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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.

songgio123
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    Type of bond (Triple > Double > Single) has higher priority than total number of single bonds. – Sam202 Oct 21 '22 at 06:49
  • 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. – Martin - マーチン Oct 21 '22 at 07:18
  • In lieu of the IUPAC Blue Book, you might visit this link. – user55119 Oct 21 '22 at 17:40
  • @Sam202 Based on what I remember, that statement is not correct. A triple bond is treated as a a carbon plus two "phantom" carbons. If you compare this to an ethyl group, the triple bond will have higher priority because in the ethyl group, the connected carbon only has one carbon attachment. But against a t-butyl group, the t-butyl group should have higher priority than an alkyne. Cf. https://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules#Double_and_triple_bonds – Zhe Oct 22 '22 at 13:08
  • @Zhe Organic Chemistry by L.G. Wade puts isopropyl's CIP priority under vinyl group. Situation should be analogous between t-butyl and ethynyl. – Sam202 Oct 22 '22 at 18:00
  • @Sam202 I don't think that's right... – Zhe Oct 22 '22 at 19:08
  • @Zhe Well, it seems 2 books support that (Wade and OP's book). Maybe you could cite one that states otherwise? – Sam202 Oct 22 '22 at 19:13
  • Looks like I gave away my copy of Elial, but I found IUPAC blue book P-92.1.4.2. You need to expand the multiple bond, duplicating the atom on the end, but the duplicates each have phantoms attached with 0 priority. So, a t-Bu is a carbon with three carbons attached, while an ethynyl is also a carbon with three carbons attached. But in the next sphere out, the t-Bu has more real hydrogens where the ethynyl has only one real hydrogen and the rest are phantoms. – Zhe Oct 22 '22 at 20:44
  • Comparing ethynyl vs t-butyl in terms of shells we have, respectively: [C,C,C][C,C,H] vs [CCC][HHH]. Ethynyl should have higher priority. – Sam202 Oct 22 '22 at 21:06
  • I don't understand what you're saying there. Ethynyl has [C][CCC][H] and tBu has [C][CCC][HHHHHHHHH] – Zhe Oct 23 '22 at 14:42
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    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]. – Sam202 Oct 23 '22 at 15:10
  • @Sam202 You're totally right. I forgot about the phantoms on the other end... – Zhe Oct 23 '22 at 17:17
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    @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. – Sam202 Oct 23 '22 at 17:28
  • @Sam202 That is correct. I also erred on my original statement. – Zhe Oct 23 '22 at 18:02

1 Answers1

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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.

user55119
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