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#)?

2 Answers
Jan 25, 2017

Answer:

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#.

Jan 25, 2017

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#.