Consider the following reactions:

'A' is:

1. Identification of Fragment 'B':

   Fragment 'B' is formed via the ozonolysis of compound 'A'. It gives a positive iodoform test (yellow precipitate with $$\text{I}_2 + \text{NaOH}$$) and a positive Tollens' test (silver mirror with $$\text{Ag}_2\text{O}$$). This uniquely identifies 'B' as acetaldehyde ($$\text{CH}_3\text{CHO}$$), a 2-carbon aldehyde containing a methyl carbonyl group.




2. Deducing the Carbon Count of Fragment 'C':

   Since the total molecular formula of 'A' is $$\text{C}_7\text{H}_{14}$$ and ozonolysis cleaves the double bond into two fragments:

   $$\text{Carbons in 'C'} = \text{Total Carbons in 'A'} - \text{Carbons in 'B'} = 7 - 2 = 5\text{ carbons}$$



3. Identification of Fragment 'C' and 'D':

   • 'C' shows a negative iodoform test, indicating it is not a methyl ketone ($$\text{R--COCH}_3$$).
   • Reduction of 'C' with $$\text{LiAlH}_4$$ ($$\text{LAH}$$) produces alcohol 'D'.
   • Alcohol 'D' reacts with Lucas reagent (anhydrous $$\text{ZnCl}_2$$ and conc. $$\text{HCl}$$) to yield white turbidity within 5 minutes, confirming that 'D' is a secondary ($$2^\circ$$) alcohol.

   A 5-carbon ketone that is not a methyl ketone must be symmetrical: 3-pentanone ($$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$$). Its reduction gives 3-pentanol (a $$2^\circ$$ alcohol).




4. Reconstructing Compound 'A':

   Combining the carbonyl carbons of acetaldehyde ($$\text{CH}_3\text{CH=O}$$) and 3-pentanone ($$\text{O=C}(\text{CH}_2\text{CH}_3)_2$$) by forming a double bond gives the structure of 'A':

   $$\text{CH}_3\text{CH}=\text{C}(\text{CH}_2\text{CH}_3)_2 \quad \text{(3-ethylpent-2-ene)}$$

Why Other Structural Options Are Incorrect:

• Isomers producing Methyl Ketones (e.g., 2-pentanone derivatives):

  If compound 'A' were structured to produce 2-pentanone ($$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$$), fragment 'C' would yield a bright yellow precipitate during the iodoform reaction. This directly contradicts the observed negative test result.




• Isomers producing Branched Aldehydes:

  If fragment 'C' were an aldehyde (such as 2-methylbutanal or 3-methylbutanal), its subsequent reduction with $$\text{LiAlH}_4$$ would produce a primary ($$1^\circ$$) alcohol. Primary alcohols do not produce turbidity with Lucas reagent at room temperature within 5 minutes, failing to match the experiment.




• Isomers producing Tertiary Alcohols:

  If 'D' were a tertiary ($$3^\circ$$) alcohol, it would react with Lucas reagent to produce white turbidity immediately (under 1 minute) rather than taking the characteristic 5 minutes required for secondary alcohol pathways.

Conclusion:

The only structure matching all chemical constraints, including carbon count, functional group classification, and targeted negative/positive tests, is 3-ethylpent-2-ene.

Answer: 3-ethylpent-2-ene