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.

  • 1
    $\begingroup$ Type of bond (Triple > Double > Single) has higher priority than total number of single bonds. $\endgroup$
    – Sam202
    Oct 21, 2022 at 6:49
  • $\begingroup$ 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. $\endgroup$ Oct 21, 2022 at 7:18
  • $\begingroup$ In lieu of the IUPAC Blue Book, you might visit this link. $\endgroup$
    – user55119
    Oct 21, 2022 at 17:40
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    $\begingroup$ 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]. $\endgroup$
    – Sam202
    Oct 23, 2022 at 15:10
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    $\begingroup$ @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. $\endgroup$
    – Sam202
    Oct 23, 2022 at 17:28

1 Answer 1


As the old adage goes, "A Picture is Worth a Thousand Words." Perhaps a picture is required. I apologize in advance for any redundancy with explanations provided in the Comments.

There are four functional groups in question. I have arranged them as substituents about a single carbon atom with arbitrarily assigned stereochemistry and atom numbers. A digraph is the best way to assign priorities to the four substituents. Duplicate atoms (ghost, phantom, etc.) are shown in parentheses and they have the same atomic number as their non-duplicate (original) atom. However, duplicate atoms have three atoms of atomic number zero attached to them and they are not displayed in the digraph. Their designation is {0,0,0} and, by default, have the lowest possible ranking.

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.


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