Question
Explain the reason for the solution of chloroform and acetone showing negative deviation from Raoult's law.
✓
Answer
Chloroform molecule cannot form hydrogen bonds with acetone molecule, hence the solution of chloroform and acetone shows negative deviation from Raoult's law:
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Name the alcohol that is used to make the following ester:
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ || \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}-\text{CH}-\text{CH}_3$
→$ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ || \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}-\text{CH}-\text{CH}_3$
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The semicarbazone of cyclobutanone
The semicarbazone of cyclobutanone
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Write the IUPAC name of the following compound:
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{CH}_3\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|\\\text{CH}_2=\text{CH}-\text{C}-\text{OH}\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{CH}_3$
→$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{CH}_3\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|\\\text{CH}_2=\text{CH}-\text{C}-\text{OH}\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }|\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{CH}_3$
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→