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What is the IUPAC name for the compound $\ce{(CH3)3CCH2CH(Br)C6H5}$?

The answer I got myself was (1-Bromo-3,3-dimethylbutyl)benzene, but from looking online most sources list the name as 1-Bromo-3,3-dimethyl-1-phenylbutane, even though the benzene group is supposed to take precedence, right? Why does butane form the suffix and not benzene?

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    $\begingroup$ Please note that names of chemicals (the 'bromo' in your suggested names) should not be capitalised unless occurring at the beginning of a sentence. $\endgroup$
    – orthocresol
    Jun 24 at 11:54
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The compound whose name is to be determined is the following.

enter image description here

As you can see, we have two options for the parent hydride.

  1. The butane chain which has 4 substituents - two methyl groups, one bromine atom, and one phenyl group.
  2. The benzene ring which has one substituent.

Now, from Section P-52.2.8 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, IUPAC Blue book, we have,

P-52.2.8 Selection between a ring and a chain as parent hydride

Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, we need to choose option for the preferred IUPAC name, therefore we now need to find the one substituent attached to the benzene ring.

enter image description here

In order to find the right numbering of locants, we use Section P-14.

Numbering

When several structural features appear in cyclic and acylic compounds, low locants are assigned to them in the following decreasing order of seniority.

(c) principal characteristic groups and free valences (suffixes);

(…)

(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

(g) lowest locants for the substituent cited first as a prefix in the name;

(…)

Note that Rule f takes precedence over Rule g.

Furthermore,

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)

Therefore in this case, we have two possible locant sets:

  1. 4-bromo-2,2-dimethylbut-4-yl [2,2,4]
  2. 1-bromo-3,3-dimethylbut-1-yl [1,1,3]

We can remove the '-1-' in option 2 using section P-29.2, which states:

P-29.2 GENERAL METHODOLOGY FOR NAMING SUBSTITUENT GROUPS

[...]
(1) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

By section P-14.4 (c), option 2 is preferred, therefore the PIN of the compound becomes (1-bromo-3,3-dimethylbutyl)benzene

enter image description here

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  • $\begingroup$ OK. One last thing, and sorry that I'm drip-feeding like this, but I notice it as we go along. The choice between butan-1-yl and butan-4-yl (note 'butan' not 'but', cf. P-29.2 again) is made not by locant set but rather by P-14.4(c), which comes before (f) and (g) and states that free valences (i.e. -yl) have priority for lower numbers. Again, I should probably have collated all these; sorry. $\endgroup$
    – orthocresol
    Jun 24 at 13:25
  • $\begingroup$ Hmm, in fact, I wonder if P-14.4 even comes into this at all; because by virtue of P-29.2/P-57.1.1.1 we already know the preferred prefix is going to be something-butyl. I get the feeling that P-14.4 is for deciding between butan-2-yl and butan-3-yl, for example. Ah, it does get pretty complicated sometimes..... $\endgroup$
    – orthocresol
    Jun 24 at 13:28
  • $\begingroup$ I would digress here, since I would say it's because of P-14.4(c) (now noticing) that the selection takes place and not via the set of locants rule. $\endgroup$ Jun 24 at 13:31
  • $\begingroup$ @Safdar Faisal Thanks for answering my question! By the way, where did you get all those diagrams you used in your answer from? $\endgroup$
    – Shud
    Jun 24 at 15:17
  • $\begingroup$ I used ChemdrawJS. Another alternative is Chemdoodle @Shud. Consider hitting the green check mark if the question is answered. $\endgroup$ Jun 24 at 15:59

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