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xxMikexx [17]
2 years ago
15

D( )-Camphor is treated with 5-methoxytryptamine in the presence of sodium cyanoborohydride to generate a new molecule, as a sin

gle diastereomer, with molecular formula: C21H30N2O. Draw the starting materials, the imine intermediate, and the final product. (No mechanism is necessary). What is the rational for the reaction being diastereoselective (yielding a single diastereomer)
Chemistry
1 answer:
Svetach [21]2 years ago
4 0

The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).

The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.

The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.

To view more about rational reaction, refer to:

brainly.com/question/20308523

#SPJ4

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3 years ago
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Using the following equation for the combustion of octane, calculate the heat associated with the formation of 100.0 g of carbon
givi [52]

Answer: Heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

Explanation:

Reaction equation will be as follows.

    2C_{8}H_{18} + 25O_{2} \rightarrow 16CO_{2} + 18H_{2}O;  \Delta H^{o}_{rxn} = -11018 kJ

Mass of CO_{2} = 100 g

Hence, moles of CO_{2} present will be calculated as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{100 g}{44.0095 g/mol}

                             = 2.27 mol

Therefore, heat produced by 2.27 mol for the given reaction will be calculated as follows.

         2.27 mol \times \frac{11018 kJ}{16 mol CO_{2}}

             = 1563.2 kJ

Thus, we can conclude that heat associated with the formation of 100.0 g of carbon dioxide is 1563.2 kJ.

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