Question

An alkene A with molecular formula C\(_6\)H\(_{10}\) on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B, and C.

Hint:

Ozonolysis of alkenes cleaves the double bond, producing carbonyl compounds such as aldehydes and ketones.

Answer 1

To solve the problem, we need to identify the alkene A with molecular formula C\(_6\)H\(_{10}\), and the compounds B and C formed upon its ozonolysis, given that B gives a positive Fehling’s test and reacts with iodine and NaOH, while C does not give a positive Fehling’s test but forms iodoform.

1. Understanding Ozonolysis:
An alkene with formula C\(_6\)H\(_{10}\) has two degrees of unsaturation, indicating one double bond (since it’s an alkene). Ozonolysis cleaves the double bond, forming carbonyl compounds (aldehydes or ketones). The products B and C together account for all six carbons and ten hydrogens, plus oxygen atoms added during ozonolysis.

2. Analyzing Compound B:
Compound B gives a positive Fehling’s test, indicating it is an aldehyde (since Fehling’s solution detects aldehydic groups, not ketones). Additionally, B reacts with iodine and NaOH, suggesting it undergoes the iodoform test, which is positive for compounds with a methyl ketone group ($ \text{CH}_3\text{CO}- $) or aldehydes like acetaldehyde ($ \text{CH}_3\text{CHO} $) that can be oxidized to such structures. Since B is an aldehyde, it is likely acetaldehyde ($ \text{CH}_3\text{CHO} $), which has the formula C\(_2\)H\(_4\)O and gives iodoform upon reaction due to its methyl group adjacent to the carbonyl.

3. Analyzing Compound C:
Compound C does not give a positive Fehling’s test, so it is not an aldehyde; it is likely a ketone. It forms iodoform, indicating it has a methyl ketone group ($ \text{CH}_3\text{CO}- $). A ketone with the iodoform reaction could be acetone ($ \text{CH}_3\text{COCH}_3 $), C\(_3\)H\(_6\)O, or a larger ketone with a methyl ketone moiety. Let’s denote C’s formula and test later.

4. Determining the Alkene A:
Since ozonolysis of A (C\(_6\)H\(_{10}\)) produces B and C, the sum of the carbons in B and C must be 6, and hydrogens must balance with oxygens added. Suppose B is acetaldehyde (C\(_2\)H\(_4\)O). Then, C must have 6 − 2 = 4 carbons. Let’s try C as a ketone with 4 carbons that gives iodoform. A likely candidate is butan-2-one ($ \text{CH}_3\text{COCH}_2\text{CH}_3 $), C\(_4\)H\(_8\)O, which has a methyl ketone group and forms iodoform.

Check the molecular balance:
- B: C\(_2\)H\(_4\)O (acetaldehyde)
- C: C\(_4\)H\(_8\)O (butan-2-one)
- Total: C\(_2\)H\(_4\)O + C\(_4\)H\(_8\)O = C\(_6\)H\(_12\)O\(_2\).
Ozonolysis adds two oxygens and two hydrogens (from reductive workup). The alkene C\(_6\)H\(_10}\) gains 2H and 2O, forming C\(_6\)H\(_12\)O\(_2\), which matches.

5. Reconstructing Alkene A:
Ozonolysis reverses to the alkene by joining the carbonyl carbons with a double bond. If B is $ \text{CH}_3\text{CHO} $ and C is $ \text{CH}_3\text{COCH}_2\text{CH}_3 $), the alkene’s double bond is between the carbonyl carbons:
- $ \text{CH}_3\text{CH}= $ (from B, losing O)
- $ =\text{C(O)CH}_2\text{CH}_3 $ (from C, losing O).
Removing the oxygens and forming the double bond gives $ \text{CH}_3\text{CH}=\text{C(O)CH}_2\text{CH}_3 $. However, we need the alkene, so the structure before ozonolysis is $ \text{CH}_3\text{CH}=\text{C(CH}_2\text{CH}_3\text{)}H $ or similar. Let’s write the alkene: $ \text{CH}_3\text{CH}=\text{C(CH}_3\text{)CH}_2\text{CH}_3 $ (2-methylpent-2-ene), C\(_6\)H\(_10\).

Verify ozonolysis:
- $ \text{CH}_3\text{CH}=\text{C(CH}_3\text{)CH}_2\text{CH}_3 $ → $ \text{CH}_3\text{CHO} $ + $ \text{CH}_3\text{COCH}_2\text{CH}_3 $.
This fits: $ \text{CH}_3\text{CHO} $ (B) and $ \text{CH}_3\text{COCH}_2\text{CH}_3 $ (C).

6. Verifying Properties:
- B ($ \text{CH}_3\text{CHO} $): Positive Fehling’s (aldehyde), positive iodoform (acetaldehyde forms iodoform).
- C ($ \text{CH}_3\text{COCH}_2\text{CH}_3 $): Negative Fehling’s (ketone), positive iodoform (methyl ketone).
- A: C\(_6\)H\(_10\), 2-methylpent-2-ene, yields B and C on ozonolysis.

Final Answer:
The compounds are:
- A: 2-methylpent-2-ene ($ \text{CH}_3\text{CH}=\text{C(CH}_3\text{)CH}_2\text{CH}_3 $)
- B: acetaldehyde ($ \text{CH}_3\text{CHO} $)
- C: butan-2-one ($ \text{CH}_3\text{COCH}_2\text{CH}_3 $).