Problem 34
Question
Iodoform test is not given by (a) 2 -pentanone (b) ethanol (c) ethanal (d) 3 -pentanone
Step-by-Step Solution
Verified Answer
3-Pentanone does not give a positive iodoform test.
1Step 1: Understanding Iodoform Test
The iodoform test is a chemical reaction used to determine the presence of a methyl ketone group, \(-COCH_3\), or a methyl carbinol group, \(-CH(OH)CH_3\). An aldehyde or ketone must have at least one of these functional groups to give a positive iodoform test.
2Step 2: Analyzing 2-Pentanone
2-Pentanone has the structure \(CH_3COCH_2CH_2CH_3\). It contains a methyl ketone group \(-COCH_3\), so it will give a positive iodoform test.
3Step 3: Analyzing Ethanol
Ethanol (\(CH_3CH_2OH\)) can be oxidized to acetaldehyde (ethanal), which contains a methyl alcohol group \(-CH(OH)CH_3\). Thus, ethanol gives a positive iodoform test.
4Step 4: Analyzing Ethanal
Ethanal has the structure \(CH_3CHO\) and contains a methyl ketone group after oxidation. Therefore, ethanal will also give a positive iodoform test.
5Step 5: Analyzing 3-Pentanone
3-Pentanone has the structure \(CH_3CH_2COCH_2CH_3\). It lacks the \(-COCH_3\) or \(-CH(OH)CH_3\) groups necessary for the iodoform test, as neither are present in the molecule. Therefore, 3-pentanone does not give a positive iodoform test.
Key Concepts
Methyl KetoneMethyl Carbinol2-PentanoneEthanolEthanal
Methyl Ketone
Methyl ketones are organic compounds characterized by a specific functional group,
COCH_3.
This group is found in carbonyl-containing molecules where a methyl group,
CH_3,
is attached to a carbonyl carbon.
The iodoform test is specifically sensitive to the presence of methyl ketones. A reaction with iodine and a base triggers the formation of a yellow precipitate of iodoform, confirming their presence.
Methyl ketones are typically found in many organic molecules, and they play a crucial role in the predictability of certain chemical reactions.
The iodoform test is specifically sensitive to the presence of methyl ketones. A reaction with iodine and a base triggers the formation of a yellow precipitate of iodoform, confirming their presence.
Methyl ketones are typically found in many organic molecules, and they play a crucial role in the predictability of certain chemical reactions.
- Common Methyl Ketones: Acetone, 2-pentanone
- Importance: Identification through chemical tests
- Test Result: Positive iodoform test
Methyl Carbinol
Methyl carbinol, also known as secondary alcohols with a specific structure,
CH(OH)CH_3,
are another group that reacts positively to the iodoform test.
These compounds contain a hydroxyl group (-OH) attached to a secondary carbon atom bonded to one methyl group.
Notably, ethanol is an example of a primary alcohol that can be oxidized to form a methyl carbinol, hence it can yield a positive result when tested.
These compounds contain a hydroxyl group (-OH) attached to a secondary carbon atom bonded to one methyl group.
Notably, ethanol is an example of a primary alcohol that can be oxidized to form a methyl carbinol, hence it can yield a positive result when tested.
- Structure: Contains one methyl and one hydroxyl group
- Example: Ethanol converts to this group
- Test Result: Positive iodoform test after oxidation
2-Pentanone
2-Pentanone is a five-carbon methyl ketone with a carbonyl group located on the second carbon.
It is represented as CH_3COCH_2CH_2CH_3.
This structure contains a methyl group directly adjacent to the carbonyl carbon, qualifying it as a methyl ketone.
As a result, when subjected to the iodoform test, 2-pentanone forms the distinctive yellow precipitate, indicating a positive outcome.
Understanding the presence of this functional group is key to predicting the iodoform test result.
It is represented as CH_3COCH_2CH_2CH_3.
This structure contains a methyl group directly adjacent to the carbonyl carbon, qualifying it as a methyl ketone.
As a result, when subjected to the iodoform test, 2-pentanone forms the distinctive yellow precipitate, indicating a positive outcome.
Understanding the presence of this functional group is key to predicting the iodoform test result.
- Structure: A carbonyl group at the second carbon
- Importance: Classified as a methyl ketone
- Test Result: Positive iodoform test
Ethanol
Ethanol, with the formula
CH_3CH_2OH,
is one of the simplest alcohols, often found in beverages and disinfectants.
Alone, ethanol does not contain a methyl ketone, yet it can be oxidized to ethanal. This conversion makes it bear a functional similarity to a methyl carbinol, amplifying its role in the iodoform test.
When ethanol is oxidized, it can give a positive iodoform test due to its ability to form ethanal, which contains the required group.
Alone, ethanol does not contain a methyl ketone, yet it can be oxidized to ethanal. This conversion makes it bear a functional similarity to a methyl carbinol, amplifying its role in the iodoform test.
When ethanol is oxidized, it can give a positive iodoform test due to its ability to form ethanal, which contains the required group.
- Formula: CH_3CH_2OH
- Importance: Can undergo oxidation to form methyl carbinol-like structures
- Test Result: Positive iodoform test after oxidation
Ethanal
Ethanal, also called acetaldehyde, has the structure
CH_3CHO.
It belongs to the class of aldehydes but can also form functional groups similar to a methyl ketone, especially upon oxidation.
Its proximity to the methyl group allows ethanal to give a positive iodoform test.
This makes ethanal a fascinating compound in chemical reactions, working as an intermediary between simple acetates and more complex polymers.
Its proximity to the methyl group allows ethanal to give a positive iodoform test.
This makes ethanal a fascinating compound in chemical reactions, working as an intermediary between simple acetates and more complex polymers.
- Structure: Simplest aldehyde and close chemical relationship to methyl ketones
- Usage: Found in synthetic organic chemistry
- Test Result: Positive iodoform test
Other exercises in this chapter
Problem 32
In the reaction of \(\mathrm{p}\)-chlorotoluene with \(\mathrm{KNH}_{2}\) in liquid \(\mathrm{NH}_{3}\), the major product is (a) o-toluidine (b) \(\mathrm{m}\)
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Freon-12 is commonly used as (a) refrigerant (b) a solvent (c) insecticide (d) a fire extinguisher
View solution Problem 37
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