Are all alkene geometrical isomers achiral?
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I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.
stereochemistry
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show 1 more comment
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I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.
stereochemistry
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4
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You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
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– Ivan Neretin
Jan 14 at 13:06
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The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
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– Karl
Jan 14 at 21:51
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@Karl Sadly, that statement isn't true. See my answer below...
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– Zhe
Jan 16 at 14:27
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@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
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– Karl
Jan 16 at 22:15
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Such questions always remind me of this one, no offence intended: xkcd.com/169
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– Karl
Jan 16 at 22:19
|
show 1 more comment
$begingroup$
I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.
stereochemistry
$endgroup$
I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.
stereochemistry
stereochemistry
asked Jan 14 at 12:49
Abhay SharmaAbhay Sharma
91
91
4
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You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
$endgroup$
– Ivan Neretin
Jan 14 at 13:06
$begingroup$
The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
$endgroup$
– Karl
Jan 14 at 21:51
$begingroup$
@Karl Sadly, that statement isn't true. See my answer below...
$endgroup$
– Zhe
Jan 16 at 14:27
$begingroup$
@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
$endgroup$
– Karl
Jan 16 at 22:15
$begingroup$
Such questions always remind me of this one, no offence intended: xkcd.com/169
$endgroup$
– Karl
Jan 16 at 22:19
|
show 1 more comment
4
$begingroup$
You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
$endgroup$
– Ivan Neretin
Jan 14 at 13:06
$begingroup$
The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
$endgroup$
– Karl
Jan 14 at 21:51
$begingroup$
@Karl Sadly, that statement isn't true. See my answer below...
$endgroup$
– Zhe
Jan 16 at 14:27
$begingroup$
@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
$endgroup$
– Karl
Jan 16 at 22:15
$begingroup$
Such questions always remind me of this one, no offence intended: xkcd.com/169
$endgroup$
– Karl
Jan 16 at 22:19
4
4
$begingroup$
You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
$endgroup$
– Ivan Neretin
Jan 14 at 13:06
$begingroup$
You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
$endgroup$
– Ivan Neretin
Jan 14 at 13:06
$begingroup$
The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
$endgroup$
– Karl
Jan 14 at 21:51
$begingroup$
The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
$endgroup$
– Karl
Jan 14 at 21:51
$begingroup$
@Karl Sadly, that statement isn't true. See my answer below...
$endgroup$
– Zhe
Jan 16 at 14:27
$begingroup$
@Karl Sadly, that statement isn't true. See my answer below...
$endgroup$
– Zhe
Jan 16 at 14:27
$begingroup$
@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
$endgroup$
– Karl
Jan 16 at 22:15
$begingroup$
@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
$endgroup$
– Karl
Jan 16 at 22:15
$begingroup$
Such questions always remind me of this one, no offence intended: xkcd.com/169
$endgroup$
– Karl
Jan 16 at 22:19
$begingroup$
Such questions always remind me of this one, no offence intended: xkcd.com/169
$endgroup$
– Karl
Jan 16 at 22:19
|
show 1 more comment
1 Answer
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While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.
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1 Answer
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1 Answer
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$begingroup$
While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.
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add a comment |
$begingroup$
While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.
$endgroup$
add a comment |
$begingroup$
While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.
$endgroup$
While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.
Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.
answered Jan 14 at 14:19
ZheZhe
12.2k12450
12.2k12450
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4
$begingroup$
You want a rule which is not there. Alkenes, like nearly all other classes of organic compounds, may or may not be chiral. Let's put it this way: having a double bond is irrelevant to chirality.
$endgroup$
– Ivan Neretin
Jan 14 at 13:06
$begingroup$
The statement in your book is a truism: Correct, but does not transport any additional insight into the underlying concepts. Except as an exercise question: Why do these isomers not show a stereoisomery?
$endgroup$
– Karl
Jan 14 at 21:51
$begingroup$
@Karl Sadly, that statement isn't true. See my answer below...
$endgroup$
– Zhe
Jan 16 at 14:27
$begingroup$
@Zhe I would say that calling a cumulated diene an alkene is also a truism: Correct, but not at all helpful. ;-) If it is correct. Because the central carbon atom is sp hybridised.
$endgroup$
– Karl
Jan 16 at 22:15
$begingroup$
Such questions always remind me of this one, no offence intended: xkcd.com/169
$endgroup$
– Karl
Jan 16 at 22:19