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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67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
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calculation:
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67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO