# Tag Info

## Hot answers tagged synthesis

11

Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol ...

9

There is indeed very little to be found in the literature. A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here. The discussion ...

5

Mostly it's a matter of economy. In most diazonium salt synthesis the counterion is not involved in the organic reactions; it's just a spectator. So we choose chloride because hydrochloric acid is a cheap and effective chemical to acidify the aromatic amine+nitrite mixture and make the diazonium ion in the first place. Ergo an aryldiazonium chloride. The ...

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 ...

4

Good answer by @Waylander and @OscarLanzi. They provided some literature evidence of the synthesis procedure of Benzenediazonium bromide and iodide and discussed its usage and worth and explained why they are not that important as compared to benzenediazonium chloride. My answer just revolves on the actual synthesis procedure(slightly abridged in my answer) ...

4

The scheme you have written will satisfy your teacher in my opinion but it can be shortened. The suggestion of @Eli Jones of the V-H formylation to give benzaldehyde is a good one. However, it can be done in 2 steps using reasonably well-known chemistry: Step 1 - Brominate with $\ce{FeBr3/Br2}$ to give bromobenzene Step 2 - Either lithiate by Li-Halogen ...

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.,...

4

As @Zhe and @Zenix have mentioned, you have to hope that there is a difference in the rate of enolate formation to differentiate the two ketones. To do this cool a rigourously dried solution of the diketone and a trapping reagent ($\ce{Me3SiCl}$ if you're doing an anhydrous workup otherwise TIPS-Cl or MeI) to $\mathrm{-75 \, ^\circ C}$ and slowly add 1 eq. ...

3

There are two basic choices here if you are trying to do this on a lab scale. The alcohol plus alkene with catalytic acid is not an approach I would consider unless I needed a lot of this. t-BuOH plus acid is a recipe for isobutylene. Sodium or potassium t-butoxide is commercially available and reacted with an excess of iPr-I should give a decent yield of ...

3

The given question is a bad question. Calculate, how much of $\ce{CO2}$ is made during the reaction: $\ce{2CO + O2}$, if we use $\pu{33,6 dm^3}$ of $\ce{O2}$, also calculate the amount of $\ce{CO}$ used. M(CO)=28, M(CO2)=44. The OP gives the answer as 132 g of CO2, and 67,2 L of $\ce{CO}$. The issues with the problems are: "...how much of $\ce{... 2 Step 1 - Form the anion of your starting material ($\ce{NaOEt/EtOH}$will do this nicely) and react with phenethyl bromide ($\ce{PhCH2CH2Br}$). Step 2 - Add the product to hot polyphosphoric acid$\ce{^{[1]}}\$ to cyclize it to the required product. Reference Cyclization of Aryl-Aliphatic Esters with Phosphorus Pentoxide in Phosphoric Acid by Richard C. ...

1

We have to distinguish between the bicarbonate in solution and the bicarbonate as a solid. While many metal cations are compatible with bicarbonate anions in dilute aqueous solution, only in a few cases can a solid bicarbonate be expected as a precipitate. In most cases, such as with lithium or with hard-water forming metals such as calcium and magnesium, ...

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