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Question #45089

Difficulty: 8/10 | 🟢 Free

The Question

An amount of 24.6 g24.6\text{ g} of nitrobenzene (\ceC6H5NO2\ce{C6H5NO2}) is treated with an excess of \ceLiAlH4\ce{LiAlH4} in anhydrous diethyl ether at 298 K298\text{ K}. The reaction goes to completion, yielding a major organic product Y\textbf{Y}. Subsequently, Y\textbf{Y} is subjected to complete reductive cleavage using \ceSnCl2\ce{SnCl2} in aqueous \ceHCl\ce{HCl}, followed by neutralization with excess \ceNaOH\ce{NaOH}, to yield a single organic compound Z\textbf{Z}. Which of the following correctly represents the theoretical yield of Y\textbf{Y} (in grams) and the number of moles of \ceBr2\ce{Br2} required for the complete bromination of one mole of Z\textbf{Z} in an aqueous medium? (Given molar masses in g/mol: \ceC=12\ce{C}=12, \ceH=1\ce{H}=1, \ceN=14\ce{N}=14, \ceO=16\ce{O}=16)

Options

  • A.

    Yield of Y=18.2 g\textbf{Y} = 18.2\text{ g}; Moles of \ceBr2=3\ce{Br2} = 3

  • B.

    Yield of Y=18.6 g\textbf{Y} = 18.6\text{ g}; Moles of \ceBr2=3\ce{Br2} = 3

  • C.

    Yield of Y=36.4 g\textbf{Y} = 36.4\text{ g}; Moles of \ceBr2=3\ce{Br2} = 3

  • D.

    Yield of Y=18.2 g\textbf{Y} = 18.2\text{ g}; Moles of \ceBr2=6\ce{Br2} = 6

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Reaction Intermediates Stability|Reduction Selectivity (Sn/HCl vs LiAlH4)|Reaction Progress Coordinate

Why This Question Is Tricky

Common Mistakes

This problem tests a critical exception in reduction selectivity and stoichiometric accounting.

Students often rote-learn that \ceLiAlH4\ce{LiAlH4} reduces \ce−NO2\ce{-NO2} to \ce−NH2\ce{-NH2}.

However, for aryl nitro compounds in anhydrous media, the reduction halts at the azo stage (azobenzene) due to a bimolecular pathway.

Failing to account for this intermediate alters the stoichiometric equivalence, halving the expected molar yield (Trap B assumes aniline is formed directly; Trap C forgets the stoichiometric coefficient division).

Furthermore, it tests the knowledge of reductive cleavage of azo bonds and the subsequent electrophilic aromatic substitution of the resulting aniline, ensuring students read carefully to calculate bromination per mole of Z\textbf{Z}, not Y\textbf{Y} (Trap D).

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Question Information

SubjectChemistry
ChapterOrganic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
ConceptReaction Intermediates Stability
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree