Procedures in experimental chemistry

Question

An experiment was carried out involving $\alpha$-bromoheptaldehyde dimethyl acetal. The experimental details are given below.

A solution of enol acetate (15.6 g, 17.7 mL) in chloroform (20 mL) was cooled to 0°C. A solution of bromine (16 g, 5.1 mL) in chloroform (5 mL) was added slowly with stirring with the rate of addition being controlled to keep the reaction temperature below 10°C. After the completion of addition, anhydrous methanol (60 mL) was added to the mixture and it was stirred for 48 hours. The mixture was diluted with water (200 mL) and ether (50 mL). The organic layer was separated, washed with water (100 mL) and with 5% aqueous sodium carbonate (100 mL). The organic layer was dried using sodium sulfate and filtered. The solvents were removed by distillation under reduced pressure. The residual oil was distilled under reduced pressure in the presence of a small amount of sodium carbonate. The fraction boiling at about 118°C at 17 mmHg was collected to give bromo acetal as a colourless oil (19.5 g).

(a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.

(b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?

(c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?

(d) Sodium carbonate is added just before the final distillation. Why?

(e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?

(f) Suggest two serious and different hazards associated with this experiment.

My attempt

(a) It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

(b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

(c) The organic layer will be at the bottom. Since it is know that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

(d) To remove any excess acid that might be present in the mixture.

(e) Rotatory evaporator

(f) Bromine is toxic if inhaled, while methanol is a carcinogen

I am unsure of my answers though. Especially part (c) as I can't decide whether the aqueous or organic layer will be at the top. Could anyone explain this.

Answer 1

I believe that when you say "a solution of enol acetate" you are refering to a solution of the corresponding enol-acetal of hetanal? You should use full names of the compounds used, if not, it might be confusing.

My opinions

(a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.
It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

I don't see a reason you would need any inert conditions for this experiment. I think you will be fine if you perform it open to the air.

(b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?
(b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

Indeed. The end point of the reaction is reached when the calculated amount is absorbed and the bromine is no longer decolorized.

(c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?
The organic layer will be at the bottom. Since it is how that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

This is wrong. The organic layer IS the one which is will be a mixture formed by chloroform, unreacted enol acetate and ether. The other phase will be the one composed mainly by water and methanol. Methanol+water solution will probably be less dense than chloroform+ether+enolacetate solution, and therefore, organic layer will be the upper one. I cannot tell for sure, since you have mixtures in high proportions of compounds with quite different densities, but and easy way to tell is: - Separate both phases using an extraction funnel. Put some tap water on the separated phase you suspect is the aqueous phase. It dissolves? That is the aqueous phase. It does not dissolve and forms a two-phased system? It was the organic phase, extract it again. You can double check adding water to the other phase you extracted, and see that the opposite effect should be observed.

(d) Sodium carbonate is added just before the final distillation. Why?
To remove any excess acid that might be present in the mixture.

Yeah. The acetal must be free of acid; otherwise decomposition takes place during distillation.

(e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?
Rotatory evaporator

Correct. Rotovap is your guy.

(f) Suggest two serious and different hazards associated with this experiment.
Bromine is toxic if inhaled, Methanol is a carcinogen

Bromine will definitely be your top priority concern. Even more if you have to prepare the solution from pure bromine. Also, it has a very strong unpleasant smell, which gives it the name it has (from Greek: βρῶμος, brómos, meaning "strong-smelling" or "stench") It is corrosive very toxic, volatile (fumes a lot, MUST be handled on a fumehood). Chloroform is also toxic; Heptanal is volatile and flammable.

Methanol, even though you should not drink it wouldn't be of an important concern.

To sum up, some of the reactants are of unpleasant odor or have lachrymatory effects it is best to use a good hood.

And also a reference: Organic Syntheses, Coll. Vol. 3, p.127 (1955); Vol. 29, p.14 (1949).