
The IUPAC name of \[C{{H}_{3}}-CH=CH-COOH\]
(A) But-2-ene-1-oic acid
(B) But-1-ene-1-oic acid
(C) But-2-ene-1-carboxylic acid
(D) Propene-1-carboxylic acid
Answer
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Hint: According to IUPAC (International Union of Pure and Applied Chemistry), whenever we are going to write the name of the compound, we have to give numbering first to functional groups. Means lower numbering should become a functional group present in the molecule or compound.
Complete step by step answer:
-We have to choose the longest chain in the given compound and label the carbons with numbers in such a way that the functional groups should get less number.
-The given compound is
\[C{{H}_{3}}-CH=CH-COOH\]
-In the given compound there are two different functional groups present. Those functional groups are alkene (double bond) and carboxylic acid.
-The carboxylic functional group is present at the terminal position of the chain. So, we have to start giving numbers from a carboxylic functional group.
\[\underset{4}{\mathop{C}}\,{{H}_{3}}-\underset{3}{\mathop{C}}\,H=\underset{2}{\mathop{C}}\,H-\underset{1}{\mathop{C}}\,OOH\]
-There are four carbons present in the main chain then we are supposed to say “but”.
-For a carboxylic functional group we have to write “oic acid”.
-According to the IUPAC name we have to write “ene” to represent the double bond, in the name of the compound.
-Carboxylic group labeled as one and alkene got a number of 2 while giving numbers to the carbons present in the given structure.
\[\underset{4}{\mathop{C}}\,{{H}_{3}}-\underset{3}{\mathop{C}}\,H=\underset{2}{\mathop{C}}\,H-\underset{1}{\mathop{C}}\,OOH\]
-Therefore, the IUPAC name of the compound is But-2-ene-1-oic acid.
-So, the correct option is A.
Note: According to IUPAC for double bond we have to write “ene” and for triple bond we have to write “yne”. While giving numbering to carbons in the longest chain we are not supposed to start giving numbering from the middle of the chain. We have to start giving numbers from the starting or from the end of the chain.
Complete step by step answer:
-We have to choose the longest chain in the given compound and label the carbons with numbers in such a way that the functional groups should get less number.
-The given compound is
\[C{{H}_{3}}-CH=CH-COOH\]
-In the given compound there are two different functional groups present. Those functional groups are alkene (double bond) and carboxylic acid.
-The carboxylic functional group is present at the terminal position of the chain. So, we have to start giving numbers from a carboxylic functional group.
\[\underset{4}{\mathop{C}}\,{{H}_{3}}-\underset{3}{\mathop{C}}\,H=\underset{2}{\mathop{C}}\,H-\underset{1}{\mathop{C}}\,OOH\]
-There are four carbons present in the main chain then we are supposed to say “but”.
-For a carboxylic functional group we have to write “oic acid”.
-According to the IUPAC name we have to write “ene” to represent the double bond, in the name of the compound.
-Carboxylic group labeled as one and alkene got a number of 2 while giving numbers to the carbons present in the given structure.
\[\underset{4}{\mathop{C}}\,{{H}_{3}}-\underset{3}{\mathop{C}}\,H=\underset{2}{\mathop{C}}\,H-\underset{1}{\mathop{C}}\,OOH\]
-Therefore, the IUPAC name of the compound is But-2-ene-1-oic acid.
-So, the correct option is A.
Note: According to IUPAC for double bond we have to write “ene” and for triple bond we have to write “yne”. While giving numbering to carbons in the longest chain we are not supposed to start giving numbering from the middle of the chain. We have to start giving numbers from the starting or from the end of the chain.
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