Has anyone done a reaction using potassium Phthalimide? How long do you react it with an alkyl bromide?
611155
You should be able to monitor this reaction by tlc, looking for the disappearance of the alkyl halide or the appearance of the alkylated product.  a tlc stain of phosphomolybdic acid is very good for nitrogen containing compounds as they stain a distinct color from others.
24993
In what solvent?\n\ndissolve 1-g of phenothalein into 99-grams of solvent.
803799
Indicator name:-3',3",5',5" - Tetrabromo - m - cresol - sulfonephthalein (Bromocresol Green)   \npH range:-3.8 - 5.4  \nColour change:- yellow to blue  \nPreparation:- 0.1 gm in 14.3 ml 0.01 M NaOH + 235.7 ml water\n\nBuddy I got another one, But I don't know whether that will help you or not,\n\nPreparation of 2.0% bromocresol Green in a stable solvent combination is obtained as follows:\n\n(1) 35 ml (2%) of propan-2-of was added to 1750 ml of silicone oil (20 mm.sup.2/ s and stirred well. The solution is left for 3 minutes and a clear colorless solution was obtained.\n\n(2) 35 ml of bromocresol green was added to the mixture and stirred.\n\n(3) The mixture is indirectly heated in a water bath. The temperature of the water bath did not exceed 100.degree. C. The mixture was left in the water bath for 10 to 12 minutes with continuous stirring. This enhances the indicator solubility.\n\n(4) The mixture was removed from the water bath and left to cool down to room temperature. A clear yellow solution was obtained by the end of this step.\n\nThis indicator works on paper pads which have been treated with sodium hydroxide and changes color from yellow to blue.\n\n\nPreparation of a mixture of two pH indicators\n\nA mixture of bromocresol green in silicone oil 20 mm.sup.2 /s and methyl red in silicone oil 20 mm.sup.2/ s at the ratio 1:1 was produced as follows:\n\n5a. Preparation of bromocresol green combination:\n\n(1) 17.5 ml of propan-2-of were added to 900 ml of silicone oil 20 mm.sup.2/ s and stirred well. The mixture was stirred for 3 minutes. A clear colorless solution was obtained.\n\n(2) 17.5 ml of bromocresol green were added to the mixture obtained from step 1 and stirred well.\n\n(3) The temperature of the water bath did not exceed 100.degree. C. The mixture was left in the water bath for 10 to 12 minutes with continuous stirring. This enhances the indicator solubility.\n\n(4) The mixture was removed from the water bath and left to cool down to room temperature. A clear yellow solution was obtained by the end of this step.\n\n\nPreparation of mixture of phenolphthalein and bromocresol green in a 3:1 ratio was obtained as follows:\n\n(1) 27 ml of octan-1-of were added to 1300 ml of silicone oil 20 mm.sup.2/ s and stirred well. The mixture was stirred for 3 minutes. A clear colorless solution was obtained.\n\n(2) 27 ml of phenolphthalein was added to the mixture of silicone oil and alcohol and stirred well.\n\n(3) The temperature of the water bath did not exceed 100.degree. C. The mixture was left in the water bath for 10 to 12 minutes with continuous stirring. This enhances the indicator solubility.\n\n(4) The mixture was removed from the water bath and left to cool down to room temperature. A colorless solution with white clouds should be obtained by the end of this step.\n\nThe bromocresol solution was prepared , except that 9 ml of bromocresol green was dissolved in 450 ml of silicone oil 20 mm.sup.2 /s. The solutions were then mixed and stirred. The final solution color was faint yellow. This combination works on paper pads that have been pre-treated with sodium hydroxide and changes color to light mauve.
151615
you can find it in most college organic chemistry textbooks. A couple of my favorites are McMurry or Solomons.\nGrignard reagents are also called organometallic compounds where you take an organic compound(usually an alkyl halide) and react it with a metal, most of the time we use magnesium.\nso for example you could combine bromobenzene reacts with magnesium metal(using an organic solvent like ether) to produce phenylmagnesium bromide.\nfor more look at those textbooks. you'll find lots more.
701112
If Br water ismixed with KI the Br will replace the I in KI and make Kbr and free I2(diatomic) It is not terribly soluble in water. It could buble out as a violet gas if the temp were high enough. \n\nIf KF is mixed with Cl water no reaction because Cl will not replace F. They will replace each oter from bottom up. Nothing replaces F.
412197
It is a nucleophilic substitution reaction, where the nucleophile -OH from KOH replaces -Cl from 1-chlorobutane; this forms KCl and 1-butanol as products.\n\nKOH + CH3-CH2-CH2-CH2Cl ---> KCl + CH3-CH2-CH2-CH2OH
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
542412
See number 28:\nhttp://elchem.kaist.ac.kr/jhkwak/OkanaganPdb97/molecule/alkane.htm
837634
It can either be a strong acid, like HCl, or a strong base, like NaOH (aq.)  The reactions are:\n2Al + 6HCl => 2AlCl3 + 3H2\n2Al + 2NaOH + 2H2O =>2NaAlO2 + 3H2
419913
It is known as 1,2-Benzenedicarboxylic anhydride; PAN; Phthalic acid anhydride.The chemical formula is C6H4(CO)2O. Phthalic anhydride is the anhydride of phthalic acid. It dissolves in alcohol and some other organic solvents.\n\nPhthalic anhydride is widely used in industry in the production of dyes (rhodamine, anthraquinone derivatives), insecticides, plasticizers, in pharmacy, in analytic chemistry, etc.
871537
It is not an redox or acid-base reaction. so you can't employ oxidation number method or ion-electron method. So the options you have left are the inspection method & algebraic method. Among them, I think this equation is way too simple to employ algebraic method.Why don't you employ the simple inspection method.\n\nC2H5OH-----------> Oh! BY THE WAY! WHAT IS HCO2. THERE IS NO SUCH COMPOUND AS THIS.\n\nHowever the possible products are:-\nC2H5OH----------------->CH3CHO + H2(Cu,300deg C)\nC2H5OH----------------->H2CCH2 + H2O(H2SO4,170 deg. C)\nC2H5OH----------------->C2H5OC2H5(H2SO4,140 deg. C)\nC2H5OH----------------->H2CCH2 + H2O(Al2O3,300 deg. C)\nC2H5OH----------------->H2CCH2 + H2O(H3PO4,250 deg. C)
561728
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
1147469
Looking at http://www.clinchem.org/cgi/reprint/22/5/679 , the suggestion I get is that the predominant reaction involved in the Kober assay involves the aromatic A ring in the oestradiol steroid. As ethinylestradiol contains this, as well as an alkyne, it will probably react in a similar fashion, but the alkyne is also likely to be reactive under those conditions. Whether it will in such a way to affect the colourmetric changes, I'm not so sure. You may want to see if you have access to this article, which may dicuss things further (unfortunately, I don't have full access to it) http://makeashorterlink.com/?C20032D3D
1058834
Let me be the first to say this is a diabolical question that has led me on a merry chase through the Internet. Not having a virtual lab with both chemicals on hand, I consulted Erik B., a chemical engineer from Purdue, to get your answer. We discovered that there are many variables with this compound; your product is determined by the quantities of each chemical that you use. On one hand, you can produce a completely colorless form of silver that has been used in the past as a discrete poison. The most common product of this compound, however, is SILVER CHLORIDE [silver chloride] chemical compound, AgCl, a white cubic crystalline solid. It is nearly insoluble in water but is soluble in a water solution of ammonia, potassium cyanide, or sodium thiosulfate ( "hypo" ). On exposure to light it becomes a deep grayish blue due to its decomposition into metallic silver and atomic chlorine. This light-sensitive behavior is the basis of photographic processes. Since silver bromide, AgBr, and silver iodide, AgI, react similarly, all three of these silver halide salts are used in making photographic films and plates. Both the bromide and iodide are less soluble in water and more sensitive to light than the chloride. The bromide forms light yellow cubic crystals; the iodide forms yellow hexagonal or yellow-orange cubic crystals, depending on the temperature. Besides use in photography, silver chloride is used in silver plating, and silver iodide is used for seeding clouds. The chloride, bromide, and iodide occur naturally as the minerals cerargyrite, bromyrite, and iodyrite, respectively. Silver fluoride, AgF, forms colorless cubic crystals; it is much more soluble in water than the other silver halides. Incidentally, should you wish to try the home-grown poison, just like for photos, silver oxide gathers in the skin and reacts with light, creating a bluish-gray tint that is a clear signal of foul play. *Don't* try this at home.\n\nHappy Answering,\n~Tia~\nLive from the big, purple brain\nhttp://advision.webevents.yahoo.com/asktheplanet/brain.html\nASK YOUR QUESTIONS, CALL THE BRAIN - 212-382-3592
302954
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.
767795
Pb2+ + 2I- <----> PbI2 (s)? \nMost often this reaction is shown as a double replacement reaction as follows:\n\nPb(NO3)2 + 2 KI ==> PbI2¯ + 2 KNO3   \nSince both solutions have to be heated before mixing, you might think it is an endothermic reaction requiring heat. However, the Lead Iodide precipitate will not form till the mixture is cooled. This would imply the reaction is actually endothermic since cooling or removing heat energy from the solution results in the precipitate.
192443
Write down the equation of reaction: K + Cl2 -> KCl (not balanced)\n\nBalance the equation of reaction: 2K + Cl2 -> 2KCl (balanced)\n\nFind the moles of Potassium: 5 g K / 39.10 g/mol K = 0.128 mol\n\nFind the moles of Chlorine gas: 7 g Cl2 / (2 x 35.45 g/mol) = 0.098 mol\n\nDetermine the stoichiometric ratios: \nCl2 -> 2 KCl\nK ->   KCl   \n\nFind the moles of product: \n\nFor potassium: 0.128 mol K x (1 mol KCl / 1 mol K) = 0.128 mol KCl\n\nFor chlorine:  0.098 mol Cl2 x (2 mol KCl / 1 mol Cl2) = 0.196 mol KCl\n\nSo potassium is the limiting reactant. \n\nMass of KCl that can be formed: 0.128 mol K x (1 mol KCl / 1 mol K) x (39.10 g/mol K +  35.45 g/mol Cl) =  9.54 g KCl