Heating 1-bromobutane with 2M NaOH ->
Butan-1-ol + NaBr
Acidification with 2M HNO3
Then Addition of 0.1 AgNO3 dropwise
What is the equation here?
Also why does the mix need to be acidic before adding AgNO3?
The equation for the given reaction sequence is as follows:
1-bromobutane + NaOH → Butan-1-ol + NaBr
The first step involves heating 1-bromobutane (C4H9Br) with a solution of 2M NaOH (sodium hydroxide). This process is called nucleophilic substitution, in which the bromine atom (Br) is replaced by the hydroxyl group (OH) from the NaOH solution. As a result, 1-bromobutane is converted into butan-1-ol (C4H9OH) and sodium bromide (NaBr).
The second step is the acidification of the reaction mixture with 2M HNO3 (nitric acid). This step is carried out to neutralize the remaining NaOH and convert the sodium salt (NaBr) into its corresponding acid (HBr).
The third step involves adding 0.1 M AgNO3 (silver nitrate) dropwise to the acidified mixture. Here, AgNO3 reacts specifically with HBr to form a precipitate of silver bromide (AgBr). The equation for this reaction is:
AgNO3 + HBr → AgBr + HNO3
Now, let's address the second part of your question. The mixture needs to be acidic before adding AgNO3 in order to selectively detect the presence of the halide ion (Br-) in the reaction mixture. This is because AgNO3 is highly selective towards halide ions and forms insoluble silver halides, such as silver bromide (AgBr) in this case.
However, if the mixture remains basic (basic pH) while adding AgNO3, other ions such as hydroxide ions (OH-) would compete with the halide ions for the silver ions (Ag+). As a result, silver hydroxide (AgOH) would precipitate instead of the silver halide, leading to a false interpretation of the reaction.
To overcome this, the acidification step is performed to convert any remaining NaOH into HBr. By doing so, there is a presence of HBr in the solution, promoting the selective formation of AgBr when AgNO3 is added. Thus, the mixture needs to be acidic before adding AgNO3 to obtain accurate results.