Question 4.9: Give the structure of an alkene that would give 2-bromopenta...
Give the structure of an alkene that would give 2-bromopentane as the major (or sole) product of HBr addition. (The numbers are for reference in the solution.)
an\> alkene + HBr \longrightarrow CH_3CH_2\overset{3}{C} H_2\overset{2}{\underset{\mid}{ C}} H\overset{1}{C} H_3 \\\hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \quad Br \\\hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \quad 2-bromopentaneThe bromine of the product comes from the H—Br. However, there are many hydrogens in the product! Which ones were there to start with, and which one came from the H—Br? First, recognize that the carbon bearing the bromine must have originally been one carbon of the double bond. It then follows that the other carbon of the double bond must be an adjacent carbon (because two carbons involved in the same double bond must be adjacent). Use this fact to construct all possible alkenes that might be starting materials. Do this by “thinking backward”: Remove the bromine and a hydrogen from each adjacent carbon in turn.
Remove Br from carbon-2 and H from carbon-3 ⇒ CH_3CH_2CH=CHCH_ 3
2-pentene (cis or trans)
Remove Br from carbon-2 and H from carbon-1 ⇒ CH_3CH_2CH_2CH= CH_2
1-pentene
(The symbol ⇒ means “implies as starting material.”) Which of these is correct? Or are they both correct? You haven’t finished the problem until you’ve mentally carried out the addition of HBr to each compound. Doing this and applying the known regioselectivity of HBr addition leads to the conclusion that the desired alkyl halide could be prepared as the major product from 1-pentene. However, both carbons of the double bond of 2-pentene bear the same number of alkyl groups. Eq. 4.17 (p. 153) indicates that from this starting material we should expect not only the desired product, but also a second product:
Furthermore, the two products should be formed in nearly equal amounts. This means the yield of the desired compound would be relatively low and it would be difficult to separate from its isomer, which has almost the same boiling point. Consequently, 1-pentene is the only alkene that will give the desired alkyl halide as the major product (that is, the one formed almost exclusively).
In solving this type of problem, it isn’t enough to identify potential starting materials. You must also determine whether they really will work, given the known characteristics—in this case, the regioselectivity—of the reaction.