Organic Chemistry Help-SN1 or E1 Reactions?
958454
To visualize (C6H5)2CHBr, you have to identify the "side groups".  Obvious Br is one side group and so is one H.  Then you have two phenyl side groups so that leaves you that carbon to build on.  The molecule one carbon bound to two phenyl groups, one hydrogen and one bromine, each with a single bond.\n\nOK, let's build on that.  You are now going to react it with ethanol and you want to know what mechanism to expect.  Since Teach didn't show you the products, you're going to have to think about this one.  First, analyze what type of molecules you have.  The first molecule we spoke of, what type of carbon is the main carbon (primary, secondary, tertiary)?  What type of reaction (Sn1 or E1) does that type of carbon favor?  How about the ethanol?\n\nActually, the alpha carbon in ethanol, being a primary carbon, would favor Sn2 but that's not an option.  OH is an awful leaving group and alkanes don't like to carry a charge.  So we can rule the carbon out.\n\nThe problem must revolve around diphenylmethylbromide.  Having two phenyls would stabilize a positive charge fairly well and that is the first step for both E1 and Sn1 mechanisms.  So, which one is it?  The big giveaway is that En reactions are called elimination for a reason.  They lose a pair of substituents to form a double or triple bond, always a higher order bond.  Can there be any higher order bonds formed in this molecule?  How about Sn1, then?  That would result in (C6H5)2CH(OC2H5) + HBr.  Is this more likely than elimination?  Would this be endothermic or exothermic?\n\nThat's as much as I'm going to do.  The rest is for you, as you only really needed to know the structure of the first molecule.  But I hope this helps you understand what you are doing.\n\nGood luck.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
963233
Lucas Reagent replaces the OH group by a Cl group.\nTherefore CH3-C5H8-OH + HCl -> CH3-C5H8-Cl + H2O
535526
a.  1-Bromobutane will undergo a faster reaction than 1-chlorobutane, as bromide is a better leaving group than chloride, due to being a weaker base.\nb.  1-Bromobutane will undergo a faster reaction than 2-bromobutane, as primary halides undergo SN2 reactions faster than secondary halides, due to steric considerations.
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
1151125
for understanding organic chem u must know all the reactions by heaart, that u can do by preparing charts of various reaction n relating each other . the more u read more u become strong in it. books by s chand publication is good
752247
Remember that the halide, which is going to eventually serve as the leaving group, is pulling electrons away from the carbon center, inducing a dipole where most of the electron density is around the halide (leaving group) and less of it around the carbon center.  The incoming nucleophile is thus even MORE drawn toward that carbon center when there is more induction of the dipole.  Therefore, the more electronegative the leaving group, the stronger the induced dipole.  The stronger the induced dipole, the more reactive is the carbon center towards SN mechanisms.
445447
1-chloropropane, because chlorine has a higher value of electronegativity.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
362907
I made flashcards while studying organic chemistry.  Break all the reactions down into the different types of reactions, then make a flashcard for each reaction within all of the types of reactions.  If you want to, you could leave certain details out, like solvents or conditions, and put them on the back to make sure you are remembering them.  I would also write for what cases the reactions would work on each card (eg. primary and secondary alcohols only).  It might be a lot of work to write them out, but it worked for me.  Also, just writing them out will probably help too.  Good luck!!
707224
E1 reactions proceed through carbonium ions. Carbonium ions resulting from particular carbon skeletons/derivatives are unique. Once the carbonium ion forms, it has a characteristic tendency to eliminate to an olefin, or to hydrate or solvate to an alcohol or an ether. So you can't control how much olefin, alcohol, or ether you'll get.\n\nOn the other hand, if the substrate is really cheap, like cyclohexanol, then an E1 reaction is the way to go, because you'll get cheap cyclohexene regardless of what ever else you have to put up with.
771541
Product 1: 3-methyl-2-pentene, HBr, and Water.\nReaction 1 is alkene formation by dehydration.\n\nProduct 2: 1-Chloro-2-ethylbutane and junk.\nRecation 2 is substition of OH with Cl.  I think this is an SN1 reaction but I'm not sure.  I'm sure about the product though.
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
293072
You could probably just dehydrate it with concentrated sulfuric acid and heat.\n\nAnother method would be to generate a alkyl halide with thionyl chloride (SOCl2) and then eliminate it in a E2 reaction with sodium ethoxide in ethanol (especially strong base to be sure you get elimination, and not SN1)
506807
I'm guessing this is a question for class. In practice, working with acetylene gas as a synthetic reagent isn't a good idea. It's a flammable, explosive gas and most synthetic labs really don't have the necessary safety equipment to handle it. It's much safer to buy the butyne you need and work from there.\n\nHowever, on paper you would first make the 2-butyne. You would react 1 eq of acetylene with 2 eq of n-BuLi followed by quenching with 2 eq of MeI and standard workup and purification. If you want the 1-butyne, you can isomerize 2-butyne to 1-butyne with an amide base like NaNH2 or KHMDS.  The reason you can't just alkylate with iodoethane is that you are going to end up with a mix of acetylene, 1-butyne and 3-hexyne since you can't selective monodeprotonate every acetylene molecule. \n\nHope this helps.
975754
I'm not sure if "catalyzes" is the right word, but what you're doing by adding iodide (as KI or NaI) to the reaction mixture is adding another nucleophile to the mix.  This would then speed up the SN2 reaction of your starting material.  You would probably run the risk of iodide becoming another (competing) nucleophile.  Since you don't specify what the initial nucleophile actually IS, this may or may not be a bother to you; the gyst of your question was whether the reaction would speed up.  It probably will, because\n\nrate of SN2 rxn = k [substrate][nucleophile]\n\nAdding I- is like adding more nucleophile, unless of course  your existing nucleophile is more reactive towards the substrate than iodide is.  And as the relation above says...more nucleophile means a faster reaction.  The question is, what's your product?\nIs it the desired one?  or do you care?  (are you content with just seeing the reaction speed up....)\n\nBest I can do with what you gave me.
749219
a) Ag2S\nb) SnCl4\nc) Na3N\nd) SrI2\n\nDo us a favor and pay attention during class. That way you'd know how to do these.