Question #f1b76

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anor277
Mar 29, 2017

#"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?

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