Draw the curved arrow mechanism to show the conversion of hex-1-ene and CH3CH2OH in acidic solution into 2-ethoxyhexane. Follow
the given instructions for each step. Draw all necessary atoms and charges; do not add any structures.
1 answer:
Answer:
The formation of ether from alkene and alcohol in the presence of an acid take place in four steps.
Step 1:
In first step the oxygen atom of alcohol abstracts the proton of acid and becomes a strong conjugate acid.
Step 2:
Double bond of alkene being electron rich and nucleophilic in nature adds the proton from alcohol across its double bond and forms a stable carbocation (i.e. in this case a 2° carbocation).
Step 3:
The oxygen atom of alcohol now attacks the electrophilic carbon (i.e. carbocation).
Step 4:
The conjugate base of acid catalyst formed in first step abstracts the proton from oxonium ion and allows the final product to form as shown in attached reaction mechanism.
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Answer:
2.5 % butane, 42.2 % pentane and 55.3 % hexane
Explanation:
Hello,
In this case, the mass balance for each substance is given by:
Whereas accounts for the fractions at the outlet distillate and
for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:
So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:
The total distillate flow:
And the total bottoms flow:
Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:
Then, the molar fraction of pentane and hexane:
Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.
NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.
Regards.