28

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-...


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

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 ...


12

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 ...


11

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: $\hspace{60 mm}$


10

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 ...


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. ...


8

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 $\...


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

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 ...


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

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 ...


7

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, ...


7

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 ...


7

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 ...


7

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 ...


7

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 (...


7

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 $(\...


6

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 ...


6

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 ...


6

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 ...


6

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. [...


6

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 ...


6

With the same reasoning as provided here, I recommend that you name this compound (2 E)-2-bromobut-2-ene. Using (cis/trans) may be useful around an isolated double bond with only two substitutents. While in the example presented by you the two methyl groups are on the same side of the double bond, they are not the only ones to be considered here. Triple and ...


6

Glucose and Galactose are best described as epimers of one another. The three structures below all have the same molecular formula and the same connectivity, but differ in the absolute configuration at a single stereocentre (this is the definition of an epimer). You can see that glucose and galactose are epimeric at the $\mathrm{C4}$ position, whilst glucose ...


6

Trans alkenes have a $C_\mathrm{2h} $ symmetry, identified as such because it has a $C_2$ rotational axis (you need to rotate 180° to have an identical molecule) and a mirror plane perpendicular to that axis $\sigma_\mathrm h$. You need to follow the symmetry flowchart to assign symmetry groups. A $C_\mathrm{2h}$ group is characterised by having: $E$: the ...


5

As I said in my comment, the mathematical way to solve this kind of problem relies on Graph Theory. This article (Applications of Graph Theory in Chemistry, preprint) reviews the problem and further more. If you google 'isomer' and 'graph theory' you will find anything you want about this exciting topic. But for generic question, there is no generic answer. ...


5

In simple molecules like what you've drawn above, see if the hydrogens are on the same side of the double bond or not. If they are on the same side, then the compound is cis; if the hydrogens are on opposite sides, then the compound is trans. In more complex cases, prioritize the four substituents on the double just as you've been doing with "R\S" ...


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