5. What is the product of reaction between cyclohexene and concentrated hydrobromic acid?
546423
Bromocyclohexane. You don't need to add the "1-" because it's assumed to be 1 by IUPAC convention.
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.
469836
(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.
788034
Once the hydroxyl group leaves the alcohol, the structure would lose its pyramidal shape and flatten into a two-dimensional shape.  The halide ion would add from either side, having a 50% chance of attaching from either side.  Once halogenated, the new molecule would again be optically active but you would have an equal abundance of both R and L enantiomers.  This follows one of the rules of stereochemistry in that optically inactive reactants (in this case HX) always yield optically inactive products.
858619
The easiest way to approach problems like these, my friend, is to look at your alkyl halide first and then look at your nucleophile or base.\n\nT-butyl bromide is a tertiary halide.  Tertiary halides undergo E1 and SN1 reactions because of their ability to form a stable carbocation intermediate.  If you recall, tertiary carbocations are more stable than primary and secondary.\n\nYour -OCH3 is a strong base known as methoxide, the conjugate base of methanol.  Strong bases favor elimination reactions and thus the elimination product is your answer.
516493
a) Toluene and Cl2 form p-chlorotoluene and HCl in presence of AlCl3\nb) nitrobenzene + Cl2 -(AlCl3)-> m-chloro-nitrobenzene\nc) chlorobenzene + Cl2 -(AlCl3)-> p-dichlorobenzene\nd) 1-(3-chlorophenyl)ethanone\ne) 3-chlorophenyl acetate\nf) methyl 3-chlorobenzoate\n\nOrganic Chem rocks!
454861
Zn + 2HCl---> zncl2 + H2\nIt is a displacement reaction. Zinc is more reactive then Hydrogen and is capable of displacing it from Hydrogen Chloride.
139601
This is an esterification, \nButanoic Acid (CH3CH2CH2CO2H)+ ethanol (HO-CH2CH3), in basic media, yields ethyl butyrate. \n\nCH3CH2CH2CO2H + HO-CH2CH3 --> CH3CH2CH2COO-CH2CH3
250756
Hydroboration is syn addition, so the product would be trans-2-phenyl-1-cyclohexanol, with both phenyl and OH groups equatorial.
793613
1,6-diaminohexane to react with 1,8-dicarboxylic acid. extract with a pair of tweezers because the film forms on top.
801483
b. Tertiary bromide. Br2 dissociates into Br. + Br. The first Br. abstracts a proton preferentially from CH3CH2-C-H to give the most stable free radical (most stable because most hindrered and so least likely to revert to hydrocarbon again). That free radical abstracts Br from a new Br2, giving a tertiary bromide and generating a new Br. to continue the chain reaction.
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
305340
A Bronsted acid is a substance which releases H+ in water.\n\nHF + H2O --> F- + H3O+\n\nNH4+ + H2O --> NH3+ H3O+\n\nH2CO3 + H2O --> HCO3- + H3O+\n\nHNO2 + H2O --> NO2- + H3O+\n\nNote that the final result is always the hydronium ion \n(H3O+)
965746
The Grignard Synthesis is usually used to convert a brominated ring compound to the corresponding phenol (Hydroxy group).\n\nThe Magnesium turnings provide metallic magnesium which inserts between the Bromine and the ring.  This is then hydrolyzed to the Hydroxy group.\n\nSee the site below for details:
799473
1\na) H2C = CH - CH2 - CH2 - CH = CH2\n ( 1,6-hexadiene )\n\nb) H2FC - CHF - CFH2\n ( 1,2,3-Trifluoropropane )\n\nc) H3C - CH2 - CH2 - CH2OH\n  ( 1-Butanol )\n\nd) H3C - CH2 - CH2 - CHO\n  ( Butanal )\n\n2.\na) Usually FeBr3 is used in addition to Br2 to brominate benzene, but for this:\n\nProduct: Bromobenzene\n\nb) \nProduct: Ethyl butanoate\n\nc) Dunno this one.\n\nd) \nProduct: 1,2-Dichloropropane
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
815271
Br2 reacts with the double bond of an alkene.    The double bond disappears and the carbon atoms of the former double bond are now bonded to a Br atom.  Since Br2 is broken apart, then it no longer has the color.