19

This is standard for purifying substances. To wash means to add your product solution to an aqueous solution (or just water, but frequently a saturated solution) to a separatory funnel. After shaking, you drain the lower layer (which is usually aqueous). This process removes water soluble impurities. This is frequently repeated. Drying is accomplished by ...


12

The current methods employed for synthesis fall into the following three categories1: Chemical synthesis Biochemical methods Isotope exchange reactions Chemical Synthesis: Most chemical syntheses involving $\ce{^14 C}$ are done with Grignard reagent and $\ce{^14 CO2}$ or $\ce{^14 CN-}$. As an example, Benzene-14 is synthesized as such2: Basic preparatory ...


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 ...


10

The solvent of this experiment should be non polar, and may be a hydrocarbon, or a chlorinated hydrocarbon. This organic phase may contain as impurities some acids and other polar substances soluble in water, that should be eliminated. This liquid is then mixed with an aqueous solution of soda $\ce{NaHCO3}$. Aqueous solutions and non polar substances are not ...


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 ...


8

This took a bit of thinking about but here is a viable route from commercially available bulk materials. Start with nitroethane, deprotonate ($\ce{t-BuOK}$ in THF would be my choice, but $\ce{NaOEt/EtOH}$ should do it) and react with the commercially available dioxane protected 3-bromopropanal. Reduce the nitro group - alkyl nitro groups are tricky to ...


7

Nicco: Of course you can't "see it". If you were given the correct structure, the solution to your post would be straightforward. There should be a double bond in the lactam ring as shown in compound 1. The reaction is conducted in acetonitrile (bp. 82oC) at reflux. These conditions allow for a Claisen rearrangement to occur to form ketolactam 2. Subsequent ...


7

Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM. Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol. Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol. Remove the THP group using ...


6

I did some literature research and I wasn't able to find any methods involving steam (as water in the gas phase) for the synthesis. This in fact makes sense as alkali metal cyanides are prone to hydrolysis and tend to decompose to formate $\ce{K(HCOO)}$ over time or in hot water: $$\ce{KCN(aq) + 2 H2O(l) -> K(HCOO)(aq) + NH3(g)}$$ which further ...


6

There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I). In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride. Method 1 Electrolysis of molten sodium chloride: $$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$ Oxidation of sodium metal to oxide by burning: $$\ce{4 Na + ...


6

Divide your alkene starting material in two portions - one twice the amount of the other. Take the larger portion and do an anti-Markovnikov addition of HBr to it using peroxide or light reference to give the terminal bromide (1-Bromo-4,4-dimethylpentane). Take the smaller portion and ozonolyse it, working up with Dimethyl Sulfide to give the aldehyde (3,3-...


5

There is one incredible simple, cheap and easy way to make nitrogen gas with near 99 percent purity if you don't need it compressed. I've used this method in my chemistry experiments when I needed just modest amounts of nitrogen or when I need to do reactions in the absence of reactive oxygen. Simply buy handwarmers from Walmart or many other sources. Use a ...


5

The complete paragraph, of which you took a particular phrase, reads: The Potassium cyanide, $\ce{KCN}$, is manufactured by methods similar to those employed for the corresponding sodium derivative. Potassium ferrocyanide is heated either in absence of air, or with potassium carbonate and charcoal, or with sodium, the potassium cyanide being extracted by ...


5

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate ...


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

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

The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever ...


5

For preparation of the Grignard reagent, use allylmagnesium bromide. $$\ce{CH2=CH-CH2-Br ->[Mg][ether] CH2=CH-CH2-Mg+Br-}$$ Now we react the Grignard reagent formed with ethylene oxide, our starting compound. $$\ce{C2H4O ->[CH2=CH-CH2-Mg+Br-][Et2O] HO-CH2-CH2-CH2-CH=CH2}$$ This gives us pent-4-en-1-ol. Now we protect the alcoholic group by ...


5

The description you provided is essentially a shorthand for a standard workup procedure. After performing a reaction, one wants to isolate one’s product while removing the catalyst or other reagents and side products that one doesn’t need. For example, in many cross-coupling reactions an inorganic halide salt is an undesired side-product and a base such as ...


4

Chiral amino alcohols such as L-valinol are generally prepared by the reduction of corresponding α-amino acids and suitable reducing reagent. Most commonly used reducing agent is $\ce{LiAlH4}$ since most common reducing reagent, $\ce{NaBH4}$ does not reduce carboxylic acid to corresponding alcohols. However, use of $\ce{LiAlH4}$ has notable disadvantages ...


4

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury". Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though). Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in ...


4

I think you're being a little too strict with your definition of what a synthesis reaction is. A better way to describe it might be building more complex molecules from simpler ones. Wouldn't you say $\ce{C6H12O6}$ is more complex than $\ce{CO2}$ and $\ce{H2O}$?


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. ...


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

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

Here is a potential solution using old-school chemistry, though with several steps. Hydroborate the alkyne and work up with $\ce{H2O2}$ to give the aldehyde. Alkylate the aldehyde enolate with $\ce{RI}$, either directly (LDA or LiHMDS) or via enamine/silyl enol ether Reduce aldehyde to alcohol ($\ce{NaBH4/EtOH}$) Form triflate/mesylate and eliminate (DBU/...


4

It is true that until recently, the syntheses of primary anilines mostly rely on the reduction of corresponding nitroarenes. However, recent publications allow the synthesis of primary anilines or their derivatives in laboratory. One such method is Buchwald–Hartwig amination: The palladium-catalyzed cross-coupling of amines and aryl (pseudo)halides, now ...


4

$\ce{NH3}$ and $\ce{SCl2}$ form $\ce{NH4Cl}$ which evolves as white fumes. $\ce{NH4Cl}$, when passed through $\ce{SCl2}$ (cherry-red liquid), gives the appearance of dense-pink fumes $$\ce{\underset{(cherry red)}{6SCl2} + 16 NH3 -> S4N4 +2S + \underset{(white)}{12NH4Cl}}$$ That is why the fumes of $\ce{NH4Cl}$ turned pink


3

Now a days, sulfonephthaleins can be prepared by reaction of readily available saccharin and the desired plenol compound (Ref.1 & Ref.2). In this method, active reagent, sulfobenzoic anhydride will be prepared in situ as depicted in following diagram: In this method, it's described the preparation of phenolsulfonephthalein (Phenol Red). If you use ...


3

The Lemieux-Johnson oxidation (details here) using catalytic $\ce{OsO4}$ or $\ce{RuO4}$ turned over with $\ce{NaIO4}$ will do the transformation you want in one step. These conditions should be selective for the exo double bond.


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