What is the structure of 5,5-diethyl-2,2,3-trimethylheptane?

1 Answer
Truong-Son N.
Oct 1, 2017

I would start from the parent name, heptane. The prefix "hept" means #7#, so the main chain contains #7# carbons:

#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#

Then I would work from right to left in the name. The #2,2,3-"trimethyl"# suggests that:

  • #"tri" -># three duplicate branching groups
  • #"methyl" -> "H"_3"C"-# groups
  • The first two are on carbon-2 and the third one is on carbon-3.

So far, we fill it in as:

#color(red)(color(white)(..........)"CH"_3color(white)(.)"CH"_3)#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3)#

Lastly, we proceed to the leftmost part of the name. The #5,5-"diethyl"# suggests that:

  • #"di" -># two duplicate branching groups
  • #"ethyl" -> "H"_3"CCH"_2-# groups
  • These are both on carbon-5.

So we get:

#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(red)(color(white)(..........)"CH"_3color(white)(.)"CH"_3color(white)(.....)"CH"_2)#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3color(white)(..............)color(red)("CH"_2))#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#

Lastly, since this is an alkane, the most saturated kind of hydrocarbon (relative to alkenes and alkynes), just fill in the rest of the branches with #"H"#.

#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(red)(color(white)(.........)"CH"_3color(white)(.)"CH"_3color(white)(.)"H"color(white)(...)"CH"_2color(white)(.)"H")#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#"H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH"_3#
#color(white)(...........)|color(white)(......)|color(white)(......)|color(white)(......)|color(white)(......)|#
#color(red)(color(white)(..........)"CH"_3color(white)(.)"H"color(white)(..)"H"color(white)(...)color(red)("CH"_2)color(white)(.)"H")#
#color(white)(..........)color(white)(..........)color(white)(...........)|#
#color(white)(..........)color(white)(..........)color(white)(..........)color(red)("CH"_3)#

And this in reality looks like:

Related questions
Impact of this question
2191 views around the world
You can reuse this answer
Creative Commons License