Certainly! Below you will find **detailed stepwise mechanisms** for both reactions, with all arrows, charges, and intermediates described in text (as I can't draw here). I will use R-CN for nitrile, R-COOR' for ester, and MeNH₂ for methylamine. --- ## 1. **Hydrolysis of Nitrile to Carboxylic Acid (Acidic Conditions)** ### **Step 1: Protonation of Nitrile** - **Reactants:** R-C≡N + H₃O⁺ - The lone pair on the nitrile nitrogen attacks a proton (H⁺) from the hydronium ion, resulting in a protonated nitrile (R-C≡N⁺H). ### **Step 2: Nucleophilic Addition of Water** - **Intermediate:** R-C≡N⁺H - A water molecule attacks the electrophilic carbon of the nitrile (now more electrophilic due to N protonation). - This forms an imidate intermediate: R-C(OH)=NH⁺. ### **Step 3: Proton Transfer** - One of the hydrogens from the attached water is transferred to the imine nitrogen, so you get R-C(OH)=N-H. ### **Step 4: Tautomerization to Amide** - The imidic acid (R-C(OH)=N-H) tautomerizes to the more stable amide (R-CONH₂). - This involves shifting a proton from the OH to the nitrogen, and the C=N double bond becomes a C=O. **Summary so far:** R-CN → [Protonated Nitrile] → [Nucleophilic Addition] → [Imidic Acid] → [Amide, R-CONH₂] --- ### **Step 5: Protonation of Amide Carbonyl** - **Reactants:** R-CONH₂ + H₃O⁺ - The carbonyl oxygen is protonated, making it a better electrophile (R-C(=O⁺H)-NH₂). ### **Step 6: Nucleophilic Addition of Water** - Water attacks the carbonyl carbon, leading to a tetrahedral intermediate (R-C(OH)₂-NH₂). ### **Step 7: Proton Transfers** - One proton from the newly added water is transferred to the amide nitrogen, converting NH₂ to NH₃⁺. ### **Step 8: Elimination of Ammonia** - The lone pair from one OH reforms the C=O, ejecting NH₃ as a leaving group. - Result: R-COOH (carboxylic acid). **Net Reaction:** R-CN → R-CONH₂ → R-COOH --- ## 2. **Aminolysis of Ester with Methylamine (MeNH₂): Formation of Amide** ### **Step 1: Nucleophilic Addition** - **Reactants:** R-COOR' + MeNH₂ - The lone pair on the nitrogen of MeNH₂ attacks the carbonyl carbon of the ester. - The C=O pi bond breaks, forming a tetrahedral intermediate: R-C(OH)(N(Me)H)-OR'. ### **Step 2: Elimination of Alcohol** - The lone pair on the oxygen reforms the C=O, kicking out the alcohol leaving group (R'OH). - The resulting intermediate is a protonated amide (R-CON(Me)H₂⁺). ### **Step 3: Deprotonation** - The alcohol leaving group (R'OH), or another molecule of MeNH₂, deprotonates the nitrogen of the amide, generating the neutral amide R-CONHMe. **Net Reaction:** R-COOR' + MeNH₂ → R-CONHMe + R'OH --- ### **Summary of Key Arrow-Pushing Steps** #### **Nitrile to Carboxylic Acid** 1. Curved arrow: N lone pair → H⁺ 2. Curved arrow: H₂O lone pair → C of nitrile (π bond shifts onto N) 3. Proton transfer from attached H₂O → N 4. Tautomerization: OH proton → N (shifts double bond from N=C to C=O) 5. Protonation of amide carbonyl O 6. Water attacks C=O 7. Proton transfer from H₂O → NH₂ 8. Lone pair on O reforms C=O, kicks off NH₃ #### **Ester to Amide** 1. N of MeNH₂ attacks ester C=O; π electrons to O 2. Tetrahedral intermediate forms 3. O reforms C=O, kicks off OR' as leaving group 4. Deprotonation of N by base (R'OH or MeNH₂) --- **Let me know if you need these mechanisms as drawn structures, or if you want this for base-catalyzed nitrile hydrolysis instead!**