9

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.)...


5

Here is what I think is going on: $Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate. $Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B $Step$ $3$ is loss of one carboxy group by ...


5

"A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible. As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language. "...


4

@Waylander has provided a well-reasoned solution to the post. I offer a different interpretation. Knoevenagel product 1a has been converted in high yield to cyano ester 2a and subsequently hydrolyzed to phenylsuccinic acid (3a) 1. In the addition of cyanide to the double bond of 1a and protonation, one of the labile ester groups is cleaved by cyanide. (e.g.,...


3

This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions Hydrazine is a ...


3

The answer is (b); majority elimination by an E2 mechanism. Quoting from chem.libretext.org: Acetylide anions are strong bases and strong nucleophiles[...] Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.


3

Yes, E2 and SN2 are competing reactions. Keep in mind, E2 reactions require a base whereas SN2 reactions require a nucleophile. Nucleophilicity corresponds to the ability to donate a lone pair of electrons ‒ here, to a carbon bearing leaving group ‒ which is a kinetic property (time does matter). Basicity corresponds to accepting a hydrogen, a thermodynamic ...


3

If you leave it to E1, you will not get the selectivity you need. The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction. In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is ...


3

That's not the right view. The $S_{N}2$ mechanism does not favor strong nucleophiles. That statement doesn't make sense at a high level. Mechanisms can't favor anything. The key here is to look at what controls reactivity in $S_{N}2$ and $S_{N}1$ reactions. The former is concerted where the bond breaking and bond forming take place at the same time. ...


2

The first thing you notice is an alcoholic base reacting with a molecule. The base looks for the most acidic hydrogen in the molecule. Since the no.of moles of $alc.KOH$ is not mentioned, we go for both the acidic hydrogens at the adjacent positions to the carbonyl functional group. Once the acidic hydrogens are abstracted, it's a simple elimination of $Cl$ ...


2

The issue in this Grob fragmentation is whether or not the C-C and C-OTs bonds are antiperiplanar. The C-OH bond is free to rotate and the unpaired electrons of the negative oxygen can align with the other two bonds. In the first example a), the red bonds are gauche to one another. Standard E2 elimination is likely the source of the mixture. Examples b), c) ...


2

The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to ...


2

An E1 mechanism is highly unlikely. I don't know what level you're at, but one could invoke the Hammond Postulate in this case, and look at the high energy cation intermediate. There are three chloride leaving groups that could generate one of two intermediates. Both of these intermediates are primary cations. A chloride substituent isn't likely to be able ...


1

The most important factor in these reactions are nucleophiles. Here, the use of strong base (Sodium methoxide) in both of the reactions favors the bi-molecular mechanisms ($\ce{S_N2}$ and $\ce{E2}$). Now, the second factor would be degree of halogen-linked carbon. Lower degree carbons ($\ce{\text{deg.} < 2}$) favor $\ce{S_N2}$, while higher degree ...


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