Deuterium (D) is the isotope of hydrogen of mass number 2, wit...

Question

Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review PROBLEM 4-57)
a. Propose a mechanism to explain each product in the following reaction.
Chemical reaction diagram showing bromopropane reacting with KOH in alcohol to yield an elimination and a substitution product.

a. Propose a mechanism to explain each product in the following reaction.

Chemical reaction diagram showing bromopropane reacting with KOH in alcohol to yield an elimination and a substitution product.

Accepted Answer

All right. Hello everyone. So this question is asking us to give the mechanism for the following reaction to rationalize each of the products formed. Now, our starting material is three chloral pentane reacting with sodium hydroxide in ethanol to create EP two E and pentane three. All right. So let's go ahead and get started. Now, in this case, we're going to have both an elimination and substitution reaction. But the question is, is elimination going to be E one or E two and is substitution going to be acid one or SN two? First, it's worth mentioning that in our starting material, our leaving group is chlorine and that leaving group is attached to a secondary carpet. Now recall that a secondary carbon can favor both a uni molecular or a bi molecular reaction, whether that's E one or E two and SN one or SN two. So we're going to need a bit more information to see what mechanisms we're going to keep in mind here because of this, we have to consider our other reagents in solution. We have sodium hydroxide, end ethanol as a solvent. Now, ethanol is simply a solvent but hydroxide in particular, does have a negative charge with sodium serving as a counter ion. Now, because hydroxide does have a negative charge, it's considered a strong nucleo file or base, which means that these two reactions are going to proceed via E two and SN two because the bi molecular reactions require a strong nucleo file or base. So first, let's start by forming our elimination product and I am going to go ahead and redraw our starting material. So here recall that the elimination mechanism requires the removal of a beta hydrogen to create a double bond between the beta carbon and the adjacent carbon that contained the leaving group initially. Now, the E two mechanism is different from E one because the E two mechanism is concerted in one single step. On the other hand, the E one mechanism is characterized by the formation of a carbo carrion. And in this conversation about elimination, we have to consider the predominance of Zits or Hoffman's product hydroxide is a relatively small base. So because of this, the Zev product is going to be favored, which means that the more substituted keen is favored if more than one is possible. So as was discussed previously, the carbon atom that contains our leaving group, chlorine is the alpha position, which means that there are two Betta carbons relative to this alpha position. However, the given starting material, the given oy Halide is symmetrical and a chiral, which means that the beta hydrogens are actually going to be equivalent and will produce the same product. So this means that I can go ahead and eliminate a hydrogen from either beta carbon. But for the sake of simplicity, I'm going to eliminate a hydrogen from the beta carbon. On the left side here, recall that the beta hydrogen and the leaving group require anti coplanar stereo chemistry, which is why hydrogen and chlorine are on the same plane but on opposite sides of each other, hence anti coplanar. So now we can proceed with our mechanism. So here's hydroxide, our base and hydroxide is going to start by removing that beta hydrogen, which is going to break the bond between the beta carbon and the beta hydrogen. Those electrons are going to go in between the alpha and beta carbons which simultaneously displaces the leaving group and creates our final all keen product. So here's our old keen which is in the E confirmation right now or the E configuration and that's more stable. And in terms of by-product, we end up with water and a free floating chloride ion. So now let's talk about the substitution product. And in a nucleic substitution reaction, the leaving group is simply replaced with the Nile on the carbon, it was initially attached to. However, in an sn two reaction, this is also a one step mechanism. The essen one reaction is characterized by the formation of a carbo cion intermediate. So heres our starting material, but this time hydroxide is directly going to attack carbon three of the parent chain where the leaving group is attached. So hydroxide attacks and that is simultaneously going to displace our leaving group. This means that we end up with our same five carbon chain. But instead of having a pie bond, chlorine is simply going to be replaced with a hydroxy group. And we have a free floating chloride ion as a by-product. Now keep in mind that the SN two mechanism normally results in the inversion of stereo chemistry. But because the carbon atom that contained our leaving group was initially a Cairo, the SN two reaction does not produce stereo isomers. So we can ignore stereo chemistry for this discussion. But there you have it here, we have the formation of both the elimination product via E two and the substitution product via SN two. So with that being said, thank you so very much for watching. And I hope you found this helpful.

Key Concepts

Kinetic Isotope Effect

Elimination vs. Substitution Reactions

Reaction Mechanisms

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