# How can I assign relative priorities to the groups or atoms in each of the following: -CH_2OH, -CH_3, -H and -CH_2 CH_2OH?

Jun 4, 2015

Step 1: Look at the atoms that are directly attached to the bond.

They are, respectively, $\text{C, C, H}$, and $\text{C}$

The three $\text{C}$ atoms are priority 1, 2, and 3. The lowly $\text{H}$ is priority 4.

To decide between the three carbon atoms, we must go to

Step 2: Look at the atoms one bond further out.

We list the atoms in order of decreasing atomic number.

• In $\text{–CH"_2"OH}$, the bonds directly attached to $\text{C}$ are $\text{(O,H,H)}$.
• In ${\text{–CH}}_{3}$, the bonds directly attached to $\text{C}$ are $\text{(H,H,H)}$.
• In $\text{–CH"_2"CH"_2"OH}$, the bonds directly attached to $\text{C}$ are $\text{(C,H,H)}$.

Step 3: Look for the point of first difference in each list.

• $\text{-CH"_2"OH}$ is priority 1, because $\text{O}$ has the highest atomic number
• $\text{–CH"_2"CH"_2"OH}$ is priority 2, because $\text{C}$ has a lower atomic number than $\text{C}$
• ${\text{-CH}}_{3}$ is priority 3, because $\text{H}$ has the lowest atomic number

The order of decreasing priority is

$\text{-CH"_2"OH" > "–CH"_2"CH"_2"OH" > "-CH"_3> "-H}$