
Iodoform Reaction Quiz
Authored by Khaleel Mohammed
Chemistry
10th Grade
Used 2+ times

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5 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the chemical formula for iodoform?
NaCl
CO2
H2O
CHI3
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Explain the mechanism of the iodoform reaction.
The iodoform reaction is a chemical reaction in which a methyl ketone is oxidized to form a carboxylic acid and a molecule of iodoform (CHI3) in the presence of iodine and a base.
The iodoform reaction is a chemical reaction in which a methyl ketone is reduced to form a carboxylic acid and a molecule of iodoform
The iodoform reaction is a chemical reaction in which a methyl ketone is hydrolyzed to form a carboxylic acid and a molecule of iodoform
The iodoform reaction is a chemical reaction in which a methyl ketone is dehydrated to form a carboxylic acid and a molecule of iodoform
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What type of compounds undergo the iodoform reaction?
Methyl ketones and alcohols with the CH3CO- group undergo the iodoform reaction.
Sodium chloride and sulfuric acid
Ethanol and water
Propanol and sodium hydroxide
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the role of hydroxide ions in the iodoform reaction?
Hydroxide ions have no role in the iodoform reaction
Hydroxide ions act as an acid in the iodoform reaction
Hydroxide ions inhibit the iodoform reaction
Hydroxide ions act as a base in the iodoform reaction, deprotonating the methyl ketone to form the iodoform precipitate.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How is the iodoform test used to identify the presence of certain functional groups?
The iodoform test is used to identify the presence of carboxylic acids or compounds with the COO- group.
The iodoform test is used to identify the presence of aldehydes or compounds with the COOH group.
The iodoform test is used to identify the presence of alcohols or compounds with the OH group.
The iodoform test is used to identify the presence of methyl ketones or compounds with the CH3CO- group. When a compound containing this group is treated with iodine and a base, it forms a yellow precipitate of iodoform, which confirms the presence of the functional group.
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