Question

Q.13) Write IUPAC name of the following: (a) (CH3)2C=CHCOCH3 CH3C=CCH=CHCOOH​

Answer 1

Answer: a)4-methyl-pent-3-ene-2-one

b) 5-methyl-hex-2,4-enoic acid

Explanation:

a) In this compound the functional group is keto group, so 'one' is added to the last. now if we count from the carbon nearest to the functional group it can be seen that keto group is in the second position and the 4 number carbon has two methyl group and so there is 4 methyl and the third carbon has a double bond so there an 'ene' is added and as there are total five carbon thus the name is 4-methylpent-3-ene-2-one.

b) So keeping the double bond and functional group in mind methyl is in the 5 position, double bond is in 2 and 4 position and there are 6 carbons so thus the name.