Page 37 - Q & A STPM 2022 Chemistry

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Thus, the more alkyl groups attached to the carbon atom that carries the
lone electron the more stable is the free radical. Take the example of the
.
butyl free radicals, C 4 H 9 below:
. . .
CH 3 CH 2 CH 2 —C—H CH 3 CH 2 —C—CH 3 CH 3 —C—CH 3
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H H CH 3
1° free radical 2° free radical 3° free radical
In the primary free radical there is one electron-donating group,
CH 3 CH 2 CH 2 !. In the secondary free radical there are two electron-
donating groups, CH 3 CH 2 ! and CH 3 !. In the tertiary free radical there
are three electron-donating groups, three CH 3 ! groups. As a result, the
stability of the free radical increases in the order: 1° < 2° < 3°.

14.6 Molecular Structure and Its Effect on Physical Properties

Section A Multiple-choice Questions

Question 1
The boiling point of pentane is higher than its isomer, 2,2-dimethylpropane.
Why?
A The bonds in pentane is more polar.
B A pentane molecule is linear while 2,2-dimethylpropane is branched.
C The molar mass of pentane is higher than 2,2-dimethylpropane.
D 2,2-dimethylpropane is optically active but pentane is not.

Answer: B
The strength of van der Waals force increases with the total number of electrons
in the molecule as well as the total surface area of the molecule.
Both pentane and 2,2-dimethylpropane are hydrocarbons. Their molecules are
non-polar.
Both the isomers have the same molar mass.
There is no chiral centre in 2,2-dimethylpropane. Hence, it is not optically
active.
Being linear, pentane has a larger surface area than 2,2-dimethylpropane,
which is more spherical. This increases the strength of the van der Waals force
between pentane molecules. Term

3

Pentane 2,2-Dimethlpropane

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14 Q&A STPM Chem T3 Layout.indd 353 4/19/19 5:29 PM