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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-...
30
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-...
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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 ...
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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 $\ce{OH}...
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I think Martin has provided an excellent answer, and I would like to supplement it with a few additional details and examples that might prove insightful.
So as I already mentioned in the comments, the definition of chirality is rooted in symmetry. A chiral compound can contain no improper axis of rotation ($S_n$), which includes planes of symmetry ($S_1 = \...
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Racemization isn't "exact," but rather very very close to equality. It is just simple probability.
Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations ...
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(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 ...
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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 ...
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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 ...
17
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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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 (...
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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 ...
14
It is neither reactant nor catalyst, and equilibrium concepts do not apply
There are processes other than photochemistry that to go in a direction away from equilibrium. They require mechanical or electrical work, and there is no established way to incorporate these into a chemical equation.
If we write a conceptual equation for charging a battery, it could ...
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 ...
13
Circularly polarized light is like a helix that twists through space. The two components are mirror images of each other.
Now, every molecule interacts with both the left-handed twisting light and the right handed twisting light. The interactions differ. Every molecule, in different orientations, interact differently with the left-handed and the right-handed ...
12
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 ...
12
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 ...
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
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 ...
11
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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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 ...
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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 ...
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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 ...
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