Question

What product, including its stereochemistry, would you expect from the reaction between [(MnMe(CO)_5] and Pph_3 we consider the reaction between [MnMe(CO)_5] and PPh_3 we can see that it is unlikely to be the simple replacement of a carbonyl ligand by the phosphine as this reaction would require a strongly bound carbonyl ligand to dissociate. A more likely reaction would be the migration of the methyl group on to an adjacent CO ligand to give an acyl group, with the phosphine filling the vacated coordination site. This reaction has a low activation barrier, and the product would therefore be expected to be c/s-(Mn(MeCO) (pph_3)(CO)_4] Explain why [Pt(Et)(CI)(PEt_3)2] readily decomposes, whereas [Pt(Me)(CI)(PEt_3)_2] does not.

This is the question I do not have any information , Just Explain Why the first complex decomposes whereas the second one is does not

Answer 1

The ethyl group in [Pt(Et)(Cl)(PEt₃)₂] is prone to β - hydride elimination.
β-hydride elimination is a reaction in which an alkyl group bonded to a metal centre is converted into the corresponding metal-bonded hydride and an alkene.The alkyl must have hydrogens on the β-carbon.
for example:


In [Pt(Me)(Cl)(PEt₃)₂] , the methyl (Me : -CH₃) group doesn't has a β-carbon, while
In [Pt(Et)(Cl)(PEt₃)₂], the ethyl (Et : -C₂H₅) group there is a β-carbon present with a hydrogen attached to it, hence β - hydride elimination readily takes place.