Step-by-step Solution:

1. Determine the structure of the intermediate compound J.

The problem provides the following information about compound J:

• Molecular Formula: $C_9H_8O_2$
• Gives effervescence on treatment with $NaHCO_3$: This is a characteristic test for a carboxylic acid group ($-COOH$). The acidic proton of the carboxylic acid reacts with bicarbonate to produce carbon dioxide gas.
• Gives a positive Baeyer's test: This indicates the presence of carbon-carbon unsaturation (a double or triple bond). Baeyer's reagent (cold, dilute, alkaline $KMnO_4$) is decolorized by compounds containing C=C or C≡C bonds.

Let's calculate the Degree of Unsaturation (DBE) for J ($C_9H_8O_2$): $DBE = C + 1 - \frac{H}{2} - \frac{X}{2} + \frac{N}{2}$ $DBE = 9 + 1 - \frac{8}{2} = 10 - 4 = 6$

A benzene ring accounts for 4 degrees of unsaturation (3 double bonds + 1 ring). A carboxylic acid group ($-COOH$) contains a carbonyl group (C=O), which accounts for 1 degree of unsaturation. The sum of DBEs from a benzene ring and a carboxylic acid group is $4 + 1 = 5$.

The calculated DBE is 6, so there is one remaining degree of unsaturation, which must be a C=C double bond, consistent with the positive Baeyer's test.

Now, let's assemble the structural components: a phenyl group ($C_6H_5-$), a carboxylic acid group ($-COOH$), and a C=C double bond (as part of a $-CH=CH-$ group).

The most stable and common structure that fits these components and the molecular formula is Cinnamic acid: $C_6H_5-CH=CH-COOH$ Let's verify the formula: $C = 6+1+1+1 = 9$; $H = 5+1+1+1 = 8$; $O=2$. The formula $C_9H_8O_2$ is correct.

Thus, the intermediate J is Cinnamic acid.

2. Determine the structure of compound K.

The question asks for compound K, which is formed from J. Although the specific reaction is not explicitly stated in the provided text (likely due to a transcription error from the original source), we can infer the reaction by examining the options. The options represent products of common reactions of cinnamic acid.

• J: Cinnamic acid ($C_6H_5-CH=CH-COOH$)

Let's analyze the options:

• A: 3-phenylpropanoic acid ($C_6H_5-CH_2-CH_2-COOH$). This would be the product of the reduction of the C=C bond in J (e.g., using $H_2/Pd$).
• B: 3-hydroxy-3-phenylpropanoic acid ($C_6H_5-CH(OH)-CH_2-COOH$). This would be the product of the hydration of the C=C bond in J (e.g., using dilute acid).
• C: Benzoic acid ($C_6H_5-COOH$). This is the product of the oxidative cleavage of the C=C bond in J. This reaction is typically carried out using a strong oxidizing agent like hot, alkaline $KMnO_4$ or ozonolysis followed by oxidative workup.
• D: Benzaldehyde ($C_6H_5-CHO$). This could be a product of ozonolysis followed by reductive workup, but the other product would be glyoxylic acid (CHO-COOH).

Among these possibilities, oxidative cleavage is a very common and important reaction for unsaturated compounds in the JEE syllabus. The reaction of cinnamic acid with a strong oxidizing agent like hot $KMnO_4$ cleaves the double bond.

The reaction is as follows: $C_6H_5-CH=CH-COOH \xrightarrow{KMnO_4, OH^-, \Delta} C_6H_5-COOH + HOOC-COOH$ The products are benzoic acid and oxalic acid. Under these strong oxidizing conditions, oxalic acid is further oxidized to carbon dioxide and water: $HOOC-COOH \xrightarrow{KMnO_4, \Delta} 2CO_2 + H_2O$

The major stable organic product, K, is benzoic acid.

3. Conclusion

Based on the analysis, compound J is cinnamic acid. A plausible and common reaction for cinnamic acid is oxidative cleavage, which yields benzoic acid. Benzoic acid corresponds to option C.

The structure of K is:

This matches option C.