How do you start from the line structure, then determine the hybridization in order to figure out the structure with explicit atoms, for 1,3,4-trimethyl-1-pentene? What is the molecular formula?
1 Answer
The idea is that organic chemists like drawing compounds to be convenient, so they "abbreviate" the structures like so:
#"C"-="C"# becomes three parallel lines (#-=# ), and each end of the lines has an implicit carbon.#"C"="C"# becomes two parallel lines#=# ), and each end of the lines has an implicit carbon.#"C"-"C"# becomes one line (#-# ), and each end of the lines has an implicit carbon.- All hydrogens on carbon atoms are implicitly there (unless the compound has only one or two carbons, in which case it would look nicer to write
#"H"_n# ). That means they are either not drawn or they are written as#"H"_n# . - All heteroatoms (non-carbon atoms) remain explicitly visible.
So, all you have to do is count atoms and tally up how many of each type you have. The challenge may come in:
- Converting from implicit to explicit sketches
- Identifying how many hydrogens are on a carbon based on its hybridization (
#sp^3# #-># three hydrogens per terminal carbon,#sp^2# #-># two hydrogens per non-terminal carbon,#sp# #-># one hydrogen per terminal carbon). - Working out the approximate bond angles
The structure in an explicit sketch looks like this:
Now, we'd count the atoms to be:
-
#"C"# :#8# #3# of these carbons are primary#sp^3# and thus have three hydrogens
#2# of these carbons are tertiary#sp^3# and thus have one hydrogen
#1# of these carbons is a secondary#sp^3# and thus has two hydrogens
#1# of these carbons are secondary#sp^2# and thus have one hydrogen
#1# of these carbons are tertiary#sp^2# and thus have no hydrogens -
#"H"# :#14# #9# of these are from the#"CH"_3# 's (bottom-left, upper-left, upper-right)
#2# of these are from the#"CH"R_1R_2# 's (bottom-left, upper-left)
#2# of these are from the#"CH"_2# (top)
#1# of these are from the#-"CH"=("C"R_1R_2)# (bottom right)
So the molecular formula is
