Give the products from the reaction of but-l-yne with:\n(a) 1 mole of bromine\n(b) excess HCl\n(c) mercury(II) sulfate and dilute sulfuric acid\n(d) sodium metal\n(e) potassium permanganate
212887
(a) 1,2-dibromobut-1-ene\n(b) 2,2-Dichlorobutane (based on Markovnikovs rule)\n(c) don't know\n(d) don't know.  Na is often used to reduce multiple bonds in a free-radical manner, however you need a source of H.\n(e) KMnO4 is a very strong oxidising agent and is commonly used as tests for unsaturation.  The side product is MnO2, showing that two atoms of oxygen are transferred to the product.  The product formed I think is 1-hydroxy-butan-2-one.
377266
1) Zinc Chloride\n2) Copper (II) Sulphate\n3) there is no such thing as NH4OH, but oh well..\n    Ammonium Hydroxide\n4) Iron (III) Sulphide\n5) Aluminium Phosphate\n6) Lead (II) Nitrate\n7) Sulphur hexafluoride\n8) Sodium Sulphate\n9) Mercury (II) oxide\n10) Potassium Clorate (V)\n11) Tin (II) Choride\n12) Iron (II) hydroxide\n13) Mercury (II) Sulphate\n14) Carbon Dioxide
462271
Li2O + CO2 ---> LiOH\nNa2O2 + CO2 ----> Na2CO3 + O2\nKO2 + CO2 ---->  K2CO3  + O2
444434
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
329417
fe 2+ , SO4 2-, Ag+, NO3 -\n\nthe products:\n\nFe(NO3)2 + Ag2SO4
426846
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
161835
CH3-CH=CH2 has an electrohilic addition of bromine.  The bromine adds to the double bond via an intermediate known as a bromonium ion.  This is essentially a polarised bromine molecule, with a small dipole.\n\n\n....Brd-\n....Brd+\n../....&#92;\nCH2-CH-CH3 (ignore the dots, only there for spacing).\n\n---->\n\n............+\nBr-CH2-CH-CH3   secondary carbocation is formed\n\nThe Br-Br bond then breaks the bromonium ion formally forms a single covalent bond with the terminal carbon, and creates a secondary carbocation.  The Br- formed from the breaking of the Br-Br bond then attacks the carbocation to form 1,2-dibromopropane.\n\nBecause you are doing the reaction in the presence of NaCl and H2O, then Cl- (from the NaCl) or -OH, from the water can also attack the formed carbocation, giving 1-bromo-2-chloropropane and 1-bromopropan-2-ol as a result.  \n\nThe reason for not generating any 2-bromo-1-chloropropane for example is that this would require the preparation of a primary carbocation.  This is far less stable than a secondary carbocation, so the three products described by far predominate in the reaction mixture
425157
I would have to guess that it involves the neutralization reaction of more than 2 substances.\n\nJust a guess
472148
a. N2O4\nb.PCl5\nc.NF3\nd.S2Cl2
310266
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
66446
the correct answer is B. A is wrong b/c it isn't an oxidation reduction reaction. Looking at a table of solubility products will confirm the answer is B.
410008
1. Methyl iodide: would involve a primary carbocation, which do not exist. Methyl iodide only reacts by an SN2 mechanism.\n2. 2-iodopropane: secondary carbocation\n3. 2-chloro-2-methylpropane--i.e. t-butyl chloride: tertiary carbocation\n4. 2-iodo-2-methylpropane--i.e. t-butyl iodide: tertiary carbocation. Iodide is a better leaving group than chloride, so this substrate is better for SN1.
319335
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
507465
1) NH(4)Cl - Ammonium chloride\n\n2) HClO(2) - Chlorous acid\n\n3) Ca(BrO(3))2 - Calcium Bromate\n\n4) BeSO(4) - Beryllium sulphate\n\n5) (NH(4)) (3)N - Ammonium nitride\n\n6) RbI - Rubidium iodide
407062
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
441596
This is a addition-elimination reaction (http://www.chem.ucalgary.ca/courses/351/Carey/Ch23/ch23-3-3.html). can be accelerated by lewis acids.
430404
Well number 10 is\n\n(Z)-3,4-dimethylhept-3-ene\n\n14,15 and 15 and 17.... Need to be drawn out.\n\nGood Luck!