When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain
1 answer:
Answer:
See explanation and image attached
Explanation:
The first step in the reaction is the attack of the hydroxide ion from the base abstracting a proton from (S)-2-methylcyclopentanone.
The abstraction of a proton from water by the substrate yields the enol form in equilibrium with the keto form.
The product formed is racemic hence the optical activity of (S)-2-methylcyclopentanone is lost.
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In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol .
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is .
(b) Rate equation for first order reaction is as follows:
now, 75% of the compound is decomposed, if initial concentration is 100 then concentration at time t
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.