Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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
Transcribed Image Text:**Transcription for Educational Website:**
**Title: Determining Major Organic Products in Haloform Reactions**
**Text:**
"Indicate the major organic products from the reaction shown."
**Reaction Illustration Explanation:**
The image depicts the structure of a ketone with the molecular formula of a methyl ketone (acetone) on the left side. The ketone features a carbonyl group (C=O) with an adjacent trimethyl-methane group (tert-butyl group).
An arrow points to the right indicating a chemical reaction, with the following reagents written above it:
- **NaOH:** Sodium hydroxide, a strong base.
- **Br₂ (excess):** Bromine, indicated to be in excess.
This reaction involves the transformation of the given methyl ketone in the presence of sodium hydroxide and an excess of bromine, suggesting a haloform reaction. This scenario typically results in the formation of a carboxylate ion, derived from the original ketone, and a haloform molecule (in this case, bromoform).
**Discussion:**
The reaction depicted represents a classic example of a haloform reaction, where a methyl ketone is subjected to halogenation followed by basic hydrolysis. This leads to the cleavage of the carbon-carbon bond adjacent to the carbonyl, forming two major products:
1. **Carboxylate Ion (Salt):** The acetyl group (part of the original ketone) gets converted into a carboxylate ion.
2. **Bromoform (CHBr₃):** Formed due to the action of excess bromine on the methyl group.
This reaction is commonly used in organic synthesis for the identification and transformation of methyl ketones.
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