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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-...

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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 ...

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Generally speaking, graph-theoretical enumeration aims at counting chemical compounds as graphs (2D structures). In other words, it is concerned with constitutional (or structural) isomers. The most famous method for combinatorial enumeration of graphs is based on Polya's theorem: G. Polya and R. C. Read, Combinatorial Enumeration of Groups, Graphs, and ...

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First, let's establish that we're all talking about octahedral complexes to eliminate any possible confusion. Fix $\ce{c}$. You can have either $\ce{b}$ trans or $\ce{a}$ trans. If $\ce{b}$ is trans to $\ce{c}$, there's only one distinct way to arrange the remaining $\ce{b}$ and three $\ce{a}$'s. So that's one isomer. If $\ce{a}$ is trans to $\ce{c}$, then ...

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Here's a pre-MOT rationalisation of the cis form being more stable than the trans form for 1,2-dihaloethylenes: The lone pair of chlorine atoms is involved in resonance with the double bond, as it does so positive charge appears on one chlorine and negative on another. A cis geometry allows for a stabilizing interaction (attraction) between these positive ...

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These two isomers are enantiomers, i.e. mirror images, of each other: $\hspace{50 mm}$ This one has a plane of symmetry through the molecule and is therefore meso: $\hspace{60 mm}$

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As it is said in the definition of IUPAC, for a carbon atom to be chiral it should be attached to 4 different groups: In case of 1,4-dimethylcyclohexane, you will see that the two groups on the right and on the left are identical. So in this case it's not Cabcd, it's Cabbc. If you alter one group (let's say the one on the right), these groups will no longer ...

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I hope the statutes of limitation haven't run out. Structure 1 is a (Z)-alkene. Using Cahn-Ingold-Prelog (CIP) protocol, the rings are deconstructed as shown in diagram 2. The larger red arrow marks the higher priority carbon (C,C,C). The smaller red arrow points to the lower priority[carbon (C,H,H). For the blue arrows, sulfur takes priority over carbon. [...

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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. ...

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The enantiomers 1a and 1b can have their double bonds defined by CIP rules 1a and 5. Rule 1a dictates that CH3>H while Rule 5 has R>S. Thus, 1a has an E-double bond and 1b is of the Z-configuration as exemplified by the red bonds. This issue has been addressed previously on this site. Addendum: While the enantiomers 1a and 1b (vide supra) have their double ...

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As you have mentioned, the basic Fischer operations are: Vertical positions are below the plane of paper and horizontal positions are Above the plane of paper, thus you have already remember that when working with Fischer projection. What you have to remember about chiral compounds are: If you switch two groups, you get the epimer and if you switch two other ...

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Geometric isomerism occurs when two structures with the same connectivity are not interconvertible. Cis-Trans isomerism is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\... 9 In this answer, I will point out an inadequacy of the reasoning based on resonance structures and I will also provided another MO perspective, which I believe is more convincing, on the issue. A flaw in the resonance argument As presented by Abcd, the phenomenon can be explained without use of MO theory and purely based on resonance structures. Please ... 8 I managed to crack the formula for optical isomers with odd chiral centers, so I'll share my attempt here. Hopefully others may innovate on it and post solutions for other formulae. Pseudo-chiral carbon atoms - an introduction The Gold Book defines pseudo-chiral/pseudo-asymmetric carbon atom as: a tetrahedrally coordinated carbon atom bonded to four ... 8 As @Zhe points out, its not possibly to definitively answer your questions without knowing the structure of the olefin, as its important what is around the olefin, as well as just how many protons are attacked to the olefin. If you read many of the original Grubbs' papers (and indeed any papers using a Grubbs' metathesis to make tri-substituted alkenes), ... 8 Hmm, you seem to be having a problem visualizing the structures in the case of 2 identical groups on one carbon. But never fear, 3D molecules are here. Check out chloroethene: It has two identical groups (hydrogens) on the left side. If you tried to make a geometrical isomer for this, you'd get: But hey! That's just the previous image flipped upside down! ... 8 Edited version: I originally wrongfully assumed these two structures are not geometrical isomers of the double bond, because it is a trisubstituted double bond with at least two identical groups (although they have two different spatial arrangements as R- and S-designations). As a result, I conclude that the two compounds cannot be distinguished by (E) and (... 8 Technically, this double bond should not be labelled as (E)- or (Z)-. It is what is known as an enantiomorphic double bond, for which the proper stereodescriptors are seqCis and seqTrans. This is described in P-92.1.1 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book). Notice that for your compound, we have$(\...

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There are two possible eliminations that can happen (see the image below) As you can see, we then have two final products depending upon which $\ce{H}$ is eliminated. From the image 1a is cis and 2a is trans. Now, how do we decide which one is the major product? This can be based on looking at the stability of the transition state. This means we have to ...

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The best way to demonstrate this kink is with pictures. Usually we have some motivation to try to represent the molecule on a single line. So for oleic acid (cis-octadec-9-en-1-oic acid), we often get pictures like the following: Elaidic acid, the trans-isomer is also sometimes depicted unhelpfully to save space: The second structure for elaidic acid is ...

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You are right, there is no difference between the two molecules that result if you rotate 180 degrees about the $\ce{C=C}$ double bond - they are the same exact molecule. So for this molecule, the terms "cis" and "trans" would not be used, they have no meaning here. The name of the molecule is simply 3-ethyl-3-hexene. If the on-line course says otherwise, ...

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The reasoning you have used is clearly faulty as there is no meaning in assigning more number of carbon to one end in a cyclic system. Just look at the both the branches you have marked using orange and blue. The logic followed is in accordance to Cahn-Ingold-Prelog(CIP) rules. The first atom in branches, marked as blue and as orange , have two hydrogen ...

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The answer in your textbook is correct. Here are all possible isomers: I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of ...

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Just adding on to Ivan's comment, you have to split the double bond "into half" such that both halves have two sets of double bonds not having "to split and remain with single bond". After that assign priorities. I don't know if I explained well, but here is a better picture: Considering each of the double-bond carbons separately, look at the two ...

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It is the same reasoning that is used to predict relative barrier heights in amines. In the example pictured below, the imine starts with the nitrogen lone pair in "more or less" an $\ce{sp^2}$ orbital. In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital. The term "more or less" is important. If we start with a molecule ...

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Contrary to the generally accepted belief, many cis isomers of olefinic compounds are more stable than their trans isomers. The stable cis form of substituted olefins such is 1–fluoro–1, 3–butadiene and l,4–difluoro–1,3–butadiene as being due to intramolecular van der Waals forces between the substituents [ Angew. Chem. , 75 , 793 (1963) ]. These forces are ...

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As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration. For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference ...

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Properly speaking, these sorts of allenes/cumulated dienes do not exhibit geometric isomerism in the way that alkenes can. Rather, when each of the two respective carbons bear two different substituents, the molecule is chiral and will exhibit enantiomerism. More specifically, these asymmetric allenes exhibit axial chirality. This is consequent to the two ...

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The lack of an octet and the empty p-orbital on boron make trialkyl boron compounds quite reactive. If we replace one of the alkyl ligands with an alkoxy group (borinate), we can now draw a resonance structure where oxygen shares one of its lone pairs with boron. In such a resonance structure the boron atom has achieved an octet and the empty p-orbital on ...

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