From this list, select the element that forms π bonds most readily. from this list, select the element that forms bonds most rea
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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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Answer : Half life and radioactive decay are inversely proportional to each other.
Explanation :
The mathematic relationship between the half-life and radioactive decay :
................(1)
where,
N = number of radioactive atoms at time, t
= number of radioactive atoms at the beginning when time is zero
e = Euler's constant = 2.17828
t = time
= decay rate
when then the number of radioactive decay become half of the initial decay atom i.e
.
Now substituting these conditions in above equation (1), we get
By rearranging the terms, we get
Now taking natural log on both side,
By rearranging the terms, we get
This is the relationship between the half-life and radioactive decay.
Hence, from this we conclude that the Half life and radioactive decay are inversely proportional to each other. That means faster the decay, shorter the half-life.