Is H2O2 present? If so, HBr and H2O2 will form Br2. That then reacts to add Br across the double bond, forming 1,2-dibromo-1,2-diphenylethane.
If H2O2 is not used, the HBr is added across the double bond. The result of this is a single C-C bond, with a new H-atom bonded to one side, and a new Br-atom bonded to the other side. Since the molecule is perfectly symmetrical, it doesn't matter which side the H and the Br are added to (although if it wasn't symmetrical, there are rules to determine which atoms goes on which side). I'm not sure about the name but I think it'd be something like this: 1-bromo-1,2-diphenylethane. Note that the carbon bonded to the Br is a stereocenter (it is bonded to a phenyl, H, Br, and C, so it has 4 unique substituents), but the product will be a racemic mixture of the two stereoisomers (I'm pretty sure anyway).
Take a look at the Web Links to the left, but note that most of these pages discuss the addition of Br2 rather than HBr (although they use HBr to prepare the Br2 with H2O2).
This equation is:HBr + LiOH = LiBr + H2O
CH3CH2CH3 + Br2 = CH3CHBrCH3 + HBr The major product formed from the reaction is 2-bromopropane. It is a free radical bromination reaction. It is called a substitution reaction because you are substituting a H an atom for a Br atom and making HBr.
HBR doesn't react with Propane, but it does with Propene. The product is either 1-bromo propane(minor product) or 2-bromo propane(major product). To determine which product will be the major product, use the Markovnikov's rule.
When 1-propene reacts with hydrogen bromide (HBr), it undergoes an electrophilic addition reaction, yielding 2-bromopropane as the major product. This reaction follows Markovnikov's rule, where the bromine atom attaches to the more substituted carbon atom of the double bond. Additionally, a minor product, 1-bromopropane, may also be formed.
To balance the reaction HBr + B → BBr3 + H2, we consider the number of atoms of each element on both sides. The balanced reaction is 6 HBr + 2 B → 2 BBr3 + 3 H2. Thus, the two substances that would have a coefficient of 2 in the balanced equation are BBr3 and H2.
The reaction is:CH3NH2 + HBr = CH3NH3Br
This equation is:HBr + LiOH = LiBr + H2O
CH3CH2CH3 + Br2 = CH3CHBrCH3 + HBr The major product formed from the reaction is 2-bromopropane. It is a free radical bromination reaction. It is called a substitution reaction because you are substituting a H an atom for a Br atom and making HBr.
The reaction mechanism for the addition of HBr to 1,3-pentadiene involves the formation of a carbocation intermediate followed by the attack of the bromide ion to form the final product.
The reaction mechanism for the addition of HBr to 2,4-hexadiene involves the formation of a carbocation intermediate followed by the attack of the bromide ion to form the final product.
When bromine reacts with water, it forms hydrobromic acid (HBr) and hypobromous acid (HOBr). The overall reaction can be represented as: Br2 + H2O → HBr + HOBr. This reaction is reversible and depends on the pH and conditions of the solution.
The reaction between CH4 (methane) and Br2 (bromine) would result in the substitution of one hydrogen atom in methane with a bromine atom, forming bromomethane (CH3Br) and hydrogen bromide (HBr).
CH4 + Br2 --> CH3Br + HBr
The reaction is:Cd + 2 HBr = CdBr2 + H2
NH3 (aq)+ HBr(aq) --> NH4+ (aq)+ Br- (aq)
The products of the reaction between hydrogen bromide (HBr) and sodium hydroxide (NaOH) are sodium bromide (NaBr) and water (H2O). This is a neutralization reaction where the acid (HBr) reacts with the base (NaOH) to form a salt (NaBr) and water.
The reaction between HBr and KOH is a 1:1 ratio. This means that the moles of HBr present in the solution will be equal to the moles of KOH used in the neutralization reaction. Using this information and the volume and concentration of KOH used, you can calculate the concentration of the HBr solution.