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OrgBio Chem > Experiment 5: Formaldehyde and Aromatic Aldehyde > Flashcards

Experiment 5: Formaldehyde and Aromatic Aldehyde Flashcards

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1
Q

In Experiment 5, what class of organic compounds are being tested?

A

Aldehydes (both aliphatic & aromatic) and Ketones

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2
Q

What do Aldehydes and Ketones have in common?

A

They are both organic compounds with the functional carbonyl group or C=O

Aldehydes: CHO
Ketones: COR

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3
Q

What is an Aromatic Aldehyde?

A

The compounds in which the aldehyde functional group is attached to the aromatic ring

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4
Q

Aliphatic Aldehydes vs Aromatic Aldehydes:

A

Aliphatic aldehydes are more reactive than aromatic aldehydes.

Aromatic aldehydes are less reactive due to their stability.

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5
Q

PREPARATION OF FORMALDEHYDE FROM METHANOL: Procedure & Equation

A

PROCEDURE:

  1. A copper spiral is heated in the burner’s oxidizing flame.
  2. It is then dipped into a solution of 1 mL methyl alcohol / CH3OH in 5-6 mL of H2O.

EQUATION:

2 CH3OH + O2 [Cu]–> 2 HCHO + 2 H2O

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6
Q

How was CH2O / HCHO produced?

A

It was produced by the oxidation of methanol.

Methanol loses an H atom and gains an O atom –> OXIDATION

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7
Q

In preparation of HCHO, what was the catalyst?

A

Cu from the copper spiral

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8
Q

Formaldehyde vs Formalin

A

Formaldehyde is a gas at room temperature. It is neutral and soluble in water.

Formalin is a solution of 37-40% formaldehyde in water

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9
Q

5.1A. SCHIFF’S TEST: Procedure

A

PROCEDURE:

1 mL of colorless Schiff’s Solution / C20H20N3 HCl is added to three different test tubes containing

  • HCHO
  • C6H5CHO / benzaldehyde
  • CH3COCH3 / acetone
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10
Q

5.1B. Why is C20H20N3 HCl colorless? What was its original color?

A

C20H20N3 HCl was originally magenta-red because it contained fuchsin which gave it its color.

However, it was decolorized by sulfurous acids / sulfur dioxide, making it lose its color

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11
Q

5.1C. SCHIFF’S REST: Results & Discussion

A
  • Schiff in HCHO: Saturated magenta color
  • Schiff in C6H5CHO: Pale pink bubble on liquid

When aldehydes react with Schiff’s reagent, they restore the fuchsin’s magenta color.

  • Schiff in CH3COCH3: Remains purple

Ketones do not give this color.

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12
Q

5.1D. Why does Schiff’s only work with aldehydes?

A

The free and uncharged amine groups in the reagent react with the aldehyde groups to form aldimine, which further reacts into a bisulfite ion. Finally, the magenta bisulfite adduct is formed.

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13
Q

5.2A. RESORCINOL TEST: Procedure

A

PROCEDURE:

  1. HCHO (formalin) is added with resorcinol solution / C6H6O2.
  2. The test tube is then inclined and added with conc. H2SO4.
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14
Q

5.2B. RESORCINOL TEST: Results & Discussion

A

A red flocculent precipitate line forms at the junction of two layers

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15
Q

5.2C. What is the Resorcinol Test specifically used to test for?

A

The presence of formaldehyde / HCHO

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16
Q

5.2D. Resorcinol is often utilized with formaldehyde to manufacture _____

A

Resins

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17
Q

These two solutions are mild oxidizing agents

A
  1. Tollen’s solution
  2. Fehling’s solution
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18
Q

5.3A. What is the other name of the Silver Mirror Test?

A

Tollen’s Test

19
Q

5.3B. SILVER MIRROR TEST: Procedure

A

PROCEDURE

  1. 2 mL of ammoniacal silver nitrate / Ag(NH3)2NO3 is added into three different test tubes containing HCHO, C6H5CHO & CH3COCH3 each.
  2. Agitate all mixtures thoroughly then warm in a water bath.
20
Q

5.3C. What formed when Ag(NH3)2NO3 was exposed to aldehydes, and why?

A

Silver mirror-like substances formed. This is because the silver ions were reduced

Ketones do not give this result.

21
Q

5.3D. Aside from the reduction of silver ions, what occurs when aldehydes are exposed to Ag(NH3)2NO3?

A

If an aldehyde is present in ammoniacal silver nitrate, it is readily oxidized by the reagent into its corresponding carboxylic acid

-CHO oxidized to -COOH

Ketones do not give this result

22
Q

5.3E. SILVER MIRROR TEST: Equations & Results

A

HCHO + Ag(NH3)2 + 3 OH –> HCOOH / Formic Acid + 2 Ag + 4 NH3

C6H5CHO + Ag(NH3)2 + 3 OH –> C6H5COOH / Benzoic Acid + 2 Ag + 4 NH3

CH3COCH3+ Ag(NH3)2 + 3 OH –> No reaction

23
Q

5.4A. HCHO ON PROTEIN SUBSTANCES: Procedure

A

PROCEDURE:

  1. Two gelatin sheets are prepared.
  2. One sheet is placed in 1 mL formalin in water, while the other is placed in water only. Allow them to stand.
  3. Test their solubility in hot water.
24
Q

5.4B. What was the purpose of the gelatin sheets?

A

They were used as protein.

25
Q

5.4C. HCHO ON PROTEIN SUBSTANCES: Results & Discussion

A
  • The first gelatin sheet exposed to the formalin-H2O solution solidified more and sank to the bottom.
  • The second gelatin sheet exposed only to water softened and did not sink to the bottom.
26
Q

5.4D. Why did the gelatin sheet harden when exposed to HCHO only?

A

Formaldehyde is known for its ability to deactivate or immobilize proteins. Crosslinking occurs between formalin and the gelatin upon interacting, leading to a decrease in the sheet’s swelling and solubility

27
Q

5.5A. FEHLING’S TEST: Procedure & Equation

A

PROCEDURE:

  1. 1 mL of Fehling’s solution / Cu(OH)2 is added to three different test tubes, each containing HCHO, C6H5CHO & CH3COCH3.
  2. All test tubes are put in a warm water bath.

EQUATION: for HCHO only

HCHO + Cu(OH)2 –> HCOOH + Cu2O + H2O

28
Q

5.5B. Fehling’s solution is a mixture of:

A
  • Copper sulfate / CuSO4
  • Sodium hydroxide / NaOH
  • Rochelle Salt or Sodium Potassium Tartrate
29
Q

5.5C. FEHLING’S TEST: Results & Discussion

A
  • HCHO in Cu(OH)2 resulted in the formation of red precipitates, indicating a positive reaction
  • C6H5CHO in Cu(OH)2 and CH3COCH3 in Cu(OH)2 yielded no reaction; thus solution remained blue.

This is because ketones and aromatic aldehydes do not work with Fehling’s solution

30
Q

5.5D. What happens when aliphatic aldehydes interact with Cu(OH)2?

A

Aliphatic Aldehydes are oxidized in the reaction with Fehling’s solution, leading to the formation of red precipitate

31
Q

5.5E. How does Cu2O become a byproduct in Fehling’s test?

A

This is because the copper (II) ions from Fehling’s reagent are reduced to Cu2O / cuprous oxide ions

32
Q

5.6A. AUTOXIDATION OF C6H5CHO: Procedure & Equation

A

PROCEDURE:

  1. Place a few drops of C6H5CHO on a watch glass.
  2. Expose to the atmosphere.

EQUATION:

C6H5CHO [O2]—> C6H5COOH / Benzoic acid

33
Q

5.6B. What did C6H5CHO oxidize to?

A

C6H5COOH or Benzoic Acid in the form of crystals

34
Q

5.6C. What is Autoxidation?

A

The spontaneous oxidation of a compound in the presence of oxygen

35
Q

5.6D. Why did C6H5CHO autoxidize easily and form crystals?

A
  • Benzaldehyde, as an aldehyde, autoxidizes easily because of its present H atom bonded to its carbonyl functional group
  • It formed crystals because of its acidic nature
36
Q

5.7A. ADDITION OF SODIUM BISULFITE: Procedure & Equation

A

PROCEDURE:

  1. 1 mL of cold conc. sodium bisulfite / NaHSO3 is added to 0.5 mL of CH3COCH3
  2. Shake vigorously.

EQUATION:

CH3COCH3 + NaHSO3 —> CH3-C-OH-SO3NaCH3

37
Q

5.7B. What formed after NaHSO3 was added?

A

Bubble layer and white precipitates

PPT is called CH3-C-OH-SO3NaCH3 or Acetone sodium bisulfite

38
Q

5.7C. How was acetone sodium bisulfite formed?

A

The carbonyl group in acetone is highly polar. This causes the nucleophile NaHSO3 to form a bond with the electrophilic C=O functional group. The alkoxide is then protonated by the addition of an acid to form an alcohol

39
Q

5.8A. FORMATION OF PHENYLHYDRAZONE: Procedure & Equation

A

PROCEDURE:

  1. 3 drops of CH3COCH3 are added with 2,4-dinitrophenylhydrazine / C6H6N4O4, then shaken.

EQUATION:

CH3COCH3 + C6H6N4O4 —> CH3-C=N2C6H6CH3

40
Q

5.8B. What was the reaction after C6H6N4O4 was added?

A

The solution’s texture thickened, and an orange-yellow precipitate formed

CH3-C=N2C6H6CH3 or Acetone 2,4-dinitrophenylhydrazone

41
Q

5.8C. What does the orangish color of acetone 2,4-dinitrophenylhydrazone precipitate indicate?

A

That the original carbonyl group in acetone was non-conjugated

42
Q

5.9A: IODOFORM TEST: Procedure & Equation

A

PROCEDURE:

  1. In a test tube with CH3COCH3, add 8 drops of 10% NaOH.
  2. Then, slowly add drop-by-drop I2.

EQUATION:

CH3COCH3 [I2 & NaOH]—> CHI3 + CH3COONA

43
Q

5.9B. What is Iodoform Test applied to?

A

The test is applied to identify methyl ketones

44
Q

5.9C. What forms in Iodoform Test, and what is it the result of?

A

The successful transformation of CH3COCH3 to CH3COONa results in the formation of yellow precipitates called Iodoform / CHI3

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