For the transformation of 1,3-dibromobenzene to bromobenzene boronic acid, different groups use different boration reagents namely trimethylborate, triethyborate and triisopropylborate).

The reaction is straightforward and involves doing a lithium-halogen exchange using nBuLi then quenching with the borate species. The addition of hydrochloric acid will then yield the boronic acid through the removal of the aliphatic chains.

Using triethylborate (reagent = 1,3-dibromobenzene), the yield is 85%. Using trimethylborate (reagent = 3,5-dibromotoluene) the yield is 65%. Using process chemistry for 1,3-dibromotoluene using triisopropylborate (same reaction with optimized parameters such as lithiation reaction time), the yield for 1,3-dibromotoluene is 91%.

Assuming the reactivity of dibromotoluene and dibromobenzene are equivalent, how exactly does the boration reagent affect the reaction?

Carbon is more electronegative than Boron and a methyl group is more electrodonating than hydrogen according to this response:

Why do larger alkyl groups show a greater inductive (+I) effect?

Hence the trimethylborate carbon (CH3) is more positively charged than the triethyl and triisopropylborate.

This higher positive charge will lead to a stronger inductive effect B->C (electron density will go from the boron to the carbon due to the difference in electronegativity).

Hence the boron will be "more positively charged/ more electron deficient" for trimethyl borate than for longer alkyl chains. Since the reaction simply involves the attack of the benzene anion onto the electron-deficient boron, one would expect a higher yield with a more electro deficient boron, which doesn't seem to be the case.

Theoretically speaking, none of this should really matter since the benzene anion is extremely reactive and that the difference in reactivity between the trimethyl borate and triethyl borate should be extremely small. Thus, yields would be expected to be equivalent for all borating reagents. Steric effects also do not seem to play a role.

Hence my question. How does one chose which borate to use for these kinds of conditions and why?


Kim, H. J., Kim, M. J., Park, H. D., Lee, J. H., Noh, S. T., Lee, Y. C., & Kim, J. J. (2010). Near-IR electromer emission from new ambipolar carbazole containing phosphorescent dendrimer based organic light emitting diode. Synthetic Metals, 160(17–18), 1994–1999. https://doi.org/10.1016/J.SYNTHMET.2010.07.024


Usutani, H., & Cork, D. G. (2018). Effective Utilization of Flow Chemistry: Use of Unstable Intermediates, Inhibition of Side Reactions, and Scale-Up for Boronic Acid Synthesis. Organic Process Research and Development, 22(6), 741–746. https://doi.org/10.1021/ACS.OPRD.8B00118/ASSET/IMAGES/LARGE/OP-2018-00118Q_0004.JPEG


Group 4 Post-Metallocenes Supported by [OCH2N,C(σ-aryl)] Auxiliaries Bearing a Seven-Membered Metallacycle: Synthesis, Characterization, and Catalysts for Olefin Polymerization Cham-Chuen Liu, Qian Liu, Shek-Man Yiu, and Michael C. W. Chan Organometallics 2019 38 (15), 2963-2971 DOI: 10.1021/acs.organomet.9b00307

  • $\begingroup$ A potential reason suggested to me by a fellow chemists is a s follows. Trimethylborate is more reactive (as explained in the question) and thus is more prone to hyrdolysis over time (degradation of starting material). Introduction of even slight amounts of boronic acid instead of pure trimethylborate would quench the anion (in this case the bromo benzene anion), forming the dehalogenated specie sideproduct thus inducing a loss in yield. $\endgroup$ Mar 28 at 2:11


Your Answer

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