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I need to write balanced equations describing the following reactions:

  • one mole of $\ce{Al2Me6}$ with two moles of water
  • excess of $\ce{Al2Me6}$ with silicon dioxide
  • excess of $\ce{Al2Me6}$ with tin(IV) chloride

My answer so far:

  • a) $\ce{2H2O + Al2Me6 -> Al2Me4(H2O)2}$?
  • b) $\ce{SiO2 + Al2Me6 -> SiMe4 + Al2Me2O2}$?
  • c) $\ce{SnCl4 + Al2Me6 -> SnMe4 + Al2Me2Cl4}$?
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    $\begingroup$ Organoaluminium compounds are very reactive. If a substance is capable of attacking one of the methyl groups in trimethylaluminium, then the process likely doesn't stop until all carbon-aluminium bonds are broken. Can you figure out the answer now? $\endgroup$ Commented Apr 21, 2014 at 13:35
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    $\begingroup$ Balance your equations. Draw structures to see what is happening. Why is trimethylaluminum dimeric? For (a), what happens when you mix a Bronsted acid with a Bronsted base given a potent oxophile? $\endgroup$
    – Uncle Al
    Commented Apr 21, 2014 at 15:10
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    $\begingroup$ I'm voting to close this question as off-topic because it's an AMIRITE question unlikely to be of help to future visitors. $\endgroup$
    – M.A.R.
    Commented Apr 8, 2017 at 16:37

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For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water.

Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.

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