Is it more reasonable to use stratification to separate similar compounds rather than using temperature?

Question

In a simple reaction between butane and chlorine, it is more likely for the secondary carbons of the butane to lose a hydrogen and gain a chlorine. The carbocation is a more stable intermediate in that situation compared to a primary carbon losing the hydrogen. So if the desired synthesis is 1-chlorobutane as opposed to 2-chlorobutane then it seems reasonable that a chemist might use stratification. The reason that I believe it is less desirable to use temperature as a method of separation is because that, at least in this example, both compounds will move between the gas and liquid phase randomly providing the possibility of still having mixed samples. Is this true, and will stratification work more reliably if purification is the desired result?

Furthermore, I never really learned why mono-chlorinated products form. Will there not also be compounds that have been chlorinated twice? Or will, over time due to entropy, all compounds be mono-chlorinated assuming that the reagents are in a ratio of 1 chlorine atom per molecule of butane?

Answer 1

The answer to your second question:

The halogenation of saturated hydrocarbons occurs by free radical substitution in the presence of UV light or heat and not by formation of carbocations. There's a very clear representation of the reaction mechanism on this Wikipedia page.

Since free radical intermediates are involved, this is a chain reaction and so what you said is correct, a large number of substituted products can be formed depending on the reaction conditions.

For example: If you consider the simple reaction between Methane and Chlorine in the presence of sunlight, Chloromethane, Dichloromethane, Chloroform, Carbon tetrachloride and a small amount of Ethane (by combination of Methyl radicals) are formed depending on the amount of chlorine available.

All these products will be formed even if the reactants are in a 1:1 ratio but they will be present in low concentration since there is less chlorine available.

In the case of higher alkanes like Butane and Pentane, the secondary carbons are substituted faster since the secondary free radical intermediates are more stable (reasons are the same as that for secondary carbocations).

Answer 2

Density stratification will only work if the substances are immiscible (mutually insoluble). For example, water and hexane could be separated this way because they are immiscible. Water will settle to the bottom (density = 1.0 g/mL) and hexane will float on top (0.65 g/mL).

Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) cannot be separated this way because they are miscible with each other. They form a homogenous solution when mixed. One of the properties of homogeneous solutions is that they do not separate into their components without a transfer of heat to or from the solution. You do not see bottles of soda separating into their components on the shelf because they are homogenous mixtures. You do see bottles of salad dressing separate because they are not homogeneous mixtures.

While the various monochlorinated, dichlorinated, trichlorinated and so forth derivatives of butane have different densities, they are likely all miscible in each other, which means the homogeneous mixture will not separate by density.

If you are looking for an alternative to the fractional distillation, you could try fractional freezing, in which you cool down the mixture, and the substances crystallize out according to the order of their freezing points. To achieve high purity, you would need to repeat the procedure several times for each substance. A good fractional distillation apparatus already has the repeated distillations built in, so it would probably be the fastest technique.