EBK ORGANIC CHEMISTRY
EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 4, Problem 4.70AP
Interpretation Introduction

Interpretation:

The alcohol that undergoes dehydration faster is to be stated. The speed of dehydration of the faster alcohol is to be sated with explanation.

Concept introduction:

A thermodynamic potential that is utilized in the calculation of the highest reversible function taking place in the thermodynamic system at the constant value pressure and temperature is known as Gibbs-free energy. It is denoted by ΔG°.

Expert Solution & Answer
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Answer to Problem 4.70AP

The rate of hydration of 2methyl2propene is 411 times faster as compared to the hydration rate of methylenecyclobutane.

Explanation of Solution

The given standard free energy of activation for hydration of 2 methylpropene to 2methyl2propanol is 91.3kJmol1 or 21.8 kcalmol1.

The given standard free energy ΔG° for hydration of 2 methylpropene is 5.56 kJmol1(1.33 kcalmol1).

The given rate of hydration of methylenecyclobutane to give an alcohol is 0.6 times the rate of hydration of 2 methylpropene.

The equilibrium constant for the hydration of methylenecyclobutane is about 250 times higher than 2 methylpropene.

The hydration of methylenecyclobutane is supposed to be reaction 1.

The hydration of 2 methylpropene is supposed to be reaction 2.

The expression for the relationship between the rate for the reaction 1, reaction and their standard free activation energies is given below.

log(rate1rate2)=ΔG°2++ΔG°1++2.3RT …(1)

Substitute the ratio of rate of reaction 1 and 2 as 0.6, ΔG°2++ as 91.3kJmol1 and 2.3RT as 5.7 kJmol1 in the above expression.

log(0.6)=91.3kJmol1ΔG°1++5.7 kJmol1ΔG°1++=91.3kJmol15.7 kJmol1×0.222=92.6kJmol1

The expression for the relationship between the rate for the reaction 1, reaction and their rate constants is given below.

log(rate1rate2)=log(K1K2) …(2)

Substitute equation (1) in equation (2) as follows.

log(K1K2)=ΔG2oΔG1o2.3RT

Substitute the ratio of rate constants of reaction 1 and 2 as 250, ΔG°2++ as 5.56kJmol1 and 2.3RT as 5.7 kJmol1 in the above expression.

log(250)=5.56kJmol1ΔG°1++5.7 kJmol1ΔG1o=5.56kJmol12.4×5.7 kJmol1=19.2kJmol1

The expression for ΔG°1++ (reverse) is given below.

ΔG°1++ (reverse)=ΔG°1++ (forward)+ΔG1o

Substitute the values for ΔG°1++ (forward) and ΔG1o in the above equation.

ΔG°1++ (reverse)=ΔG°1++ (forward)+ΔG1o=19.2kJmol1+92.6kJmol1=111.8kJmol1

The expression for ΔG°2++ (reverse) is given below.

ΔG°2++ (reverse)=ΔG°2++ (forward)+ΔG2o

Substitute the values for ΔG°2++ (forward) and ΔG2o in the above equation.

ΔG°2++ (reverse)=ΔG°2++ (forward)+ΔG1o=91.3mol1+5.56kJmol1=96.9kJmol1

Substitute the values of ΔG°1++ (reverse) and ΔG°2++ (reverse) in equation (1).

log((rate1(reverse)rate2(reverse)))=ΔG°2++ (reverse)ΔG°1++ (reverse)2.3RT=111.896.95.7=14.95.7=2.614

The further simplification of the above equation is given below.

log(rate1(reverse)rate2(reverse))=2.614(rate1(reverse)rate2(reverse))=antilog(2.614)=411

Therefore, the above equation suggest that the hydration rate of 2methyl2propene is about 411 times faster as compared to the hydration rate of methylenecyclobutane.

Conclusion

The hydration rate of 2methyl2propene is about 411 times faster as compared to the hydration rate of methylenecyclobutane.

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