The unknown compound had the same functional group as one of the given known compounds. To determine which functional group the unknown compound had, the unknown was subjected to the same tests as each of the five known compounds. The results of each test were then compared to the results with the known compounds.
- One drop of the unknown compound was added to one millilitre of Tollen’s reagent in a test tube and to react for less than 10 minutes before the result was noted.
- 2 drops of the unknown were placed in a test tube with about 1 ml of the 2,4-dinitrophenylhydrazine reagent, which was then sealed and shaken to accelerate the reaction. The observations were then noted down.
- 2 ml of Lucas reagent were placed in a test tube; five drops of the unknown were then added. The test tube was shaken and the contents were allowed to react. The observations and the time it took for the mixture to react were then noted down.
- Since the test for the primary and the secondary alcohols is the same, only producing different results, it was performed only once.
- The odour of the unknown was then noted and compared to the odour of the ester sample.
The results of the tests of the unknown were then compared to each of the five tests performed on each of the five known substances. The results that showed the greatest similarity indicated the functional group of the unknown to be the same as on the known compound with the similar observations.
To find the molar mass of the unknown compound Dumas method was used. The molar mass of a volatile liquid is equal to the molar mass of its vapour, assuming that the liquid neither dissociates nor associates in the progress. To find the molar mass of the vapour, a known volume was captured at known temperature and pressure. The vapour was then condensed and its mass measured. A 125 ml Erlenmeyer flask was used as a container to trap the vapour. 4 ml of the unknown compound were placed in the Erlenmeyer flask; the liquid was then placed in boiling water and boiled off. A piece of aluminium foil with a pinhole was placed over top the flask to allow the liquid to evaporate but to preserve the vapour once all the liquid had boiled away. The flask was then taken out of the boiling water, and the vapour allowed to condense. The flask with the condensate was then weighed and the mass compared to the mass of the mass of the empty flask measured before the experiment, giving the mass of the condensate.
Observations:
Part 1
- The reaction of the aldehyde with Tollen’s reagent produced a black mixture with silvery substance appearing on the walls of the test tube.
- The reaction of the 2,4-dinitrophenylhydrazine with the ketone caused yellow crystals to precipitate.
- The primary alcohol showed no reaction.
- The secondary alcohol with Lucas reagent produced a cloudy mixture after 5 to 7 minutes.
- The ester had a sharp distinctive odour to it.
The only reaction that proceeded when the unknown was reacted with each of the reagents was the reaction with Lucas reagent, producing a cloudy mixture. The unknown had a different smell than the ester sample.
Part 2
The weight of the empty flask, foil and band 65.8270g ± 0.0001g
Weight of flask, foil, band, and liquid after volatizing 66.2126g ± 0.0001g
Weight of volatized liquid 0.3856g
Temperature 369ºK
Volume of flask 152ml ± 1.0ml = 0.152 L ± 0.001 L
Atmospheric pressure 757.3 torr
Discussion:
Since the only reaction that proceeded to completion, involving the unknown and each of the three reagents was the reaction with the Lucas reagent, that indicated the unknown to be a secondary alcohol. No other reaction ensued and the odour of the unknown did not correspond to the odour of the ester sample.
Using the ideal gas law PV = nRT the molar mass M of the unknown sample could be determined.
The atmospheric pressure, the measured volume and temperature were used in the calculations. An R of 62.36 torr L mol-1 ºK-1 was used.
PV = nRT
(757.3 torr)(0.152L) = n (62.36)( 369ºK)
n = 0.005 mol M = m/n
M = (mass of volatized liquid) / (number of moles)
M = (0.3856g) / (0.005 mol)
M = 77.12 g/mol
The relative uncertainty in the molar mass was found using the following formula:
∆z / z = √ [(∆x/x)2 + (∆y/y)2]
The relative uncertainty came out to be:
∆M / M = √ [(0.001L / 0.152 L)2 + (0.5K / 369ºK)2 + (0.0002g / 0.3856g)2]
= 6.7 x 10-3
Only a four carbon secondary alcohol could produce a molar mass similar to that obtained.
The closest molar mass is the one of the butan-2-ol which is 74 g/mol.
Since butan-2-ol has no structural isomers the unknown compound must have had the same molecular formula.
H OH H H
| | | |
H – C – C – C – C – H M = 74 g/mol
| | | |
H H H H
Conclusions:
The unknown compound was found to be a secondary alcohol with a formula C4H9OH, and name butan-2-ol. The functional group of the unknown was determined to be the hydroxide because it reacted with the alcohol reagent. The time of reaction proved it to be a secondary alcohol. The molar mass was found using Dumas method and found to be 77 g/mol. That molar mass corresponds to the closest molar mass of a secondary alcohol butan-2-ol, with a molar mass of 74 g/mol.
References:
The only source used for this lab was the course lab manual.
Prof. J. Poe. Chemistry 140Y Laboratory Manual; University of Toronto :Toronto, 2004
