Nur Anis Arisya Binti Mohd Ramli

Izzah Binti Shamsol

Ethanal

the grey precipitate dissolves

1. The functional group for aldehydes and ketone is the carbonyl group.

2. Aldehydes and ketones can be further classified as aliphatic or aromatic, depending on whether they contain a straight-chain or aromatic ring structure.

3. The test that differentiates aldehydes and ketones from other classes of compounds is the 2,4-Dinitrophenylhydrazine (DNPH) test, which forms a yellow or orange precipitate.

4. The test that can be used to differentiate between aldehydes and ketones is the Tollens' test, where aldehydes produce a silver mirror, while ketones do not react.

5. The test that can be used to differentiate between aliphatic aldehydes and aromatic aldehydes is the Fehling's test, where aliphatic aldehydes form a red precipitate of copper(I) oxide, while aromatic aldehydes generally do not react.

6. The purpose of the iodoform test is to identify methyl ketones or compounds with a CH3CO group, as they produce a yellow precipitate of io

doform (CHI3).

Dropper

Test tube

5mL measuring cylinder

1. Disinfectants (Formaldehyde) – Formaldehyde (an aldehyde) is used as a disinfectant and preservative due to its ability to kill bacteria, viruses, and fungi by denaturing proteins and disrupting cell structures.

2. Nail Polish Remover (Acetone) – Acetone (a ketone) is a strong solvent that dissolves nail polish, making it easy to remove by breaking down the chemical bonds in the polish.

3. Food Flavoring (Vanillin) – Vanillin (an aldehyde) is responsible for the sweet, vanilla aroma and flavor in food products. It enhances the sensory appeal of foods like baked goods, chocolates, and beverages.

1. Formaldehyde (Aldehyde) – Disinfectant and Preservative

• Use: Formaldehyde is used as a disinfectant to sterilize medical instruments and preserve biological specimens.

• Effectiveness: It works by denaturing proteins and disrupting microbial cell structures, making it effective in killing bacteria, viruses, and fungi.

2. Acetone (Ketone) – Pharmaceutical Solvent

• Use: Acetone is used as a solvent in medicines and in the production of certain drugs.

• Effectiveness: Its ability to dissolve a wide range of substances makes it useful in drug formulation and topical treatments.

3. Cortisone (Ketone) – Anti-Inflammatory Drug

• Use: Cortisone, a ketone-based steroid, is used to treat inflammation in conditions like arthritis and asthma.

• Effectiveness: The ketone functional group plays a role in binding to steroid receptors, reducing inflammation and immune response.

• Atom Economy: In pharmaceutical synthesis, reactions are designed to maximize reactant use, reducing unwanted byproducts.

Bio-Based Production of Vanillin

Traditional vanillin production involves chemical synthesis from petroleum-based precursors, which generates waste. A green alternative uses fermentation of lignin (from plant biomass) by engineered microbes, producing vanillin in a sustainable and renewable way.

1. Approximately 1 mL of 2,4-dinitrophenylhydrazine was placed into 4 separate test tubes.

2. Five drops of ethanol, benzaldehyde, propanone, and unknown Y were added to the 4 test tubes accordingly.

3. The test tubes were shaken.

4. The formation of a precipitate was observed.

(B) Fehling’s test

1. Approximately 1 mL of ethanol, benzaldehyde, propanone, and unknown Y were placed in 4 separate test tubes.

2. Two mL of Fehling’s solution was added to each test tube.

3. The test tubes were gently shaken.

4. The mixtures were heated in a hot water bath for 15–20 minutes.

5. Observations were recorded.

(C) Tollens’ test

1. Tollens’ reagent was prepared by adding one drop of 10% NaOH to 2 mL of 0.3 M AgNO₃ in a boiling tube.

2. Five percent ammonia was added dropwise until the precipitate dissolved.

3. Approximately 1 mL of ethanol, benzaldehyde, propanone, and unknown Y were placed in separate test tubes.

4. One mL of Tollens’ reagent was added to each test tube, and the mixtures were gently shaken.

5. The mixtures were allowed to stand for 3 minutes. If no change occurred, the mixtures were warmed in a water bath at 60–70°C for 5 minutes.

6. Observations were recorded.

(D) Iodoform test

1. Ten drops of iodine in potassium iodide solution were placed into 3 mL of distilled water in a boiling tube.

2. Five drops of ethanol were added to the boiling tube and shaken gently.

3. Ten percent NaOH was added drop by drop until the color of the iodine faded.

3. The mixture was allowed to stand for 2–3 minutes. If no precipitate formed, the boiling tube was warmed in a water bath at 60–70°C.

4. If the color of iodine disappeared, more iodine in potassium iodide solution was added until the color was retained, and Step 4 was repeated.

5. Observations were recorded.

6. The above steps were repeated with benzaldehyde, propanone, and unknown Y.

2. Suitable tests to differentiate aldehydes and ketones are:

• tollens’ test

• fehling’s test

3. Unknown Y is aldehydes

1. Ten drops of iodine in potassium iodide solution were placed into 3 mL of distilled water in a boiling tube.

1. b) The functional group present in M is a carbonyl group (C=O), specifically in an aldehyde or ketone.

2. One chemical test to distinguish between propanone and propanal is the Tollen's test. Propanal, being an aldehyde, will react with Tollen's reagent (ammoniacal silver nitrate) to form a silver mirror on the inner surface of the test tube. Propanone, being a ketone, will not give a positive result with Tollen's reagent.

3. Both propanol and propanone are very soluble in water because they can form hydrogen bonds with water molecules. Propanol has a hydroxyl group (-OH) which can engage in hydrogen bonding, while propanone has a carbonyl group (CO) that can also form hydrogen bonds with water.

When M was warmed with alkaline iodine, a yellow precipitate was obtained.

M reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate.

(a) Name the yellow precipitate and write its molecular formula.

The yellow precipitate is tri-iodomethane (iodoform).

Its molecular formula is CHI₃.

(b) What is the functional group present in M?

The reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH) confirms the presence of a carbonyl functional group (C=O).

Since it also reacts with alkaline iodine, it must be a methyl ketone (R-CO-CH₃).

The functional group present in M is ketone (-C=O).

2. Suggest one chemical test to distinguish between propanone from propanal. Explain.

Test: Tollens' reagent (ammoniacal silver nitrate solution)

Explanation:

Propanal (an aldehyde) reacts with Tollens' reagent, producing a silver mirror on the test tube walls due to the oxidation of propanal to propanoic acid.

Propanone (a ketone) does not react with Tollens' reagent because ketones cannot be easily oxidized.

3. Explain why both propanol and propanone are very soluble in water.

Propanol (CH₃CH₂CH₂OH) contains a hydroxyl (-OH) group, which can form hydrogen bonds with water molecules, making it highly soluble.

Propanone (CH₃COCH₃) contains a polar carbonyl (-C=O) group, which also interacts with water molecules via dipole-dipole interactions and hydrogen bonding.

Since both molecules can form strong hydrogen bonds with water, they are highly soluble in water.

1. b) The functional group is a carbonyl group (C=O), in an aldehyde or ketone.

2. One chemical test is the Tollen's test. Propanal, being an aldehyde, will react with Tollen's reagent (ammoniacal silver nitrate) to form a silver mirror on the inner surface of the test tube. Propanone, being a ketone, will not give a positive result with Tollen's reagent.

3. Propanol (CH₃CH₂CH₂OH) contains a hydroxyl (-OH) group, which can form hydrogen bonds with water molecules, making it highly soluble.

Propanone (CH₃COCH₃) contains a polar carbonyl (-C=O) group, which also interacts with water molecules via dipole-dipole interactions and hydrogen bonding.

Since both molecules can form strong hydrogen bonds with water, they are highly soluble in water.

2. Suitable tests to differentiate aldehydes and ketones are:

• tollens’ test

• fehling’s test

3. Unknown Y is aldehyde

(a) The yellow precipitate is tri-iodomethane (iodoform).Its molecular formula is CHI₃.

(b) The reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH) confirms the presence of a carbonyl functional group (C=O).

Since it also reacts with alkaline iodine, it must be a methyl ketone (R-CO-CH₃).The functional group present in M is ketone (-C=O).

2. Test: Tollens' reagent (ammoniacal silver nitrate solution)

Propanal (an aldehyde) reacts with Tollens' reagent, producing a silver mirror on the test tube walls due to the oxidation of propanal to propanoic acid.

Propanone (a ketone) does not react with Tollens' reagent because ketones cannot be easily oxidized.

3.Propanol (CH₃CH₂CH₂OH) contains a hydroxyl (-OH) group, which can form hydrogen bonds with water molecules, making it highly soluble.

Propanone (CH₃COCH₃) contains a polar carbonyl (-C=O) group, which also interacts with water molecules via dipole-dipole interactions and hydrogen bonding.

Since both molecules can form strong hydrogen bonds with water, they are highly soluble in water.