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$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions.
$\alpha$-D-glucose
$\beta$-D-glucose
Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is down, and, in the $\beta$ form, the $\ce{OH}...
27
The correct definition of chirality is given in the IUPAC gold book as follows:
chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane, σ = S1, a centre of inversion, i = S2, a rotation-...
25
Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested. However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding trans isomer.
trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-...
24
(1) As for the number of alkanes ($\ce{C_nH_{2n+2}}$), Table 1, which is extracted from the data reported in
S. Fujita, MATCH Commun. Math. Comput. Chem., 57,
299--340 (2007) (access free),
shows the comparison between two enumerations based on
Polya's theorem and on Fujita's proligand method.
The number of alkanes ($\ce{C_nH_{2n+2}}$)
as ...
24
I think Martin has provided an excellent answer, and I would like to supplement it with a few additional details and examples that might prove insightful.
So as I already mentioned in the comments, the definition of chirality is rooted in symmetry. A chiral compound can contain no improper axis of rotation ($S_n$), which includes planes of symmetry ($S_1 = \...
23
It isn't easy but it is an interesting research topic
Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for ...
21
Final update, all earlier edits incorporated.
Groundrules: Considering compounds with:
only carbon and hydrogen
only 4 bonds to carbon
There are 37 isomers without considering trans isomers; 49 when trans isomers are included. Also, many of these compounds seem extremely unstable and therefore unlikely to exist.
Note to self: check back in 20 years and ...
21
There are two things we need to understand before we can answer the question.
1) More highly substituted double bonds are generally more stable than less substituted double bonds.
This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$ hybridized carbon in the double bond. Electron density likes ...
16
No, knowing the mass ratios is not sufficient by itself. In the absence of additional information (for example, molar mass) that would only be enough information to determine the empirical formula, which is the formula that contains the smallest integral ratios of atoms. (E.g., the empirical formula of $\ce{C2H6}$ would be $\ce{CH3}$.) There are (...
16
take 2,3-dimethylbutane
In the case of different substituents on different or the same atom, we use a combination of the position and the name of the substituent, such as in
2-chloro-3-methoxy...
3-cyano-4-bromo..., etc.
Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier di, together with the positions, just ...
answered Apr 16 '15 at 14:06
Klaus-Dieter Warzecha
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14
The compound 2,3,7,8-tetramethylspiro[4.4]nonane has five potential stereogenic or pseudoasymmetric centres: the four C atoms with the methyl substituents and the spiro atom. In total there are $2^5=32$ possible configurations. However, many of these combinations are actually identical compounds.
A simple and obviously exhaustive (but also exhausting) ...
13
If we have one double bond in a hydrocarbon compound we have an olefin or alkene. Ethylene is the simplest example of this class of compounds. The carbons in the double bond and the 4 atoms attached to them lie in the same plane. One pair of cis-trans isomers is possible in compounds with a single double bond.
If we add a nother double bond directly on ...
12
In mass spectrometry, ions with the same nominal mass (but possibly differing exact masses) are called isobaric.
12
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),
The locants o, m, p are no longer recommended; the numerical locants ‘1 ,2-‘, ‘1 ,3-‘, and ‘1,4-‘ must be used in substitutive names. However, as an exception, the three isomers of xylene are still recognized as o-, m-, and p-...
11
How many different organic structures (from the pure theoretical
viewpoint) can be drawed with only 4 (exact) carbon atoms and
with/without hydrogen?
We could make strict rules like each carbon has exactly 4 bonds and get a specific answer, but this is not reality. There can be lone pair electrons and unpaired electrons. The octet rule is not stictly ...
11
Take 2,3 dimethylbutane for example. Why does one add the numerical multiplier di when one has already explicitly mentioned two locations? Why doesn't one just use 2,3 methylbutane?
Actually, it is the other way round. Multiplicative prefixes (di, tri, tetra, etc.) are always used in names to denote multiplicity of identical features in structures, whereas ...
11
In your example, the stereodescriptors R and S are already given. The remaining question is about the double bond which gives rise to cis-trans isomerism.
In the hierarchical Sequence Rules, we finally find
P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.
...
11
This scheme I just drew up specifically for you should answer your question.
Diastereomers are stereoisomers that are not enantiomers of each other. That includes conformers (geometric isomers that derive from single bond rotation; usually interconverting rapidly) and atropisomers (under which I would subsume E/Z isomers; they derive from hindered rotation ...
11
Real molecules surely do rotate around their single C-C bonds all the time(*). So yes, if two models can be superimposed after an internal rotation of such sort, this means they represent the same molecule, as is the case with your B and C.
This, BTW, is the reason why you should orient all substituents in a standard manner when drawing your Fischer ...
11
Gaurang's answer deals well with the theory part I just wanted to give a visual aid.
(Taken from Atropisomerism of naphthyl alcohol)
Let's call the rotating substituent on the naphtahlene a fan.
The fan has two black fins one red fin.
Now let us consider two conformations
Red fin at 3'o clock
And
Red fin at 9'o clock
If you were to take ...
10
Let me give you two helpful hints.
Propane
If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products ...
10
There are 12 different constitutional isomers:
There is also the further possibility of stereoisomerism in some of the compounds above.
3: Two diastereomers (cis- and trans-) are possible.
7, 8, 9: Geometrical isomerism is possible in these alkenes.
11: C-3 (the carbon with the chlorine) is a chiral centre, which leads to two possible enantiomers.
That ...
10
epimers as compounds which differ by configuration at only one carbon
Yes, at only one centre.
However isn't that the same thing as diastereomers?
No, not completely. Diastereomers differ at least at one, but at less than all stereocentres. If two compounds would differ at all stereocentres, they would be enantiomers.
Update
As far as epimerism in ...
answered Apr 21 '16 at 7:47
Klaus-Dieter Warzecha
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10
If you search a library for a copy of Inorganic Chemistry by Miessler and Tarr, they have an excellent diagram showing which explains the phenomenon.
$\ce{Cl-}$ ligands have lone electrons that can donate into the metal, as you said yourself. The trick is to recognize what effect this has on the $t_\mathrm{2g}^*$ and $t_\mathrm{2g}$ orbitals. Remember that ...
9
There are a few widespread misconceptions about this topic. The original question, edits, and comments give the impression that such misconceptions may also contribue to the confusion in this case. Therefore, we should like to clear up some of the misconceptions first.
High-octane fuel does not contain a high concentration of octane.
The composition of ...
9
I was taught that geometric (cis-trans) isomers are considered isomers because there is a high energy barrier to breaking the double bond...
There's a problem here. Geometrical isomers are isomers because the arrangement of different groups around the pi bond is different. Consider the following molecule:
It also has a pi bond, we might be tempted to say ...
9
Here is another approach to @loong's solution to the seven stereoisomers of this tetramethyl spiro[4.4.0]nonane. [This question is akin to one recently asked about tetramethylspiropentanes].
Using the arbitrary numbering scheme shown above and assigning C1 as having the R-configuration, the eight permutations for C1-4 are listed on the left of the diagram. ...
9
The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long ...
9
Consider 2,3-butanediol.
This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.
A carbon center is ...
8
I see a couple of problems here:
Beam diameter vs volume of the beverage
The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source ...
answered Mar 10 '15 at 19:43
Klaus-Dieter Warzecha
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