Question

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

Answer 1

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.