When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain
this observation and draw a mechanism that shows how racemization occurs. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.
1 answer:
Answer:
See explanation below
Explanation:
In this case, we need to see which is the structure of this compound. Now, racemization occurs basically because we are in an aqueous basic medium, and the ketone can reacts again with water in the medium to form the starting reagent.
First, the base will take out the Alpha hydrogen from the ketone, then, the negative charge goes down and opens up the carbonile group, forming a double bond in there. Later, with the water of the medium, it reacts and substract a proton, and then, with keto enolic equilibrium, forms again the ketone, but this ketone is different from the start, it will be the R isomer which is not optically active.
See picture below for mechanism
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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.
Be sure your answer has the correct number of significant digits.
Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 3.16 L
n = number of moles = ?
R = gas constant =
T =temperature =
According to stoichiometry:
4 moles of chlorine produces = 1 mole of carbon tetrachloride
Thus 0.141 moles of methane produces = moles of carbon tetrachloride
volume of carbon tetrachloride =
Thus volume of carbon tetrachloride that would be produced is 0.788 L