Before you can start anything you need to know the equation you will be using to complete this problem. The equation is C6H6 + 1 Br2 --> 1 C5H6Br + 1 HBr . Fill out the equation and work it through to get the answer.
35g C6H6* 1mol C6H6/78.108g C6H6 * 1 mol C6H5Br/ 1 mol C6H6 * 157g C6H5Br/1mol C6H5Br = 70.35g of Bromobrezene.
Yes, the dark stops the formation of a bromine radical
The substitution reactions of phenol are easier than benzene, phenol directly reacts with bromine and gives tribromo phenol while benzene requires FeCl3 as a catalyst and gives mono bromo phenol.
Adding halogens to alkene groups (X2) requires that the product adopt an anti configuration. Hexene will also lose its double bond upon bromination. Benzene is energetically unfavorable when a reaction attempts to break its double bond. The resonance benzene has makes it very stable, and thus very hard to break.
due to extensive delocalization of pie-electrons of benzene , it do not undergoes Bayer's and bromine test. 6 carbon nuclei hold the pie electronic cloud which make it difficult for an electrophile to attack.
When Bromine is mixed with Benzene it becomes bromobenzene.
You are trying to make a para compound, so the trick here is to recognize that bromine is an ortho-para director (albeit a weak one) and nitro is a meta director. Therefore, you want to add the bromine first and then the nitro. Doing the reaction in reverse order will result in the meta product. Your reaction pathway is: 1) Benzene + Br2 + FeBr3 => Bromobenzene 2) Bromobenzene + HNO3 + H2SO4 (catalytic) => 1,2 bromonitrobenzene + 1,4 bromonitrobenzene
You can add the sodium hydroxide solution(NaOH). Because NaOH reacts with Br2(Bromine) and generates NaBr. NaBr is dissolved in water while bromobenzene is layered with water, then you could remove Bromine by a liquid separation. The reaction is Br2 + 2NaOH = NaBr + NaBrO + H2O.
No!! Benzene wont de colourise bromine water although it is an unsaturated compound ,as it is an aromatic compound and it does not undergo addition reaction.
35g C6H6* 1mol C6H6/78.108g C6H6 * 1 mol C6H5Br/ 1 mol C6H6 * 157g C6H5Br/1mol C6H5Br = 70.35g of Bromobrezene.
Yes, the dark stops the formation of a bromine radical
Meso-stilbene dibromide is an organic molecule. Its structure is a benzene ring bonded to a carbon with a hydrogen and a bromine. That carbon is bonded to another carbon with a bromine that is ANTI to the first bromine. This carbon is then also bonded to a benzene ring.
The substitution reactions of phenol are easier than benzene, phenol directly reacts with bromine and gives tribromo phenol while benzene requires FeCl3 as a catalyst and gives mono bromo phenol.
Toluene forms from the reaction of bromine with acetophenone.
Adding halogens to alkene groups (X2) requires that the product adopt an anti configuration. Hexene will also lose its double bond upon bromination. Benzene is energetically unfavorable when a reaction attempts to break its double bond. The resonance benzene has makes it very stable, and thus very hard to break.
no reaction
The reaction between methane and bromine is a photochemical reaction. Refer to the related link below.