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

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

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## pent-2-one (or 2-pentone)

##CH_3-CH_2-CO-CH_2-CH_3## pent-3-one (or 3-pentone)

##CH_3-CO-CH(CH_3)-CH_3## 3-methyl but-2-one (or iso-pentone). Since the places of the ##=O## group in relation to the ##CH_3-## branch are fixed in relation to each other (i.e. ther is only one way), you may leave out the numbers: methyl butone