Three isomers A, B and C (mol. formula C??H??O?) gives the following results:
B or C →Decarboxylation→ D(C?H??)
A →Decarboxylation→ E(an isomer of D)
Both D and E are non-resolvable.
D or E →H?/KMnO?→ Benzoic acid
B is more reactive than C towards decarboxylation reaction.
A, B, C respectively are:

Option 1 - <p>Ph-CH₂-CH(CH₃)-COOH, Ph-CH(COOH)-CH₂CH₃, Ph-CH(CH₃)-CH₂COOH<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>Ph-CH(CH₃)-CH₂COOH, Ph-CH(COOH)-CH₂CH₃, Ph-CH₂-CH(CH₃)-COOH</p>
Option 3 - <p>Structures involving a benzene ring with different alkyl and carboxyl groups.<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>Ph-CH₂-CH(CH₃)-COOH, CH₃-C(Ph)(COOH)-CH₃, Ph-CH(CH₃)-CH₂-COOH</p>
5 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
R
5 months ago
Correct Option - 4
Detailed Solution:

D and E are non-resolvable.

D = P h - C H 2 - C H 2 - C H 3 ? H + / K M n O 4 P h - C O O H   (Benzoic Acid)

Or 

 

 

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Chemistry Ncert Solutions Class 12th 2023

Chemistry Ncert Solutions Class 12th 2023

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