Question

What is the limiting reagent for meso-hydrobenzoin? (Benzil + NaBH4 ------> meso-hydrobenzoin)

Benzoin= 1.01g MW= 212.24 g/mol

Sodium Borohydride= .11g MW= 46.07g/mol

Recrystalized meso-hydrobenzoin= .91 g MW= 214.26g/mol

(When in class my teacher said something about there being 4 hydrogens on Sodium Borohydride but only one attacks the carbon so you will either have to divide or multiply by four? I'm not really sure what he meant by this)

Answer 1

The balanced reaction is

Benzil + 2H-   ---> Meso -Hydrobenzoin

Here 1 NaBH4 gives 4H- , hence 2H- is given by ( 2/4) NaBH4 ie 1/2 NaBH4

Hence balanced eq becomes

Benzil + 1/2 NaBH4 ---> Meso-hydrobenzoin

Benzoin moles = mass/ Molar mass of benzoin

       = 1.01 / 212.24

     = 0.00476

NaBH4 moles = 0.11 /46.07   = 0.002388

As per reaction Benzoin moles = 2 times NaBH4 moles

Hence for 0.002388 moles NaBH4 we need benzoin moles = 2 x 0.002388 = 0.004776

but we had only 0.00476   moles .

Hence benzoin is relatively less , hence it is Limiting reagent