How can I assign relative priorities to the groups or atoms in each of the following: $- C H_{2} O H$ $-CH_2OH$ , $- C H_{3}$ $-CH_3$ , $- H$ $-H$ and $- C H_{2} C H_{2} O H$ $-CH_2 CH_2OH$ ?

Question

8984 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-04T01:54:46

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

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

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

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

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

The order of decreasing priority is

${-CH}_{2} OH > {-CH}_{2} {CH}_{2} OH > {-CH}_{3} > -H$ $"-CH"_2"OH" > "–CH"_2"CH"_2"OH" > "-CH"_3> "-H"$

How can I assign relative priorities to the groups or atoms in each of the following: -CH_2OH−CH2OH-CH_2OH, -CH_3−CH3-CH_3, -H−H-H and -CH_2 CH_2OH−CH2CH2OH-CH_2 CH_2OH?