Start Learning Free

Before You Begin

Question #44036

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following two independent reaction sequences aimed at synthesizing sterically hindered primary amines:

\textbf{Reaction I:} Potassium phthalimide + 2-Bromo-2-phenylpropane →DMF,353 K\xrightarrow{\text{DMF}, 353\text{ K}} Major nitrogen-containing product PP + Major non-nitrogenous organic product QQ

\textbf{Reaction II:} 2-Methyl-2-phenylpropanamide →Br2,NaOH(aq),343 K\xrightarrow{\text{Br}_2, \text{NaOH(aq)}, 343\text{ K}} Major organic product RR

Identify PP, QQ, and RR.

Options

  • A.

    P=PhthalimideP = \text{Phthalimide}, Q=2-PhenylpropeneQ = \text{2-Phenylpropene}, R=2-Phenylpropan-2-amineR = \text{2-Phenylpropan-2-amine}

  • B.

    P=N-(2-Phenylpropan-2-yl)phthalimideP = \text{N-(2-Phenylpropan-2-yl)phthalimide}, Q=NoneQ = \text{None}, R=2-Phenylpropan-2-amineR = \text{2-Phenylpropan-2-amine}

  • C.

    P=PhthalimideP = \text{Phthalimide}, Q=2-PhenylpropeneQ = \text{2-Phenylpropene}, R=2-Phenylpropan-1-amineR = \text{2-Phenylpropan-1-amine}

  • D.

    P=N-(2-Phenylpropan-2-yl)phthalimideP = \text{N-(2-Phenylpropan-2-yl)phthalimide}, Q=NoneQ = \text{None}, R=2-Phenylpropan-1-amineR = \text{2-Phenylpropan-1-amine}

Check Your Answer

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

Hoffmann Bromamide Rearrangement|Gabriel Phthalimide Synthesis|Ammonolysis Complexity

Why This Question Is Tricky

Common Mistakes

The question tests the critical limitations of primary amine synthesis methods.

Students often blindly apply the Gabriel phthalimide synthesis to any primary amine target.

However, Gabriel synthesis relies on an SN2 mechanism, which completely fails with tertiary alkyl halides due to severe steric hindrance, leading exclusively to E2 elimination.

Conversely, the Hoffmann Bromamide Rearrangement bypasses this steric limitation because the C-N bond is already formed in the amide precursor, and the subsequent 1,2-alkyl shift to the electron-deficient nitrogen accommodates bulky tertiary groups perfectly.

The trap is assuming both methods are universally interchangeable for primary amine synthesis.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAmines - Class 12
ConceptHoffmann Bromamide Rearrangement
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree