How do you name #"C"_5"H"_11"OH"# if it is a straight-chained alcohol?

1 Answer
Mar 12, 2017

Count the number of carbons to find that there are five. That means, from the prefixes you were taught, you can match up with one of the following:

  • #"meth"-# = one carbon
  • #"eth"-# = two carbons
  • #"prop"-# = three carbons
  • #"but"-# = four carbons
  • #"pent"-# = five carbons
  • #"hex"-# = six carbons

Since this is an alkane (the main chain contains only #"C"# and #"H"#, and no double or triple bonds. That is, the compound is completely saturated with hydrogen atoms), the middle of the name contains #-"an"-#.

So far we have #bb("pentan"-)#.

Then, you should note that #-"OH"# connected to a #-"CH"_2-# is called a hydroxyl group. Any hydrocarbon with a #-"CH"_3# (a terminal methyl group) replaced with a hydroxyl group is called an alcohol. Those end in the suffix #-"ol"#.

#=> "pent" + "an" + "ol" = [ . . . ]#

Therefore, this is called #bbcolor(blue)("pentanol")#.