1 ADDITIONS ACROSS C=C Π BONDS. We’ve learned a number of different substitution and elimination reactions of alkyl halides. We’ve learned that compounds with lone pairs are nucleophiles, and alkyl halides are electrophiles. Now we’re going to learn the reactions of a different class of nucleophiles: alkenes. 9.1. Reactivity of Alkenes. The C=C π bond of an alkene is higher in energy than the C—H or C—C σ bond. As a result it can act as a nucleophile towards various electrophilic species. When a C=C π bond acts as a nucleophile, one of the C atoms takes the π electrons away from the other C and uses them to make a bond to the electrophile. The other C atom either becomes electron-deficient: or it gets a pair of electrons from the electrophile at the same time as the electrophile is getting a pair of electrons from the π bond Note the balance of charge in both examples! Because the C=C π bond is lower in energy than a lone pair (it is, after all, a bond, not an unshared, nonbonding pair of electrons), it reacts only with very strong electrophiles and not at all with alkyl halides (except under circumstances you will learn in CHE 232). 9.2. Addition of Strong Acids HX to Alkenes to give Alkyl Halides. Alkenes C=C react with strong acids HX to give alkyl halides H—C—C—X. HX can be HF, HCl, HBr, or HI, although HCl and HBr are used most often. Alkenes also react with carboxylic acids to give esters H—C—C—O2CR. When alkenes react with HX, they do so in a two-step mechanism. Alkenes are nucleophiles, so the first step in their reaction with HX must involve reaction with an electrophile. Acids are electrophiles because they give up H+. The first step in the reaction of an alkene with HX, then, is addition of H+ to the alkene to give a2 carbocation intermediate. The carbocation is an electrophile, of course, so the next step is combination with a nucleophile. The nucleophile is X–, the conjugate base of the acid. In fact, the second step of the mechanism is the same as the second step in the SN1 mechanism. Here you just have a different way of generating the carbocation intermediate. We can look at the reaction coordinate diagram for HX addition to an alkene (see Jones Figure 9.3). Remember the Hammond postulate states that the structure of a transition state resembles the structure of the nearest stable species. Hence transition states for endothermic reaction steps structurally resemble product, and transition states for exothermic steps structurally resemble starting materials. Thus according to the Hammond postulate, any factor that stabilizes a high energy intermediate should also stabilize the transition state leading to that intermediate. So, factors that lead to a more stable carbocation intermediate will also stabilize the transition state of HX addition to alkene! Suppose an alkene that has different substituents on each C. In principle we can obtain two products. The major product is the one where the nucleophile has added to the C better able to bear a positive charge, in many cases the more substituted C. This rule, called Markovnikov’s rule, derives directly from the fact that the reaction proceeds through a high-energy carbocation, so the faster reaction is the one that gives the lower energy carbocation intermediate. Another way of formulating Markovnikov’s rule is that the C with more H’s gets another one (them that has, gets more). If the two carbocations are about equal in energy, then a mixture of regioisomers will be formed (See Jones Figures 9.26 and 9.27).3 The carbocation intermediates in HX addition reactions are the same as the carbocation intermediates in SN1 and E1 reactions, and as such they can undergo rearrangement to generate a more substituted carbocation. A rearrangement can occur via 1,2-shifts of alkyl or H- groups. For example, 3,3-dimethyl-1-butene reacts with HBr to give 2-bromo-2,3-dimethylbutane as product, via a tertiary carbocation formed by 1,2-shift of a methyl group. H3CH3CCH3HBrH3CCH3H3CBrCH3H3CCH3H3CCH3H3CCH3H3CCH3+H1,2-methyl shift+Br- Similarly, a 1,2- hydride shift will produce 2-bromo-2-methylbutane from 3-methyl-1-butene:4 HH3CCH3HBrH3CCH3H3CBrHCH3H3CCH3H3CCH3H3C+H1,2-hydride shift+Br- Again, the rearrangements occur to give more substituted (hence more stable) carbocations, so again, 2° carbocations are most prone to undergo rearrangement, 3° carbocations tend not to undergo rearrangement, and 1° carbocations can’t be formed in the first place. Alkenes are of course flat, whereas the product alkyl halides are not, so the question arises, do the H and the X add to the same face of the alkene (called syn addition), to opposite faces (anti addition), or is a mixture of products obtained? When the alkene reacts with H+, a carbocation is obtained. Suppose the H+ adds to the bottom face of the alkene. The carbocation is planar, and the top face of the carbocation is no different from the bottom face, so the X– can add to either face. As a result, a mixture of products is obtained. The addition, therefore, is nonstereospecific. We will shortly see additions that are stereospecific. HX also adds to alkynes to give haloalkenes, but the intermediate C(sp) carbocations derived from alkynes are much higher in energy than the C(sp2) carbocations derived from alkenes, so the reaction is much more difficult and less widely used than the addition to alkenes. You have already learned how to make alkyl halides from alcohols (SN1 substitution). Now you know how to make alkyl halides from alkenes. (You also know how to make alkenes from alkyl halides – elimination reactions!) 9.3. Retrosynthetic analysis Retrosynthetic analysis refers to the skill of figuring out what starting materials can be used to make a desired product. We use a5 double-bodied arrow, ⇒, to show a retrosynthetic step leading from product to starting materials. The arrow should be read as “can be made from”. To identify the simpler starting materials, we need to identify the bond that
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