Dear Chandan Kumar Tiwari

Many thanks for posting an interesting technical question on RG. LiAlH4 is a strong reducing agent which can reduce polar multiple bonds. Alkynes have non polar C-C triple bond and hence this should not be reduced with LiAlH4. However, your ligand has C-C triple bond which is further bonded to trimethylsilane in one side, and the other side is bonded to R2MCl2 unit which may result the C-C triple bond to be polarized. So it may also be reduced with LiAlH4. I have no practical experience working with LiAlH4. So do not regard my answer as an expert's one. You can carry out the reaction with special care. Good luck with your work.

Dear Chandan,

I think you should just give it a try. In this case a quick experiment will be better than any speculations. In general, I assume that the chloride ligands in your complex will be more reactive towards LiAlH4 than the alkyne, so that the hydrogenation reaction will take place. As an internal alkyne, bis(trimethylsilyl)acetylene should be more or less unreactive with LiAlH4.

For some potentially useful general information about the chemistry of alkynylsilanes please have a look at the following interesting review article:

Some Aspects of the Chemistry of Alkynylsilanes

Article Some Aspects of the Chemistry of Alkynylsilanes

This review is freely available as public full text on RG. Good luck with your work and best wishes!

Hello Chandan,

P.S. I should add that in the presence of catalysts the hydroalumination of internal alkynes with LiAlH4 is possible. For more information please search the internet for the term "hydroalumnation of internal alkynes". A typical publication about this topic is the following:

Hydroalumination of Alkenes and Alkynes with LiAlH4 Catalyzed by [C5(CH3)5]2TiCl2

Article Hydroalumination of Alkenes and Alkynes with LiAlH4 Catalyze...

This paper has also been posted as public full text on RG. The difference to your case is that the organotitanium complex is used only in catalytic amounts. With best wishes, Frank Edelmann

I do not think LiAlH4 attacks acetylene. Certainly it can convert M-Cl to M-H bond.

Also think about using NaBH4 or NaH. I hope you are referring to M4+ system like Ti or Zr? They are bit more reactive and form several other products via radical mechanism. Temperature is very crucial here. Alcoholic-KOH can give monohydride. Another important, mild and ideal reagent is Li[BEt3H].

Best wishes

Balakrishna

You will need a milder reducing agent than the LAH, trimethoxy derivative can also be tried.

I would suggest sodium bis(methoxyethoxy)aluminum hydride (Red-Al, Vitride) as an alternative. It is easier to handle than LAH and in many cases gives better results. The triple bond should be stable, but one never knows until the reaction is tried. No matter how the reaction is carried out, workup will be problematic.

Are the chlorines co-ordinated with the metal or counter ions? What is the oxidation state of the metal, 0 or +2? The given formula is a little ambiguous. If the chlorines are in the co-ordination sphere NaH or KH may work, but if they are counter ions I would think not. Interesting problem.

Dear Chandan, as a synthetic inorganic chemist my personal suggestion would be to just go ahead and try a reaction with lithium aluminum hydride. Use a clear solution of high-quality LiAlH4 in diethyl ether and see what comes out. In this case I think that a quick try is better than all speculation. Good luck with your experiment!