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17 Nov 2019

Hi! I need the following:

1. Write the mechanism of the reaction of reduction of cyclohexanone with NaBH4 and CH3OH.

2. Why in the reduction key step, does cyclohexanone take hydrogen from sodium borohydride and not from methanol?

3. What do you expect to observe in the 1 H NMR spectrum of cyclohexanol compared to that of cyclohexanone? Discuss the differences in relation to the number of signals and the number of protons responsible for them.

Thanks a lot!

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Trinidad Tremblay
Trinidad TremblayLv2
17 Nov 2019

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Related questions

16.23

What is the purpose of this lab??

Introduction

Metal hydrides (sources of H-) such as LiAlH4 (LAH) and NaBH4, are widely used in reducing carbonyl groups such as aldehydes and ketones. The somewhat reduced reactivity of NaBH4 allows it to be used in both alcoholic and aqueous media in contrast to LAH. For each of these reagents, four hydrogen atoms are available for reaction, and thus one mole of hydride reagent can reduce four moles of ketone.

The stereochemistry of hydride reductions is very interesting. It appears that the alcohol derived from the reaction below appears to result from hydride attack on the carbonyl group on the least hindered side of the molecule. Such effects are called steric approach control (control of stereoselectivity of a reaction by steric hindrance towards attack of the reagent, which is directed to the less hindered face of the molecule). However, in the reduction of simple ketones, the alcohol obtained appears to be the thermodynamically more stable product. Such product formation is called product development control (this term is used of reaction under kinetic control where the selectivity parallels the relative thermodynamic stabilities of the products). In this experiment, approach of the hydride from the bottom (endo) leading to isoboreol appears to be less hindered than that from the top (exo) which leads to the isomer, borneol. In this experiment, the technique of NMR spectroscopy will be used to ascertain the composition of the reduction product in the reaction mixture. Before coming to lab, review the meaning of the terms "exo" and "endo" in your organic lecture text.

Procedure

In a 25-mL Erlenmeyer flask, dissolve 0.40g of racemic camphor in 2 mL of cold methanol. Cautiously add 0.30 g of sodium borohydride, in portions, to the solution. If necessary, cool the flask in an ice bath to keep the reaction mixture at room temperature. When all the borohydride is added, in the hood, heat the methanolic solution to boiling for 2 minutes.

Pour the hot reaction mixture into about 15 g of chipped ice, using small portions of methanol to aid the transfer (if you add to much methanol, the product will not reprecipitate). When the ice melts, collect the white solid by vacuum filtration (rinse with cold water), transfer the solid to a small flask and in the hood add about 20 mL of ether to dissolve the product. Add a spatula full of anhydrous magnesium sulfate to dry the solution, gravity filter to remove the drying agent, and evaporate the solvent from a Petri dish in a hood.

Purify the dry isoboreol by sublimation. The product in a Petri dish is placed on a hot plate with the cover on, and a beaker of ice and water is placed on the cover. Heat the dish on a low setting until the sublimation is complete. If water condenses inside the lid during the sublimation, remove the Petri dish from the heat and wipe away the water. Weigh the purified material and calculate the theoretical and percent yield.

Determine the melting point (sealed tube); pure racemic isoborneol melts at 212°C.

The product will be analyzed by H-NMR in CDCl3. Analyze the spectrum by comparing to the spectra of the pure alcohols obtained from a spectral database. Use the integration to determine the ratio of isoborneol to borneol. Based upon the results obtained, what conclusions can one make concerning the sterochemistry of the reduction?

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