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Iupac Name to Structure

The various steps involved are :

(i) Locatethe root-word and write the skeleton of principal chain accordingly (straight chain in case of acyclic or open chain and closed in case theroot-word carries the prefix cyclo) and number this skeleton carbon chain from right end to left hand.

(ii) Locate the primary functional group, i.e., —ane, —ene or -yne and fill the multiple carbon bond, if present, at the proper place(s) in the parent skeleton chain according to the positions given.

(iii) Attachthe secondary functional group according to the position mentioned. [In case the chain terminates in functional group, the carbon of the functional group must be considered as the part of the principal chain and numbered as one]. Secondary functional group(s), substituents and side chain(s) indicated by suffixes and prefixes in the name are also introduced to given specific number.

Example : Let us write down the structure of compound with IUPAC name; 7-Hydroxy-3-methoxy-5-methyl octa-5-enoic acid.

(i) Root-word is oct, hence write skeleton chain of eight carbon atoms and number them from right end to left.

8 7 6 5 4 3 2 1


(ii) The primary suffix is ene, so introduce a double bond starting from the carbon five to six as specified in the name.

8 7 6 5 4 3 2 1

c—c—c =c—c—c—c—c

(iii) The primary functional group is oic, i.e., carboxylic group, since no number is specified it is at number one carbon atom, further, the carbon of the —COOH group is the carbon one of the chain itself.

8 7 6 5 4 3 2 1


(iv) The substituents and side chains include hydroxy (—OH) at 7C — atom, methoxy (- OCH3) at 3C — atom and methyl group (- CH3) at 5C - atom,

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OH CH3 OCH3 (iv)Finally the hydrogen atoms are attached to satisfy the tetravalency ofeach carbon leading to following structure of the given compound.



H—C—C—C =C — C—C--C—COOH

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H OH CH3 H OCH3 H or


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