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.
When Bromine is mixed with Benzene it becomes bromobenzene.
Liquid bromine is the Real Bromine, while Bromine water is a mixture of Bromine and Water
Liquid bromine is the Real Bromine, while Bromine water is a mixture of Bromine and Water
Bromine has less valence shells than lead making the distance between its valence electron and its nucleus less than that of lead. This means that there is greater attraction between the nucleus and electron for bromine and it requires a higher ionisation energy to remove its electron.
Its a double elimination. Since these carbons are secondary they can undergo E1 or E2, depending on the strength of the base used, if the adjacent hydrogen is stericly hindered (blocked by a bulky group). Using a strong base such as -OH (from K+OH-) You have the Bromine attack the K+, making a carbocation, then have the H of the adjacent carbon leave (donate) its electron to the carbocation creating a double bond, and have the -OH attach to that leaving Hydrogen; forming water as a result. This leaves you with a bromine attached to one carbon on the double bond, and a Hydrogen on the other. So you just repeat those steps again. Leaving you with the Diphenlacetlyene. Which is just PhC(triple bond)CPh + HOH + 2KBr
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
well maybe because coca-cola remove gum from hair and well their is bromine in their
It is a liquid.
The simplest method would be distillation, because bromobenzene has a higher boiling point than chlorobenzene.
Bromobenzene is a clear, colorless liquid with a pleasant odor. It is used as a solvent and motor oil additive, and in making other chemicals.
When a mixture of carbonmonoxide and HCl is added to Bromobenzene (Gatterman's reaction) a mixture of ortho and para product is obtained which may be separated on the bases of solubilities.
Beryllium chloride is a nonpolar molecule.
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.
The Bromination of benzene in presence of Ferric chloride produces Bromobenzene the nitration of bromobenzene with dilute nitric acid gives ortho and para products which may be separated by physical means.
By solvent extraction with water. By fractional distillation.
Liquid bromine is the Real Bromine, while Bromine water is a mixture of Bromine and Water