Question

There are three possible products from this experiment: the Markovnikov addition product (1methylcyclohexanol), and the two anti-Markovnikov addition products ((15,2R)-2-methyl- 1cyclohexanol and (1R 2R)-2-methyl-1-cyclohexanol). 1. BH "THE 2. H2O2. NaOH or 3 or 3 OH 1-methyl-1-cyclohexene 1. What is your predicted product? Why? 2. Calculate the theoretical yield. Show calculations including those for the limiting reactant. 0.5 mL of 1-methyl-1-cyclohexene 2.5 mL of 1 M BH, THF solution 0.7 mL of 3 M NaOH followed by 1 mL of 30% H,O. 3. Why is 1H NMR not useful for identifying the product from this reaction?

Answer 1

H3CH OH the addition is syn addition. Considering the given alkene to be 1M solution, no of millimales of alkene are 0.5. the number of millimdes of BH, are 2.5 and that of NaOH are 2.1 3 mole of alkene reacts with 1 mole of BH, SO BH3 as well as NaOH is in excess. So 0.5 mM of alkene requires 1.5mM of BH3 which is available to the extent of 2.5mm 30% H, O, means 30 gram H2O2 in 100 mL of water. So 1 mL of water contains 0.3 gram of H,O,. i.e 8.8 mm so the limiting reagent is alkene. Theoretical yield is 100% 3. 1H NMR cannot be used to distinguish between streoisomers as the products are streoisomers, NMR 1H is not useful.