Question

Write IUPAC names of the following
i. $C{H_3} - C{H_2} - C{H_2} - C{H_3}$
ii. $C{H_3} - C{H_2} - COOH$
iiii. $C{H_3} - CO - C{H_2} - C{H_3}$

Answer 1

Hint: In short the IUPAC name of the compound is written out with the substituent in alphabetical order followed by the base name (derived from the number of carbons in the parent chain, mostly the longest chain is taken). Commas are used between numbers and dashes are used between letters and numbers. There are no spaces in the name.

Complete Step by step answer: Let us derive the name of each of the following compounds one by one.
First is $C{H_3} - C{H_2} - C{H_2} - C{H_3}$, in this compound the chain is having 4 carbons so we will use but in starting of the name. Now as the compound is alkane or we can say is unsaturated and having no substituent group attached to it, so we will take –ane to be the suffix. So this implies that IUPAC name of the compound $C{H_3} - C{H_2} - C{H_2} - C{H_3}$ is Butane.
Second is $C{H_3} - C{H_2} - COOH$, in this compound there is a carboxyl group attached to it. And Carboxylic acids are named by counting the number of carbons in the longest continuous chain including the carboxyl group and by replacing the suffix -ane of the corresponding alkane with -enoic acid. So in this case the chain is having three carbon atoms so we will use prop and suffix is –oic acid. So the name of the compound $C{H_3} - C{H_2} - COOH$ is Propanoic acid.
Third one is $C{H_3} - CO - C{H_2} - C{H_3}$, in this compound ketone group. And for ketones, the suffix is replaced by –one. And in this case the longest chain is having four carbons so we will use prop. Now as the ketonic group is present on the second carbon so we will name it according to the rule. So the name of the compound $C{H_3} - CO - C{H_2} - C{H_3}$ will be 2-butanone.
So IUPAC names of the compounds given are butane, propanoic acid and 2-butanone.

Note: Always take the longest chain as a parent chain. Always keep in mind that if there is a group which is having higher priority than previous then always take that group in the parent chain which is having higher priority.