Preparation, purification and combined analytical evidence
Introduced| English | Meaning |
|---|---|
| infrared/ˌɪnfrəˈred/ | Electromagnetic waves absorbed by bonds at characteristic wavenumbers, used in IR spectroscopy. |
| distil | To boil and recondense a liquid to purify it. |
| reflux/ˈriːflʌks/ | Heating a reaction mixture while condensing its vapour and returning the condensate to the same vessel. |
| anhydrous/ænˈhaɪdrəs/ | Containing no water, especially no water of crystallisation. |
Can one peak prove which liquid you made?
- Propanal and propanone both have formula C₃H₆O, molecular mass 58 and a carbonyl group. One molecular ion or one C=O absorption cannot distinguish them.
- Combine reaction conditions, purification, mass spectra, infrared 红外 data and chemical tests. Give a supported candidate and explain which competing candidate each observation rejects.
Choose preparation conditions for the product
- Acidified potassium dichromate(VI) changes orange to green when reduced. For a primary alcohol, distil 蒸馏 the aldehyde as it forms; excess oxidant under reflux 回流 gives the carboxylic acid. A secondary alcohol gives a ketone under these conditions.
- For propan-1-ol: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O; with further oxidation, CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O. Propan-2-ol gives CH₃COCH₃ + H₂O with [O]. Distillation removes vapour; reflux returns condensed vapour and allows prolonged heating.
Which conditions favour collecting propanal from propan-1-ol with acidified dichromate?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Distillation removes the aldehyde as it forms, reducing further oxidation. Excess oxidant under reflux favours propanoic acid. The controlled apparatus must be used under approved supervision.
Purify without confusing bulk water and residual water
- In solvent extraction, use a separating funnel to separate immiscible layers. If the organic layer has density $0.87\,\dfrac{\mathrm g}{\mathrm{cm}^3}$ and the aqueous layer $1.00\,\dfrac{\mathrm g}{\mathrm{cm}^3}$, the organic layer is upper; identify it from the stated densities, not an always-upper rule.
- After bulk separation, a suitable anhydrous 无水的 salt removes residual water. Remove the drying salt, then distil as appropriate. Measure boiling temperature with the thermometer bulb at the vapour take-off. A narrow range near the expected value at the stated pressure supports purity but is not unique structural proof.
A dried liquid boils at 85–86°C at a stated pressure. This is consistent with a relatively pure liquid having that boiling range. It does not prove a single named isomer: compare independent spectral or chemical evidence, and consider whether the apparatus measures vapour temperature correctly.
Two immiscible layers have densities organic 0.87 g/cm³ and aqueous 1.00 g/cm³. Which statement follows?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The lower-density organic liquid is upper in this stated case. A separating funnel removes bulk aqueous liquid; drying salt removes residual water after separation.
A narrow boiling range at the expected temperature uniquely proves the molecular structure.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A narrow range supports relative purity under the stated pressure and measurement conditions. Different structures can have similar boiling temperatures; independent identification evidence is needed.
Keep mass and fragment evidence distinct
- A singly charged molecular ion at m/z 58 supports molecular mass 58. C₃H₆O gives $3(12)+6(1)+16=58$. The base peak is the most intense peak; it need not be the molecular ion.
- A peak at m/z 29 could be CHO⁺, mass 12 + 1 + 16, or C₂H₅⁺, mass 24 + 5. Check plausible fragmentation and the full pattern. A positive warmed Fehling's/Benedict's test gives a brick-red precipitate and supports propanal over propanone in this two-candidate comparison.
Use C = 12, H = 1 and O = 16. What nominal m/z does singly charged CHO⁺ have?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
12 + 1 + 16 = 29 for charge +1. C₂H₅⁺ also has nominal mass 29, so this one peak is not a unique fragment formula.
Combine infrared bands and tests
- Use the supplied wavenumber table: C–H bands differ in alkanes, alkenes and aldehydes; C=C supports an alkene; alcohol O–H is broad while acid O–H is very broad. C=O can occur in aldehydes, ketones and acids; C–X supports a halogenoalkane and N–H an amine. Predict all relevant bonds from a candidate structure before comparing its spectrum.
- For example, supplied broad O–H at $3200–3600\,\dfrac{1}{\mathrm{cm}}$ and C=O at $1680–1750\,\dfrac{1}{\mathrm{cm}}$ can fit HOCH₂CHO; an acid has a much broader O–H region around $2500–3300\,\dfrac{1}{\mathrm{cm}}$. A carbonate test releasing CO₂ supports an acid. For inorganic unknowns use clean aliquots: warming NH₄⁺ with OH⁻ releases NH₃; acidified barium reagent gives white BaSO₄ with sulfate. These observations are not universally unique outside a stated candidate list.
An absorption shows evidence for a bond, not the whole structure. Overlapping bands and ambiguous fragments require more evidence. These are written method and interpretation checks; they do not certify practical performance or replace approved supervised procedures.
Control the preparation, explain each purification step, and combine molecular mass, plausible fragments, infrared bonds and selective tests. State exactly how far the evidence supports the identification.
The candidate list is propanal or propanone. Both fit C₃H₆O and C=O. A warmed Fehling’s test gives a brick-red precipitate. Which is supported?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Propanal is an aldehyde and gives this positive test. Propanone is a ketone and does not under these conditions. Formula and carbonyl evidence alone cannot distinguish them.