Question

Ru(PPh3)4H2 (0.5 mol%) NaBH4 (0.5 eq.) H2O (2 eq.) Toluene, 100 °C, 2.5 h 100% (71%)
please show me the mechanism of this reaction with clarification of side products in a clear picture?   


and what is the meaning of 100% (71%) under the product?
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Answer #1

The mechanism is as follows: Phy Phun mappha Phan Rupph Phap r vacancy Phype R uppha/nucleophilis attack 7 PP - Hinteraction

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