D( )-Camphor is treated with 5-methoxytryptamine in the presence of sodium cyanoborohydride to generate a new molecule, as a sin
1 answer:
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.
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The overall reaction is given by:
The fast step reaction is given as:
The slow step reaction is given as:
(slow step
)
Now, the expression for the rate of reaction of fast reaction is:
The expression for the rate of reaction of slow reaction is:
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as takes place in this reaction.
The expression of rate of formation is:
= (1)
Now, consider that the fast step is always is in equilibrium. Therefore,
Substitute the value of in equation (1), we get:
=
=
Thus, rate law of formation of in terms of reactants is given by
.
Answer:- The natural abundance of is 0.478 or 47.8% and
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for and
as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of as n then the abundance of
would be 1-n .
Let's plug in the values in the formula:
151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:
negative sign is on both sides so it is canceled:
The abundance of is 0.478 which is 47.8%.
The abundance of is =
= 0.522 which is 52.2%
Hence, the natural abundance of is 0.478 or 47.8% and
is 0.522 or 52.2% .