Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

17. Amines

Amine Reactions

GOB Chemistry

17. Amines

Amine Reactions

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3:11 minutes

Problem 38b

Textbook Question

Write the balanced chemical equations for the (1) reaction of each of the following amines with water and (2) neutralization with HBr:
b. propylamine

Verified step by step guidance

Step 1: Understand the problem. Propylamine (C₃H₇NH₂) is a primary amine. When it reacts with water, it acts as a weak base, accepting a proton (H⁺) from water to form its conjugate acid and hydroxide ions (OH⁻). In the neutralization reaction with HBr, the amine reacts with the acid to form an ammonium salt.

Step 2: Write the reaction of propylamine with water. The general reaction for an amine (RNH₂) with water is: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. For propylamine, substitute R with C₃H₇: C₃H₇NH₂ + H₂O ⇌ C₃H₇NH₃⁺ + OH⁻.

Step 3: Balance the reaction of propylamine with water. Ensure that the number of atoms on both sides of the equation is equal. In this case, the equation is already balanced as written: C₃H₇NH₂ + H₂O ⇌ C₃H₇NH₃⁺ + OH⁻.

Step 4: Write the neutralization reaction of propylamine with HBr. The general reaction for an amine (RNH₂) with a strong acid (HX) is: RNH₂ + HX → RNH₃⁺X⁻. For propylamine, substitute R with C₃H₇ and X with Br: C₃H₇NH₂ + HBr → C₃H₇NH₃⁺Br⁻.

Step 5: Balance the neutralization reaction. Ensure that the number of atoms and charges are balanced. In this case, the equation is already balanced as written: C₃H₇NH₂ + HBr → C₃H₇NH₃⁺Br⁻.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Balanced Chemical Equations

A balanced chemical equation represents a chemical reaction where the number of atoms for each element is the same on both sides of the equation. This is crucial for adhering to the law of conservation of mass. Balancing involves adjusting coefficients to ensure that reactants and products have equal numbers of each type of atom.

Amines and Their Reactions

Amines are organic compounds derived from ammonia by replacing one or more hydrogen atoms with alkyl or aryl groups. They can act as bases, reacting with water to form ammonium ions and hydroxide ions. Understanding the behavior of amines in aqueous solutions is essential for predicting their reactions with acids like HBr.

Neutralization Reactions

Neutralization reactions occur when an acid reacts with a base to produce water and a salt. In the case of amines, when they react with strong acids like HBr, they form an ammonium salt. This concept is fundamental in understanding how amines behave in acidic environments and the products formed from such reactions.