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18

Here is a 3-D conformer from PubChem As you can clearly see, a plane of symmetry can be sent along the black line perpendicular to the plane of the screen. Hence, the molecule is achiral. If you take a mirror image, you can ultimately super-impose it again on the parent form Here is an illustrative 3D image(courtesy of andselisk) which clearly shows the ...


11

Yes, both inner and outer placements are possible with some nuances. From Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) [1, p. 1926]: ST-1.3.2: Stereogenic centers at ring fusion atoms Stereogenic centers at ring fusion atoms should be drawn with hashed wedged or solid wedged bonds to the exocyclic substituent ...


9

In this complex there are two different 31P environments which are not related by symmetry: The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted. As extra proof, consider that the green P is cis to both carbonyl ...


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

Biphenyl 3 is the only optically active compound here. These stereoisomers are due to the hindered rotation about the 1,1'-single bond of the compound (Ref.1). Biphenyl 2 is noty optically active, because partially allowed rotation about the 1,1'-single bond of the compound (rotation is only partially restricted). To illustrate this phenomenon, I depicted ...


7

To answer this you need to consider the hybridization and geometry around each atom but also think about all of the single bonds present. Single bonds will (generally) allow rotation along their axes as this can happen to a sigma bond without breaking the orbital overlap. If rotation about a single bond moves other atoms in space then you must consider ...


7

Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$. $\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure). $\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically active compounds formed after ozonolysis is $\ce{= 1}$.


6

They can, and they do. The secret is to have a chiral cation and separately a chiral anion. Letting $D^+$ be the dextrorotatory form of the cation, $L^+$ be the levorotatory form, and anologously for the anion, we then have four isomeric salts: $D^+D^-$ $D^+L^-$ $L^+D^-$ $L^+L^-$ The first and second are diastereomers because only the anion is mirror-...


6

Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite,which gives various stereochemical products as shown here. Extending this to the current question , alkyl chloro sulfite 2 is formed.${Cl^-}$ attacks in a ${S_N^2}$ mode to give 3. Base deprotonates 3 to give 4. Alkoxide and chloride in 4 are in correct trans configuration for ...


6

The given answer is correct. The product scheme you drew is correct as well. However, as I marked in your scheme (see below), products A and B are essentially the same enanthiomer (both have (2S)-configuration). The compounds you have drown in right-hand side are also an identical compound, which is not optically active. Therefore, ozonolysis has given only ...


6

(3) is not chiral, so maybe it is (2).


5

Not necessarily. Although the change you suggest would not alter the absolute stereochemistry (S in both cases) of the asymmetric center starred, the two molecules are diastereomers and therefore will have different physical and optical properties. Even if they did both rotate light the same way, they probably won’t have the same degree of rotation for the ...


5

@Waylander provided a correct written explanation for your question. I'll expand on his comments by considering the effect of temperature and provide diagrams. Using M+ = Sc+3 to save some space, the Lewis acid coordinates with the more basic of the two acetate oxygens in 1 to provide the reactive oxonium species 2. [The oxonium species is planar about the ...


5

The number of isomers are ${9}$ as shown below. AlaAla has ${4}$ isomers. AlaGly has ${2}$ isomers. GlyAla has ${2}$ isomers. GlyGly has ${1}$ isomer.


4

Azobenzenes are among the most widely studied molecular photoswitches, yet, I'm not sure this is a project capable of doing in a high school lab. It probably need sophisticated instruments (e.g., azobenzene needs to be covalently bound to carbon nanotubes (CNTs)) and has done previously, notably by Department of Materials Science and Engineering in ...


4

You don't need a digraph for this one. Your structure is A. Remove the double bond and add duplicate carbon atoms to each of the double bond carbons (B). These duplicate carbon atoms each bear three phantom atoms of atomic number zero. Clearly at C5 hydrogen has the lowest priority and methyl is the next lowest. The methylene groups at C4 and C6 are a tie. [...


4

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too? Many molecules exist in both forms in nature. One fun example are the enantiomeric terpenoids R-(–)-carvone and S-(+)-carvone. The R-form smells like spearmint while the S-form smells like caraway. The difference in smell shows that properties other than the ...


4

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4. So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical ...


4

It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper ...


4

It is achiral. The cyclopropane ring is planar. The substituted C will be $\mathrm{sp^3}$ hybridised (tetrahedral). If you consider two of the C's in one plane and the H and Br in the other plane, these two planes will make 90° angle between them and one can draw a mirror plane along the plane containing H and Br. Thus, it will be achiral.


3

In organic chemistry bulkiness isn't just a matter of geometry; one can find the volume of the group using vdW radii, but it's a crude approximation. Quantitatively steric bulk relies on energy measurements and can be described with a cyclohexane A-value — energy required for a group to switch between axial and equatorial positions in cyclohexane. From the ...


3

Sameer. The key to understanding this is that the two asymmetric carbons are indistinguishable because of the symmetry in the molecule. When you do your inversions on them, the molecule looks different, and should be an enantiomer, but after you do the 180 deg rotation, you have essentially switched them over (now the molecule looks the same as the first ...


3

I agree with you but in a slightly different way. The question is correct, i.e. there are two structural isomers, but as given in your answer, the two isomers are not cis and trans dichlorocyclopropane, because as you mentioned correctly, cis and trans isomers are definitely stereoisomers and not structural isomers. To answer your question about two ...


3

Am I on the right track? Generally, yes. For this question, yes but you have over thought this question. For starters, any answer with Y can be totally eliminated as Y is neither cis nor trans. Now you are down to answers 'A.' and 'D.' Here you correctly determined the E and Z configurations of molecules with identical functional groups, but could have ...


3

I offer a different perspective to your question about the inositols and how the chirality of the "top" carbons in D-chiro-inositol 2 (C2; S-configuration) and scyllo-inositol 3 (C2; r-configuration) are determined. Generic inositol 1 is numbered the same way as D-chiro-inositol and scyllo-inositol. The method for assigning the CIP descriptor for each ...


3

For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds. To confirm, I have optimized the molecular geometry on different levels of quantum ...


3

The literature reveals that cis-1,2-cyclohexanediol 1 cleaves about 20-times faster than trans-1,2-cyclohexanediol 2 although the equilibrium between the diol and the cyclic periodate intermediate slightly favors the trans-isomer. The rates of cleavage of trans-1,2-cyclohexanediol 2, (-)-(R,R)-2,3-butanediol 3 and meso-2,3-butanediol 4 are comparable. [The ...


3

Because of the absence of a lone pair, C undergoes usual $\mathrm{S_N2}$ transformation under the conditions. However, the availability of a lone pair in A undergoes the nucleophilic substitution of cyanide ion via neighboring-group participation (see above comments by Waylander and J. Deans) by the sulfur atom under the conditions (Ref.1). A very reliable ...


3

A search of SciFinder will also produce the formation of the (E)-dibromide 3. The conditions described for the formation of the (Z)-dibromide 2 are taken from a paper by Myers1 as reported earlier by Tippett.2 Rossi3 has reported the formation of the (E)-dibromide 3 using pyridinium hydrobromide perbromide. Given the more vigorous conditions for the ...


3

Almost everything that needs to be said has been said. But I do have a few comments about configurations. In enantiomeric, trans-dibromides 1 and 2, the methyl groups are chirotopic and non-stereogenic. Inversion of the stereochemistry of the methyl group in either 1 or 2 returns the same structure, respectively. Thus, there is no designation for ...


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