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According to the Cahn-Ingold-Prelog rules, a double bond has to be considered as two single bonds, and a triple bond has to be considered as three single bonds. Considering the triple bond in your question, as three single bonds, the compound would be correctly classified as Z.


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From here The SN1 reaction - A Nucleophilic Substitution in which the Rate Determining Step involves 1 component. -SN1 reactions are unimolecular, proceeding through an intermediate carbocation. -SN1 reactions give racemization of stereochemistry at the reaction centre. -The first step is slower and therefore determines the rate. -Neighbouring group ...


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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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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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You can also assign based on the highest priority. After assigning all priorities, if the highest priority group is forward and the direction from 1->2->3 is clockwise, the configuration is R. If counterclockwise, the configuration is S. In this case, the absolute configuration of d is straightforward. You do not switch from S to R because the low ...


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Because Jmol may assign stereogenic centers, it is likely other programs (perhaps including Gaussian) are equally capable to perform this task. For example, on a hypothetical bromofluoromethanol: you select the carbon atom in question, and type on Jmol's console the commands set labelalignment right; set labeloffset -20 0; label %[chirality]; to obtain ...


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In the case of cis-trans isomerism, you can refer to carbon molecules forming the carbon-carbon double bond as stereogenic centers. An example of cis-trans isomers is taken from here: The carbon atoms that form the C=C double bond in 2-butene are called stereocenters or stereogenic atoms. A stereocenter is an atom for which the interchange of two groups ...


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D and L configuration of sugars: Draw the given sugar (aldose) in the Fischer projection with the most oxidized carbon at the top (i.e. aldehyde carbon). If it is a ketose, make the closest terminal carbon to the keto group at the top. Now: if the $\ce{OH}$ on the bottom chiral center points to the right hand side, the sugar is referred to as D-sugar. if ...


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It is worth pointing out what exactly enantiomers are: exact mirror images of each other. Most of what enantiomers do and don't do can be understood by remembering that basic definition and applying it logically. First, let's compare a chiral compound to an achiral one. Imagine an achiral molecule in front of a (molecular-sized) mirror. The orientation does ...


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I think a simplified version of the enzyme example is helpful here. Consider this image of binding a molecule to a receptor: Using a model of enantiomers where the attached groups are just differently shaped blocks, it is clear to see that if one enantiomer fits, the other one can not, regardless of the way it is oriented. The different rotation of light is ...


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In this case, there are 3 stereogenic centers, they are $\ce{C-3}$, $\ce{C-4}$, $\ce{C-5}$ $\ce{C-3}$,$\ce{C-5}$ are straightforward. Using the main CIP rules (1-3), we get $\ce{C-3}$ to be $R$ and $\ce{C-5}$ to be $S$ The term 'pseudoasymmetric' is defined in Subsection P-92.1.1 (d) of the IUPAC Blue Book as follows. Stereogenic units are called ...


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As noted by Safdar, the two chiral compounds are meso-compunds. Like for meso-tartaric acid, there a is a mirror plane in the molecule. It is possible to name them unambiguously, too. You may use a page like the one about e.g., ChemDraw to train yourself in this part of chemical nomenclature: While the compound is chiral, it is better to speak about the ...


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The central $\ce{C}$ is said to be pseudochiral. This means that its chirality will depend upon the configuration of the branches attached to it. For example, in the compounds you've drawn, the central $\ce{C}$ indeed is, in fact, achiral. They are diastereomers and not enantiomers, and both are meso compounds, hence optically inactive. However, if you had ...


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I'm hazarding an educated guess here and going to say that the O-C bond is longer than the O-H bond and that there are more interaction between the axial substituents of the H from the hydroxyl group compared to the O-C-H of the longer distant from the other substituent methoxy group.


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Your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, does not have line symmetry (plane of symmetry), and hence $\ce{C}$3 and $\ce{C}$6 are chiral centers. Thus, maximum possible stereoisomers are $2^2 = 4$. However, there is a symmetry element in the structure, which is order 2 ($180^\circ$) rotational symmetry (also called point symmetry). Therefore, one ...


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While this rule is certainly not infallible, it is often used in organic compounds, especially for Fischer projections. The odd/even rule implies that no change of configuration occurs if an even number of swaps are made at a single chiral center and change of configuration occurs if an odd number of swaps are made at a single chiral center. Here, you have ...


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