what is the mechanism of acetonide formation?
323945
Try http://www.scirus.com/srsapp/
947946
water hydrolyzes acetic anhydride to promote peptide formation with the primary amine.
480971
acetates are ionic, and will more easily dissolve in water than an oil. thats your best bet.
605888
There are many different ways of proving a chemical mechanism.  The methods can be different depending on how fast the reaction is, and the conditions of the reaction.\n\nI personally have used deuterium labelling studies.  This is where you prepare your substrate with a deuterium label in place of an hydrogen.  This product undergoes the chemical reaction and you identify in your product which hydrogen in your product has been replaced with a deuterium.  This then tells you how atoms are interacting.  You may need more than one experiment.  You can use 13C, 19F, 31P labelling etc also if necessary.  The trans-elimination of water from an alcohol to give and alkene was further proven using these techniques.\n\nOther methods are to use Mass Spectrometry to obtain the MS data on your intermediates.  By using accurate mass analysis you can determine the molecular formula of your intermediates and the postulate structures, based on experience.  This has been used for example to demonstrate some palladium coupling mechanisms, and proving that the oxidative addition process takes place.\n\nOther methods use NMR (nuclear agnetic resonance) data.  This is where you look at the environment for all H atoms in your molecule.  If your intermediates are stable enough you can see the environments change as the reaction progresses and from experience with NMR interpretation you can make postulates as to what the intermediates and hence mechanisms are.\n\nIn some cases the nature of the conditions you use for the reaction to be successful are enough of an indication as to what mechanism is going on.  \n\nBefore trying to prove a mechanism though you need some information.  You must know what your starting materials are, and you need to know what your product is.  So correct identification of the product is crucial.  NMR/MS/HPLC-MS/X-ray etc are all used to accomplish this, although you wouldn't need to use all the techniques for every molecule.\nYou must also have created a postulate of what the mechanism is that you are going to prove or disprove.  If you don't then you don't know where to look and which technique to use most effectively.\n\nI hope this helps.
962900
The main mechanism of energy production in the body is through the formation and oxidation of adenosine triphosphate, ATP.  Cells convert carbohydrates (which we eat) into ATP, and then oxidize the molecules later to access the energy in their chemical bonds.
794459
The 3-hydroxypropionate cycle has been proposed to operate as the autotrophic CO2 fixation pathway in the phototrophic bacterium Chloroflexus aurantiacus. In this pathway, acetyl coenzyme A (acetyl-CoA) and two bicarbonate molecules are converted to malate. Acetyl-CoA is regenerated from malyl-CoA by l-malyl-CoA lyase. The enzyme forming malyl-CoA, succinyl-CoA:l-malate coenzyme A transferase, was purified. Based on the N-terminal amino acid sequence of its two subunits, the corresponding genes were identified on a gene cluster which also contains the gene for l-malyl-CoA lyase, the subsequent enzyme in the pathway. Both enzymes were severalfold up-regulated under autotrophic conditions, which is in line with their proposed function in CO2 fixation. The two CoA transferase genes were cloned and heterologously expressed in Escherichia coli, and the recombinant enzyme was purified and studied. Succinyl-CoA:l-malate CoA transferase forms a large (αβ)n complex consisting of 46- and 44-kDa subunits and catalyzes the reversible reaction succinyl-CoA + l-malate → succinate + l-malyl-CoA. It is specific for succinyl-CoA as the CoA donor but accepts l-citramalate instead of l-malate as the CoA acceptor; the corresponding d-stereoisomers are not accepted. The enzyme is a member of the class III of the CoA transferase family. The demonstration of the missing CoA transferase closes the last gap in the proposed 3-hydroxypropionate cycle.
852453
The previous answer is partially correct.\n\nWater reacts with the excess acetic anhydride which is converted to acetic acid.\n\nSince the product is only slightly soluble in water, this also makes it possible to collect the product as a solid.
43485
Well, I'm not gonna try to explain this or anything, but check this website out:\n\nhttp://www.usm.maine.edu/~newton/Chy251_253/Lectures/Esterification/Esterification.html\n\nScroll down to "Scheme 2: Saponification of an Ester," and that explains the reaction process fairly well.  I don't know if this is what you're looking for, but I hope it helps!
151412
1) It's readily soluble in water\n\n2) It solvates to become the ammonium cation. The ammonium cation and its conjugate base the amine group is seen in most organic compounds, such as nucleotides, amino acids, etc. So it's a source by which plants can build the molecules they need via enzymetic action.
901337
After glycolysis is finished 2 molecules of acetic acid are "loaded" into Kreb's on the enzyme Acetyl Co-A.  Co-A then returns to glycolysis in the cytoplasm to get more acetic acid.\n\nMean time the acetic acid adds to oxaloacetice acid (A 4 carbon structure) and gives a 6 carbon structure called citric acid. (2+4=6).\n\nThen CO2 and NADH2 is formed and we drop down to a 5 carbon structure called alpha ketgluteric acid and then lose CO2 and NADH2.  We also have created enough energy at this point to form an ATP and a low energy "hydrogen bus" FADH2.\n\nWhich brings us back to the 4 carbon structure oxaloaceitic acid, and we start all over again. \n\nThe purpose of Kreb's is to load what I call hydrogen busses (NADH2) because they are what unload electrons in the electron transport system which gives us a huge amount of ATP's (34).\n\nHope this helps.
635418
CH3CH2CH2CH3 + CL2 -hV--> \nCH3CH2(CL)CHCH3 + HCL ..\n\nwhat are the ways to get the CL at the end of the chain..so that\n\nCH3CH2CH2CH2CL + 2NA + CLCH2CH2CH2CH3\n--->CH3--------CH3\n\nThat's a radical reaction, the chlorination of an alkane with chlorine in UV light. The intermediate is a carbon radical and secondary radicals are more stable than primary. There are also more secondary carbons on this molecule. So probability and reaction kinetics dictates that the primary alkyl halide will be a minor product.\n\nI guess, if this were my synthesis, I'd:\n\n1. Put the glucose in HCl to cleave the anomers it likes to make.\n\n2. Esterify the carboxlic acid with whatever's available (Fisher esterification with MeOH, for example)\n\n3. Reduce the ester with Diisobutyl aluminium hydride (DIBALH)\n\n4. React the aldehyde with 1 equivalent of Ethyl Magnesium Bromide (Grignard)\n\n5. Reduce the hydroxy groups with a Barton-McCombie deoxygenation, although I've never seen it done in a case like this.\n\nThen again:\n\nI'd go to the stockroom and save my research advisor a lot of money. :)
337759
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.
115234
It's going to be hard to show a mechanism using this website, so I suggest you look up "esterification" and learn the mechanism.
394756
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.
959671
The rate detemining step in SN1 reactions is the first step. Step one in SN1 reaction involves carbocation formation by ionisation. Any positive feature of the alkyl halide or leaving group will increase the reaction rate:\nIn general the rate of reaction is:\nMethyl > 1° > 2° >>>3°\nSimilarly leaving group (LG) ability is a function of stability of the LG. The weaker the conjugate base, the better the LG\nFor the acids: HF, HCl, HBr, HI the acidity increases as you go down the period, so then the anion stability also increases as you go down the period. Then the general stability is:\nI- > Br- > Cl- >> F-\n\nPutting it all together you get (slowest to fastest):\nCH3Br < CH3CH2CH2I < CH3CHBrCH3 < CH3CHICH3 < (CH3)3Cl
373250
C3H7NO2 \n\nwww.chemfinder.com is a useful resource.  \n\nI hope this helps :)
677813
An amide is the product of the reaqction between a carboxylic acid and ammonia, by the mechanism:\n\nR-C(O)-OH  + NH3 ===> R-C(O)O-NH4 (an ammonium salt)\n\nR-C(O)-ONH4 + HEAT ===> R-C(O)-NH2 (an amide) + H20\n\nA second mechanism is the reaction of ammonia with an acid chloride:\n\nR-C(O)OH  + SOCl2  =======>  R-C(O)-Cl\n\nR-C(O)-Cl + NH3 ==========> R-C(O)-NH2