Can somebody tell me about bis-phenol A and if there is any reasone for useing later A in that name.\n\nis that related to a position of the functoal group or what?\nand do we have another type of bis-phenol ????
123972
http://www.bisphenol-a.org/index.html\n\nThere is a Bisphenol B.  It is a butyl group versus the propyl in A.\n\nhttp://www.fujita-hu.ac.jp/JSCD/all_folder/text_folder/vol1/bisb.htm
16661
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!
254441
the OH group is para in the 4-Hydrobezoic acid, and the OH group is ortho in the salicylic, para bonds to a withdrawing group will cause greater stability in the ring due to resonance and therefore the boiling/melting points will be higher because greater stability in the ring will make it harder to break bonds. If you have more questions jsut get back to me :)
240352
Simple\nFirst Benzene is reacted with AlCl3 and chloroethane, performing the Friedel Crafts reaction.\nThen secondly, now youll be having a compound toluene (methyl benzene). Now start oxidsing it with say Potassium dichromate (any oxidising agent),toluene is oxidised to Benzyl alcohol (Ph-CH2OH), which is further oxidised to benzaldehyde(Ph-CHO), which can be further oxidised to Ph-COOH)----- Ph stands for phenyl group\n\nPlease make it the best answer
219725
I don't see B6 or B12 in NOX-CG3, which I find really helps with focus during workouts. I use BSN's N.O. Explode, which is essentially the same thing (competing product). I have noticed my endurance and strength substantially increasing. Definitely helps with recovery, not just after a workout, but even during sets. I give N.O supplements a thumbs up.
263856
Grignard Synthesis of Benzoic Acid\n\n\nPreparation:  McMurry, sections 10.8 and 20.6.\n\nConversion of an organic halide into a Grignard reagent completely transforms the reactive character of the organic compound. Carbon bears partial positive charge when bonded to halogen but partial negative charge when the carbon is bonded to magnesium. The Grignard reagent is a very reactive nucleophile and base. \n\nIn this experiment you will make the Grignard reagent phenyl magnesium bromide; this reagent will then be treated with carbon dioxide to form benzoic acid. The sequence (Equation I) is explained in McMurry Sect 20.6.\n\n\nAs in any work, this experiment may not succeed the first time; be prepared to repeat it from the beginning. On the second try, you should be able to work much more efficiently and with more understanding. \n\n\nProcedure: Obtain a 150 mm piece of magnesium ribbon. Hold one end with forceps and with the other hand remove the residue on the ribbon by scraping it with a spatula, scissors or pliers to obtain a shiny surface. Cut at least 0.12 g of the shiny ribbon into 2-3 mm pieces (do not handle the pieces) and place on a folded piece of paper to obtain the mass on the balance. Transfer the magnesium pieces to the 5 mL conical vial, add the dry spin vane, add one small crystal of iodine and add 300 µL (0.3 mL) of anhydrous ether. Be prepared to add more ether as necessary to maintain the liquid level at a minimum of 300 µL.\n\nAssemble the rest of the apparatus as in Mayo, et al., Fig 3.24. Clamp the apparatus firmly to a stand.\n\nIn the hood, prepare a solution of 510 µL of bromobenzene in 700 µL of anhydrous ether in the 3 mL vial using the delivery pipettes provided. Cap this vial with a polyethylene top. \n\n(*Note:  Be certain the dry ice is ready before you initiate the reaction below.)\n\nDraw half of the bromobenzene solution into the syringe and insert the needle through the polyethylene cap in the Claisen head as in the figure. While stirring the mixture below, add 6-8 drops of the bromobenzene solution. Evidence of reaction is the formation of tiny bubbles. When the reaction has started, add the remainder of the bromobenzene solution slowly over 3-5 min. The temperature should be maintained at about 40 oC to allow gentle refluxing (Mayo, et al., p. 29) of the ether. To the 3 mL vial add an extra 300 µL of ether, then transfer this to the reaction via the syringe. Adjust the total volume to 3 mL with the addition of more anhydrous ether. Reflux the mixture for 10 minutes, then allow the gray-brown solution of the Grignard reagent to cool to room temperature. What do you observe?\n\nThe Dry Ice (at least 1.0 g) is added in small chunks to the Grignard solution. Note the reaction as this is added. To convert the C6H5C(=O)-OMgBr, to benzoic acid (last step, Equation I), use a pipet to VERY SLOWLY add 1 mL of 6 N HCl and again observe the result. Test the aqueous layer with pH paper; it should be acidic. Remove the ether by gently heating the vial for 5 minutes on the steam bath. Cap the vial, mark it and set it in your drawer until the next laboratory period. \n\nPreparation for the 2nd laboratory period:  The benzoic acid has been synthesized, but it must be purified. Water soluble magnesium salts are removed by extracting away the water layer; but, a number of organic side products remain with the benzoic acid.  These are side products of the Grignard reaction; small amounts of benzene are formed (Equation II), unreacted bromobenzene remains, and biphenyl, H5C6-C6H5, is formed from the reaction (Equation III). \n\nThe benzoic acid can be isolated from this mixture by taking advantage of its property as an acid. Only it can dissolve in aqueous NaOH (Equation IV) while the side products remain in the ether. After the ether-soluble side products have been removed, strong acid is again added to reverse this reaction in order to restore the benzoic acid.\n\nRecrystallization of the crude product in
64980
how about something that has to do with what the best powder for lifting prints is.\nthe hypothesis could be:\n______ powder is the best for lifting prints. does that make sense?  that way you could talk about all the advancements that have been made on the powders over the years.\nfor example lead powder vs. carbon powder.
253609
Hmm, kinda hard to help you here, since we can't see all your options for the first two.  The ellipses (. . .) indicate that you have more in your message than is being displayed.  Perhaps if you retyped them, it would be easier for us to help.\n\nHere's what I *can* help with:\n\nHalogens:Reactive as Noble Gas:STABLE\n(the halogens are very reactive gases, while the noble gases are extremely stable, almost to the point of being inert)\n\nOxygen:molecule as chloride:ION\nOxygen gas is a molecule, while chloride is the ionic form of chlorine\n\nRatio:subscript as oxidation number:CHARGE\nthe ratio of components in a mixture is denoted by the subscripts, as the oxidation number of an atom is denoted by its ionic charge
514761
Use a retro synethic method.\n\nTarget Molecule <-----  1,2-dibromo- 1,3-dibenzylpropane   (M2) (use 3eq. of NaNH3 in ammonia)\nM2 <---- 1,3-diphenyl-1-propene (M3) (use Br2 in CCl4)\nM3 <---- 1,3-diphenyl-2-propanol (M4) (use KOH/EtOH)\nM4 <---- 2-phenylethanal (M5) + benzyl bromide (M6) (Grignard rxn)\nM5 <---- 2-phenylethanol (M7) + Cr6+ (use chromic acid)\nM7 <----  Styrene (starting material #1) + i)BF3 ii) H2O2, HO-, H2O (anti-Markovnikov hydroboration)\nM6 <---- Benzyl alcohol (starting material #2) + Mg(s) in Et2O (make your Grignard to be condensed in a step later on)\n\nNow, you do the connections. Note my nomenclature is rusty and I'm not 100% sure about the phenyl nomenclature as Benzen side chains to a higher priority parent chain.
612797
i find  your question, interesting but quite inexplicable .\n this is something new to me . I was not  very attentive in my chemistry studies ..\n\n My memory centre is all clogged up with other facts , just now ; \n this must be the reason for me , to draw a blank here\n \n i will go hang my head in shame .\n \n or hunt up that  answer,  LIT UP for all to see , on Google or some other wondrous search engine .\n\n good luck \n \n.
220586
every per bottle has a marking at its bottom...a triangle with the numbers 1 2 or 3....the ones marked with 1 should not be used more than once due to the change in chemical compositions taking place in it.how ever the ones marked with higher numbers can be reused...but not 4 very loong.
355408
Your starting material, 2,6-dimethyl-4-hexanol and your product 2,3,6-trimethyl-3-hexanol have no existence in organic chemistry.\n\nIn the second part of your question, 2-methyl-2-butanol;, a tertiary alcohol, reacts. According to the Saytzeff rule, the most highly substituted olefin results. 2-methyl-2-butene
339846
when there are two hydroxyl groups, it is named ending in "diol".\n\nif you say u have somthg which is like cyclohexanol but with 2 hydroxyl groups, you can call it cyclohexandiol.\nnotice the position of the hydroxyl group, it can be 1,2-cyclohexandiol or 1,3-cyclohexandiol or 1,4-cyclohexandiol.\nfind the number for the position of the hydroxyl group that make it the smaller number possible.\n\nfor example, instead of 1,6-cycloexandiol, it must be named 1,2-cyclohexandiol.\ninstead of 1,5-cyclohexandiol, it must be named 1,3-cyclohexandiol.\n\nthe maximum (furthest) position for 2 hydroxyl groups in a cyclohexanol is 1,4-cyclohexandiol.\n\nhope u do the best in yr exam.
280342
This is a pretty good site for naming.  I gave you the one that goes straight to the fused ring system\nhttp://www.chem.qmul.ac.uk/iupac/fusedring/index.html
305449
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. :)
51015
I have clicked on the link and I read about a chemical substance(filter), that is called PBSA and is contained in some sunscreens. \nIt's impressing!\nWe put on our skin a product in order to protect it and we do exactly the opposite! Thanks for the information. I haven't heard about this, but BBC site is a trustworthy site, so this notice must be valid.\nI suppose that this special case is the result of a European Union program about chemicals, their detection and their elimination from our daily life.\n\nKaterina
126443
the previous reaction will provide 0% yield of n-heptyl benzene.  it'll all isomerize to the 2-heptyl, not "some of." Friedel-Craft is the right reaction, but you want acylation not alkylation with heptanoic acid (acid chloride, or prepare in situ).  then reduce the resulting ketone by any of a zillion ways ... zinc/HCl being the classic.\n\nand where did you get the idea that the longer the synthesis the better? that's just nutty.