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My previous Chemistry teacher while teaching IUPAC nomenclature said that the double bond is always given greater priority over the triple bond and the numbering of the carbon atom chain is done in such a way that the double bond gets the lowest possible number while my present Chemistry teacher says that the numbering of the carbon atoms is done in such a way that the multiple bond gets the lowest number whether it is the double bond or the triple bond (whichever multiple bond is nearest to the end when both the ends are considered, numbering is done from that end).

On surfing through the various sites on the Internet, i came across the lowest sum of locants rule.

I am confused and cannot understand how to number the carbon atoms for IUPAC nomenclature purposes when many multiple bonds are present in the molecule. What is the actual rule?

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The individual rules that have been mentioned by your teachers are correct; however, you have to pay attention to the priorities. For any given compound, there may be other relevant rules (e.g. concerning the principal characteristic group that is expressed as suffix, or the longest chain) that determine the parent chain and the numbering of locants first. Only if there is still a choice, the rules concerning multiple bonds are taken into account.

In this case, the principal chain is the chain with the greater number of multiple bonds. If there are more than one chain with the greatest number of multiple bonds, the principal chain is the one with the greater number of double bonds.

The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)

The senior ring, ring system, or principal chain:

(a) has the greater number of multiple bonds (P-44.4.1.1);

(b) has the greater number of double bonds (P-44.4.1.2);

(…)

Likewise, with regard to numbering of locants, low locants are given first to multiple bonds, all considered together as a set. If there is still a choice, low locants are given first to double bonds:

P-14.4 NUMBERING

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

(…)

(e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

(…)


The “lowest sum of locants rule” that you have mentioned does not exist in the IUPAC Recommendations and can lead to wrong results. The corresponding section in the current IUPAC Recommendations actually reads as follows:

P-14.3.5 Lowest set of locants

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; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

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The simple way which I use for solving such question is here.

(1). The numbering of the parent chain should always be done from that end which gives lowest sum for the multiple bonds. For example,

enter image description here

In this case if you will start numbering from LHS then it is $1+3=4$. And If you start from RHS then $2+4=6$. As lowest sum is obtained from LHS, hence name will be 3-penten-1-yne and NOT 2-penten-4-yne.

(2). If, however, there is a choice for numbering, i.e. numbering from both LHS & RHS gives same sum then, double bond is always given preference over the triple bond. For example:

enter image description here

In this case, numbering from both LHS & RHS gives same sum i.e. $1+4=5$. So here you should prefer numbering from that side which will give priority to double bond. Thus name will be 1-penten-4-yne and NOT 4-penten-1-yne.

This is a easy way for answering such questions , just you need to keep those above two points in your mind. I hope this will help you.

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