Question

Question #f1b76

Answer 1

`"H"_3"CCH"_2"CH(CH"_3)"C(=O)H"`

Explanation:

So draw the parent hydrocarbyl chain (four carbons long for butyl):

`C_1-C_2-C_3-C_4`

Substitute in the functional groups at the appropriate position. It is an aldehyde so it has a carbonyl at the head of the line. At `C_2` we substitute a methyl group as required.

`H(O=)C_1C_2(CH_3)C_3-C_4`

Satisfy carbon's quadrivalency (add hydrogens):

`H(O=)C-CH(CH_3)-CH_2-CH_3`

The exercise is a hell of a lot easier on paper than by using an editor like this one. You will find these sorts of problems trivial in a week or so.

Would this molecule have left and right handed isomers? Why?