Question

How would you classify the following compounds as saturated, unsaturated, or substituted hydrocarbons: hexane (`C_6H_14`), isopropyl alcohol (`C_3H_7OH`), 2-chlorobutane (`C_4H_9Cl`), pentene (`C_5H_10`), and butyric acid (`C_3H_7COOH`)?

Answer 1

Saturated organic molecules have formula of `C_nH_(2n+2)` OR the equivalent.

Explanation:

A saturated organic molecule can be substituted with halide or oxygen to give (saturated formula of) `C_nH_(2n+1)X`.

And thus `"hexane"` is saturated.

`"Isopropyl alcohol"` is saturated.

`"2-chlorobutane"` is saturated. Because a halide counts as one hydrogen atom.

`C_5H_10` is `"UNSATURATED"` and has 1 degree of unsaturation; as does `"butyric acid, "C_4H_8O_2`.

Answer 2

Saturated open chain hydrocarbons are classified as ALKANE. The general molecular formula of alkane is `C_nH_(2n+2)`. A saturated compound does not contain any `C=C or CequivC` in its structure. An unsaturated compound must posses atleast one `C=C or CequivC`. The general molecular formula of open chain hydrocarbon with one `C=C` is given by `C_nH_(2n)`

When a compound is produced replacing one H-atom of alkane molecule by a univalent group or atom G, then the molecular formula of the compound formed becomes `C_nH_(2n+1)"-"G`.This compound also belongs to saturated class. This type of compound is classified as substituted hydrocarbons.

So hexane having molecular formula `C_4H_14=C_4H_(2*4+2)` follows the general formula of alkane `C_nH_(2n+2)` for `n=4`.

Isopropyl alcohol having MF

`C_3H_7OH=C_3H_(2*3+1)"-"OH`,

2-chlorobutane having MF

`C_4H_9Cl=C_4H_(2*4+1)"-"Cl`,

and butyric acid having MF

`C_3H_7COOH=C_3H_(2*3+1)"-"COOH`

belong to substituted hydrocarbon class as they follow general molecular formula `C_nH_(2n+1)"-"G` .

And finally the open chain hydrocarbon pentene `(C_5H_10)` is classified as unsatrated hydrocarbon as its molecular formula follows general mf `C_nH_(2n)` of alkene with one `C=C` for `n=2`.