Detailed mechanism for the preparation of methyl salicylate?
Preparation of methyl salicylate, Here’s a simple synthesis of methyl salicylate (oil of wintergreen), for those that just started (organic) chemistry (though of course, even if you are a bit more experienced, nothing beats a minty smelling flask/test tube). Necessary reagents Methanol, CH3OH Salicylic acid, OHC6H4COOH Concentrated sulfuric Acid, H2SO4 Outline of procedure The procedure is a Fischer esterification and is quite simple, and can be followed below. Note that I did this on a microscale, but if you have the right apparatus, this can easily be scaled up. You could also get more complicated by using a Dean-Stark trap for esterification, etc. 1) Dissolve 0.2 grams salicylic acid into 2.0 mL of methanol in a 5 mL round bottom flask. 2) Add 4-5 drops concentrated H2SO4 3) Add a few boiling chips to promote smooth boiling. 4) Reflux for 40 minutes. 5) After 40 minutes, remove and isolate product. Discussion In step 4, I used a hot water bath and not a sand bath, which is not the best method. The water vapours that come off from the water bath cause the condenser to heat up much too quickly, and some of the product vapour managed to escape the condenser. In addition, I didn’t have a pipe cleaner handy (to wrap around the condenser to keep 1 (paper towel taken away for better image) it cool) so I improvised and used some small strips of wet paper towel in an attempt to keep the condenser cool. I also did not bother isolating the product (i.e. separating from the water, methanol that is left over, etc.) as I don’t have the necessary reagents. If you want to purify your product, you can do the following: 1) Pour the mixture into a test tube and add 4 mL diethyl ether. 2) Remove the aqueous layer with a (Pasteur) pipette. 3) Pour the organic layer into another (dry) test tube and remove any residual water. 4) Add 2 mL NaHCO3 solution and mix. This will neutralize any remaining acid. 5) Transfer the ether layer to another (dry) test tube and add CaCl2 pellets until the product is dry.
Why is concentrated sulphuric acid used for the preparation of other acids like nitric acid and hydrochloric acid?
Sulphuric acid is cheap and plentiful. When it is put on the common salts of these acids like sodium chloride (table salt) and ammonium nitrate (fertilizer) acid gases are generated that are easily dissolved in water to make pure acids and leaving any impurities behind. Nitric acid is only made in this way for laboratory use. Industrial production uses a completely different process.
When carboxylic acid reacts with ethanol in the presence of H2SO4, what is the product?
Here what you need to know is that Sulfuric acid (H2SO4) is a dehydrating agent. So it will remove water (H2O) from a compound/s.Carboxylic acids have a OH group, a R group (attached to another hydrocarbon) and an oxygen. Let's me take an example of propanoic acid. (3 carbon)Ethanol on the other hand has an OH group.As I have stated earlier H2SO4 is a dehydrating agent. So the OH group from ethanol and the H (OH group) get eliminated to form an Ester. This process is also known as esterification. This is a quite important reaction to know in your A levels/ high school. The result is:Th end product for any reaction between a primary alcohol and a carboxylic acid will be an Ester.
What is the multi-step synthesis of ethyl ethanoate (starting from ethanol)?
The OP wants a multistep synthesis.OK no problem:Reagent 1: trifluoroperoxyacetic acid.First of all we need to convert our ethanol to trifluoroacetic acid.This can be done by initially oxidising it to acetic acid and make the acetyl chloride with e.g. thionyl chloride. The acetyl chloride is reacted with Hydrogen fluoride to form a trifluoroacetyl fluoride that can be hydrolysed to trifluoroacetic acid.CH[math]_3[/math]CH[math]_2[/math]OH + 2[O] → CH[math]_3[/math]COOH + 2H[math]_2[/math]OCH[math]_3[/math]COOH + SOCl[math]_2[/math] → CH[math]_3[/math]COCl + SO[math]_2[/math] + HClCH[math]_3[/math]COCl + 4HF → CF[math]_3[/math]COF + 3 H[math]_2[/math]O + HClCF[math]_3[/math]COF + H[math]_2[/math]O → CF[math]_3[/math]COOH + HFIn the next step the trifluoroacetic acid is to be converted to the corresponding trifluoroperoxyacetic acid with e.g. hydrogen peroxide.CF[math]_3[/math]CO[math]_2[/math]H + H[math]_2[/math]O[math]_2[/math] → CF[math]_3[/math]CO[math]_3[/math]H + H[math]_2[/math]OReagent 2: ethyl methyl ketone (aka MEK or butan-2-one)Part A: make CH[math]_3[/math]CH[math]_2[/math]MgBr by initially converting the ethanol to the bromide with e.g. PBr[math]_3[/math] and subsequent reaction with Mg turnings.Part B: convert the Ethanol to acetaldehyde by partial oxidation. (e.g. using Collins reagent.)If we combine the Grignard with the acetaldehyde we obtain butan-2-ol. Just oxidise this to butan-2-one.Finish it by a Baeyer-Villiger Oxidationmethyl ethyl ketone (see 2) is oxidised using the trifluoroperoxyacetic acid of step 1. This ‘inserts’ an extra oxygen in the ketone.The methyl group is a poor migrating group so the ethyl group will migrate preferrably.Voilà we’ve made Ethyl acetate. (aka ethyl ethanoate) from ethanol.That was fun!Obviously you can use a simple Fischer esterification using ethanol and acetic acid, but that would be just one step. The OP clearly stated a multistep procedure.
Why ethanol is used in the nucleophilic substitution of halogenoalkane?
The factsIf a halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide, the halogen is replaced by -OH and an alcohol is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met.For example, using 1-bromopropane as a typical primary halogenoalkane:You could write the full equation rather than the ionic one, but it slightly obscures what's going on:The bromine (or other halogen) in the halogenoalkane is simply replaced by an -OH group - hence a substitution reaction. In this example, propan-1-ol is formed.The mechanismHere is the mechanism for the reaction involving bromoethane:This is an example ofnucleophilic substitution.Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an SN2 reaction.Note: Unless your syllabus specifically mentions SN2 by name, you can just call it nucleophilic substitution.If your examiners want you to show the transition state, draw the mechanism like this:
What's the mechanism for acid hydrolysis of ester?
Ethyl ethanoate is heated under reflux with a dilute acid such as dilute hydrochloric acid or dilute sulphuric acid. The ester reacts with the water present to produce ethanoic acid and ethanol.Because the reaction is reversible, an equilibrium mixture is produced containing all four of the substances in the equation. In order to get as much hydrolysis as possible, a large excess of water can be used. The dilute acid provides both the acid catalyst and the water.Step 1The actual catalyst in this case is the hydroxonium ion, H3O+, present in all solutions of acids in water.In the first step, the ester takes a proton (a hydrogen ion) from the hydroxonium ion. The proton becomes attached to one of the lone pairs on the oxygen which is double-bonded to the carbon.The transfer of the proton to the oxygen gives it a positive charge, but the charge is actually delocalised (spread around) much more widely than this showsOne way of representing this delocalisation is to draw a number of structures called resonance structures or canonical forms joined by double-headed arrows. None of these formulae represents the true structure of the ion - but each gives you some information about it. For example, notice where the positive charge is in the three structures.Step 2The positive charge on the carbon atom is attacked by one of the lone pairs on the oxygen of a water molecule.Step 3What happens next is that a proton (a hydrogen ion) gets transferred from the bottom oxygen atom to one of the others. It gets picked off by one of the other substances in the mixture (for example, by attaching to a lone pair on a water molecule), and then dumped back onto one of the oxygens more or less at random.Eventually, by chance, it will join to the oxygen with the ethyl group attached. When that happens, the net effect is:Step 4Now a molecule of ethanol is lost from the ion. That's one of the products of the reaction.Step 5The hydrogen is removed from the oxygen by reaction with a water molecule.And there we are! We have produced the ethanoic acid (the other product of the reaction) and the hydroxonium ion catalyst has been regenerated.
Making ester results in poor amount why ?
It would help if you told which ester, and the reaction/procedure(s) you are using. It is hard to give specific help with your experiment otherwise. If you are mixing the organic acid and the alcohol together with a small amount of acid catalyst and heating, the trouble is that the acid also causes the reverse reaction as well, so unless you have a way to drive the reaction to the right, you won't get a good yield (depends on the nature of the reactants and conditions used). Many times the anhydride or the acid chloride of the organic acid is used. In the first case, the byproduct water is tied up in converting the anhydride to the corresponding acid; in the second case, the leaving group is HCl which will leave the mixture and drive the reaction. Sometimes, if the ester is high enough Bp you can drive the water out directly. Sometimes the ester, such as ethyl acetate, is lower bp than water or staring materials, so you can evaporate the formed ester to drive the reaction. Sometimes the reaction can be driven by using a large excess of the cheaper (or, sometimes, easier to separate) reactant. Hope this is of use to you. Added: I think, for a Dean-Stark trap to work, you need the most volatile component of the reactant mixture to be relatively immicible with water, or the water just dissolves and washes back into the reaction. Depending on the reactants used, this may or may not be the case. You might be able to get it to work by the addition of an inert solvent that is insoluble in water, but forms an azeotrope with water that boils lower than either of the reactants; perhaps toluene or hexane etc.
What method of extraction is used to extract caffeine from tea leaves?
The question initially asked for “What are the “other methods” of extracting Caffeine from TEA leaves! “How do you extract Caffeine from Tea leaves”?======For the benefit of “others” let us see what are the chemical and natural methods by which, the naturally existing Caffeine is removed from the tree, made it impotent, and we drink the “Tea” by adding the residue of such process(Like the sugar cane pith ,after removing the sugar cane juice ),and we call it Premium. Like Zero alcohol Beer/Cheat beer!Ironically the Caffeine isolated in such process is SOLD for other purposes like Butter from the Milk!!Any way, they have Two chemical methods and they are by use of ETHYL ACETATE and the Other by use of METHYLENE CHLORIDE!Natural methods are like Water processing or by Carbon di oxide processing.It is very simple like use of TEA recycled.What we can call if some tea powder is used once, filtered out/washed Dried and sold as Decaffeinated?Basically the process is similar. may not be exact! One method of reuse.Hotels and restaurants are banned from selling used Coffee powder or tea leaves / powder residues for anything other than compost for this reason!!Reputed manufacturers do it scientifically others use crude ways,but both reach the market shelves. And there is no illegality. What is sold is TEA with less caffeine!!