for electrophilic addition of ethene with Br2, CCl4 is sometimes used as an inert solvent together with Br2.?
921966
Br2 is extremely soluble in CCl4 - much more soluble than in water.\nLook up "partition".
889156
The answer is BrF3.\n\nand the balance equation is \n\nBr2  +  3F2  --->  2BrF3\n\nhow? get the moles of each using this formula:\nn = PV/RT\n\nyou'll get:\nmol Br2 = 0.041mol\nmol F2 = 0.123 mol\nmol pdt = 0.082 mol\n\nbase from this, you'll found that ratio of Br:F is 1:3. Hence, the product is BrF3. you'll also found that ratio of Br2:F2:pdt is 1:3:2 which also agrees with our balance equation above.
220779
For benzene to undertake this addition reaction, it must be subject to heat (323K) and approx 400atm pressure.  In direct sunlight alone, there may be no reaction.  Addition of a catalyst drives electrophilic substitution towards C6H5CL + HCL
897106
I don't know if this is what you're looking for but... E1 reactions are caused by the solvent pulling off the leaving group, thus a carbo-cation is produced. The leaves the door open for rearrangement of the carbon skeleton, you'll often see alkyl and hydride shifts in the intermediate (carbons and hydrogens moving around to stabilize the atom). E1 is usually preformed in reflux conditions (boiling point of the solvent), which also makes carbon rearrangement more likely. E1 is also in competition with Sn1 (which isn't an elimination at all), so you'll get rearmaments at chiral centers, alkyl halides, and other trash. So E1 kinda messes with your product. \n\nBut, your other option, E2, requires antiperiplanar geometry when removing the other hydrogen, so rings are a whole mess by themselves. And E2 is not great for removing OH groups, you usually have to protonate the OH (usually with acid) and make an oxonium ion (positive oxygen) and get the OH to leave as H2O, which is an E1 mechanism. Man, organic chem is a pain.
130958
The equation is correct, if I perceive it right. You have 3 ethoxy groups on the boron atom, which does correspond to that formula. It looks like what might happen if you reacted borane (BH3), possibly tethered in a THF complex, with ethanol (EtOH) to get the product and H2 gas, but a lot of heat too. Such a substituted Borane would be rather reactive, as a Lewis acid.\n\nB(OC2H5)3 + 6O2 ---> H3B03 + 6CO2 + 6H2\n\nis my best guess
400328
Since both HCl and KCl are strong elecrolyes, both should remain totally dissassocitated in solution.  I dont think any HCl will react with KCl in this case.\n\nHowever, have you ever heard the rule "Always Add Acid"?\nWhenever you dilute an acid with water, the resulting solution gives off energy (dissolving HCl in water is an exothermic reaction).  This when you dilsute 1 M HCl to .5 M HCl, energy will be released and will cause a temperature change in the solution.\n\na)  The number of moles of HCl used in this experiement is constant. This value will be equal to the number of moles of HCl present in the initial 100 ml of 1 M solution.\n1 M = x moles / .1 L, thus you have .1 moles of HCL.\n\nb)  The heat of the reaction will be the heat given off when the HCl solution is diluted.\n\nc)  Once you know the heat energy given off by the reaction, you can calculate the change in temperatutre if you know the specific heat of the final solution.  I am not sure if you are suppose to estimate the final solution's specific heat as being equal to that of water (4.186 J/g C), but that would be my guess.\nQ = m*c*(delta T)\n\n\nEDIT:\n"davidchingwong13" is correct in his application of the common ion effect.  By dissolving additional Cl- ions in the solution by adding HCl, the equilibrium between solid and aqeuous KCl is disturbed.  However, KCl is a strong electrolyte, it readily dissolves and breaks in to ions in water.  I doubt the addion of so few moles of Cl- ions to an already unsaturated KCl solution willl effect it that much.\n\nHowever....dissolving KCl in water is a endothermic reaction, so if, indeed, the KCl did precipitate out, it would be exothermic and give off additional energy.
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. :)
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
733447
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
889022
sterics help the water to attack the central carbonyl, as the other 2 are crowded by the phenyl ring. and with the electronics of the system the central carbonyl is the most electropositive making it a great target for nucleophilic addition
971292
as it involves leaving group at primary carbon so the carbocation if formed will be highly unstable ......so neither  E1 nor SN1.\n\nIt will be SN2,.......because ..EtO-  is a good Nu- and also Br- is a good leaving group.......and also C--O bond which will be formed will be stronger than C--Br bond ,so energetically also favourable.
193729
most of the chemical properties of alkyl halides are related to the polar bond.  this bond strengthens with decreased length.  therefore, as fluorine is the smallest halogen, the carbon-fluorine bond will be the shortest and strongest.  the boiling points of alkyl halides are generally higher than those of the corresponding alkane, and the boiling point rise with with increasing size of the halogen as a result of greater London dispersion forces.\n\ni have studied alkyl halides in respect to their ability to undergo both nucleophilic subsititution and elimination reactions.\n\nhope that helps!
1134694
Add a second carbon atom.  The molecular formula for methans is CH4.  The molecular formula for ethene is C2H4.
431100
further explanation - when certain compounds are mixed, for instance an amine (primary or secondary) and an acid chloride, they will  form an amide and an equivalent of HCl  which will form an equilibrium with the starting materials.   Using pyridine (or another tertiary amine such as triethylamine or diisopropylethylamine) will trap the HCl and drive the reaction forward.
1121888
Well you got C3H3Br + CH3S,  I'd guess something like ethane and bromosulfide.. C2H6 + BrS..
72715
CH3CH2Cl is an example.  An alkyl group (R) with at least (an usually) one halogen (X), so R-X
979779
hmmm... acid chlorides can easily react with AlCl3 to form  RCO+ that will serve as the electrophile during acylation. \n\nacid anhydrides may have the difficulty in forming the electrophile because of its O=C--O--C=O system where electrons can freely move around. in effect, site of nucleophilic attack for AlCl3 becomes weak - weaker than that of acid chlorides.