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In chemical analysis, the presence of a ...

In chemical analysis, the presence of a methyl ketonic group is tested with `I_(2)` in presence of an alkali like NaOH or KOH. In another chemical analysis, aldehyde group is tested either by Tollens' reagent `(AgNO_(3)NH_(3)OH)` which gives silver mirror or by Fehling's solution `(CuSO_(4)+NaOH+` Roschelle salt) which gives red ppt. of copper (I) oxide.
Q. Consider the following compounds:
I. `CH_(3)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)`
II. `CH_(3)-overset(O)overset(||)(C)-overset(O)overset(||)(C)-CH_(3)`
III. `CH_(3)-overset(O)overset(||)(C)-OH`
IV. `CH_(3)-overset(O)overset(||)(C)-Cl`
Which will give iodoform test?

A

Only I

B

Both I and II

C

Only II

D

All

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given compounds will give a positive iodoform test, we need to identify the presence of a methyl ketone group (–C(=O)–CH3) in the structures of the compounds. The iodoform test is specifically positive for compounds that contain a methyl ketone group or ethanol. Let's analyze each compound step by step: ### Step 1: Analyze Compound I **Compound I:** `CH3-C(=O)-CH2-C(=O)-CH3` - This compound has two ketone groups. The first ketone group is attached to a methyl group (–C(=O)–CH3). - Therefore, this compound contains a methyl ketone group. **Conclusion:** Compound I will give a positive iodoform test. ### Step 2: Analyze Compound II **Compound II:** `CH3-C(=O)-C(=O)-CH3` - This compound has a ketone group (–C(=O)–) that is directly attached to two methyl groups (–C(=O)–CH3). - Thus, it also contains a methyl ketone group. **Conclusion:** Compound II will give a positive iodoform test. ### Step 3: Analyze Compound III **Compound III:** `CH3-C(=O)-OH` - This compound is a carboxylic acid (specifically acetic acid). - It does not have a methyl ketone group. **Conclusion:** Compound III will NOT give a positive iodoform test. ### Step 4: Analyze Compound IV **Compound IV:** `CH3-C(=O)-Cl` - This compound is an acyl chloride. - It does not have a methyl ketone group. **Conclusion:** Compound IV will NOT give a positive iodoform test. ### Final Conclusion The compounds that will give a positive iodoform test are: - Compound I - Compound II ### Summary of Results: - **Compound I:** Positive iodoform test - **Compound II:** Positive iodoform test - **Compound III:** Negative iodoform test - **Compound IV:** Negative iodoform test

To determine which of the given compounds will give a positive iodoform test, we need to identify the presence of a methyl ketone group (–C(=O)–CH3) in the structures of the compounds. The iodoform test is specifically positive for compounds that contain a methyl ketone group or ethanol. Let's analyze each compound step by step: ### Step 1: Analyze Compound I **Compound I:** `CH3-C(=O)-CH2-C(=O)-CH3` - This compound has two ketone groups. The first ketone group is attached to a methyl group (–C(=O)–CH3). ...
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Knowledge Check

  • The order of stability of the following tautomeric compound is I. CH_(2)=overset(OH)overset("| ")"C "-CH_(2)-overset(O)overset(||)C-CH_(3) II. CH_(3)-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CH_(2) III. CH_(3)-overset(OH)overset("| ")"C "=CH-overset(O)overset(||)C-CH_(3)

    A
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    B
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    C
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    D
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    A
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    B
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    C
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    D
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    A
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    B
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    C
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    D
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