Question

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`?

Answer 1

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

They are, respectively, `"C, C, H"`, and `"C"`

The three `"C"` atoms are priority 1, 2, and 3. The lowly `"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 `"–CH"_2"OH"`, the bonds directly attached to `"C"` are `"(O,H,H)"`.
• In `"–CH"_3`, the bonds directly attached to `"C"` are `"(H,H,H)"`.
• In `"–CH"_2"CH"_2"OH"`, the bonds directly attached to `"C"` are `"(C,H,H)"`.

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

• `"-CH"_2"OH"` is priority 1, because `"O"` has the highest atomic number
• `"–CH"_2"CH"_2"OH"` is priority 2, because `"C"` has a lower atomic number than `"C"`
• `"-CH"_3` is priority 3, because `"H"` has the lowest atomic number

The order of decreasing priority is

`"-CH"_2"OH" > "–CH"_2"CH"_2"OH" > "-CH"_3> "-H"`