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

1 Answer
Jul 27, 2015

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.