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Iodoform on heating with KOH gives (a) \(\mathrm{CH}_{3} \mathrm{COOK}\) (b) HCOOK (c) HCHO (d) \(\mathrm{CH}_{3} \mathrm{CHO}\)

Short Answer

Expert verified
Iodoform on heating with KOH gives potassium acetate, \\( \mathrm{CH}_{3} \mathrm{COOK} \\\), which is option (a).

Step by step solution

01

Understand Iodoform Reactivity

Iodoform, also known as triiodomethane (CHI鈧), reacts with a strong base like KOH in a reaction known as the iodoform test. This reaction involves the formation of carboxylate ions.
02

Identify the Reaction Process

In the presence of KOH, the iodoform undergoes a basic reaction. This involves the release of iodide ions and the cleavage of the carbon-hydrogen bond, ultimately forming potassium salts of carboxylic acids.
03

Determine the Product

When iodoform is heated with KOH, the reaction leads to the formation of potassium acetate (CH鈧僀OOK). The methyl group from CHI鈧 contributes to the formation of the acetate ion (CH鈧僀OO鈦), which combines with K鈦 from KOH.
04

Confirm Final Answer

The correct product of heating iodoform with KOH is potassium acetate, not any other compound listed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iodoform Test
The Iodoform test is a fascinating chemical reaction that is used to identify methyl ketones or secondary alcohols with a methyl group adjacent to the carbonyl group. It revolves around the reactivity of iodoform, known chemically as triiodomethane (CHI鈧). When CHI鈧 interacts with a strong base like potassium hydroxide (KOH), it undergoes a distinct transformation. This is due to the ability of the iodine to easily leave the compound, facilitating significant structural changes. During the test, if the compound being tested contains the necessary methyl group adjacent to the carbonyl, iodoform 鈥 a yellow crystalline solid 鈥 will form as precipitate. This visible change often confirms the presence of such functional groups. However, in the scenario given in the exercise, iodoform itself takes part in the reaction and produces different compounds by yielding carboxylate ions and becoming potassium acetate.
Carboxylate Ions Formation
The formation of carboxylate ions is a crucial aspect of various organic reactions, including the iodoform test. Carboxylate ions are formed when a carboxylic acid loses a proton (H鈦), leaving behind the anion form of the acid, denoted as RCOO鈦. In the scenario of iodoform reacting with KOH, the carbon-hydrogen bond in triiodomethane breaks, facilitating the formation of these carboxylate ions. Here's how it works:
  • The hydroxide ions (OH鈦) from the KOH attack the carbon atom in the iodoform molecule.
  • This reaction leads to the removal of iodide ions and results in the formation of carboxylate ions.
These ions play a pivotal role in the creation of acetates, as they enable the binding of potassium (from KOH) with the acetate ion to form potassium acetate. Understanding carboxylate ion formation is essential for mastering many reactions involving carboxylic acids and their derivatives.
Potassium Acetate
Potassium acetate (CH鈧僀OOK) is the end product of heating iodoform with KOH. This compound is formed when the acetate ion (CH鈧僀OO鈦) binds with the potassium ion from KOH. Potassium acetate is widely used in various fields, including medicine and food additives, due to its properties and ease of formation. In chemical reactions such as the one noted in the exercise, it's important to recognize how the specific ions come together:
  • The methyl group from the iodoform provides the necessary component for the acetate structure.
  • Simultaneously, the iodide ions are displaced, leaving space for the formation of potassium acetate.
The reaction highlights the role of strong bases in facilitating the formation of stable salts like potassium acetate from more reactive substances like iodoform. This transformation is a perfect example of organic synthesis processes capable of producing useful compounds.

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Most popular questions from this chapter

Consider the following compounds: I. DDT II. gammexane III. carbon tetrachloride IV. chlorobenzene The correct sequence of these compounds in the increasing order of percentage of chlorine in them is (a) I, II, III, IV (b) IV, I, II, III (c) IV, II, I, III (d) III, I, II, IV

Match List I (compound) with List II (use) and select the correct answer using the codes given below the lists: List I List II (Compound) (Use) I. acetylsalicylic A. insecticide II. DDT B. drug III. naphthalene C. moth repellent IV. carbon tetrachloride D. fire extinguisher E. refrigerant (a) I-E, II-A, III-C, IV-D (b) I-B, II-C, III-D, IV-A (c) I-E, II-C, III-D, IV-A (d) I-B, II-A, III-C, IV-D

Which of the following represents the correct order of densities? (a) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{Cl}\) (c) \(\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CHCl}_{3}>\mathrm{CCl}_{4}\) (d) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CH}_{3} \mathrm{Cl}\)

Which of the following compounds is not chiral? [2004] (a) 2-chloropentane (b) 1 -chloropentane (c) 3 -chloro-2-methylpentane (d) 1 -chloro-2-methylpentane

Compound (A), \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{9} \mathrm{Br}\), gives a white precipitate when warmed with alcoholic \(\mathrm{AgNO}_{3}\). Oxidation of (A) gives an acid (B), \(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}\). (B) easily forms anhydride on heating. Identify the compound (A).

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