# geometrical and opitcal isomer exam question pls help

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#1
Q2.Which one of the following can exhibit both geometrical and optical isomerism?
A (CH3)2C=CHCH(CH3)CH2CH3
B CH3CH2CH=CHCH(CH3)CH2CH3
C (CH3)2C=C(CH2CH3)2
D CH3CH2CH(CH3)CH(CH3)C=CH2

why is the answer b? ive drawn them all out but I still find it hard to see the chiral carbon and geometrical isomerism? how do I draw C
0
3 years ago
#2
Q2.Which one of the following can exhibit both geometrical and optical isomerism?
A (CH3)2C=CHCH(CH3)CH2CH3
B CH3CH2CH=CHCH(CH3)CH2CH3
C (CH3)2C=C(CH2CH3)2
D CH3CH2CH(CH3)CH(CH3)C=CH2

why is the answer b? ive drawn them all out but I still find it hard to see the chiral carbon and geometrical isomerism? how do I draw C
Geometric isomerism requires that each of the carbon atoms of the double bond have different substituents.
Optical isomerism requires that a carbon atom has four different groups attached.
Only B has both.
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#3
(Original post by charco)
Geometric isomerism requires that each of the carbon atoms of the double bond have different substituents.
Optical isomerism requires that a carbon atom has four different groups attached.
Only B has both.
thanks is the chiral centre on B on
CH3CH2CH=CHCH(CH3)CH2CH3
0
3 years ago
#4
thanks is the chiral centre on B on
CH3CH2CH=CHCH(CH3)CH2CH3
0
1 year ago
#5
why is it not A then ? I think it does have a chiral center with it and is also an optical isomer.
Last edited by az2004583; 1 year ago
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1 year ago
#6
(Original post by az2004583)
why is it not A then ? I think it does have a chiral center with it and is also an optical isomer.

A. (CH3)2C=CHCH(CH3)CH2CH3

does indeed have a chiral carbon (shown above)

But it has no stereoisomers (cis, trans) due to lack of rotation about the double bond as there are two identical groups attached to the first carbon of the double bond.
0
1 year ago
#7
(Original post by charco)

A. (CH3)2C=CHCH(CH3)CH2CH3

does indeed have a chiral carbon (shown above)

But it has no stereoisomers (cis, trans) due to lack of rotation about the double bond as there are two identical groups attached to the first carbon of the double bond.
okay yes , Thank you
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