# Cahn-Ingold Prelog rules concerning tie breaking on second level

comparing the cahn-ingold prelog priority of the above groups, I should first compare the atoms attached to the first carbon. So we get:

(A): {O,C,H}

(B): {O,C,H}

(C): {O,C,H}

(D): {O,C,H}

therefore we have to consider the second level, and this is where I get confused: Do I compare it this way:

(A):{C,H,H,H}

(B):{O,H,H,H}

(C):{F,C,H,H}

(D):{C,H,H,H}

I did this by rearranging all the "second level atoms" in descending order of atomic numbers.

since A and D are still in a tie, I rearrange all the "third level atoms" in descending order of atomic numbers:

(A):{H,H,H}

(D):{F,H,H}

so the priority is C>B>D>A Is that correct?

• In your examples you have made the decision because you have{CHO}. Three different groups.. If you had{CCO) for example, you would have to go to a second level to break the C C tie. That is if I understand your question correctly. Take a look here: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… May 8 at 2:50

look, all i can say is that $$D > C > A$$
why ?
Lets compare A and C
$$OCH$$3 and $$H$$ are common to both. one has CH3 and the other has CH2F. of course, C > A.
compare A and D in the same way and find that D > A
compare D and C now.
there is clear proof that D > C because it has F in two places while C has H in those two exact positions.

thus it is seen that D > C > A

now see B and C
in B : after O we have CH3 and after C we have { H , H , F }
in C : after O we have H and after C we have { H , H , O }
looking at second layer after oxygens : B > C
looking at second layer after carbons : C > B.

thus even i am confused when it comes to B

but definitely D > C > A