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Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
