Question: Why do we insert a numeric multiplier after group positions when naming an isomer?

Take 2,3 dimethylbutane for example. Why do we add the numerical multiplier di when we've already explicitly mentioned two locations? Why don't we just use 2,3 methylbutane? This just doesn't feel intuitive.


2 Answers 2


take 2,3-dimethylbutane

In the case of different substituents on different or the same atom, we use a combination of the position and the name of the substituent, such as in

  • 2-chloro-3-methoxy...
  • 3-cyano-4-bromo..., etc.

Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier di, together with the positions, just shortens and simplifies the name and still is unambiguous.

The multiplier di explicitly denotes that there are two identical substituents at the given positions and not just one that bridges two centres.

enter image description here

An example for such a situation would be 1,4-methanonaphthalene (1). Note that even in this context, di can be used, as in 1,4:5,8-dimethanonaphthalene (2).

  • $\begingroup$ Why not just 1,2-methoxyethane? $\endgroup$ Apr 16, 2015 at 14:08
  • 1
    $\begingroup$ @VatsalManot Because that does not explicitly state that there are two substituents and not one, that is bridging two centres, such as in 1,4-methanonaphthalene. $\endgroup$ Apr 16, 2015 at 14:14

Take 2,3 dimethylbutane for example. Why does one add the numerical multiplier di when one has already explicitly mentioned two locations? Why doesn't one just use 2,3 methylbutane?

Actually, it is the other way round. Multiplicative prefixes (di, tri, tetra, etc.) are always used in names to denote multiplicity of identical features in structures, whereas locants (1, 2, 3, etc.) may be omitted under certain conditions when there is no ambiguity, e.g.
‘trichloromethane’ (instead of ‘1,1,1-trichloromethane’) or
‘heptafluorobutanoic acid’ (instead of ‘2,2,3,3,4,4,4-heptafluorobutanoic acid’).

  • $\begingroup$ Why did you omit 2,2,3,3,4,4,4 ? I thought only 1 can be omitted. $\endgroup$
    – user31607
    Feb 8, 2017 at 8:50
  • 2
    $\begingroup$ @A---B because it is not ambiguous. Only one molecule with the name heptafluorobutanoic acid exists. In other words, those numbers are implicitly understood. $\endgroup$
    – Yashas
    Feb 8, 2017 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.