
What will be the final product of the following reaction sequence?
\[C{H_3}C{H_2}CN\xrightarrow{{{H^ + }/{H_2}O}}A\xrightarrow[\Delta ]{{N{H_3}}}B\xrightarrow{{NaOBr}}C\]
A. $C{H_3}C{H_2}CON{H_2}$
B. $C{H_3}C{H_2}OBr$
C. $C{H_3}C{H_2}N{H_2}$
D. $C{H_3}C{H_2}C{H_2}N{H_2}$
Answer
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Hint: The compounds containing a cyanide group act as an electron rich species in a protic medium or in aqueous conditions in which there are a lot of protons available. The carboxylic acid on reaction with ammonia forms an intermediate ionic complex and dehydrates on the application of strong heat and loses a molecule of water. Bromine in the presence of sodium hydroxide acts as a strong reducing agent.
Complete answer:
We have been given a reaction sequence in the question along with certain reagents that have different functions. It is because of the difference in the activity of these reagents that yield different products. Let us determine the unknown products one by one.
(i) When propyl cyanide ($C{H_3}C{H_2}CN$ ) hydrolyses with the help of water or in an aqueous environment, it yields propionic acid. The reaction can be determined as follows:
$C{H_3}C{H_2}CN\xrightarrow{{{H^ + }/{H_2}O}}C{H_3}C{H_2}COOH$(A)
(ii) When the propionic acid reacts with ammonia, the product that is yielded in the reaction is an intermediate salt of ammonium propionate which on further heating loses a water molecule in order to yield propionamide. The reaction can be determined as follows:
$C{H_3}C{H_2}COOH\xrightarrow{{N{H_3}}}C{H_3}C{H_2}COONH_4^ + \xrightarrow{\Delta }C{H_3}C{H_2}CON{H_2}$(B)
(iii) $NaOBr$ acts as a strong reducing agent and is produced by $B{r_2}/NaOH$ and reduces the propanamide into ethyl amine. The reaction that takes place in this final step of the reaction yields us the final product and the reaction can be determined as follows:
$C{H_3}C{H_2}CON{H_2}\xrightarrow{{NaOBr}}C{H_3}C{H_2}N{H_2}$(C)
Thus option C is the correct answer.
Note:
Ethylamine is an organic compound with the formula \[C{H_3}C{H_2}N{H_2}\] . This colorless gas has a strong ammonia-like odor. It condenses just below room temperature to a liquid miscible with virtually all solvents. It is a nucleophilic base, as is typical for amines. Ethylamine is widely used in the chemical industry and organic synthesis.
Complete answer:
We have been given a reaction sequence in the question along with certain reagents that have different functions. It is because of the difference in the activity of these reagents that yield different products. Let us determine the unknown products one by one.
(i) When propyl cyanide ($C{H_3}C{H_2}CN$ ) hydrolyses with the help of water or in an aqueous environment, it yields propionic acid. The reaction can be determined as follows:
$C{H_3}C{H_2}CN\xrightarrow{{{H^ + }/{H_2}O}}C{H_3}C{H_2}COOH$(A)
(ii) When the propionic acid reacts with ammonia, the product that is yielded in the reaction is an intermediate salt of ammonium propionate which on further heating loses a water molecule in order to yield propionamide. The reaction can be determined as follows:
$C{H_3}C{H_2}COOH\xrightarrow{{N{H_3}}}C{H_3}C{H_2}COONH_4^ + \xrightarrow{\Delta }C{H_3}C{H_2}CON{H_2}$(B)
(iii) $NaOBr$ acts as a strong reducing agent and is produced by $B{r_2}/NaOH$ and reduces the propanamide into ethyl amine. The reaction that takes place in this final step of the reaction yields us the final product and the reaction can be determined as follows:
$C{H_3}C{H_2}CON{H_2}\xrightarrow{{NaOBr}}C{H_3}C{H_2}N{H_2}$(C)
Thus option C is the correct answer.
Note:
Ethylamine is an organic compound with the formula \[C{H_3}C{H_2}N{H_2}\] . This colorless gas has a strong ammonia-like odor. It condenses just below room temperature to a liquid miscible with virtually all solvents. It is a nucleophilic base, as is typical for amines. Ethylamine is widely used in the chemical industry and organic synthesis.
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