Carbon Reactivity

The goal of this unit is to introduce one aspect of carbon reactivity that plays a very important role in the chemical transformations that occur in living organisms. Previously, we learned about the common functional groups that exist on hydrocarbons. These functional groups will eventually allow us to see how proteins, DNA, sugars, and lipids are synthesized in the body. Before we can understand that, we first need to consider a couple ideas that profoundly influence chemical reactions: energy and mechanisms. Reaction Energies To begin this discussion, let's first think about how chemical reactions take place. The simplest conceptual pathway is one where every bond in the reactants break apart resulting in individual atoms. Once all bonds are broken, new covalent bonds can be formed. For example, check out the example below which examines the synthesis of HCl from H2 and Cl2. In the first step, the H-H and Cl-Cl bonds are broken to form an intermediate combination of atoms. Since these atoms all lack a satisfied octet, they are not very stable. Consequently, we call this a high energy intermediate or transition state. The term transition state suggests that we are in transition between two stable states (reactants and products). From the transition state, the atoms can recombine into new compounds that satisfy the octet rule, regenerating a stable electron configuration. H2 (g) + Cl2 (g) ⇌ 2 HCl (g) Step 1: H2 + Cl2 --> 2 H + 2 Cl Step 2: 2 H + 2 Cl --> 2 HCl Untitled picture.png 00 In the past, we've learned that we can uses Hess' law to determine the net DG, DH, and DS of this reaction by adding up the individual steps (as long as they sum to the desired reaction, which they do in this case). So, to figure out the change in enthalpy of this reaction, we can look up the H-H, Cl-Cl, and H-Cl bond enthalpies and apply them appropriately. A table of bond enthalpies is shown below. Reaction DH Step 1 H2 + Cl2 --> 2 H + 2 Cl 435 + 243 = 678 kJ/mol Untitled picture.png
Step 1 H2 + Cl2 --> 2 H + 2 Cl 435 + 243 = 678 kJ/mol Step 2 2 H + 2 Cl --> 2 HCl 2*-431 = -862 kJ/mol Overall H2 (g) + Cl2 (g) ⇌ 2 HCl (g) 678 + -862 = -184 kJ/mol Breaking bonds are endothermic (this process requires energy) and creating a bond is exothermic (energy is released - it is energetically favorable to create a bond which is why they form). Don't forget stoichiometry! 2 HCl bonds are formed. A couple things to note here: Untitled picture.png Table of Bond Enthalpies (kJ/mole) at 25 oc Bond H-CI H-Br c-c Enthalpy 435 569 431 364 297 414 460 465 377 368 724 963 Bond C-N c-o C-Br C-CI NEN N-N O-O Si-Si Enthalpy 301 352 532 234 331 950 297 498 213 159 339 Bond PEP Br-Br CI-CI H-Se H-Te H-Si H-P C-Si 1-1 Enthalpy 490 193 243 276 243 427 260 393 318 289 151 So we see that overall this reaction does release heat. We would also expect it to be spontaneous because there should not be a significant change in entropy (2 gasses become 2 gasses). You could calculate this value from a table of standard molar entropies (expect to do this in your problem set). So this reaction is spontaneous. Does this mean that every time an H2 molecule bumps into a Cl2 molecule, HCl is formed? Not necessarily. For this reaction to happen as we have explained, step 1 must happen prior to step 2. As you can see in the image below, this is troublesome because step 1 requires quite a bit of energy! Untitled picture.png Energy
Untitled picture.png Energy This demonstrates a very important concept in chemical reactions - all (well, nearly all) reactions must proceed through a high energy state (transition state) before products can be formed. The energy requiring step absolutely necessary if products are to be formed. This big energy step is known as the activation energy (Ea). It corresponds to the energy that the reactants must "find" to reach the highest energy state and ultimately be converted to products. As you might imagine, the more energy that is required to get to the transition state, the less likely it is for that reaction to happen. We will explore this concept and how it relates to chemical and enzyme kinetics later in the term. For now, let's explore how this idea influences the chemistry that our common functional groups can participate in. Hydrolysis Reactions So first let's clarify - we'll be focused almost exclusively on identifying what bonds we can break with water. The chemical term for a reaction that breaks a bond is "lysis" - because we are lysing a bond (where lyse is to break). When this process is done by water, this is called "hydrolysis" (hydro = water). In general terms, we are always going to be thinking about breaking a C-X bond, where X can theoretically be any atom. We say that water is "added across the bond" such that we always end up with C-OH and H-X (the three atoms from water are shown in green). Untitled picture.png Another way of looking at this would be to arrange water "across" the bond so that it's more clear where the OH goes and where the H goes: Untitled picture.png So in this case we have two bonds that need to be broken: the H-O bond of water and the C-X bond. The identity of X will have a large impact on the activation energy and therefore the likelihood of a reaction being successful. For example, let's consider the hydrolysis of an ether: Untitled picture.png 3 In this example, we need to break an O-H bond and a C-O bond to reach the transition state (these bonds are highlighted in the example below). This transition state is very unstable because there are unpaired electrons present on four different atoms. Untitled picture.png 3
Untitled picture.png 3 As we did above, we can calculate the amount of energy needed to reach the transition state (Ea) by adding up the bond enthalpies: Ea = 465 + 362 = 827 kJ mol-1 This is a tremendous amount of energy; to put it into context, it would take a photon with l = 145 nm (far into the UV) to accomplish this for one reaction. Alternatively, it would take a golf ball travelling at 13481 mph to provide enough energy to break these bonds. Punchline, it is unlikely that this reaction will happen very frequently. The same can be said for nearly all hydrolysis reactions that look like this; it takes special reaction conditions for hydrolysis of C-X bonds - this will be studied in detail in Organic Chemistry. For our purposes, the large activation energy that we discussed above will prevent these types of reactions from occurring. Hydrolysis of Carbonyls If there was a way to decrease the activation energy for a hydrolysis reaction, it would become much more practical. One way for this to happen is when a carbon double bonded to an oxygen (so a carbonyl) is part of the bond being broken. In this case, a different pathway can be followed; instead of breaking the C-O and O-H bonds directly. The figure directly below shows the two reactants (water and an ester) with solid arrows connecting the two compounds - each solid arrow indicates movement of a pair of electrons. The NEW C-O bond can form first (1st in figure below). This happens when a lone pair of electrons on the oxygen of water become a covalent bond between the C and O - this is called an "attack". Since carbon cannot break the octet rule, a pair of electrons in the double bond "flow" up to the oxygen (2nd in figure below). Untitled picture.png 工 0 工 0 ○ 工 The carbonyl carbon is on the positive end of multiple polar bonds. This gives it a strong positive character - we call this an electrophile = an atom that is susceptible to attack because of its high degree of positive character. The oxygen on water is very electronegative and has a high degree of negative character. We call this a nucleophile. Nucleophiles must be at least partially negative. The electrophile carbon is part of a carbonyl. This means that we can "push" electrons up to the oxygen. This is a low energy step because oxygen is highly electronegative and doesn't mind being an anion. The incoming oxygen ALWAYS ends up with the carbonyl. This step is possible for three main reasons: The resulting transition state is high energy because we have created a cation on a oxygen, the second most electronegative atom. However, it is much more stable than generating four unpaired electrons as we saw in the hydrolysis reaction above. Consequently, this reaction pathway has a significantly lower activation energy.
The resulting transition state is high energy because we have created a cation on a oxygen, the second most electronegative atom. However, it is much more stable than generating four unpaired electrons as we saw in the hydrolysis reaction above. Consequently, this reaction pathway has a significantly lower activation energy. By moving the proton from one oxygen to another, we can draw a second transition state that is equally stable; these two forms of the molecule exist in equilibrium. This is notable because the final step of the reaction pathway begins from the 2nd equilibrium form. Untitled picture.png OH In the final step of this reaction, electrons from the oxygen anion flow back to create a double bond between C and O and carbons octet is preserved by breaking the C-O bond. This is an energetically favorable step because it minimizes formal charge and gets rid of the unstable oxygen cation. Untitled picture.png ○ 工 You can recognize a C-X bond that is susceptible to hydrolysis by identifying an electrophile - this is not always carbon. If the electrophile is adjacent to an oxygen or nitrogen, the bond can be hydrolyzed. If it is adjacent to a carbon, that bond will not be hydrolyzed. Perhaps you can propose a reason for this. When the nucleophile attacks (the oxygen from water in the example above), a new bond is formed and the electrons in the double bond flow to the oxygen to form an oxyanion. When the oxyanion collapses, the desired bond is broken. So let's summarize some important aspects of this: Here is one more example - hydrolysis of an amide. First, let's predict what the product will be. We identify the hydrolysable bond finding a carbonyl adjacent to an oxygen or nitrogen. Untitled picture.png Now we add water across this bond. When this happens, the OH ends up with the carbonyl and the H ends up with the nitrogen. Untitled picture.png ○ 工 Ink Drawings  
Untitled picture.png ○ 工 Finally, let's make sure that this can actually happen by showing the electron flow steps: Step 1 Untitled picture.png Step 2 Untitled picture.png Step 3 Untitled picture.png OH Condensation Reactions One of the reactants is a carboxylic acid. One of the reactants is an amine or alcohol. Condensation is the exact opposite of hydrolysis: when we create a C-X bond by releasing water. Let's not go into as much detail here and instead focus on recognizing when a condensation reaction can happen and what the products will be. To recognize IF condensation is possible, think critically about the reactants. If both of these criteria are met, then a condensation reaction that generates water is possible: So let's consider the reaction below between an amine and a carboxylic acid. As we have seen in the past, the nucleophile (electronegative atom) is aligned with the electrophile (electropositive atom). You can see a "box" of bonds - two that need to be broken (shown highlighted in yellow) and two that need to be created (shown as red arrows). When both of these steps are completed, the two new compounds are water and an amide. Ink Drawings    
So let's consider the reaction below between an amine and a carboxylic acid. As we have seen in the past, the nucleophile (electronegative atom) is aligned with the electrophile (electropositive atom). You can see a "box" of bonds - two that need to be broken (shown highlighted in yellow) and two that need to be created (shown as red arrows). When both of these steps are completed, the two new compounds are water and an amide. Untitled picture.png We could think about this reaction using exactly the same three-step reaction process: Nucleophile attacks the electrophile and electrons flow from the double bond to the oxygen. An equilibrium is established between a protonated nitrogen and a protonated oxygen. The oxyanion collapses and the double bond is reformed. The C-X bond that has a positive charge is broken. Ink Drawings Ink Drawings Ink Drawings Ink Drawings

 

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