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Patrick A.
Chemistry and mathematics tutor for twenty years
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Organic Chemistry
TutorMe
Question:

Heating a solution of 3-methylbutan-2-ol with a catalytic amount of acid produces 2-methylbut-2-ene. Please suggest a method that would yield 3-methylbut-1-ene from 3-methylbutan-2-ol.

Patrick A.
Answer:

What makes this question tricky is that you have to draw out the structures of the reactant and the two different products in order to visualize it easily. Take some time and draw out the structures of all three compounds. You should see that the production of 2-methylbut-2-ene results from the elimination of the hydroxyl with the beta-hydrogen on the adjacent tertiary carbon from 3-methylbutan-2-ol. 2-methylbut-2-ene is what we would call the thermodynamic elimination product or Zaitsev product. 2-methylbut-2-ene is the favored product under heated conditions with catalytic acid because its double bond is more substituted and more thermodynamically stable. Producing 3-methylbut-1-ene would require the formation of the less substituted double bond through elimination of the hydroxyl with the beta-hydrogen on the adjacent primary carbon from 3-methylbutan-2-ol. This product is actually known as the kinetic elimination product or Hoffman product. To produce the Hoffman product, one could convert the hydroxyl to a p-toluenesulfonate (p-tosylate) by reacting 3-methylbutan-2-ol with p-tosyl chloride to form 3-methylbutan-2-yl p-tosylate. The resulting tosylate is a good leaving group. Reacting 3-methylbutan-2-yl p-tosylate with a bulky base like potassium tert-butoxide will produce 3-methylbut-1-ene. Potassium tert-butoxide is a bulky enough base that the tertiary beta-hydrogen is sufficiently sterically hindered that it is inaccessible by the tert-butoxide ion. The primary beta-hydrogen is more easily removed in the elimination by tert-butoxide, thus producing our desired product.

Chemistry
TutorMe
Question:

A friend of mine once told me that cold beer pours better, giving less froth. I have definitely experimented with this myself and indeed the colder the beer, the better the pour. As we *should* know, frothing is caused by the rapid formation of carbon dioxide from the beer's carbonated solution. Considering the carbonation equilibrium and thermodynamics, propose a reasonable theory as to why colder beer "pours better."

Patrick A.
Answer:

The carbonation equilibrium in water is as such: H2CO3(aq) <----> H2O(l) + CO2(g) One can see that the forward direction is an endothermic bond-breaking process and the reverse direction is an exothermic bond-making process. A colder temperature should hinder the forward endothermic process due to less available energy from the surroundings, thus lowering the rate at which carbon dioxide is produced.

Calculus
TutorMe
Question:

Calculus finds application throughout the physical sciences. A neat application can be found in chemical kinetics. Experimentation can easily give us an expression of the rate at which a chemical entity reacts such as (1) dC/dt=kC, where k is a proportionality constant unique to the reaction and C is the concentration of the reacting entity. Please determine the expression for the concentration of the reacting entity as a function of time.

Patrick A.
Answer:

Algebraic rearrangement of equation (1) can give (2) dC/C=kdt. One can easily integrate both sides of equation (2) to give (3) lnC=kt+A. Now all that needs to be done is the algebraic isolation of C in equation (3) using the natural logarithm base, e. This yields the following equation: (4) C=e^(kt+A). Equation (4) can also be expressed as C=(e^A)(e^(kt)). Believe it or not, (e^A) is actually the initial concentration of the reacting entity and we can give it an arbitrary constant assignment like B. Therefore, the equation can simply stated as such: C=Be^(kt)

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