# What are the states of HX when reacting with alcohol and alkenes respectively, and Br2 when reacting with benzene?

I couldn't find this in any textbooks, so I guess I will ask this here. What is the physical state of $\ce{HX}$ when reacting with alcohol (nucleophilic substitution) and alkenes (electrophilic addition) respectively?

For electrophilic substitution of benzene with bromine, what is the state of bromine? (e.g. aqueous, liquid, solid)

If possible, can you explain why they have the particular state?

• By the theory (not practically) what I know, $HX$ and $Br_2$ must be gases (at room temperature) because of lower molecular weights and the covalent bonds. Go through this link http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt, you will get idea about which compound or element will be gas. Usually, $HCl$ is gas at room temperature, but while performing experiments in our labs, we use however solutions of $HCl$, so I think both $HX$ and $Br_2$ will be in liquid state, which is most preferred. – Immortal Player Nov 17 '13 at 15:47

The first reaction ($\ce{HX}$+Ethanol) happens in solution. High concentration of HX mixed with ethanol will produce the product. However, for $\ce{X}$ other than $\ce{Cl}$, more often one put concentrated phosphoric or sulfuric acid with alcohol and a salt of X (for example, $\ce{KBr}$ or $\ce{KI}$). Concentrated phosphoric/sulfuric acid are easy to obtain and store in high concentration, as opposed to $\ce{HX}$. It is, therefore, a liquid phase reaction.
For ethene + $\ce{HX}$, it can be done in gas-phase, in an aqueous environment, or you can also add $\ce{HX}$ gas to an ethene solution in a solvent such as $\ce{CCl4}$. So, it can be either gas phase or liquid phase. It is almost always more preferable to do a reaction in solution in a laboratory set up. (So much that we call the experimental part of the chemistry wet chemistry.) This constraint is, however, lifted in the industrial world, with the availability of expansive specialized reactors.
And for benzene + $\ce{Br2}$, both reactants are liquid and $\ce{Br2}$ is also soluble in benzene, so the reaction is in also the liquid phase.