# What determines the higher priority in CIP rules, if both groups have the same atoms?

Which of the following groups has the highest priority in the Cahn-Ingold-Prelog sequence rules?

a) CH2CH3

b) CH=CH2

c) C≡CH

d) C(CH3)3

I narrowed it down to either C or D. The carbon in answer C is triple bonded to another carbon. The carbon on answer D is also bonded to three carbons. How do I decide between the two? My book gives C as the answer.

• Type of bond (Triple > Double > Single) has higher priority than total number of single bonds. Oct 21, 2022 at 6:49
• I don't think the type of bond part has been formalised. The rule are in the blue book to which I currently have no access. An answer should include this reference. Oct 21, 2022 at 7:18
• In lieu of the IUPAC Blue Book, you might visit this link. Oct 21, 2022 at 17:40
• Let X* represent a duplicate atom and X a real atom. In ethynyl, the first C is bonded to C*, C*, and C. The second C is bonded to C*, C*, and H. On the other hand, in t-butyl, the first C is bonded to C, C, and C. Then, whichever C you choose as the second layer will be bonded to H,H, and H. [CCH] beats [HHH]. Oct 23, 2022 at 15:10
• @Zhe just a final point for clarification. Your source describes two different types of imaginary atoms: "duplicate" and "phantom". While phantoms have 0 priority, duplicates keep the priority of the real atom they're representing. Distinction between the two is important if other sources name these imaginary atoms differently. Oct 23, 2022 at 17:28

In sphere 2 (red numbers) all of the carbons are equivalent. No priorities may be assigned. Progressing to sphere 3, one has $$\ce{C2}$${3,H,H}, $$\ce{C4}$${5,(4),(4)}, $$\ce{C6}$${7,8,9} and $$\ce{C10}$${11,(10),H}. With the atoms in braces ranked in descending order of priority and looking for a first difference, it is clear that $$\ce{C2}$${3,H,H} is the lowest priority atom with $$\ce{C10}$${11,(10),H} the next to lowest priority. To distinguish between $$\ce{C4}$$ and $$\ce{C6}$$, one must proceed to sphere 4. Here $$\ce{C4}$$ displays {(5),(5),H} which is higher in priority than that which is offered by $$\ce{C7}$$, $$\ce{C8}$$ or $$\ce{C9}$$--namely--{H,H,H}. Therefore, the priorities of the carbon atoms in sphere 2, and the groups of which they are comprised, have the ranking $$\ce{C4}$$>$$\ce{C6}$$>$$\ce{C10}$$>$$\ce{C2}$$. In other words, yne>t-but>ene>ane. The compound has the (S)-configuration.