and forms tin tetrachloride and two ions. As \[S{{n}^{+2}}\] changes into \[S{{n}^{+4}}\] when two electrons are released.
Mechanism of the reaction:
\[R-CN+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}R-CH=NHHCl\xrightarrow{{{H}_{2}}O}R-CHO+N{{H}_{4}}Cl\]
Example of Stephen reaction:
\[C{{H}_{3}}-C\equiv N+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}C{{H}_{3}}-CH=NHHCl\xrightarrow{{{H}_{2}}O}C{{H}_{3}}-CHO+N{{H}_{4}}Cl\]
In methyl cyanide (an alkyl cyanide), nitrogen in the cyanide group is more electronegative in comparison to the carbon. Therefore, it develops a partial negative charge and carbon develops a partial positive charge. In the reaction, one pie bond breaks between the carbon and nitrogen of the cyanide group and electrons move to the nitrogen atom, at the same time two electrons released are accepted by the carbon atom of cyanide. As a result, giving both carbon and nitrogen atoms negative charge. Further, it reacts with two ions and forms a minimum. Then iminium undergoes hydrolysis with water and gives final product methyl aldehyde.
Note: The reaction is more efficient in the case of aromatic nitriles as compared to the aliphatic nitriles. Ammonium chloride is formed as a byproduct of the reaction.
\[{{C}_{6}}{{H}_{5}}-C\equiv N+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}{{C}_{6}}{{H}_{5}}-CH=NHHCl\xrightarrow{{{H}_{2}}O}{{C}_{6}}{{H}_{5}}-CHO+N{{H}_{4}}Cl\]
Complete step by step solution:
Alkyl nitriles undergo reduction with the help of tin (II) chloride (\[S{{n}_{2}}Cl\]) and hydrochloric acid and form an intermediate imine salt which undergoes hydrolysis with water, giving the final result as aldehydes. Tin(ll) chloride reacts with hydrochloric acid and forms tin tetrachloride and two ions. As \[S{{n}^{+2}}\] changes into \[S{{n}^{+4}}\] when two electrons are released.
Mechanism of the reaction:
\[R-CN+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}R-CH=NHHCl\xrightarrow{{{H}_{2}}O}R-CHO+N{{H}_{4}}Cl\]
Example of Stephen reaction:
\[C{{H}_{3}}-C\equiv N+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}C{{H}_{3}}-CH=NHHCl\xrightarrow{{{H}_{2}}O}C{{H}_{3}}-CHO+N{{H}_{4}}Cl\]
In methyl cyanide (an alkyl cyanide), nitrogen in the cyanide group is more electronegative in comparison to the carbon. Therefore, it develops a partial negative charge and carbon develops a partial positive charge. In the reaction, one pie bond breaks between the carbon and nitrogen of the cyanide group and electrons move to the nitrogen atom, at the same time two electrons released are accepted by the carbon atom of cyanide. As a result, giving both carbon and nitrogen atoms negative charge. Further, it reacts with two ions and forms a minimum. Then iminium undergoes hydrolysis with water and gives final product methyl aldehyde.
Note: The reaction is more efficient in the case of aromatic nitriles as compared to the aliphatic nitriles. Ammonium chloride is formed as a byproduct of the reaction.
\[{{C}_{6}}{{H}_{5}}-C\equiv N+2[H]\xrightarrow{SnC{{L}_{2}}/dil.\,HCl}{{C}_{6}}{{H}_{5}}-CH=NHHCl\xrightarrow{{{H}_{2}}O}{{C}_{6}}{{H}_{5}}-CHO+N{{H}_{4}}Cl\]
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