Question

If a reduction reaction uses 0.50 g of camphor and has a (2.72x10^1) % yield of borneol and isoborneol, how many g's isoborneol does it produce? Refer to the previous question(s) for MW data. Consider borneol and isoborneol are formed in a 3/1 ratio.

Answer 1

Ans: Given data:

Camphor MW = 152.24 g/mol

Borneol & Isoborneol MW = 154.25 g/mol (isomers having the same mol. wt.)

Camphor wt. = 0.50 g

Product Yield = 2.72x10^1 = 27 %

Camphor --Reduction---- > Product (Borneol + Isoborneol mixture)

152.24 g camphor (one mole) after reduction produces 154.25 g (isomer mixture) of borneols

So, 0.50 g camphor will produce      = (154.25 g / 152.24) * 0.50 g

= 0.5066 g of reduction products (mix borneols)

Practical Yield = 27.2 % (of the theoretical yield)

So, the obtained weight of the reduction products is = 0.5066 * 27/100 = 0.1368 g

It is given that borneol and isoborneol are formed in a 3/1 ratio.

So, the above obtained mix product weight is to be proportioned in 3:1 (borneol : isoborneol) percentage

Borneol           = 3/(3+1) *100 = 75 %

Isoborneol      = 1/(3+1) *100 = 25 %

Component weight g = Total weight g * component %

Borneol           = 0.1368 * 75% = 0.1026 g

Iso-borneol    = 0.1368 * 25% = 0.0342 g

Final answer: The 0.50 g of camphor produces 0.0342 g of iso-borneol

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