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

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:

$C {H}_{3} - C O - C {H}_{2} - C {H}_{2} - C {H}_{3}$
pentan-2-one (or 2-pentanone)

$C {H}_{3} - C {H}_{2} - C O - C {H}_{2} - C {H}_{3}$
pentan-3-one (or 3-pentanone)

$C {H}_{3} - C O - C H \left(C {H}_{3}\right) - C {H}_{3}$
3-methylbutan-2-one. Since the places of the $= O$ group in relation to the $C {H}_{3} -$ branch are fixed in relation to each other (i.e. there is only one way), you may leave out the numbers:
methylbutanone.