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Question #44139

Difficulty: 8/10 | 🟢 Free

The Question

Consider the compound \ceX\ce{X} = 4-nitrophenylglyoxal (\ceC8H5NO4\ce{C8H5NO4}). \ceX\ce{X} undergoes the following reaction sequence:

  1. \ceX→2. H3O+1. conc. NaOH, Δ\ceY\ce{X} \xrightarrow[2. \text{ H}_3\text{O}^+]{1. \text{ conc. NaOH, } \Delta} \ce{Y} (Major organic product)
  2. 1 mole of \ceY→excess Sn/HCl\ceZ11 \text{ mole of } \ce{Y} \xrightarrow{\text{excess Sn/HCl}} \ce{Z1}
  3. 1 mole of \ceY→2. H2O1. excess LiAlH4, THF, Δ\ceZ21 \text{ mole of } \ce{Y} \xrightarrow[2. \text{ H}_2\text{O}]{1. \text{ excess LiAlH}_4\text{, THF, } \Delta} \ce{Z2}

Let n1n_1 be the total decrease in the oxidation state of the substrate per molecule of \ceY\ce{Y} in the formation of \ceZ1\ce{Z1}. Let n2n_2 be the total decrease in the oxidation state of the substrate per molecule of \ceY\ce{Y} in the formation of \ceZ2\ce{Z2}.

Calculate the value of n1+n2n_1 + n_2.

Options

  • A.

    10

  • B.

    12

  • C.

    14

  • D.

    16

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Electron Transfer Accounting|Cannizzaro Mechanism|Reduction Selectivity (Sn/HCl vs LiAlH4)

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to track oxidation states through complex multi-step transformations, specifically contrasting metal-catalyzed electron transfer (Sn/HCl) with nucleophilic hydride transfer (LiAlH4).

The primary trap lies in the chemoselectivity of LiAlH4.

Students frequently memorize that 'reducing agents turn nitro groups into amines.' While true for Sn/HCl or H2/Pd, LiAlH4 anomalously reduces aromatic nitro compounds to azo compounds (Ar-N=N-Ar) due to the condensation of intermediate nitroso and hydroxylamine species in the basic reaction medium.

Failing to account for this specific exception leads to an incorrect electron count (calculating 10 instead of 8 for n2, leading to the distractor 16).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterRedox Reactions - Class 11 Part 2
ConceptElectron Transfer Accounting
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree