Question

The IUPAC name of ${{C}_{6}}{{H}_{5}}-CH=CH-COOH$ is :
(A) Cinnamic acid
(B) 1-phenyl-2-carboxyethane
(C) 3-phenylprop-2-enoic acid
(D) Dihydroxy-3-phenyl propionic acid

Answer 1

Hint: Before finding the IUPAC name, we first need to identify the functional groups present in the compound. After identifying all of the functional groups, we need to classify them as primary and secondary and then go for the nomenclature.

Complete step by step solution:
-To find the name, first we must find the functional groups present in the given structure. Here we see that there is presence of double bond(=), acid group(-COOH) and a phenyl group$\left( {{C}_{6}}{{H}_{5}} \right)$
-The general method of writing the IUPAC name is
Secondary prefix+primary prefix+root word+primary suffix+seconday suffix.
The word root depends on the number of carbon atoms present in the longest chain. Eg:

No. of C atoms in chain	Word root	No. of C atoms in chain	Word root
1	Meth	8	Oct
2	Eth	9	Non
3	Prop	10	Dec
4	But	11	Undec
5	pent	20	Icos
6	Hex	30	Triacont
7	Hept	100	cent

-Primary suffix includes the saturation of compounds. Single bond is alkane, double bond alkene and triple bond alkyne.
-Secondary prefix and suffix is added to indicate the functional groups. They have priorities assigned to them. Higher priority functional group is written as a suffix and all functional groups with lower priority are written as a prefix. Some of the groups can be shown as

CLASS	NAME	SUFFIX	PREFIX
R-COOH	Alkanoic acid	Oic acid	Carboxy
$R-S{{O}_{3}}H$	Alkane sulphanoic acid	Sulfonic acid	Sulfo
R-(CO)-O-(CO)-R	Alkanoic anhydride	Oic anhydride	-----
R-COOR	Alkyl alkanoate	oate	Alkoxy carbonyl
R-(CO)-X	Alkanoyl halide	Oyl-halide	Halocarbonyl
$R-\left( CO \right)-N{{H}_{2}}$	Alkanamide	Amide	Carbamoyl
R-CN	Alkanenitrile	Nitrile	Cyano
R-CHO	Alkanal	-al	Oxo
R-CO-R	Alkanone	-one	oxo
R-OH	Alkanol	-ol	Hydroxy
R-SH	Alkanethiol	-thiol	mercapto
$R-N{{H}_{2}}$	alkanamide	-amine	amino

-Numbering is done such that double/triple bonds are given the highest location. Here, the location of the double bond is in between. So numbering is decided by the priority of the functional group present on the compound.
-We see that there are 2 functional groups, carboxylic acid and phenyl group. Acid is given higher priority. So the numbering starts from the right hand side. The number of carbon atoms are 3 in the chain and so the root word is prop and suffix is –oic acid. Phenyl group is located at the third C atom and so the final name of the compound is 3-phenylprop-2-enoic acid.
${{C}_{6}}{{H}_{5}}-CH=CH-COOH$
             3 2 1


Therefore the correct answer of the question is option C. 3-phenylprop-2-enoic acid


Note: Always follow all the steps of naming a compound in proper sequence. The sequence is mandatory in the naming. The priorities of different functional groups need to be memorised as it plays an important role when there are multiple number of functional groups.