[jadelamlam]

When we perform fractional distillation of petroleum in the laboratory,

1.Why should we use a small flame to heat the rocksil(soaked with petroleum)
and avoid using a stationary flame for heating?(it is one of the safety precautions to be taken in this experiment but I don't understand)

2.When we heat the rocksil in the test tube,it will not burn,why?

diagram:http://upload.lsforum.net/users/public/s234141u144.jpg

[fledarmus]

You move the flame to keep form getting "hot spots" on the test tube you are heating. You want the entire base of the tube to be heated at the same rate. You can only get consistent results with your distillation if the entire mass of material in the tube is at the same temperature; otherwise the hot places in your tube will be boiling off high boiling materials before the colder spots have finished boiling off lower boiling materials. Also, spot-heating glassware can cause stresses that will fracture the glass and coat the lab, your classmates, and yourself with hot, sticky, and (since you are using an open flame) possibly burning tars.

As for using a smaller flame, you are looking for a slow rise in temperature in your pot so that the lower boiling materials will be gone before you start boiling off higher boiling materials. This is how you achieve a good separation by distillation. If you plunge your pot rapidly into a very high flame, you boil off everything volatile and get no separation.

[jadelamlam]

Yes,thanks a lot.
I still have a question.
At the start,the petroleum soaked in rocksil is heated very gently.The fraction with the lowest boiling point range distills out first.
As we know,lower alkanes are extremely flammable,so why it will not burn when we heat it?

[Arkcon]

To help you, help yourself, you should first ask your self -- Why do things burn?  What is burning anyway?  Maybe if you look up some definitions such as these, you will have an A-ha moment, and be able to answer this question for your self.

[jadelamlam]

To start a fire:
fuel,O2,high temperature is required.
Put it into this case,
fuel:lower alkanes(say,CH4)
O2:present
high temperature:yes
But in fact nothing burns 

[fledarmus]

There are a couple of different temperatures associated with materials to determine whether they will burn. One is the flash point, which is the temperature at which you have enough vapor in the air that a spark will ignite it. For the example that you gave, methane (CH₄), that temperature is -188°C. So yes, with oxygen, methane will burn if you have a spark.

The other temperature that you will be concerned with is the autoignition temperature. This is the temperature at which a substance will spontaneously combust in air even if you don't have a spark. For methane, this temperature is 537°C. If you get the inside of the tube that hot, methane will burn even if you don't have a spark inside the tube.

However, the odds are there isn't any methane in that tube, because the boiling point of methane is -164°C - any methane present will evaporate rapidly at room temperature, before you even start heating. The smallest simple alkane that you might have in your sample would be pentane, with a boiling point of 36°C. This compound has a fairly low autoignition temperature, however, at 260°C, and you should be very careful about heating it. Of course, with a flash point at -49°C, you do not want a spark anywhere near it. Branched chain alkanes have higher autoignition temperatures, while longer straight chain alkanes have lower autoignition temperatures. Fortunately, the boiling points of all of these compounds are much lower than their autoignition temperature, so if you are careful with how fast you heat the pot, the most hazardous compounds will be gone before the temperature in the pot reaches the autoignition temperature.

Bottom line - this is not a particularly safe experiment. You are working with compounds that can be easily ignited, and heating them using a flame. You are relying on your products being cool enough by the time they have come out of the condenser and the collector being far enough away from the pot that there will not be enough vapor drifting out of your collector and over your flame to blow up the whole experiment. You will need to be very careful not to allow any spark or flame inside your apparatus and to keep your collector very far from your flames. This would be far safer if you used an electric heat source, either a heating mantle, electric blower, or oil bath and avoid any source of spark or flame in the lab while you were doing it. Even then, you would need to be careful not to exceed the autoignition temperatures of any of the components of your mixture

[Arkcon]

To start a fire:
fuel,O2,high temperature is required.
Put it into this case,
fuel:lower alkanes(say,CH4)
O2:present
high temperature:yes
But in fact nothing burns 

In a corked test tube, with a thermometer plugging the hole, while actively distilling organic compounds out the only other hole, is there O2 left?  Look here: http://en.wikipedia.org/wiki/Destructive_distillation

[jadelamlam]

Thx,but why branched chain alkanes have higher autoignition temperatures than the straight chain alkanes 
Because there are more atoms?

[fledarmus]

No, there are not more atoms, you can compare compounds that have the same number of atoms. For example, n-octane (C₈H₁₈) has an autoignition temperature of 220°C, while isooctane (also C₈H₁₈) has an autoignition temperature of 396°C.

As for why they autoignition temperature increases with the amount of branching, that takes a pretty deep understanding of what chemical reactions are occurring at ignition. How much organic chemistry have you had? Doing a Google Scholar search on chemistry of autoignition will lead you to several studies and explanations of the process.

In short, though, burning is a free radical process. As you raise the temperature of a molecule, you increase the possibility that you will see some homolytic bond cleavage, producing two free radicals. Once free radicals form, there is a lot of chemistry that the free radical species can undergo. The simplest reaction is that two free radicals can combine to form a covalent bond, and if it is the same two free radicals you formed in the beginning, there is no net energy change. The amount of energy absorbed to break the bond is the same as the amount of energy released when the bond is reformed. But there are other things that can happen. In particular, if there is oxygen present, the free radicals can react with oxygen, forming C-O and H-O bonds which are much stronger than the C-C and C-H bonds that were broken to form the free radicals. This releases more energy, and the temperature increases, forming more radicals, and so on. Runaway reaction. Ignition.

The free radical formed in the initiation reaction described above may also rearrange to form a more stable free radical, giving off more energy. This rearrangement creates branched structures, and since it is an intramolecular process rather than an intermolecular process, it can happen more rapidly than the processes described above, and at a lower temperature. However, the more highly branched the chain is to begin with, the less available this pathway becomes. This means that for less branched alkanes, such as n-octane, a lot of energy can be released by rearrangement, increasing the temperature more than for more branched alkanes, such as iso-octane, which have almost no rearrangement pathways available. This rearrangement process adds energy to the the system, starting the increase in temperature and the runaway reaction process at a lower temperature than is available for more branched structures.

I hope some of that made at least a little bit of sense.