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Section
III
Unit 7
Questions 31 - 35:
Question 31
D In the first example the group with highest priority on the left
carbon is CH3 and on the right carbon is the CH2CH3. As these are on the same
side of the double bond, this is a Z isomer. In the second example the F has
greater priority than the CH3 and the Br is greater than the Cl. As the F and
Br are on the same side, this is again a Z isomer.
Question 32
A Don't worry about the numbers here. Look at each double bond and
assign priority as before. In this case the carbon groups are E to each other
on each double bond.
Question 33
C Again the numbers are
irrelevant. Only one of the double bonds here can show isomerism (on the left
of the molecule). The double bond on the right has 2 H's attached to one carbon
and so can not show geometric isomerism, as stated in the question.
Question 34
C This is the only bond where the geometry
changes. As seen it goes from Z to E.
Question 35
B The two
rings in the first example are opposite each other and hence this is the E
isomer. In the second example the Cl has greater priority than the methyl group
and this is opposite the other carbon group. Hence both are E.
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