tert-Butyl chloride

Tertiary butyl chloride, also known as 2-chloro-2-methylpropane (IUPAC) is a tertiary alkyl halide used industrially as a precursor to pharmaceuticals, agrochemicals and fragrances. It also finds some use in organic chemistry as a reagent for introducing the tert-butyl group into molecules. Under standard conditions, tert-butyl chloride is a clear colourless mobile liquid with a boiling point of 51 C. I don't really have any particular use for the compound, although I might try synthesizing a tert-butyl ether sometime in the future.

tert-Butyl chloride can be easily prepared in the lab from tert-butanol and hydrochloric acid. I've included a description of the reaction mechanism at the end of this post. A large excess of hydrochloric acid is required to shift the equilibrium in favor of the product. Here is what I did.

To a 500ml round-bottom flask, I added 50ml (0.5228 moles) of tert-butanol and 150ml (1.3549 moles) of 33% hydrochloric acid (159% excess). I loosely sealed the top of the flask with some cling wrap and a few rubber bands, then gently shook the flask for around 1 minute, which resulted in a cloudy and completely homogeneous mixture. After shaking, I allowed the mixture to stand and separate into two layers for 20 minutes. A clear upper layer separated almost immediately and gradually became larger. After the 20 minutes, about two thirds of the mixture were carefully poured into a 150ml beaker and a syringe was used to remove the upper tert-butyl chloride layer, which was added to a 250ml flask. Once no more tert-butyl chloride could be syringed off, the remaining liquid was discarded and the remaining one third of liquid in the round-bottom flask was added to the 150ml beaker and the process was repeated (using a syringe to transfer the upper tert-butyl chloride to the 250ml flask. The tert-butyl chloride was then thoroughly washed with 90ml of saturated sodium bicarbonate solution to remove all traces of hydrochloric acid. After removing and discarding the lower aqueous layer, I washed the tert-butyl chloride with two 50ml portions of water in succession. The crude product was then added to a 500ml round-bottom flask containing an excess of anhydrous calcium chloride.

After a bit of swirling, the tert-butyl chloride lost its cloudiness and became clear as the calcium chloride removed water. I set the flask up for simple distillation, using a 250ml flask as the receiver. I then began the distillation. Only one fraction passed over, which was the tert-butyl chloride at 48-49 C. After distillation, the product still looked a bit cloudy, so I added a bit of anhydrous calcium chloride, whereupon it became crystal clear. Finally, I transferred the product to a bottle for storage. I got 34ml (0.3126 moles) of tert-butyl chloride as a colourless liquid which works out to a respectable yield of 60%. The product had a density of 0.782g/cm³ which is very close to the established 0.851g/cm³ of pure tert-butyl chloride. This shows my product is relatively pure.

Brief description of the SN1 reaction: a hydrogen ion from the hydrochloric acid first attacks the hydroxyl group of a tert-butanol molecule, forming an intermediate compound containing a positively charged OH2 group. The OH2 group is not very stable and acts as a leaving group, breaking off the molecule as H2O and transferring its positive charge to the alpha carbon. This leaves a carbocation which is then attacked by the chloride anion left over by the hydrochloric acid, forming the final product, tert-butyl chloride.


HCl <==> H (+) + Cl (-)

(CH3)3COH + H (+) ==> (CH3)3CO(+)H2

(CH3)3CO(+)H2 ==> (CH3)3C (+) + H2O

(CH3)3C (+) + Cl (-) ==> (CH3)3CCl

Boron, revisited

In a previous post (link), I described how I prepared a very crude sample of the element boron. I decided to have another go at boron using a different method to see if I could get a nicer sample. In my post on making silicon, I used a side reaction of aluminium and sulphur to thermally sustain the reaction. This gave nice large beads of silicon and I hoped that it might do the same for boron. Although the melting point of boron is significantly higher (over 2000 C) this didn't seem to be a problem.

To begin, I added 8g of finely powdered boron trioxide and 10g of powdered sulphur to a steel can. I then added in 11.2g of 325 mesh aluminium powder and stirred the mixture thoroughly to intimately mix the chemicals. Next, I poured the mixture onto a brick and placed a small amount of potassium nitrate/magnesium flash powder on top, then inserted a strip of magnesium metal. The flash powder and magnesium ribbon serve to kick start the thermite reaction. Using a torch, I carefully ignited the flash powder which immediately set off the thermite. Large amounts of heat and flame were given off and near the end, I noticed the fire was tinted green, presumably due to the presence of boron. Once the thermite was finished and the residue had cooled sufficiently, I used a hammer to remove as much of the residue as I could from the brick. I then added the residue to a 500ml beaker containing 200ml of water. Rapidly after addition, lots of bubbling was observed, which continued for over an hour. This is aluminium sulphide formed in the helping reaction reacting with the water to form aluminium hydroxide and hydrogen sulphide gas. After about 2 hours, the rate of gas evolution had decreased significantly and I decided to move on to the next step. I swirled the beaker, waited a few seconds, then quickly decanted as much of the gray aluminium hydroxide suspension as I could without losing any of the other material on the bottom of the beaker.

I repeated this process until almost all the aluminium hydroxide had been removed. To the material remaining in the beaker, I slowly added 50ml of 33% hydrochloric acid. A very vigorous reaction was observed with some foaming and lots of bubbling. I allowed the mixture to stand for about 2 hours to make sure the hydrochloric acid had dissolved all that it could, then decanted off the acid. At this point, I got my first proper look at what seemed to be the boron. Most of it seemed to be present as a black powder mixed in with lots of coloured impurities and hopelessly unrecoverable. However, there were a few beads of boron which looked fairly easy to separate. I washed the boron with 20ml of ethanol and after decanting the ethanol, then carefully picked out as many pieces of the black metallic boron as I could.

After drying, I was left with 0.7g of what I presumed to be boron. It certainly resembles many of the pictures I've seen on the net. In conclusion, this method doesn't seem very viable to produce large amounts of the element, but it's certainly a good way to get a nice sample.

2 Al + B2O3 ==> 2 B + Al2O3

2 Al + 3 S ==> Al2S3

Al2S3 + 6 H2O ==> 2 Al(OH)3 + 3 H2S

Bromine, preparation and isolation

Bromine is a fascinating element that exists as a strongly fuming dark red, almost black liquid at room temperature. It finds extensive use as a reagent, particularly in organic chemistry and is used commercially in flame retardants and as a disinfectant for pools.

There are many ways to make bromine. After considering a few different methods, I decided the oxidation of bromide with chlorine (generated in situ from hydrochloric acid and TCCA) was the best. So I went ahead and tried it out.

Unfortunately, the handing of pure bromine poses some extreme safety hazards, being incredibly toxic, corrosive and volatile. It can easily kill you if you're not prepared, and that's not an exaggeration. This procedure should only be performed by a chemist with lots of experience, in a fume hood or outside with a respirator.

To a 500ml round-bottom flask, I added 36g (0.1549 moles) of trichloroisocyanuric acid (TCCA) and a solution of 95g (0.9233 moles) of sodium bromide in 200ml of water. With addition of the bromide, a  yellow-red colour was observed, likely due to a small amount of chlorine being given off by the TCCA and liberating some bromine. Anyway, I then carefully poured in 34ml (0.1549 moles) of 16% hydrochloric acid, swirled the flask, then quickly attached a stillhead greased with 98% sulphuric acid. The mixture rapidly became very red with addition of the acid. I then attached the rest of the distillation equipment (condenser, vacuum adapter and receiving flask), greasing all the joints with 98% sulphuric acid. A tube was attached to the vacuum adapter and led into a bubbler trap to contain escaping bromine fumes and the receiving flask was placed in an ice bath. I then commenced distillation on medium heat. After a minute or two, I noticed red-orange bromine fumes had filled the whole apparatus and the mixture in the distilling flask became intensely dark red. A few minutes later, bromine began distilling over at a rate of about 1 drop per second. As the distillation continued, more and more water came over with the bromine. I kept distilling, until no more red bromine fumes could be seen in the distilling flask, gradually increasing the heat to keep the distillate coming over.

Once the apparatus had cooled to room temperature, I removed the receiving flask which contained a two layer system of bromine water floating on bromine. I carefully removed the top water layer with a pipette and discarded it. I then washed the bromine thoroughly with 15ml of 98% sulphuric acid to remove any water that might still present . After leaving the layers to separate for a few seconds, I removed the lower bromine layer using a pipette and transferred it to an amber glass vial for storage. A small amount of 98% sulphuric acid was also added to the vial to form a layer above the bromine and help prevent evaporation. The lid was sealed thoroughly with a generous amount of PTFE tape.

I got about 42g (0.2628 moles) of anhydrous bromine, representing a 57% yield. To test the purity, I placed one drop of the product on top of an upside-down beaker and allowed to evaporate. No residue remained, indicating my bromine to be fairly pure.

I should probably mention that my method of storage is not suitable for keeping bromine indefinitely. For permanent storage, ampoules are definitely the best option.

Also, this procedure was by know means designed by myself, I was simply following a brilliant video by the youtube chemist, NileRed (link). It's worth noting that only 1/3 of the total amount of hydrochloric acid needed stoichiometrically is actually required. NileRed seemed to think that this might possibly be because the TCCA is able to function as a source of chlorine directly to some extent.

C3N3O3Cl3 + 3 HCl ==> 3 Cl2 + C3N3O3H3

Cl2 + 2 NaBr ==> 2 NaCl + Br2

Sodium silicate and the chemical garden

Sodium silicate, often referred to as waterglass or liquid glass, is not one compound, but rather a series of compounds with the general formula (Na2SiO2)nO. It is used in high temperature glues for furnaces, in water treatment and in the past, perserving eggs as well as many other things. Sodium silicate also finds use in famous chemical demonstrations known as chemical gardens. In these demonstrations, lumps of transition metal salts are added to sodium silicate solution. The corresponding transition metal silicates gradually begin to form in such a way that an incredible plant-like structure results.

I decided to make some sodium silicate and try the demonstration.

To begin, I added 3.5g (0.0875 moles) of sodium hydroxide to a 125ml beaker. I poured in 12ml of water and swirled the flask. The sodium hydroxide rapidly dissolved, giving a clear solution. I then began adding 5.25g (0.0874 moles) of powdered silica gel in small portions with good stirring, heating the beaker between additions to dissolve everything. After all the silica gel had been added, a small amount of white solid remained which did not dissolve with heating. The solution was filtered to remove this. The resulting slightly yellow liquid should be a fairly concentrated solution of sodium silicate.

The sodium silicate solution prepared above was transferred to a beaker and diluted with 90ml of water. I added a few medium sized pieces of chromium(iii) chloride hexahydrate to the beaker and left the mixture to stand for 12 hours. After the first few hours, I noticed the start of the garden. The chromium(iii) chloride crystals expanded into a plant-like structure of green-black chromium(iii) silicates and a few columns of  the  the material rose up through the liquid, then later collapsed. This was the result after approximately 12 hours:

This is a fairly basic chemical garden. Often a whole range of transition metal salts are added to give colour variation.

2 n NaOH + n SiO2 ==> (Na2SiO2)nO + n H2O   

2 CrCl3 + 3 Na2SiO3 ==> 6 NaCl + Cr2(SiO3)3

8 CrCl3 + 6 Na4SiO4 ==> 24 NaCl + Cr8(SiO4)6

2 CrCl3 + Na6Si2O7 ==> 6 NaCl + Cr2Si2O7

Isopropylamine extraction from glyphosate herbicide

Isopropylamine, or 2-aminopropane, is an organic amine used mainly in herbicides and as a chemical intermediate. It is a colourless, highly volatile liquid with the strong rotten-fish smell typical of many amines. There are several reactions I plan to try out with isopropylamine in the future.

Isopropylamine can be fairly easily extracted from glyphosate herbicides. The herbicides contain the amine in the form of the glyphosate salt. The glyphosate is precipitated by adding hydrochloric acid, which leaves the isopropylamine in solution as the hydrochloride salt. Sodium hydroxide is then added to the solution to freebase the amine, which is distilled out of the mixture in fairly pure form.

To a 1000ml beaker, I added 500ml of Number 8 Ready To Go Glyphosate 360 Weedkiller (0.789 moles of isopropylamine glyphosate). With stirring, I poured in 88ml of 33% hydrochloric acid (0.789 moles of hydrogen chloride) and left the mixture to stand for 10 minutes. With addition of the hydrochloric acid, the mixture rapidly became cloudy as a dense white precipitate began to form. After 1 minute, a dense mass of crystals had precipitated. The crystals should in theory consist of N-(phosphonomethyl)glycine (glyphosate), however it's possible that the hydrochloric acid may cause the surfactants in the herbicide to precipitate as well. Anyway, I filtered the heavy white precipitate off, collecting the filtrate in a 1000ml conical flask. 40g (1.00 moles) of sodium hydroxide were then added to the mixture in small portions, with swirling of the flask between additions. A fairly violent reaction was observed with each addition and immediately an intense aroma of rotten fish was noticed. Once the addition was complete, I stoppered the flask and allowed it to cool to room temperature. I then poured the contents of the flask into a 1000ml flat-bottom boiling flask. Thankfully isopropylamine does not form an azeotrope with water, so distillation can be used to extract the amine. A stillhead, condenser, vacuum adapter and receiving flask were attached and a distillation was performed. Boiling water was used as the heat source. The isopropylamine gradually came over at 40-50 C as a colourless liquid.

After a while, I decided that just hot water was not a satisfactory heating source, so I placed a hotplate below the water bath to maintain a temperature of around 90 C. The heat was occasionally adjusted so the vapour temperature stayed at 40-52 C. When no more isopropylamine came over at this temperature, the distillation was ended. I dried the isopropylamine over sodium hydroxide, then redistilled it, collecting the fraction boiling at 30-41 C. Finally, I transferred the product to an amber vial for storage. I got 18ml (0.209 moles) of isopropylamine as the dry freebase. This represents a 26% yield. The density of the product was 0.601g/cm³ which is pretty close to the 0.988/cm³ of pure isopropylamine.

The distasteful smell of isopropylamine wasn't quite as bad as I was expecting. However, the foul odour was still quite unpleasant whenever I was working directly with the compound. Fishy ammoniacal, with a slight rancid note is how I would personally describe it.

The addition of hydrochloric acid at the beginning of the procedure is probably not necessary. I only added this step because I wanted to recover the glyphosate in solid form for easy disposal.


C3H9N.C3H8NO5P + HCl ==> C3H8NO5P + C3H9N.HCl

C3H9N.HCl + NaOH ==> NaCl + H2O + C3H9N