Problem 47
Question
A compound that gives a positive iodoform test is (a) 1-pentanol (b) 2 -pentanone (c) 3-pentanone (d) pentanal
Step-by-Step Solution
Verified Answer
2-pentanone (b) gives a positive iodoform test.
1Step 1: Understanding the Iodoform Test
The iodoform test is used to identify methyl ketones or alcohols that can be oxidized to methyl ketones. It involves the presence of a CH₃CO group adjacent to a hydrogen atom.
2Step 2: Analyzing 1-Pentanol
1-Pentanol is a primary alcohol with the structure CH₃(CH₂)₄OH. Since it does not have a methyl ketone group or form one upon oxidation, it does not give a positive iodoform test.
3Step 3: Analyzing 2-Pentanone
2-Pentanone is a ketone with the structure CH₃COCH₂CH₂CH₃. This contains the CH₃CO group necessary for a positive iodoform test, as the iodoform test specifically detects methyl ketones.
4Step 4: Analyzing 3-Pentanone
3-Pentanone is a ketone with the structure CH₃CH₂COCH₂CH₃, lacking the CH₃CO group necessary for a positive iodoform test, as it does not have a methyl group next to the carbonyl.
5Step 5: Analyzing Pentanal
Pentanal is an aldehyde with the structure CH₃(CH₂)₃CHO. Aldehydes only give a positive iodoform test if it forms a methyl ketone upon oxidation, which pentanal does not, so it does not pass the iodoform test.
6Step 6: Conclusion
Based on the analysis, 2-pentanone is the compound that possesses the necessary CH₃CO group for the iodoform test, confirming a positive result.
Key Concepts
Methyl KetonesOrganic Chemistry ReactionsFunctional Group Identification
Methyl Ketones
Methyl ketones are a type of organic compound characterized by the presence of a carbonyl group (C=O) flanked by a methyl group (CH₃) and an alkyl group. This unique arrangement gives methyl ketones distinct chemical properties. A classic example is 2-pentanone, which is a compound that readily gives a positive iodoform test due to its structure.
Methyl ketones are significant because they react with iodine in the presence of a base to form iodoform, which is a solid yellow precipitate.
Methyl ketones are significant because they react with iodine in the presence of a base to form iodoform, which is a solid yellow precipitate.
- They are often used as a diagnostic tool in chemistry to identify unknown compounds.
- This test is particularly useful in distinguishing between different types of alcohols and ketones.
Organic Chemistry Reactions
Organic chemistry reactions involve the transformation of organic molecules through chemical processes. These reactions form the basis of studying how functional groups within molecules interact.
The iodoform test is a notable example within this scope. It specifically involves oxidation and substitution reactions to indicate the presence of methyl ketones.
Here's how the reaction proceeds:
The iodoform test is a notable example within this scope. It specifically involves oxidation and substitution reactions to indicate the presence of methyl ketones.
Here's how the reaction proceeds:
- The compound first reacts with iodine in a base like sodium hydroxide (NaOH).
- The methyl ketone group is oxidized, and the iodine substitutes hydrogen to form iodoform, resplendent in its recognizable yellow precipitate.
Functional Group Identification
Functional groups are distinctive groupings of atoms within molecules that determine the chemical behavior of those molecules. In organic chemistry, identifying these groups is crucial for predicting the reactivity and properties of compounds.
The iodoform test is a simple method used to identify the presence of certain functional groups, specifically methyl ketones. It hinges on detecting the CH₃CO group within a compound.
For example:
The iodoform test is a simple method used to identify the presence of certain functional groups, specifically methyl ketones. It hinges on detecting the CH₃CO group within a compound.
For example:
- Compounds like 2-pentanone test positive because they contain this functional group.
- Other compounds, such as 3-pentanone and 1-pentanol, lack a CH₃CO group adjacent to a methyl group, resulting in a negative test.
- The test isn't just for methyl ketones but can also apply to specific secondary alcohols that can be oxidized into methyl ketones.
Other exercises in this chapter
Problem 46
When acetaldehyde is heated with Fehling's solution it gives a precipitate of (a) \(\mathrm{Cu}\) (b) \(\mathrm{CuO}\) (c) \(\mathrm{Cu}_{2} \mathrm{O}\) (d) \(
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The boiling point of propionic acid is less than that of \(n\)-butyl alcohol, an alcohol of comparable molecular weight.
View solution Problem 48
The reagent with which both acetaldehyde and acetone react easily is [1981 - 1 Mark] (a) Fehling's reagent (b) Grignard reagent (c) Schiff's reagent (d) Tollen'
View solution Problem 52
Consider the following reactions \(\mathrm{A} \frac{(\mathrm{i}) \mathrm{CH}_{3} \mathrm{MgBr}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{B} \frac{\
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