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Question #44462

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from hept-6-enoic acid (\textbf{X}):

\textbf{X} $\xrightarrow{\text{i) } \ce{SOCl2} \text{, ii) sodium salt of N-hydroxypyridine-2-thione}}$ \textbf{Y}
\textbf{Y} $\xrightarrow{h\nu,\ \ce{[Co(NH3)5(NO2)]Cl2} \text{ (0.5 M)}}$ \textbf{Z} (Major organic product)
\textbf{Z} $\xrightarrow{\ce{H3O+},\ \Delta}$ \textbf{W} (Major organic product)

Given: $k_{\text{cyclization}}$ of the intermediate radical $\approx 1 \times 10^5 \text{ s}^{-1}$ ; $k_{\text{trapping}}$ by the Co(III) complex $\approx 1 \times 10^8 \text{ M}^{-1}\text{s}^{-1}$ .

Identify \textbf{W} and the linkage of the ambidentate ligand in \textbf{Z}.

Options

A.
\textbf{W} = hex-5-en-1-ol; \textbf{Z} contains a $\ce{C-O-N=O}$ linkage
B.
\textbf{W} = cyclopentylmethanol; \textbf{Z} contains a $\ce{C-O-N=O}$ linkage
C.
\textbf{W} = 6-nitrohex-1-ene; \textbf{Z} contains a $\ce{C-NO2}$ linkage
D.
\textbf{W} = (nitromethyl)cyclopentane; \textbf{Z} contains a $\ce{C-NO2}$ linkage

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Free Radical Substitution Selectivity|Ambidentate Linkage Isomerism|Electrophile vs Nucleophile

Why This Question Is Tricky

Common Mistakes

The problem tests two hidden variables.

First, the 'isolated radical logic' trap: students often memorize that the 5-hexenyl radical always cyclizes.

However, calculating the pseudo-first-order trapping rate ( $5 \times 10^7 \text{ s}^{-1}$ ) reveals it outcompetes cyclization ( $10^5 \text{ s}^{-1}$ ).

Second, the ambidentate linkage trap: students might assume the radical forms the thermodynamically stable $\ce{C-NO2}$ bond.

However, in the $\ce{[Co(NH3)5(NO2)]^2+}$ complex, the nitrite is N-bonded, sterically shielding the nitrogen.

The radical is forced to attack the exposed oxygen, forming an alkyl nitrite ( $\ce{C-O-N=O}$ ), which subsequently hydrolyzes to an alcohol.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Hydrocarbons - Class 11 Part 2
Concept	Free Radical Substitution Selectivity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free