Q.An alkene 'A' (molecular formula $C_5H_{10}$) on ozonolysis gives a mixture of two compounds 'B' and 'C'. Compound 'B' gives a positive Fehling's test and also forms iodoform on treatment with $I_2$ and NaOH. Compound 'C' does not give Fehling's test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and the formation of iodoform from B and C.

Concept: Structural Isomerism – specifically, ozonolysis of alkenes to locate the double bond, combined with functional group tests (Fehling’s and iodoform) to identify aldehydes, ketones, and methyl carbonyls.

Reasoning:

1. Ozonolysis clue: Alkene $C_5H_{10}$ cleaves at the double bond to give two carbonyl compounds. Since both B and C give iodoform, each must contain a $CH_3CO-$ group (methyl ketone) or $CH_3CH(OH)-$ group (but here, ozonolysis gives carbonyls directly, so methyl ketone).

2. Fehling’s test: B gives positive Fehling’s → B is an aldehyde. C does not → C is a ketone. Both form iodoform → B is $CH_3CHO$ (acetaldehyde) and C is $CH_3COCH_3$ (acetone). …

Why this approach works

The problem tests your understanding of two key reactions: ozonolysis of alkenes and iodoform reaction. Ozonolysis cleaves the C=C double bond, giving carbonyl compounds (aldehydes or ketones). The iodoform test is positive for methyl ketones ($\ce{CH3CO-}$) and for ethanol ($\ce{CH3CH2OH}$) or acetaldehyde ($\ce{CH3CHO}$) — compounds with a $\ce{CH3CO-}$ group or a $\ce{CH3CH(OH)-}$ group that can be oxidised to a methyl ketone.

The molecular formula $\ce{C5H10}$ suggests an alkene with one degree of unsaturation. The ozonolysis products must add up to five carbons, and both must be capable of giving iodoform. That immediately narrows the possibilities.

Step-by-step reasoning

1. Analyse the iodoform clues.

   Compound B gives a positive Fehling’s test (so it’s an aldehyde) and forms iodoform. The only aldehyde that gives iodoform is ethanal ($\ce{CH3CHO}$) — because it has a $\ce{CH3CO-}$ group. So B must be ethanal.

   Compound C does not give Fehling’s test (so it’s a ketone) but does form iodoform. That means C is a methyl ketone ($\ce{CH3COR}$). Since the total carbons are 5 and B has 2 carbons, C must have 3 carbons. The only 3-carbon methyl ketone is propanone (acetone, $\ce{CH3COCH3}$).

2. Reconstruct the alkene from ozonolysis products.

   Ozonolysis of an alkene $\ce{R1R2C=CR3R4}$ gives two carbonyl compounds:

   $\ce{R1R2C=O}$ and $\ce{R3R4C=O}$.

   Here, one product is ethanal ($\ce{CH3CHO}$) and the other is propanone ($\ce{CH3COCH3}$).

   Ethanal comes from a $\ce{CH3CH=}$ fragment. Propanone comes from a $\ce{(CH3)2C=}$ fragment.

   Joining these two fragments across the double bond gives the alkene:

   $\ce{CH3CH=C(CH3)2}$ — that is 2-methylbut-2-ene.

3. Confirm the molecular formula.

   $\ce{CH3CH=C(CH3)2}$ has formula $\ce{C5H10}$ — matches perfectly.

4. Write the ozonolysis reaction. …

Method: Retrospective Analysis via Ozonolysis and Functional Group Tests

This method works backwards from the chemical tests to deduce the structure of the alkene.

Step 1: Interpret the tests on products B and C

Compound B:

• Positive Fehling’s test → It is an aldehyde (not a ketone).
• Forms iodoform with $I_2 + NaOH$ → It has a $\ce{CH3CO-}$ group or a $\ce{CH3CH(OH)-}$ group. Since it’s an aldehyde, the only way to get iodoform is if it is acetaldehyde ($\ce{CH3CHO}$).

Compound C:

• Does NOT give Fehling’s test → It is a ketone (not an aldehyde).
• Forms iodoform → It has a $\ce{CH3CO-}$ group. So it is a methyl ketone.

Step 2: Determine the carbon skeleton from ozonolysis

Ozonolysis of an alkene cleaves the double bond into two carbonyl compounds.

If alkene A ($\ce{C5H10}$) gives B ($\ce{C2H4O}$) and C ($\ce{C3H6O}$), then:

• B = $\ce{CH3CHO}$ (acetaldehyde, 2 carbons)
• C = a methyl ketone with 3 carbons → $\ce{CH3COCH3}$ (acetone)

Step 3: Reconstruct the alkene A

The double bond in A must be between the carbons that become the carbonyl groups in B and C.

• B comes from: $\ce{CH3CH=}$
• C comes from: $\ce{=C(CH3)2}$

Joining them gives:

Name: 2-methylbut-2-ene

Step 4: Write the reactions …

Here are the common mistakes students make with this exact type of structural isomerism + ozonolysis problem, and how to avoid each.

1. Misinterpreting the Fehling’s Test Result

Mistake:

Thinking that any carbonyl compound gives a positive Fehling’s test.

Students often assume that if a compound gives a positive Fehling’s test, it must be an aldehyde — but that’s only half the story.

Why it’s wrong:

Fehling’s test is positive for aldehydes (except aromatic ones) and α-hydroxy ketones. However, in this problem, compound B gives a positive Fehling’s test and forms iodoform. That combination points specifically to acetaldehyde ($CH_3CHO$) or a methyl ketone with an α-hydroxy group — but here, the iodoform test narrows it down.

How to avoid:

• Remember: Fehling’s positive = aldehyde (or α-hydroxy ketone).
• Iodoform positive = methyl ketone ($CH_3COR$) or ethanol/ secondary alcohol with $CH_3CH(OH)-$ group.
• The only small carbonyl that gives both is acetaldehyde ($CH_3CHO$).
• So, B must be acetaldehyde.

2. Confusing the Iodoform Test Conditions

Mistake:

Thinking that any aldehyde or ketone gives iodoform.

Why it’s wrong:

Only compounds with a methyl group attached to the carbonyl carbon ($CH_3CO-$) or a methyl carbinol group ($CH_3CH(OH)-$) give the iodoform test. For example, acetone ($CH_3COCH_3$) gives iodoform, but propanal ($CH_3CH_2CHO$) does not.

How to avoid:

• Memorise the structural requirement: $CH_3CO-$ or $CH_3CH(OH)-$.
• In this problem, C does not give Fehling’s test but does give iodoform → C must be a methyl ketone (not an aldehyde).
• The only methyl ketone possible from ozonolysis of $C_5H_{10}$ is acetone ($CH_3COCH_3$).

3. Forgetting Ozonolysis Cleavage Pattern

Mistake:

Assuming ozonolysis always gives two different carbonyl compounds, or that the alkene must be symmetric.

Why it’s wrong:

Ozonolysis cleaves the C=C double bond completely, giving two carbonyl fragments. If the alkene is unsymmetrical, the two products are different. If the alkene is symmetric, you get two identical molecules.

How to avoid:

• Draw the alkene with the double bond in the middle.
• Ozonolysis:

• Here, the products are acetaldehyde (B) and acetone (C).
• That means the alkene A must be:

Actually, the correct alkene is:

Ozonolysis gives:

4. Writing the Wrong Ozonolysis Reaction

Mistake:

Writing ozonolysis with $H_2O_2$ or wrong reducing agent, or forgetting to show both products.

Why it’s wrong:

Ozonolysis can be done with reductive workup (Zn/H₂O or dimethyl sulfide) to give aldehydes/ketones, or oxidative workup (H₂O₂) to give carboxylic acids. The problem implies aldehydes/ketones (since Fehling’s and iodoform are tested), so reductive workup is correct.

How to avoid:

• Write the reaction clearly:

• Label A, B, C correctly.

5. Writing the Iodoform Reaction Incorrectly

Mistake:

Writing incomplete or unbalanced equations for iodoform formation.

Why it’s wrong:

The iodoform reaction involves triiodomethyl anion formation and cleavage. Students often forget the base ($NaOH$) or write the product as $CHI_3$ without showing the carboxylate salt.

How to avoid:

• For acetaldehyde (B):

• For acetone (C): …