Use equation with curved arrows to show the reaction between a three-carbon alkene and
hydronium ion to give isopropyl carbonium ion and water.
Answers
The first step (which involves electrophilic attack by bromine on the double bond) produces a bromide ion and a carbocation, as shown in Equation 10-1.1
As we know from our study of SN1 reactions (Section 8-4), carbocations react readily with nucleophilic reagents. Therefore in the second step of the bromine-addition mechanism, shown in Equation 10-2, the bromoethyl cation is expected to combine rapidly with bromide ion to give the dibromo compound. However, if other nucleophiles, such as Cl⊖ or CH3OH, are present in solution, they should be able to compete with bromide ion for the cation, as in Equations 10-3 and 10-4, and mixtures of products will result:
To account for the
All of the hydrogen halides HF, HCl, HBr, and HI) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of 70% anhydrous hydrogen fluoride and 30% of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent I2 addition products resulting from iodine formed by oxidation reactions such as
4HI+O2→2I2+2H2O
With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled asonably reactive alkenes, although by no means as effectively as does concentrated sulfuric acid. The carbocation formed then is attacked rapidly by a nucleophilic water molecule to give the alcohol as its conjugate acid,2 which acts readily with water in a second step to give ethanol:

10-3F Aqueous versus Nonaqueous Acids. Acid Strengths
One of the more confusing features of organic chemistry is the multitude of conditions the H3O⊕ concentration, because H3O⊕ is where the protons are.
Now, if we use poorer proton acceptors as solvent we find the proton-donating powers of various "strong" acids begin to spread out immensely. Furthermore, new things begin to happen. For example, ethene is not hydrated appreciably by dilute aqueous acid; it just is too hard to transfer sulfuric acid is virtually all converted to H3O⊕, which is non-nucleophilic!
H2SO4+H2O→H3O⊕+HSO−4
id is a very good proton donor) and SbF−6 is an extremely poor nucleophile. If we add ethene to such a solution, a stable solution of CH3CH+2SbF−6 is formed. The reason is that there is no better proton acceptor present than CH2=CH2 and no nucleophile good enough to combine with the