# 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?

May 16, 2016

The idea is that organic chemists like drawing compounds to be convenient, so they "abbreviate" the structures like so:

• $\text{C"-="C}$ becomes three parallel lines ($\equiv$), and each end of the lines has an implicit carbon.
• $\text{C"="C}$ becomes two parallel lines $=$), and each end of the lines has an implicit carbon.
• $\text{C"-"C}$ 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 ${\text{H}}_{n}$). That means they are either not drawn or they are written as ${\text{H}}_{n}$.
• 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 ($s {p}^{3}$ $\to$ three hydrogens per terminal carbon, $s {p}^{2}$ $\to$ two hydrogens per non-terminal carbon, $s p$ $\to$ 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:

• $\text{C}$: $8$

$3$ of these carbons are primary $s {p}^{3}$ and thus have three hydrogens
$2$ of these carbons are tertiary $s {p}^{3}$ and thus have one hydrogen
$1$ of these carbons is a secondary $s {p}^{3}$ and thus has two hydrogens
$1$ of these carbons are secondary $s {p}^{2}$ and thus have one hydrogen
$1$ of these carbons are tertiary $s {p}^{2}$ and thus have no hydrogens

• $\text{H}$: $14$

$9$ of these are from the ${\text{CH}}_{3}$'s (bottom-left, upper-left, upper-right)
$2$ of these are from the $\text{CH} {R}_{1} {R}_{2}$'s (bottom-left, upper-left)
$2$ of these are from the ${\text{CH}}_{2}$ (top)
$1$ of these are from the -"CH"=("C"R_1R_2) (bottom right)

So the molecular formula is ${\text{C"_8"H}}_{14}$.