
The IUPAC name of $\text{C}{{\text{H}}_{2}}(\text{OH})\text{ - CH(N}{{\text{H}}_{2}}\text{) - COOH}$ is:
(A) 2 - amino - 3 - hydroxy propanoic acid
(B) 1 - hydroxy - 3 - amino propan-3-oic acid
(C) 1 - amino - 2 - hydroxy propanoic acid
(D) 3 - hydroxy - 2 - amino propanoic acid
Answer
573.3k+ views
Hint: For this problem, we have to study about the IUPAC naming, its rules for the naming of the compound and then we will do the numbering of the long-chain according to the priority orders of the functional groups.
Complete step by step solution:
-In the given question, we have to choose the correct IUPAC name of the given organic molecule.
-As we know that the IUPAC naming is the method of naming the organic molecules in a standard way which is accepted worldwide.
-Now, for the naming of any molecule, firstly we have to recognize the longest chain that is present in the molecule.
-For example, in the given compound, there are three carbons so the longest will be named as 'prop'.
-Now, after identifying the longest chain, we have to do the numbering of the carbon atoms which is done according to the functional groups present on them.
-The priority order of the functional groups is given below in the table:
-Here, as carboxylic acid is prior so the suffix oic will be used whereas other groups i.e. amine and alcohol are not prior so for them the prefix amino and hydroxy will be used respectively. -So, the numbering of the compound will start from carboxylic acid as it comes first in the priority order.
$^{3}\text{C}{{\text{H}}_{2}}(\text{OH})\text{ -}{{\text{ }}^{2}}\text{CH(N}{{\text{H}}_{2}}\text{) -}{{\text{ }}^{1}}\text{COOH}$
- So, the name of the compound will be: 2 - amino - 3 - hydroxy propanoic acid.
Therefore, option (A) is the correct answer.
Note: While doing the naming of the compound which consist of more than one functional group then the suffix will be written only for the most prior functional group whereas prefix is written for the less prior functional group.
Complete step by step solution:
-In the given question, we have to choose the correct IUPAC name of the given organic molecule.
-As we know that the IUPAC naming is the method of naming the organic molecules in a standard way which is accepted worldwide.
-Now, for the naming of any molecule, firstly we have to recognize the longest chain that is present in the molecule.
-For example, in the given compound, there are three carbons so the longest will be named as 'prop'.
-Now, after identifying the longest chain, we have to do the numbering of the carbon atoms which is done according to the functional groups present on them.
-The priority order of the functional groups is given below in the table:
| S. No. | Functional group | Prefix | Suffix |
| 1 | Carboxylic groups | -carboxy | -oic acid |
| 2 | ester | -oxycarbonyl | -oate |
| 3 | Acid halide | -halo carbonyl | -oyl halide |
| 4 | Amide | carbonyl | Amide |
| 5 | Aldehyde | formyl | -al |
| 6 | Ketone | -oxo | -one |
| 7 | Alcohol | -hydroxy | -ol |
| 8 | Amine | amino | -amine |
| 9 | Alkene | alkenyl | -ene |
| 10 | Alkane | alkyl | -ane |
-Here, as carboxylic acid is prior so the suffix oic will be used whereas other groups i.e. amine and alcohol are not prior so for them the prefix amino and hydroxy will be used respectively. -So, the numbering of the compound will start from carboxylic acid as it comes first in the priority order.
$^{3}\text{C}{{\text{H}}_{2}}(\text{OH})\text{ -}{{\text{ }}^{2}}\text{CH(N}{{\text{H}}_{2}}\text{) -}{{\text{ }}^{1}}\text{COOH}$
- So, the name of the compound will be: 2 - amino - 3 - hydroxy propanoic acid.
Therefore, option (A) is the correct answer.
Note: While doing the naming of the compound which consist of more than one functional group then the suffix will be written only for the most prior functional group whereas prefix is written for the less prior functional group.
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