I was told by my teacher that $\ce{RMgX}$ forms $\ce{PbR4}$ when it reacts with either $\ce{PbCl2/PbCl4}$ the later case is clear to me but why does $\ce{PbCl2}$ forms a compound of the form $\ce{PbR4}$.

  • 2
    $\begingroup$ What the teacher said when you had asked him/her? If one says something it need not to be true, even if it is said by a teacher. $\endgroup$
    – Poutnik
    Jun 11, 2023 at 15:27
  • $\begingroup$ PbCl4 "is best stored under pure sulfuric acid at -80 °C in the dark." (Wikipedia) $\endgroup$
    – Poutnik
    Jun 11, 2023 at 15:34
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    $\begingroup$ Note with very bulky side groups Pb(II) compounds can be isolated, e.g. pubs.acs.org/doi/10.1021/om990614g pubs.acs.org/doi/10.1021/om9708337 $\endgroup$
    – Ian Bush
    Jun 12, 2023 at 11:15
  • $\begingroup$ @IanBush "The compound is in effect the first dialkylead(II) species to be structurally characterized". Cool! It took way longer than it should have been given the first organolead compound was synthesized in 1858 ;) $\endgroup$ Jun 13, 2023 at 3:33

2 Answers 2


More is involved here than the simple nucleophilic exchange normally associated with Grignard reagents. Essentially, the proposed dialkyl lead disproportionates with displacement of metallic lead. According to bionity com, the main reactions with both organolithium and Grignard reagents bring about this disproportionation:

Lead(II) chloride is the main precursor for organometallic derivatives of lead. The usual alkylating agents are employed, including Grignard reagents and organolithium compounds[1]:

$\ce{2 PbCl2 + 4 RLi -> R4Pb + 4 LiCl + Pb}$

$\ce{2 PbCl2 + 4 RMgBr -> R4Pb + Pb + 4 MgBrCl}$

$\ce{3 PbCl2 + 6 RMgBr → R3Pb-PbR3 + Pb + 6 MgBrCl}$

Notice how these reactions produce derivatives that are more similar to organosilicon compounds, i.e. that Pb(II) tends to disproportionate upon alkylation.

The "inert pair" effect in heavy $p$-block elements becomes less prominent when the heavy atom is attached to less electronegative ligands that favor covalent instead of ionic bonding (in this case, by exchanging chlorine for alkyl groups), so the higher oxidation states ($+4$ in tetraalkyl lead $+3$ in hexaalkyl dilead) become more favorable in such situations.

Cited Reference

  1. Housecrift, Catherine and Sharpe, A.G.. Inorganic Chemistry (2nd edition). ‎Pearson College Div (January 1, 2005), p. 524.

Note that the dialkyl/diaryl lead(II) derivatives are still an intermediate products before getting disproportionated.

Simple dialkyl- and diaryl-lead(II) derivatives cannot be isolated because of their facile disproportionation to elemental lead and hexaorganodilead compounds. Nevertheless, they are presumably the initial products of the reaction of lead(II) chloride with organolithium or Grignard reagents. A deep brown–black reaction mixture which exhibits reactions attributable to the presence of diphenyllead can be obtained from the reaction of lead(II) chloride with exactly two moles of phenylmagnesium bromide in ether at −20 °C (the reported isolation of diphenyllead as a red solid by Krause and Reissaus is most probably in error). Thus addition of water to the freshly prepared "diphenyllead" solution produces lead(II) oxide and benzene, whilst iodine affords lead(II) iodide and iodobenzene together with a small amount of diphenyllead diiodide. Reaction with phenyllithium gives triphenyllead lithium. Disproportionation occurs on allowing the solution to warm, and hexaphenyldilead and lead metal result.

Source: P.G. Harrison, Comprehensive Organometallic Chemistry, 1982 (Sciencedirect Link)


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