
Ethyl cyanide and ethyl isocyanide are _______ isomers.
A. Position
B. Functional
C. Chain
D. None of these
Answer
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Hint:: Isomers have the same number of atoms and a similar chemical formula but they differ in the 3-dimension representation of atoms in space. Ethyl is an alkyl group having 2 carbons in it, and if we hydrolyze the functional (cyanide) group it will convert into the corresponding carboxylic acid
Complete answer:
To understand the type of isomer we will first write the chemical formula for both of the compounds, and since they are isomer they both would have the same chemical formula: ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{CN}}$.
The functional group cyanide in this compound ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{CN}}$ is not a typical functional group, Cyanide (-CN) has two binding sites, meaning it can form a bond with both the atoms (i.e., carbon and nitrogen). And both cases have different nomenclature, when cyanide binds to a compound with carbon atom the suffix cyanide is added, and when it binds with the nitrogen atoms suffix isocyanide is added.
Therefore, we can say that these two isomers i.e., Ethyl cyanide and Ethyl isocyanide shows structural isomerism of linkage, since the binding or linking sides are different in both the cases viz., one has a carbon binding side, ${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CN}}$ (ethyl cyanide) and the other has nitrogen binding side ${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{NC}}$ (ethyl isocyanide).
Hence, the correct answer is option (D)i.e., None of these.
Note:
All the mentioned options belong to structural isomers, and one may get confused with the position isomer, assuming the position of the Carbon and nitrogen is different in both case, but this is not correct, since in the position isomer the functional group is bonded to a different carbon atom in the hydrocarbon chain. E.g., 2-bromopropane and n-bromopropane. Hence a doubt free concept of all types of isomerism is advised, with proper examples.
Complete answer:
To understand the type of isomer we will first write the chemical formula for both of the compounds, and since they are isomer they both would have the same chemical formula: ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{CN}}$.
The functional group cyanide in this compound ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{CN}}$ is not a typical functional group, Cyanide (-CN) has two binding sites, meaning it can form a bond with both the atoms (i.e., carbon and nitrogen). And both cases have different nomenclature, when cyanide binds to a compound with carbon atom the suffix cyanide is added, and when it binds with the nitrogen atoms suffix isocyanide is added.
Therefore, we can say that these two isomers i.e., Ethyl cyanide and Ethyl isocyanide shows structural isomerism of linkage, since the binding or linking sides are different in both the cases viz., one has a carbon binding side, ${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CN}}$ (ethyl cyanide) and the other has nitrogen binding side ${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{NC}}$ (ethyl isocyanide).
Hence, the correct answer is option (D)i.e., None of these.
Note:
All the mentioned options belong to structural isomers, and one may get confused with the position isomer, assuming the position of the Carbon and nitrogen is different in both case, but this is not correct, since in the position isomer the functional group is bonded to a different carbon atom in the hydrocarbon chain. E.g., 2-bromopropane and n-bromopropane. Hence a doubt free concept of all types of isomerism is advised, with proper examples.
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