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Hey all,
=20
Thanks to everybody that responded! I am always amazed at how helpful =
this
group is. Thanks to Alan Brown, Theodore Burkey, Craig Butts, John =
Decatur,
Pete Dormer, Charlie Fry, Szalontai Gabor, Gerd Gemecker, Steve Graham, =
Bob
Hanson, Mike Harmata, Patrick Hays, Neil Jacobsen, Bernhard Jaun, Alan
Kenwright, Ivan Keresztes, Tadeusz Molinski, Martha Morton, Guillermo =
Moyna,
Brian Myers, Peter Rinaldi, Dave Scott, Andy Soper, Jane Strouse, Raju
Subramanian, and Ed Waali.
=20
There was enough interest for me to summarize anonymized responses. I
include that below, but also want to say that I was led astray initially =
by
an overly restrictive definition of diastereotopic which limits it to
molecules with chiral centers only (in Merriam-Webster=92s online =
Medical
Dictionary =96 I=92m going to avoid making a disparaging comment about =
the
medical profession). My distillation of what I received plus what I =
have
since read is:=20
=20
Stereoisomers are two or more molecules with the same empirical formula =
in
which corresponding atoms between the stereoisomers exist in the same
bonding or chemical environment. The difference between two =
stereoisomers
lies in how these atoms are arranged in space with respect to each =
other.
=20
Homotopic atoms or groups always give the same molecule if one of the =
atoms
or groups in question in labeled and the molecule is rotated about an =
axis
or if rotation about a single bond occurs (or a combination thereof). =20
=20
Aside: there are rotational isomers, but you need low temperatures or =
high
barriers to rotation to resolve these isomeric states. =20
=20
The presence of homotopic atoms or groups does not imply the existence =
of
multiple stereoisomers (consider labeling one of the H=92s in methane, =
CH4).
=20
Stereoisomers can be either enantiomers (mirror images of one another) =
or
diastereomers (everything else).
=20
Enantiotopic atoms or groups give, upon labeling, stereoisomers that are
mirror images of one another (or that require rotation plus reflection).
Enantiotopic groups must necessarily occur in (at least) pairs, as
reflection is always required. A cool molecule I just thought of that =
has
two sets of four enantiotopic H=92s is [3.3.0]bicyclooct-7-ene (number
analogously to naphthalene, put the double bond in the middle of the
molecule). =20
=20
Diastereotopic atoms or groups give, upon labeling, stereoisomers that =
are
not mirror images of one another. This includes compounds that contain
non-mirrored chiral centers (we can=92t have an R and an S center that =
reflect
in the same plane of symmetry, i.e., a meso compound), but also includes
cis-trans isomers. Therefore, a stereogenic center can be pro-R or =
pro-S or
pro-E and pro-Z. This type of stereoisomerism is also called =
geometrical
isomerism (the IUPAC strongly discourages this terminology, instead =
favoring
=93cis-trans isomerism=94).
=20
Stereogenic means there is a locus that generates a stereoisomer and =
does
not necessarily imply a chiral center. =20
=20
If you have multiple (non-reflecting) chiral centers, inverting just one
center but preserving the handedness of the other chiral center(s) gives =
an
epimer. There is a more narrow definition of epimers in the literature
pertaining to sugars.
=20
There are other types of chirality such as helical chirality. Imagine
fusing benzene rings so that the ring on one side is fused to C1 and C2,
while on the other side the ring is fused to C3 and C4. If you build a =
model
with six rings fused like this, you can arrange it in one of two ways.
There is a good example in the literature with some functionalized =
biphenyls
with the ortho sites containing bulky groups. The barrier to rotation =
about
the center bond in the molecule is too great to overcome if bulky groups =
are
ortho to the ring attachment points.=20
=20
So the answer to my original question is that yes, the axial and =
equatorial
1H=92s in t-Bu-cyclohexane are indeed diastereotopic, so one can be =
labeled
pro-E and the other pro-Z.
=20
Jeff
=20
Jeff Simpson, Director
Department of Chemistry Instrumentation Facility=20
Massachusetts Institute of Technology
77 Massachusetts Avenue, 18-0090
Cambridge, MA, 02139
617-253-2016 (1806 lab)
_____ =20
From: Jeff Simpson [mailto:jsimpson@MIT.EDU]=20
Sent: Friday, February 08, 2008 4:31 PM
To: ammrl@ammrl.org
Subject: AMMRL: diastereotopic?
=20
Hey all,
=20
I am teaching a class in NMR and I was pondering the differences between
homotopic, enantiotopic, and diastereotopic atoms and groups.
=20
If I have t-butyllcyclohexane (with the t-butyl group equatorial at the =
1
position of the ring), are the H=92s on C4 diastereotopic? The =
definition I
am getting for diastereotopicity is that there has to be a chiral center =
in
the molecule somewhere else for diastereotopicity to occur.
=20
It is clear that the axial and equatorial H=92s on C4 should have =
different
chemical shifts, but are they diastereotopic? Is there another, perhaps
better, term to describe this proton pair? Do I just say they are
=93chemically distinct=94 because of the ring=92s chair conformation but =
NOT
diastereotopic? Clearly the two cannot be separated into pro-R and =
pro-S.
=20
Send me your vote/explanation, and I=92ll post a summary if there is
sufficient interest.
=20
Jeff
=20
Jeff Simpson, Director
Department of Chemistry Instrumentation Facility=20
Massachusetts Institute of Technology
77 Massachusetts Avenue, 18-0090
Cambridge, MA, 02139
617-253-2016 (1806 lab)
=20
RESPONSES BELOW:
=20
The H's on C4 give diastereomers (in this case cis/trans isomers) if
replaced individually, thus they're diastereotopic. Enantiotopic atoms =
or
groups are related by symmetry operations of the molecule, =
diastereotopic
atoms or groups aren't.
Hmmm. Been a bit since I thought a lot about this, but the situation =
you
give seems a classic example of a diastereotopic pair. The 2nd chiral
center is generated when you replace one of the protons in the pair with =
D.
A simpler condition for diastereotopic relation is the non-existence of =
a
plane of symmetry among the objects you are looking at. The existence of =
a
chiral center is obviously a good reason for that but not a "sine qua =
non"
condition. When you cool down the cyclohexene itself you will find
diastereotopic protons too.
The topicity of two constitutionally equivalent ligands (=3Disomorphous
ligands =3D atoms, groups etc. that are bound to equivalent binding =
partners
with the same connectivity etc.) is a consequence of symmetry (group
theory):
=20
If the point group of the molecule contains a symmetry element that
transforms one of the constitutionally equivalent groups into the other, =
the
following rules apply:
=20
Rule 1. If the symmetry element is a Cn axis, the two are homtopic.
=20
Rule 2. If the symmetry element is a mirror plane (sigma) or Sn axis
(including the center of inversion i=3DS2), the two are enantiotopic.
=20
Of course, rule 1 has precedence over rule 2, i.e. if both, a Cn and a
mirror plane transform one group /atom into the other, the two are
homotopic.
=20
Rule 3. If no symmetry element is present, or the two constitutionally
equivalent ligands are not transformed into each other by one of the
symmetry elements of the molecule, the two are diastereotopic.
=20
Diastereotopicity is therefore not dependent on the presence of =
chirality
centers: classical cis/trans ligands on double bonds or rings can be
diastereotopic although the molecule is achiral.
=20
For the same reason, when comparing molecules instead of =
constitutionally
equivalent groups within a molecule, we also call the cis/trans
stereoisomers (sometimes called configurational isomers) on double bonds =
and
rings diastereoisomers.
=20
Rule 3 above applies to your CH2 at C(4) in t-butylcyclohexane: there is =
no
symmetry element of the molecule that transforms the two H into each =
other
(they are both sitting in the mirror plane and transformed into =
themselves
by it). Whether you look at the conformer with the t-butyl group =
equatiorial
or axial does not matter: topicity is a property that is independent of
conformation because you are always allowed to consider the conformation =
of
highest possible symmetry when analyzing topicity.
There is no need for a chiral centre to engender diastereotopic =
methylene
protons (although if there IS a chiral centre, they will always exist).
=20
A simple example is a geminal diethyl group (say 1,1-diethyl =
phenylethane).=20
The methylene protons on the ethyl groups are diastereotopic, as (from =
their
'perspective') the adjacent quaternary centre has three different groups
attached, hence each methylene proton occupies a different magnetic
environment. Hence they are magnetically and chemically inequivalent.
=20
The same argument applies to the C2 and C3 protons of t-butyl =
cyclohexane
(the fact that a ring exists in the structure changes nothing here).=20
Unfortunately C4 is a special case, as there is a line of symmetry =
through
the molecule, so I do not believe they are diastereotopic as such (but I
could be wrong!)..... The important feature is that they still occupy
different magnetic and chemical environments (i.e. cis or trans to the
t-butyl), there is no dependence of the ring conformation (even flat, =
they
would still be chemically distinct). The fact they are attached to the =
same
carbon is really neither here nor there (consider the terminal protons =
on an
alkene!).
=20
I learned that you had to have at least 2 chiral carbons for a =
diasteriomer
to occur. Your molecule will always be the same molecule if you put the
butyl group up or down since all you have to do is spatially rotate it =
180
and voila, the same molecule. But that is as you said, only a chemistry
point of view.
What complicates this is that your molecule could be boat or chair or =
both
(in various proportions). If the conditions froze it in only one of =
those
positions (all boat for example), then putting the butyl axial or =
equitorial
does make them different molecules But only in an NMR sense, not =
chemically
and the spectra would be different. Freeze it in the chair mode gives
another spectrum and having both gives another. So you actually have 4
different forms of the molecule available in solution (boat axial, boat
equitorial, chair axial, chair equitorial), each with different chemical
shifts. =20
To make things even more interesting, what do you call the molecule =
which
has a hydrochloride ion pair on a fused ring molecule (a secondary or
tertiary amine base) that produces two distinct NMR spectra because the =
acid
can be alpha or beta to the amine group (like above or below an =
N-methyl)?=20
The test for diastereotopic groups is to replace each group (H in this =
case)
with a new group (isotopic substitution is the least ambiguous method) =
and
determine if two different stereoisomers are formed. Chiral centers are =
not
a required for compounds to be enantiomers or diastereomers. For =
example,
the two terminal hydrogens at the 1 position of propene are =
diastereotopic.
Check an organic text or better yet IUPAC definitions for more =
information.
Wikipedia has a reasonable discussion under diastereomers.
I would call them diastereotopic. If replacement of the atoms =
individually
with an isotope results in a pair of diastereomers, then the atoms are
diastereotopic. One does not need a chiral center. The two vinylic H =
atoms
on C1 of propene, for example, are also diastereotopic, because cis- and
trans- alkene isomers are diastereomers. The ring you have works exactly
like an alkene to place the C4 Hs cis or trans to the t-butyl group.
That's a good one, and I also question myself about this everytime I =
teach
it. You are right, these two protons are not different form 'real'
diastereotopic protons. Even if there was a 1:1 equilibrium between the =
two
chairs (i.e., the same ratio of equatorial vs. axial, which is =
impossible
with t-Bu), the protons would still be different (just as when you have =
a
chiral center next to a CH2).
=20
However, diastereotopic is 'reserved' for when you have chiral centers. =
This
is due to the fact that replacing one of the protons in molecule with =
one
chiral center for something else will generate diastereomers. In the =
case of
tbutylcyclohexane, replacing one of the
C4 protons with something else will still leave you with an optically-
inactive molecule (as you said, they cannot be labeled as pro-R or pro-
S...).
=20
Thus, these are just chemically (and magntically) non-equivalent =
protons.
The H4 hydrogens on t-butylcyclohexane are indeed diastereotopic =
due to
cis/trans (or Z/E) relationships. Likewise, the two H1 alkene hydrogens =
on
propene are also diastereotopic. Cis/trans isomers meet the definition =
for
diastereomers (stereoisomers which aren't even mirror images). =20
The H4=92s in t-butylcyclohexane are most assuredly diastereotopic. =
There are
two ways to talk about this: symmetry rules and replacement tests.
=20
The replacement test is easier to show. Say you replace the axial H4 =
with
D; t-Bu and D are now cis. Now replace the equatorial H4 with D; t-Bu =
and D
are now trans. Well, that=92s a pair of diastereomers!
=20
Using the symmetry rule argument, the two H=92s in a CH2 will be
diastereotopic if they CANNOT be exchanged by a mirror plane.
=20
The confusion arises because a stereocenter is only ONE way to get a
diastereotopic CH2. 2-bromobutane is a classic example; the H3=92s are
diastereotopic.
=20
Non-chiral cases are (1) any monosubstituted alkene (e.g. chloroethene).
The H=92s in the CH2 are diastereotopic (try the replacement test) or =
(2)
prochiral molecules (of which glycerol is the classic example; see =
Sanders
and Hunter). The replacement test helps here too.
By all means 'NO"! Achiral molecules (with or without 'chiral centers') =
and
chiral molecules alike may have diastereotopic H's.=20
=20
Rather than remembering rules for each and every case (C1, C2, meso,
pseudo-meso, etc) I always apply the 'substitution' test.=20
=20
Separately substitute each of the two H's at C4 (call then Ha and Hb) =
with
an arbitrary group (say 'X'), examine the two products and make 'the =
test':
=20
Are the two products...
=20
i) diastereomers =96> Ha and Hb are diastereotopic (this is the case =
for H's
at C4 of 4-t-butylcyclohexane - independent of the property that the
cyclohexane ring conformation is fixed)
=20
ii) enantiomers =96> Ha and Hb are chemically equivalent* and Ha and =
Hb are
enantiotopic.
=20
iii) identical =96> Ha and Hb are chemically equivalent and homotopic=20
=20
This is also very useful for examining other topicity properties, e.g. =
H-H J
coupling, etc.
=20
* in an achiral environment (e.g. CDCl3)=20
well ... it IS a CH2 group, where we normally would use the term
"diastereotopic".
However, this is not a flexible molecule, but rather a rigid
(ring) system with a fixed geometry. Therefore the axial and equatorial
positions are clearly chemically distinct. Not as extreme as, e.g., the
CH3 and CH2 group in ethanol (clear case!), but like the CH2 group in
ClHC=3DCH2. Again, this IS a CH2 group, but due to the fixed geometry =
(here
double bond instead of ring) the cis/trans (or E/Z) positions are more
"distinct" than "diastereotopic".
As a definition, I (!) would use "diastereotopic" in cases where =
chirality
is involved (i.e., stereoisomers in a narrower sense), and not use it in
matters of mere "configurational isomers".
I am attaching an html file which I believe will help you. It is from
Wikipedia on the www.
=20
I am sure you are aware that all material on the internet is subject to
error and ignorance, without the checking and review to which books and
other publications should be subject.
=20
I strongly suggest that you confirm all 3 definitions from a reputable
organic chemistry book. I recommend McMurray.
It's my understanding that if protons are not interchangable by a =
symmetry
operation, such as Cn or mirror plane or Sn2, or some fast rotation that
exchanges them, then the protons are not chemical shift equivalent and =
I've
always presumed this is equivalent to diastereotopic. I believe
diastereotopic simply means two molecules are non-mirror images and are
non-super-imposable. =C5nd for protons, then that fits the symmetry =
rules
above.
=20
There are plenty of cases where a chiral carbon is not necessary. Even =
in
glycerol, the CH2 protons are diastereotopic. So, if a cyclohexane ring =
is
locked in place, then there is no symmetry between axial and equatorial
protons. =20
=20
Homotopic protons are equivalent through an axis of rotation. =
Enantiotopic
are equivalent through some other symmetry operation and are only =
equivalent
in non-chiral environments.
t-butylcyclohexane is NOT chiral since it has a mirror plane through C-1 =
and
C-4. The hydrogen atoms on C-4 are different due to the chair =
conformation
of the ring (axial vs. equatorial) and are thus is different magnetic
environments.
My vote is chemically inequivalent because their oriantation relative to =
the
t-Bu group will always be different, regardless of the conformation of =
the
ring. I think that even if the ring were boat, half chair, twist boat =
etc.
the two hydrogens will always remain distinguishable from each other.
They are non-diastereotopic because the molecule is not chiral.
tert-butylcyclohexane doesn't have a stereocenter, but the equtorial and
axial protons in the molecule will have different chemical shifts =
because
they are diastereotopic per the definition...see the attached slide.
Since I am not an organic chemist, I became confused on the topic =
by
your question. So I just had a look at an organic spectroscopy text =
that
says =93It is not necessary that a molecule contain a chirla center for
methylene protons to be diastereotopic. =85.. The term diastereotopic =
may be
applied to a broad range of cases since diastereomers are defined as
stereoisomers, including geometric, that are not enantiomers.=94 Now I =
think
I am straightened out again, but no guarantees. The book I am looking =
at is
=93Introduction to Organic Spectroscopy=94 by Lambert, Shurvell, =
Lightner, and
Cooks, MacMillan Publishing, 1987, 70-71.
I would say they were not diastereotopic, but are clearly chemically
distinct by virtue of the fixed geometry set by the equatorial t-butyl
group. In fact the inequivalence of all the CH2's can be explained the =
same
way so it isn't necessary to invoke diastereotopism(?) at all in this =
case.
Rather than being pro-R and pro-S, the hydrogens on C4 are cis and trans =
to
the t-butyl on C1 (same for the hydrogens on C3/5 and C2/6).
=20
I would suggest that arguments relying on hydrogens being diasteriotopic
should only be used when free rotation makes arguments based on geometry
(cis/trans) inapplicable.
I would define these as diastereotopic for lack of a better term. The
molecule is "locked" at room temperature with the t-butyl equatorial. =
These
2 protons are then part of a mirror plane. Clearly we need another term =
to
define this type of chirality. This is not the only molecule with this
characteristic. Maybe we could add "apparent" or "axially" to =
diastereotopic
to define these without a new word. We would create a subcategory.=20
Assuming you have no other substituent(s) in the cyclohexane molecule, =
C1 is
only a pro-chiral center. Therefore, protons at C4 are not =
diastereotopic:
you can always draw a plane across C1-C4 such that C1-t-butyl and C1-H =
are
either part of (or become symmetric) the mirror plane. However, protons =
in
C2/C3 can be diastereotopic since a substitution would make C1 a chiral
center.
=20
Protons in C4 have distinct chemical shift simply because the ring =
adopts a
conformation.
They are diastereotopic. Simple substitution test. Replace each C4
hydrogen with deuterium. The relationships between the two resulting
structures is diastereomeric, therefore the hydrogens are =
diastereotopic.
A chiral center does not have to exist on the molecule. In the example =
you
gave, replacement of H with D on the carbon adjacent to the CH-tbu =
creates a
chiral center at that site (CH2 on one side and CHD on the other). They
would also be diastereotopic in the case of Et2CH-t-bu.
=20
Thus, the methylenes on C-2, C-6 and for similar reasons C-3 and C-5 are
diastereotopic. The fact that these protons are locked into equatorial =
and
axial positions by the ring and the t-bu group makes them chemically
non-equivalent. =20
=20
Replacement of either H on C-4 produces a molecule with a plane of =
symmetry,
so these are not diastereotopic, however, they are chemically =
non-equivalent
since the t-bu locks these into equatorial and axial positions.
I guess its just another example of the inadequacies of language.
Enantiomers with its mirror image/handedness concept great, but it =
doesn't
work when there are two sites in a molecule. So we invent another word.
Disastomer ? :).
We'd probably be better off calling them double enantiomers.
I'd probably vote to stop such name calling, but if backed into a corner =
I'd
guess it was diastereomeric because replacing one of the protons could
generate an enantiomer.=20
It might be clearer to call them conformational isomers.
The nonequivalent protons of a methylene group are often referred to as
"diastereotopic" protons.
=20
What is more important is that they are chemically non-equivalent (i.e.,
they are expected to have different chemical shifts). To be chemically
non-equivalent, the two protons are not interconverted by any symmetry
element of the molecule. In your example, the only symmetry element is =
a
mirror plane passing through both C-4 protons and C-1. This plane does =
not
interconvert the two protons, so they are not equivalent. They COULD =
have
the same chemical shift by coincidence, but they are not REQUIRED BY
SYMMETRY to have the same chemical shift. I find this simple test (for
symmetry) much easier to explain than words like "diastereotopic".
=20
A more difficult concept is "magnetically equivalent". This is only
relevant if two protons are chemically equivalent (symmetry-related) but
don't have the same J coupling to any THIRD proton you can pick in the
molecule. Then they are magnetically non-equivalent.
=20
=20
=20
=20
=20
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Content-Type: text/html;
charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
Hey =
all,
Thanks
to everybody that responded!=A0 I am always amazed at how helpful this =
group is.=A0
Thanks to Alan Brown, Theodore Burkey, Craig Butts, John Decatur, Pete =
Dormer, Charlie
Fry, Szalontai Gabor, Gerd Gemecker, Steve Graham, Bob Hanson, Mike =
Harmata, Patrick
Hays, Neil Jacobsen, Bernhard Jaun, Alan Kenwright, Ivan Keresztes, =
Tadeusz
Molinski, Martha Morton, Guillermo Moyna, Brian Myers, Peter Rinaldi, =
Dave
Scott, Andy Soper, Jane Strouse, Raju Subramanian, and Ed =
Waali.
There was enough interest for me to
summarize anonymized responses.=A0 I include that below, but also want =
to say
that I was led astray initially by an overly restrictive definition of
diastereotopic which limits it to molecules with chiral centers only (in =
Merriam-Webster’s
online Medical Dictionary – I’m going to avoid making a =
disparaging
comment about the medical profession).=A0 My distillation of what I =
received plus
what I have since read is:
Stereoisomers are two or more =
molecules with
the same empirical formula in which corresponding atoms between the
stereoisomers exist in the same bonding or chemical environment.=A0 The
difference between two stereoisomers lies in how these atoms are =
arranged in
space with respect to each other.
Homotopic atoms or groups always =
give the
same molecule if one of the atoms or groups in question in labeled and =
the
molecule is rotated about an axis or if rotation about a single bond =
occurs (or
a combination thereof).=A0
Aside: there are rotational =
isomers, but
you need low temperatures or high barriers to rotation to resolve these
isomeric states.=A0
The presence of homotopic atoms or =
groups
does not imply the existence of multiple stereoisomers (consider =
labeling one
of the H’s in methane, CH4).
Stereoisomers can be either =
enantiomers
(mirror images of one another) or diastereomers (everything =
else).
Enantiotopic atoms or groups give, =
upon
labeling, stereoisomers that are mirror images of one another (or that =
require rotation
plus reflection).=A0 Enantiotopic groups must necessarily occur in (at =
least) pairs,
as reflection is always required.=A0 A cool molecule I just thought of =
that has
two sets of four enantiotopic H’s is [3.3.0]bicyclooct-7-ene =
(number analogously
to naphthalene, put the double bond in the middle of the =
molecule).=A0=A0=A0=A0=A0=A0=A0=A0=A0
Diastereotopic atoms or groups =
give, upon
labeling, stereoisomers that are not mirror images of one another.=A0 =
This
includes compounds that contain non-mirrored chiral centers (we =
can’t
have an R and an S center that reflect in the same plane of symmetry, =
i.e., a
meso compound), but also includes cis-trans isomers.=A0 Therefore, a =
stereogenic
center can be pro-R or pro-S or pro-E and pro-Z.=A0 This type of =
stereoisomerism
is also called geometrical isomerism (the IUPAC strongly discourages =
this
terminology, instead favoring “cis-trans =
isomerism”).
Stereogenic means there is a locus =
that
generates a stereoisomer and does not necessarily imply a chiral =
center.=A0
If you have multiple =
(non-reflecting)
chiral centers, inverting just one center but preserving the handedness =
of the other
chiral center(s) gives an epimer.=A0 There is a more narrow definition =
of epimers
in the literature pertaining to sugars.
There are other types of chirality =
such as
helical chirality.=A0 Imagine fusing benzene rings so that the ring on =
one side is
fused to C1 and C2, while on the other side the ring is fused to C3 and =
C4. If
you build a model with six rings fused like this, you can arrange it in =
one of
two ways.=A0 There is a good example in the literature with some =
functionalized
biphenyls with the ortho sites containing bulky groups.=A0 The barrier =
to
rotation about the center bond in the molecule is too great to overcome =
if
bulky groups are ortho to the ring attachment points. =
So the answer to my original =
question is
that yes, the axial and equatorial 1H’s in t-Bu-cyclohexane are =
indeed
diastereotopic, so one can be labeled pro-E and the other =
pro-Z.
Jeff
Jeff Simpson, =
Director
Department of Chemistry =
Instrumentation
Facility
Massachusetts Institute of =
Technology
77 Massachusetts =
Avenue
Cambridge , MA , 02139
617-253-2016 (1806 =
lab)
From: Jeff =
Simpson
[mailto:jsimpson@MIT.EDU]
Sent: Friday, February =
08, 2008
4:31 PM
To: ammrl@ammrl.org
Subject: AMMRL: =
diastereotopic?
Hey all,
I am teaching a class in NMR and I was pondering the
differences between homotopic, enantiotopic, and diastereotopic atoms =
and
groups.
If I have t-butyllcyclohexane (with the t-butyl group
equatorial at the 1 position of the ring), are the H’s on C4
diastereotopic? The definition I am getting for diastereotopicity =
is that
there has to be a chiral center in the molecule somewhere else for
diastereotopicity to occur.
It is clear that the axial and equatorial H’s =
on C4
should have different chemical shifts, but are they =
diastereotopic? Is
there another, perhaps better, term to describe this proton pair? =
Do I
just say they are “chemically distinct” because of the =
ring’s
chair conformation but NOT diastereotopic? Clearly the two cannot =
be separated
into pro-R and pro-S.
Send me your vote/explanation, and I’ll post a =
summary
if there is sufficient interest.
Jeff
Jeff Simpson, Director
Department of Chemistry Instrumentation Facility =
Massachusetts Institute of =
Technology
77 =
Massachusetts
Avenue
Cambridge, MA , 02139
617-253-2016 (1806 lab)
=A0RESPONSES =
BELOW:
The
H's on C4 give diastereomers (in this case cis/trans isomers) if =
replaced
individually, thus they're diastereotopic. Enantiotopic atoms or groups =
are
related by symmetry operations of the molecule, diastereotopic atoms or =
groups
aren't.
Hmmm.=A0 Been a bit =
since I
thought a lot about this, but the situation you give seems a classic =
example of
a diastereotopic pair.=A0 The 2nd chiral center is generated when you =
replace one
of the protons in the pair with D.=A0
A
simpler condition for diastereotopic relation is the non-existence of a =
plane
of symmetry among the objects you are looking at. The existence of a =
chiral
center is obviously a good reason for that but not a "sine qua =
non"
condition. When you cool down the cyclohexene itself you will find
diastereotopic protons too.
The topicity of two constitutionally equivalent ligands =
(=3Disomorphous
ligands =3D atoms, groups etc. that are bound to equivalent binding =
partners with
the same connectivity etc.) is a consequence of symmetry (group =
theory):
If the point group of the molecule contains a symmetry element =
that
transforms one of the constitutionally equivalent groups into the other, =
the
following rules apply:
Rule 1. If the symmetry element is a Cn axis, the two are =
homtopic.
Rule 2. If the symmetry element is a mirror plane (sigma) or Sn =
axis
(including the center of inversion i=3DS2), the two are =
enantiotopic.
Of course, rule 1 has precedence over rule 2, i.e. if both, a Cn =
and a
mirror plane transform one group /atom into the other, the two are =
homotopic.
Rule 3. If no symmetry element is present, or the two =
constitutionally
equivalent ligands are not transformed into each other by one of the =
symmetry
elements of the molecule, the two are =
diastereotopic.
Diastereotopicity is therefore not dependent on the presence of =
chirality
centers: classical cis/trans ligands on double bonds or rings can be
diastereotopic although the molecule is =
achiral.
For the same reason, when comparing molecules instead of
constitutionally equivalent groups within a molecule, we also call the
cis/trans stereoisomers (sometimes called configurational isomers) on =
double
bonds and rings diastereoisomers.
Rule 3 above =
applies to
your CH2 at C(4) in t-butylcyclohexane: there is no symmetry element of =
the
molecule that transforms the two H into each other (they are both =
sitting in
the mirror plane and transformed into themselves by it). Whether you =
look at
the conformer with the t-butyl group equatiorial or axial does not =
matter: topicity
is a property that is independent of conformation because you are always
allowed to consider the conformation of highest possible symmetry when
analyzing topicity.
There is no need =
for a
chiral centre to engender diastereotopic methylene protons (although if =
there
IS a chiral centre, they will always =
exist).
A simple example is =
a
geminal diethyl group (say 1,1-diethyl phenylethane). =
The methylene =
protons on the
ethyl groups are diastereotopic, as (from their 'perspective') the =
adjacent
quaternary centre has three different groups attached, hence each =
methylene
proton occupies a different magnetic environment. Hence they are =
magnetically
and chemically inequivalent.
The same argument =
applies to
the C2 and C3 protons of t-butyl cyclohexane (the fact that a ring =
exists in
the structure changes nothing here).
Unfortunately C4 is =
a
special case, as there is a line of symmetry through the molecule, so I =
do not
believe they are diastereotopic as such (but I could be wrong!)..... The
important feature is that they still occupy different magnetic and =
chemical
environments (i.e. cis or trans to the t-butyl), there is no dependence =
of the
ring conformation (even flat, they would still be chemically distinct). =
The
fact they are attached to the same carbon is really neither here nor =
there
(consider the terminal protons on an =
alkene!).
I learned
that you had to have at least 2 chiral carbons for a diasteriomer to
occur. Your molecule will always be the same molecule if you put =
the
butyl group up or down since all you have to do is spatially rotate it =
180 and
voila, the same molecule. But that is as you said, only a =
chemistry point
of view.
What complicates this is that your molecule could be boat or chair or =
both (in
various proportions). If the conditions froze it in only one of =
those
positions (all boat for example), then putting the butyl axial or =
equitorial
does make them different molecules But only in an NMR sense, not =
chemically and
the spectra would be different. Freeze it in the chair mode gives =
another
spectrum and having both gives another. So you actually have 4 =
different
forms of the molecule available in solution (boat axial, boat =
equitorial, chair
axial, chair equitorial), each with different chemical shifts. =
To make things even more interesting, what do you call the molecule =
which has a
hydrochloride ion pair on a fused ring molecule (a secondary or tertiary =
amine
base) that produces two distinct NMR spectra because the acid can be =
alpha or
beta to the amine group (like above or below an =
N-methyl)?
The test for
diastereotopic groups is to replace each group (H in this case) with a =
new
group (isotopic substitution is the least ambiguous method) and =
determine if
two different stereoisomers are formed. Chiral centers are not a =
required
for compounds to be enantiomers or diastereomers. For example, the =
two
terminal hydrogens at the 1 position of propene are diastereotopic. =
Check
an organic text or better yet IUPAC definitions for more information. =
Wikipedia
has a reasonable discussion under =
diastereomers.
I would call them diastereotopic. If =
replacement of
the atoms individually with an isotope results in a pair of =
diastereomers, then
the atoms are diastereotopic. One does not need a chiral center. The two
vinylic H atoms on C1 of propene, for example, are also diastereotopic, =
because
cis- and trans- alkene isomers are diastereomers. The ring you have =
works
exactly like an alkene to place the C4 Hs cis or trans to the t-butyl =
group.
That's a good one, =
and I
also question myself about this everytime I teach it. You are right, =
these two
protons are not different form 'real' diastereotopic protons. Even if =
there was
a 1:1 equilibrium between the two chairs (i.e., the same ratio of =
equatorial
vs. axial, which is impossible with t-Bu), the protons would still be =
different
(just as when you have a chiral center next to a =
CH2).
However, =
diastereotopic is
'reserved' for when you have chiral centers. This is due to the fact =
that
replacing one of the protons in molecule with one chiral center for =
something
else will generate diastereomers. In the case of tbutylcyclohexane, =
replacing
one of the
C4 protons with =
something
else will still leave you with an optically- inactive molecule (as you =
said, they
cannot be labeled as pro-R or pro- S...).
Thus, these are =
just
chemically (and magntically) non-equivalent =
protons.
=A0=A0=A0=A0
The H4 hydrogens on t-butylcyclohexane are indeed diastereotopic due to
cis/trans (or Z/E) relationships.=A0 Likewise, the two H1 alkene =
hydrogens on
propene are also diastereotopic.=A0 Cis/trans isomers meet the =
definition for
diastereomers (stereoisomers which aren't even mirror images).=A0 =
The H4’s in =
t-butylcyclohexane are
most assuredly diastereotopic. There are two ways to talk about =
this:
symmetry rules and replacement tests.
The replacement test is easier to
show. Say you replace the axial H4 with D; t-Bu and D are now cis.
Now replace the equatorial H4 with D; t-Bu and D are now =
trans.
Well, that’s a pair of diastereomers!
Using the symmetry rule argument, =
the two
H’s in a CH2 will be diastereotopic if they CANNOT be exchanged by =
a
mirror plane.
The confusion arises because a
stereocenter is only ONE way to get a diastereotopic CH2. =
2-bromobutane
is a classic example; the H3’s are =
diastereotopic.
Non-chiral
cases are (1) any monosubstituted alkene (e.g. chloroethene). The
H’s in the CH2 are diastereotopic (try the replacement test) or =
(2)
prochiral molecules (of which glycerol is the classic example; see =
Sanders and
Hunter). The replacement test helps here =
too.
By all means 'NO"! Achiral molecules (with or without =
'chiral
centers') and chiral molecules alike may have diastereotopic =
H's.
Rather than remembering rules for each and every case (C1,
C2, meso, pseudo-meso, etc) I always apply =
the 'substitution' =
test.
Separately substitute each of the two H's at C4 (call then Ha =
and Hb)
with an arbitrary group (say 'X'),
examine the two products and make 'the =
test':
Are the two products...
i) diastereomers –> Ha and Hb =
are diastereotopic (this is the case =
for H's
at C4 of 4-t-butylcyclohexane
- independent of the property that the cyclohexane ring conformation is =
fixed)
ii) enantiomers –> Ha and Hb are chemically equivalent* and Ha
and Hb are enantiotopic.
iii) identical –> Ha and Hb
are chemically equivalent =
and homotopic
This is also very useful for examining other topicity =
properties, e.g.
H-H J coupling, =
etc.
* in an =
achiral
environment (e.g. CDCl3)
well ... it IS a =
CH2 group,
where we normally would use the term =
"diastereotopic".
However, this is =
not a
flexible molecule, but rather a rigid
(ring) system with =
a fixed
geometry. Therefore the axial and equatorial positions are clearly =
chemically
distinct. Not as extreme as, e.g., the
CH3 and CH2 group =
in ethanol
(clear case!), but like the CH2 group in ClHC=3DCH2. Again, this IS a =
CH2 group,
but due to the fixed geometry (here double bond instead of ring) the =
cis/trans
(or E/Z) positions are more "distinct" than
"diastereotopic".
As
a definition, I (!) would use "diastereotopic" in cases where
chirality is involved (i.e., stereoisomers in a narrower sense), and not =
use it
in matters of mere "configurational =
isomers".
I am attaching an =
html file
which I believe will help you.=A0 It is from Wikipedia on the =
www.
I am sure you are =
aware that
all material on the internet is subject to error and ignorance, without =
the
checking and review to which books and other publications should be =
subject.
I
strongly suggest that you confirm all 3 definitions from a reputable =
organic
chemistry book.=A0 I recommend McMurray.
It's my understanding that if protons are not interchangable by =
a
symmetry operation, such as Cn or mirror plane or Sn2, or some fast =
rotation
that exchanges them, then the protons are not chemical shift equivalent =
and
I've always presumed this is equivalent to diastereotopic. I believe
diastereotopic simply means two molecules are non-mirror images and are
non-super-imposable. =C5nd for protons, then that fits the =
symmetry
rules above.
There are plenty of cases where a chiral carbon is not =
necessary.
Even in glycerol, the CH2 protons are diastereotopic. So, if a
cyclohexane ring is locked in place, then there is no symmetry between =
axial
and equatorial protons.
Homotopic protons are equivalent through an axis of rotation.
Enantiotopic are equivalent through some other symmetry operation and =
are only
equivalent in non-chiral environments.
t-butylcyclohexan=
e
is NOT chiral since it has a mirror plane through C-1 and C-4. The
hydrogen atoms on C-4 are different due to the chair conformation of the =
ring
(axial vs. equatorial) and are thus is different magnetic =
environments.
My vote is =
chemically
inequivalent because their oriantation relative to the t-Bu group will =
always
be different, regardless of the conformation of the ring. I think that =
even if
the ring were boat, half chair, twist boat etc. the two hydrogens will =
always
remain distinguishable from each other.
They are =
non-diastereotopic
because the molecule is not chiral.
tert-butylcyclohexane
doesn't have a stereocenter, but the equtorial and axial protons in the
molecule will have different chemical shifts because they are =
diastereotopic
per the definition...see the attached =
slide.
=
;
Since I am not an organic chemist, I became confused on the topic by =
your
question. So I just had a look at an organic spectroscopy text =
that says
“It is not necessary that a molecule contain a chirla center for =
methylene
protons to be diastereotopic. ….. The term diastereotopic =
may be
applied to a broad range of cases since diastereomers are defined as
stereoisomers, including geometric, that are not =
enantiomers.” Now
I think I am straightened out again, but no guarantees. The book I =
am
looking at is “Introduction to Organic Spectroscopy” by =
Lambert,
Shurvell, Lightner, and Cooks, MacMillan Publishing, 1987, =
70-71.
I would say they =
were not
diastereotopic, but are clearly chemically distinct by virtue of the =
fixed
geometry set by the equatorial t-butyl group.=A0 In fact the =
inequivalence of all
the CH2's can be explained the same way so it isn't necessary to invoke =
diastereotopism(?)
at all in this case.=A0 Rather than being pro-R and pro-S, the hydrogens =
on C4
are cis and trans to the t-butyl on C1 (same for the hydrogens on C3/5 =
and
C2/6).
I would suggest =
that
arguments relying on hydrogens being diasteriotopic should only be used =
when
free rotation makes arguments based on =
geometry
(cis/trans) =
inapplicable.
I
would define these as diastereotopic for lack of a better term.=A0 The =
molecule
is "locked" at room temperature with the t-butyl =
equatorial.=A0 These 2
protons are then part of a mirror plane.=A0 Clearly we need another term =
to
define this type of chirality.=A0 This is not the only molecule with =
this
characteristic. Maybe we could add "apparent" or =
"axially"
to diastereotopic to define these without a new word.=A0 We would create =
a
subcategory.
Assuming you have no other =
substituent(s)
in the cyclohexane molecule, C1 is only a pro-chiral center. Therefore, =
protons
at C4 are not diastereotopic: you can always draw a plane across C1-C4 =
such
that C1-t-butyl and C1-H are either part of (or become symmetric) the =
mirror
plane. However, protons in C2/C3 can be diastereotopic since a =
substitution
would make C1 a chiral center.
Protons in C4 have distinct =
chemical shift
simply because the ring adopts a =
conformation.
They
are diastereotopic.=A0 Simple substitution test.=A0 Replace each C4 =
hydrogen with
deuterium.=A0 The relationships between the two resulting structures is
diastereomeric, therefore the hydrogens are =
diastereotopic.
A chiral center does not have to =
exist on
the molecule. In the example you gave, replacement of H with D on the =
carbon
adjacent to the CH-tbu creates a chiral center at that site (CH2 on one =
side
and CHD on the other). They would also be diastereotopic in the case =
of
Et2CH-t-bu.
Thus, the methylenes on C-2, C-6 =
and for
similar reasons C-3 and C-5 are diastereotopic. The fact that these =
protons are
locked into equatorial and axial positions by the ring and the t-bu =
group makes
them chemically non-equivalent.
Replacement
of either H on C-4 produces a molecule with a plane of symmetry, so =
these are
not diastereotopic, however, they are chemically non-equivalent since =
the t-bu
locks these into equatorial and axial =
positions.
I guess its just =
another
example of the inadequacies of language. Enantiomers with its mirror
image/handedness concept great, but it doesn't work when there are two =
sites in
a molecule.=A0 So we invent another word.=A0 Disastomer ? =
:).
We'd probably be =
better off
calling them double enantiomers.
I'd probably vote =
to stop
such name calling, but if backed into a corner I'd guess it was =
diastereomeric
because replacing one of the protons could generate an enantiomer. =
It
might be clearer to call them conformational =
isomers.
The nonequivalent =
protons of
a methylene group are often referred to as "diastereotopic" =
protons.
What is more =
important is
that they are chemically non-equivalent (i.e., they are expected to have
different chemical shifts).=A0 To be chemically non-equivalent, the two =
protons
are not interconverted by any symmetry element of the molecule.=A0 In =
your
example, the only symmetry element is a mirror plane passing through =
both C-4
protons and C-1.=A0 This plane does not interconvert the two protons, so =
they are
not equivalent.=A0 They COULD have the same chemical shift by =
coincidence, but
they are not REQUIRED BY SYMMETRY to have the same chemical shift.=A0 I =
find this
simple test (for symmetry) much easier to explain than words like =
"diastereotopic".
A more difficult =
concept is
"magnetically equivalent".=A0 This is only relevant if two =
protons are
chemically equivalent (symmetry-related) but don't have the same J =
coupling to
any THIRD proton you can pick in the molecule.=A0 Then they are =
magnetically
non-equivalent.
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