what is the mechanism of esterification?
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Fischer esterification is a special type of esterification and the process of forming an ester by refluxing a carboxylic acid and an alcohol in the presence of an acid catalyst. Commonly used catalysts for a Fischer esterification include sulfuric acid or lewis acids such as scandium triflate.\n\nDirect acylations of alcohols with carboxylic acids is preferred over acylations with anhydrides (poor atom economy) or acid chlorides (moisture sensitive). The main disadvantage of direct acylation is the unfavorable chemical equilibrium that must be remedied by removal of water either by Dean-Stark distillation or the use of molecular sieves. In one study [1] it is found that Tetrabutylammonium tribromide (TBATB) is a very effective catalyst. For example the acylation of 3-phenyl propanol with glacial acetic acid and TBATB at reflux generates the ester in 15 minutes in a 95% chemical yield without the need to remove water. It is believed that hydrobromic acid released by TBATB protonates the alcohol over the acid making the carboxylate the actual nucleophile in a reversal of the standard esterification mechanism.\n\nThe reaction mechanism for this reaction has several steps:\n\n   1. Proton transfer from acid catalyst to carbonyl oxygen increases electrophilicity of carbonyl carbon.\n   2. The carbonyl carbon is then attacked by the nucleophilic oxygen atom of the alcohol.\n   3. leading to the formation of an oxonium ion.\n   4. Proton transfer from the oxonium ion to a second molecule of the alcohol gives an activated complex\n   5. Proton transfer from one of the hydroxyl groups of the activated complex gives a new oxonium ion.\n   6. Loss of water from this oxonium ion gives the ester and water\n\nAn example of an esterification is in the formation of benzocaine.
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The site pointed to above does a good job explaining the mechanism. Just remember that with any unimolecular reaction, there must be a rate determining step.  Since it is always harder to form a carbocation, the ionization step will be your rate determinant.\n\nA few other pointers on carbocation formation:\n\nYou want an ionizing solution.  This basically means a polar protic solvent like water will do the job much better than a polar aprotic one like an ether.  Hydrogen-bonds will help stabilize the carbocation intermediate really well so you're much more likely to produce a C+ in H-bonding solutions.\n\nYou will also form the most substituted product.  So look at all the elimination possibilities and pick the one that's got the fewest H's on your SP2 Carbons.  And anytime a carbocation is involved, you gotta watch out for skeletal rearrangement.
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The ester, methyl propionate is formed in this reaction.\nA sodium carbonate soln will neutralize any unreacted acid...the salt formed and methanol are soluble in water, but the ester is not...so it so it will rise to the top as a separate layer.\nYou will also note the fruity odor of the ester.
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Effect of enzyme on reaction
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the catalyst helps the reaction proceed by allowing a separate path to reaction occur.  nature uses enzymes to do this, humans use organic and inorganic chemicals
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So you acetylate the amine, thereby creating an amide. What happens, I think anway, is that the carbonyl is drastically reducing the initial nucleophilicity of the lone pair of the amine, because it's pulling electron density from the nitrogen.
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it`s very long. I`ll give u a simple explanation!\n\nfirst; glycolisis.\nthe suger is broken into 2 molecules of pyruvic acid\n\nsecond;\nthe pyruvic acid is changed into acethyl co-A\n\nthird; the krebs cycle\nthe acethyl co-A is reacted with oxaloacetate acids, and during the cycle CO2 is released, and those compound returned back in oxaloacetat acids\n\nfourth; electron chain systems
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I'll talk through it, since I can't draw structures.  Presumably, this is an acid catalyzed process.  The first step is protonation of the oxygen on acetone.  This creates a more electrophilic carbonyl.  One of the hydroxyls of the starting material will attack the carbonyl carbon, generating a hemiacetal.  Protonation of the hydroxyl of this hemiacetal, followed by elimination of water with assistance from the lone pairs on the neighboring oxygen will generate an oxonium ion suited for another nucleophilic attack by the other hydroxyl of the hydrobenzoin.  A final proton transfer will generate the acetal.
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phosphoric acid is the catalyst used in the synthesis.    Its a one to one molar ratio reaction.
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Esterification is the general name for a chemical reaction in which two chemicals (typically an alcohol and an acid) form an ester as the reaction product. These are commonly used as fragrance or flavour agents.\n\nExamples of esterification reactions include:\n\nHeating to reflux an acid (usually, but not always a carboxylic acid) and an alcohol in the presence of an catalyst forms the ester and water byproduct which can be removed to force the equibilibrium across to the desired products. This method is called Fischer esterification \nFor example, esterification of acetic acid in excess ethanol in the presence of sulfuric acid results in an ester (ethyl acetate). \nH3C-COOH + HO-CH2-CH3     →     H3C-COO-CH2-CH3 + H2O (with the presence of conc.sulfuric acid)\n\nthe reaction of an alkali carboxylate and an alcohol. For example, the reaction of sodium acetate with ethanol. \nH3C-COO- Na+ + HO-CH2-CH3 + H+ → H3C-COO-CH2-CH3 + H2O\n\nThe reaction of a silver carboxylate with a halogenoalkane (presumably driven by the insolubility of the silver halide). \nH3C-CO2-Ag+ + Cl-CH2-CH3 → H3C-COO-CH2-CH3 + AgCl\n\nThe reaction of a carboxylic acid halogenide (which is also called acyl halide) with an alcohol/phenol. \nH3C-COCl + HO-CH2-CH3 → H3C-COO-CH2-CH3 + H-Cl\n\nThe reaction of a carboxylic acid anhydride with an alcohol. (This method is used to synthesize aspirin) \nH3C-CO-O-CO-CH3 + HO-CH2-CH3 → H3C-COO-CH2-CH3 + H3C-COOH
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http://en.wikipedia.org/wiki/Reaction_mechanism\nhttp://en.wikipedia.org/wiki/Bromomethane
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if your qouestion is that how is  soap made ?\ni must say that with mixing the an ester such az R-COOR and NaOH we can get soap plus alcohol.\nNaOH+RCOOR------------RCOONa +ROH
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Formula: you more or less answered your question yourself. It is any combination of glycerine esters with any combination of these acids.\n\nReaction with NaOH is saponification, resulting in glycerine and the free acids (or more specifically, acid rests as dissolved salts).\n\nReaction with hydrogen in the presence of Pt or Raney-Ni as catalysts will ultimately produce the triglycerides of the corresponding saturated acids. This is known as "hardening" of fats since saturated fatty acids tend to produce solid fats whereas unsaturated fatty acid triglycerides of comparable mass remain liquid at the same temperatures.\n\nReaction with bromine will result in electrophile addition of Bromine on the double bonds of the unsaturated acids.
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water hydrolyzes acetic anhydride to promote peptide formation with the primary amine.
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Chemisorption of propionic acid vapors by an iron impregnated MCM-41 porous silica leads to the formation of iron-propionate.\n\nSource is from American Chemical Society Journal.
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Kreb cycle and glycolysis - is what I initially answered\n\nHowever, if it is a standard test/textbook question the questioner is asking, then the answer is probably going to be 2 different mechanisms/pathways. It probably is going to be AEROBIC and ANAEROBIC metabolism.
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Well, the kinetics are actually pretty complicated. \n\nAll of the catalysts I'm aware of for this type of reaction are heterogeneous. This complicates things for several reasons. First, the rate is going to be much more dependent on reactor design since how well the catalyst is suspended and dispersed plays a big role in the rate of reaction. Second, the rate will depend on the active surface area of the catalyst, which can vary from batch to batch of catalyst and must be determined empirically for each sample.  \n\nAnother tricky issue is that in the course of the reaction you are going from a liquid (oil) to solid (fat). This will change the physical properties of your mixture, which changes the way the catalysts is distributed and will certainly change the kinetics. Solution viscosity could potentially become an issue, especially on the industrial scale. It really just depends on how you run the reaction.\n\nIn terms of amount of H2 needed, it's going to depend on precisely what kind of oil you are starting with. Most vegetable oils are mixtures of several oils made of the triacylgylcerides of unsaturated fatty acids such as oleic acid and linoleic acid. Since the degree of unsaturation varies from acid to acid (and therefore oil to oil), it depends entirely on which specific oil you have.\n\nIn terms of catalyst, in principle it's always regenerated and never lost.\n\nI wish I could give you more specifics, but the kinetics here are really messy and not easily amenable to a traditional analysis.