CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
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Question
Chapter 17, Problem 1ASA
Interpretation Introduction
Interpretation:
The properties of a substance that definitely establish that it is a molecular material are to be stated.
Concept introduction:
Substances are classified on the basis of the kinds of bonding present in it. There are three types of substances.
- Ionic substances
- Molecular substances
- Metallic substances
Molecular substances have covalent bonds between the atoms, ionic substances have electrostatic forces between the ions, and metallic substances have metallic bonds.
Expert Solution & Answer
Answer to Problem 1ASA
Molecular substances are formed by the sharing of the electrons. They are volatile in nature, and do not conduct electricity either in solid or in the molten state.
Explanation of Solution
The properties of a substance that is molecular material are listed below.
- Molecular substances are formed by the sharing of the electron (covalent bond) between the atoms.
- Molecular substances are liquid at room temperature.
- Molecular substances are volatile in nature.
- Molecular substances do not conduct electricity either in solid or in the molten state.
- Molecular substances are soluble in a few organic solvents.
- Some molecular substances are polar in nature and they all are soluble in water.
Conclusion
Molecular are liquid at room temperature, volatile in nature, and do not conduct electricity either in solid or in the molten state. A few molecular substances are polar in nature.
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Complete the reaction in the drawing area below by adding the major products to the right-hand side.
If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead.
Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule
to represent each pair of enantiomers, using line bonds at the chiral center.
More...
No reaction.
my
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1. Na O Me
Click and drag to start
drawing a structure.
2. H
+
Predict the intermediate 1 and final product 2 of this organic reaction:
NaOMe
H+
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1
2
H
H
work up
You can draw 1 and 2 in any arrangement you like.
Note: if either 1 or 2 consists of a pair of enantiomers, just draw one structure using line bonds instead of 3D (dash and wedge) bonds at the chiral center.
Click and drag to start drawing a structure.
X
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dm
Predict the major products of this organic reaction:
1. NaH (20°C)
2. CH3Br
?
Some notes:
• Draw only the major product, or products. You can draw them in any arrangement you like.
• Be sure to use wedge and dash bonds where necessary, for example to distinguish between major products that are enantiomers.
• If there are no products, just check the box under the drawing area.
No reaction.
Click and drag to start drawing a structure.
G
Cr
Chapter 17 Solutions
CHM 111/112 LAB MANUAL >C<
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