[sodium.dioxid]

In one instance, my textbook says:

(tert-butyl OH) + (HCl) ---> (tert-butyl Cl) + (H2O)

In another instance:

(tert-butyl Cl) + (H2O) ---> (tert-butyl OH) + (HCl)

One of these reactions should disqualify the other one from occurring. Otherwise, it would be a never-ending circular reaction.

[fledarmus]

It is a never-ending circular reaction; only the rates are different. It's called a reversible reaction, and depending on the conditions you can establish an equilibrium which contains either mostly t-butyl chloride or mostly t-butanol, or any mixture in between. An equilibrium doesn't mean that the compounds are finished reacting, it just means that the number of reactions going one direction is the same as the number of reactions going the other direction.

[sodium.dioxid]

If it is a reversible reaction, then how come the textbook does not use the reversible reaction notation (the one with the two arrows pointing in opposite directions)? Instead, only the one way arrow is used. Take a look at the attachment.

[Arctic-Nation]

The reversibility of many reactions is strongly dependent on the reaction conditions. Halogenation of a tertiary alcohol using HX is done using concentrated acid, and of the two nucleophiles present (water and halide), halide is the stronger one, and will react faster. Reaction with water will just regenerate the starting alcohol, which then can react again. The inverse reaction, the conversion of a tertiary alkyl halide into the alcohol, is done simply by mixing with water. It works because the concentration of halide is so small that the halide is virtually never captured again, and the reaction conditions do not allow for the easy creation of a carbocation from the alcohol.

Thus, the reaction is reversible, but by using the correct reaction conditions you can shift the equilibrium far to either the alcohol or the alkyl halide. In practical terms, your reversible reaction behaves like an irreversible one.

(Pedantry note: all chemical reactions are technically reversible, but that doesn't mean that a given reaction actually will reverse or equilibrate.)

[sodium.dioxid]

Arctic-Nation,

so what you are saying is that if I had equal amounts of reactants, then the reaction would equilibrate with no substantial favoring of one over the other; however, having concentrated water will favor the alcohol and having concentrated hydrogen chloride will favor the alkyl halide. Correct?

[fledarmus]

Arctic-Nation,

so what you are saying is that if I had equal amounts of reactants, then the reaction would equilibrate with no substantial favoring of one over the other; however, having concentrated water will favor the alcohol and having concentrated hydrogen chloride will favor the alkyl halide. Correct?

Not quite - without some understanding of either the reaction rates of the forward and backward reactions or the equilibrium constant of the reaction you can't say "with no substantial favoring of one over the other"; the equilibrium constant tells you which one is favored.

The second half of your statement is true - that is Le Chatelier's principle. Whatever the equilibrium concentration is, if you increase the amount of HCl or remove the water as it is formed, you will drive the reaction in the direction of the chloride - if you add water or remove HCl you will drive the reaction in the direction of the alcohol.

[sodium.dioxid]

Arctic-Nation,

so what you are saying is that if I had equal amounts of reactants, then the reaction would equilibrate with no substantial favoring of one over the other; however, having concentrated water will favor the alcohol and having concentrated hydrogen chloride will favor the alkyl halide. Correct?

Not quite - without some understanding of either the reaction rates of the forward and backward reactions or the equilibrium constant of the reaction you can't say "with no substantial favoring of one over the other"; the equilibrium constant tells you which one is favored.

The second half of your statement is true - that is Le Chatelier's principle. Whatever the equilibrium concentration is, if you increase the amount of HCl or remove the water as it is formed, you will drive the reaction in the direction of the chloride - if you add water or remove HCl you will drive the reaction in the direction of the alcohol.

This makes sense.

However, I can't help but to think why this can't be done for every reaction. For example, in the reaction below, why is in not viable to work backwards (by adjustment of concentrations) using H3PO3 and an alkyl halide to form an alcohol. Does it have to do with ease of reversibility? If so, what determines the ease of reversibility?

[sodium.dioxid]

Then how come we don't do this for other reactions in organic chemistry?

[Mobius1988]

In the above case I believe that youre forming very strong P=O bonds which are much stronger than C-O bonds and therefore reversing the reaction is unfavourable.

C-O bond energy = 358 kj mol^-1
P=O bond energy = 544 kj mol^-1

This is a well known driving force for many reactions where a P=O bond is formed as a result and this principal also applies to reaction where N₂ is formed, since the triple bond between nitrogen atoms is the strongest bond there is.

For the reaction between the alcohol and the alkyl halide the bond energies are much closer together and so reversibility is a much bigger factor.

C-O = 358 kj mol^-1
C-Cl = 327 kj mol^-1

I believe that may answer your question though if anyone more experienced than me wishes to correct me please feel free.

[fledarmus]

Exactly. The larger the difference between stabilities of starting material and products, the higher the rate difference between forward and backward reactions will be, and the higher the equilibrium constant for the reaction will be, and the less effect any change in concentration will have on the progress of the reaction. Reversibility and irreversibility is a continuum, like metals and non-metals. Some reactions, the equilibrium constant is so close to 1 that the process is totally reversible and very small effects can drive it one way or the other; in other reactions, the equilibrium constant is so large or so small that the reaction is irreversible in one direction or the other regardless of what you do to the concentrations; and other reactions can fall anywhere along the line between the two.

[orgopete]

The larger the difference between stabilities of starting material and products, the higher the rate difference between forward and backward reactions will be, and the higher the equilibrium constant for the reaction will be, and the less effect any change in concentration will have on the progress of the reaction. …

I think the rate depends on the activation energy, low is fast, high is slow. The differences in stability will affect the equilibrium. The greater the difference, the larger the equilibrium constant.

[Mobius1988]

I think the rate depends on the activation energy, low is fast, high is slow. The differences in stability will affect the equilibrium. The greater the difference, the larger the equilibrium constant.

But surely if adding excess reagent can push the equilibrium towards the product this is a kinetic effect? Adding extra reagents doesnt change the thermodynamic stability of the reactants/products but it does increase the concentration and therefore increase the rate of reaction??

[orgopete]

Mobius, good point. I was referring to the effect the energy level has on reactions. If a reaction will succeed, then greater concentrations will increase the rate. That rate increase in independent of the activation energy. I was simply pointing out that even with a great differential in stability, a reaction may be slow or no reaction at all if the activation energy is high. For example, you can mix hydrogen and oxygen together and it does not react. The activation energy is too high at standard temperatures and pressures. If you give it a spark, it will react explosively. Its failure to react is not a function of hydrogen and oxygen being more stable than water.

For other reactions, they may be slow under different conditions. Heating them will accelerate the reaction. I think if you measure a reaction rate as a function of the time for it to reach its equilibrium, then increasing the concentrations can increase the number of molecules being converted per unit time, but will not shorten the time to reach equilibrium. If the activation energy could be lowered or the reaction is heated, this will increase the rate and reduce the time.