Based on the way the molecular formula is written — $\ce{Br(CH2)4Br}$ — I'm assuming this means that there are $4\ \ce{-CH2}$ groups attached to $\ce{Br}$. Am I correct? I've always thought that $\ce{Br}$ can't accept so many groups though.

  • 2
    $\begingroup$ Organic chemistry is starkly different from inorganic, even in the ways of writing the molecular formula. So in fact this is just Br-CH2-CH2-CH2-CH2-Br. $\endgroup$ Sep 6 '17 at 7:58
  • $\begingroup$ Really? I always thought adding the brackets signified another chain so as to differentiate between isomers. $\endgroup$ Sep 6 '17 at 9:10
  • 2
    $\begingroup$ Brackets may serve multiple purposes. $\endgroup$ Sep 6 '17 at 9:31

When you see a formula like the one in the question in organic chemistry, the person writing it down either already assumes a lot of chemical knowledge on the reader’s side or wants to teach them how to interpret this. So to a practising organic chemist, one inline formula may mean one thing while a similar one (with brackets in the same positions etc) may mean something completely different. Sometimes even, inline formulae are (and must be, due to their nature) potentially ambiguous. Therefore, organic chemistry relies a lot on actually drawing out the structures, whether it be each atom explicitly or in the sum formula formalism.

In the case of $\ce{Br(CH2)4Br}$, organic chemists will know that bromine will typically only form one single bond. Additionally, each $\ce{CH2}$ unit is known to be able to form two bonds — either two single bonds or one double bond. Therefore, an organic chemist will automatically expand the formula to $\ce{Br-CH2-CH2-CH2-CH2-Br}$ or 1,4-dibromobutane.

However, you might also sooner or later encounter inline formulae such as $\ce{BrC(CH3)2CH2Br}$ or $\ce{(CH3)3COH}$. Here, the brackets do actually mean repeated addition of said group to a single common atom. so the first of these two formulae describes 1,2-dibromo-2-methylpropane and could also be written $\ce{H3C-C(CH3)(Br)-CH2Br}$ while the second one represents tert-butanol or 2-methylpropan-2-ol.

Immediately realising which type of brackets is used requires practice and experience.


Although denoting linear repeating unit by parentheses (round brackets) in organic compound formulae is very very common, it's not the correct, or recommended way.

From Nomenclature of Organic Chemistry, IUPAC Recommendations and Preferred Names 2013, P-16.5 Name writing: Enclosing Marks:

P- Parentheses are used to enclose groups attached to a chain in linear formulas.


$\ce{\overset{3}{C}H3-\overset{2}{C}H(SH)-\overset{1}{C}H3}$   propane-2-thiol (PIN)

$\ce{\overset{4}{C}H3-\overset{3}{C}H(OH)-\overset{2}{C}O-\overset{1}{C}H3}$   3-hydroxybutan-2-one (PIN)

$\ce{\overset{5}{C}H3-\overset{4-3}{[CH2]}_2-\overset{2}{C}(CH3)2-\overset{1}{C}H3}$   2,2-dimethylpentane (PIN)

P- Brackets are employed in formulae to indicate repetition of a group in a chain.

Example: $\ce{CH3-[CH2]68-CH3}$ heptacontane (PIN, P-21.2.1)

Or, Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), GR-2.3 Atom labels and other chemically significant text: Atom labels representing more than one non-hydrogen atom:

GR-2.3.4 Interpretation of contracted labels

Divalent structural fragments may be enclosed in brackets and followed by a repeat count to represent repeating fragments concisely.

structures diagram


So the correct recommended short representation of, obviously 1,4-dibromobutane, should be $\style{font-weight:bold}{\ce{Br[CH2]4Br}}$ (or $\ce{Br-[CH2]4-Br}$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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