A ketone has the molecular formula C5H10O. Write the structural formulae of the isomers to show positional isomerism?

1 Answer

A ketone has one double bonded oxygen atom, but not at the end of a (sub) chain (or it would be an aldehyde).

Explanation:

Step 1:
Make a straight chain of #C#'s. You can place the #=O# at the second or third #C# (number 1 and 5 are not allowed, and 4 would be the same as two). Of course, you fill the rest of the valencies with #H# atoms.
Step 2:
Make a chain of 4 #C#'s and put a branch-#C# at number 3.
Number 3 has but one valence left, so the #=O# goes to number two (again the other ones are end-#C#'s so not allowed).

So you are left with altogether 3 isomers:

#CH_3-CO-CH_2-CH_2-CH_3#
pentan-2-one (or 2-pentanone)

#CH_3-CH_2-CO-CH_2-CH_3#
pentan-3-one (or 3-pentanone)

#CH_3-CO-CH(CH_3)-CH_3#
3-methylbutan-2-one. Since the places of the #=O# group in relation to the #CH_3-# branch are fixed in relation to each other (i.e. there is only one way), you may leave out the numbers:
methylbutanone.