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ProblomItatamont_the thermuully fully developed region thcintcnor sunaco As"(x)Ilaw thra ugn * smooth Circular [ube with dianicler 0.4, Lhe [ube surljce naled ...

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SYNTHETIC APPLICATIONS OF FRIEDEL-CRAFTS ACYLATIONS: THE CLEMMENSEN AND WOLFF-KISHNER REDUCTIONS Rearrangements of the carbon chain do not occur in Friedel-Crafts acylations. The acylium ion, because it is stabilized by resonance, is more stable than most other carbocations. Thus, there is no driving force for a rearrangement. Because rearrangements do not occur, Friedel-Crafts acylations followed by reduction of the carbonyl group to a $\mathrm{CH}_{2}$ group often give much better routes to unbranched alkylbenzenes than do FriedelCrafts alkylations. The carbonyl group of an aryl ketone can be reduced to a CH $_{2}$ group. (FIGURE CANNOT COPY) As an example, let us consider the problem of synthesizing propylbenzenc. If we attempt this synthesis through a Friedel-Crafts alkylation, a rearrangement occurs and the major product is isopropylbenzene (see also Practice Problem 15.4 ): (FIGURE CANNOT COPY) Isopropylbenzene (major product) Propylbenzene (minor product) By contrast, the Fricdel-Crafts acylation of benzene with propanoyl chloride produces a ketone with an unrearranged carbon chain in excellent yield: (FIGURE CANNOT COPY) Propanoyl chloride Ethyl phenyl ketone $(90 \%)$ This ketone can then be reduced to propylbenzenc by scveral methods, including the Clemmensen reduction (Sect. $15.7 \mathrm{A}$ ) and the Wolff-Kishner reduction (Sect. $15.7 \mathrm{B}$ ). A The Clemmensen Reduction One general method for reducing a ketone to a methylene group-called the Clemmensen reduction- consists of refluxing the ketone with hydrochloric acid containing amalgamated zinc. [Caution: As we shall discuss later (Scction $20.4 \mathrm{B}$ ), zinc and hydrochloric acid will also reduce nitro groups to amino groups.] A Clemmensen reduction (FIGURE CANNOT COPY) Ethyl phenyl ketone Propylbenzene $(80 \%)$ In general, (FIGURE CANNOT COPY) B The Wolff-Kishner Reduction Another method for reducing a ketone to a methylene group is the Wolff-Kishner reduction, which involves heating the ketone with hydrazine and base. The Wolff-Kishner reduction complements the Clemmensen reduction in that it is conducted under basic conditions, whereas the Clemmensen reduction involves acidic conditions. The WolffKishner reduction proceeds via a hydrazone intermediate (Section $16.8 \mathrm{B}$ ) that is not isolated during the reaction. Ethyl phenyl ketone can be reduced to propylbenzene by the Wolff-Kishner reduction as follows, for example. A Wolff-Kishner reduction
(FIGURE CANNOT COPY) Hydrazone intermediate (see Section $16.8 B)$ When cyclic anhydrides are used as one component, the Fricdel-Crafts acylation providcs a means of adding a new ring to an aromatic compound. One illustration is shown here. Note that only the ketone is reduced in the Clemmensen reduction step. The carboxylic acid is unaffected. The same result can be achicved using the Wolff-Kishner reduction. (FIGURE CANNOT COPY) Benzene (excess) Succinic anhydride 3-Benzoylpropanoic acid (FIGURE CANNOT COPY) 4-Phenylbutanoic acid 4-Phenylbutanoyl chloride $\alpha$ -Tetralone Starting with benzene and the appropriate acyl chloride or acid anhydride, outline a synthesis of each of the following: a. Butylbenzene b. (FIGURE CANNOT COPY) c. (FIGURE CANNOT COPY) Diphenylmethane d. (FIGURE CANNOT COPY) 9,10 -Dihydroanthracene THE CHEMISTRY OF... DDT Aryl Halides as Insecticides Insects, especially mosquitoes, fleas, and lice, have been responsible for innumerable human deaths throughout history. The bubonic plague or "black death" of medieval times that killed nearly one-third of Europe's population was borne by fleas. Malaria and yellow fever, diseases that were responsible for the loss of millions of lives in the twentieth century alone, are mosquito-borne diseases. One compound widely known for its insecticidal properties and environmental effects is DDT $[1,1,1$ -trichloro-2, 2-bis(4-chlorophenyl)ethane]. (FIGURE CANNOT COPY) DDT $[1,1,1$ -trichloro-2, 2- bis(4-chlorophenyl)ethane] From the early 1940 s through the early 1970 s, when its use was banned in the United States, vast quantities of DDT were sprayed over many parts of the world in an effort to destroy insects. These efforts rid large areas of the world of disease-carrying insects, especially those responsible for malaria, yellow fever, sleeping sickness (caused by tsetse flies), and typhus. Though it has since re surged, by $1970,$ malaria had been largely eliminated from the developed world. According to estimates by the National Academy of Sciences, the use of DDT during that time had prevented more that 500 million deaths from malaria alone. (IMAGE CANNOT COPY) DDT Eventually it began to become clear that the prodigious use of DDT had harmful side effects. Aryl halides are usually highly stable compounds that are only slowly destroyed by natural processes. As a result they remain in the environment for years; they are what we now call "persistent insecticides" or "hard insecticides." The U.S. Environmental Protection Agency banned the use of DDT beginning in 1973. Aryl halides are also fat soluble and tend to accumulate in the fatty tissues of most animals. The food chain that runs from plankton to small fish to birds and to larger animals, including humans, tends to magnify the concentrations of aryl halides at each step. The chlorohydrocarbon DDT is prepared from inexpensive starting materials, chlorobenzene and trichloroacetaldehyde. The reaction, shown here, is catalyzed by acid. (FIGURE CANNOT COPY) DDT [1.1,1-trichloro-2,2bis(4-chlorophenyl)ethane] Estimates indicate that nearly 1 billion pounds of DDT were spread throughout the world ecosystem. One pronounced environmental effect of DDE, after conversion from DDT, has been in its action on eggshell formation in many birds. DDE inhibits the enzyme carbonic anhydrase that controls the calcium supply for shell formation. As a consequence, the shells are often very fragile and do not survive to the time of hatching. During the late 1940 s the populations of eagles, falcons, and hawks dropped dramatically. There can be little doubt that DDT was primarily responsible. DDE also accumulates in the fatty tissues of humans. Although humans appear to have a short-range tolerance to moderate DDE levels, the long-range effects are uncertain. Study Problem 1
The mechanism for the formation of DDT from chlorobenzene and trichloroacetaldehyde in sulfuric acid involves two electrophilic aromatic substitution reactions. In the first electrophilic substitution reaction, the electrophile is protonated trichloroacetaldehyde. In the second, the electrophile is a carbocation. Propose a mechanism for the formation of DDT. Study Problem 2 What kind of reaction is involved in the conversion of DDT to DDE?

This is a problem. # 29 We are given a set of data regarding regarding the per capita disposable income for each of the 50 States and District Columbia. First we will construct a frequency distribution Given that the lower bound of our class is 20,000 and our class with our 2500. We will get the following for our income. Mhm. I'll just round both of these classes to make it easier to fit. So we have 20- 25 K. 22.5. Mhm. 22 5-25, 25-27 5 27.5-30 30 to 32.5 32 5-35, 35 to 37.5 37 5 to 40 and finally 40- 42.5. Now these are not inclusive of the bounds. This is obviously this will be like 22499, I just ordered like this to make it easier to ride out for our frequencies. If we count each of these data points, we get the following three, 10, 14, 12, seven, two, two, zero and one. And when we calculate our relative frequency we get 5.9%,, 19.6%,, 27.5%,, 23.5%,, 13.7%,, 3.9%,, 3.9%,, 0% And 2.0%. So as we can see most of our data is in the top part of our data set. And when you construct your hissed a gram, it does seem to follow this so it is very slightly right skewed. If we are then to change our class sizes to be 4000 wide you will find that the data is significantly more skewed left as our relative frequencies will drop to 19.6%,, 43 0.1% 27.5%,, 5.9 And 2% for the last two classes that we would have. So as you can see, this data is significantly more right. Skewed. This set of data with these smaller classes is a much more precise way of representing this data.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.

For this question. We're looking at the nuclear pore complex specifically at the poor structure. We're being told we have F G repeats and but in vitro at a concentration at least 50 million more, please form a gel. Our first part is to ask if it's reasonable to expect them to form a gel in vivo as well. And for this we're given some data. So first thing you want to do is write down the information you've been given, we've been told, is 5000 repeats and that they are found in poor but is 35 9 m in diameter and 30 nanometers any length. So how are we going to work out the concentration of these repeats in this this volume? Well, we're going to work out the belonging we're working with first, and it says it's cylindrical. So we're going to use the formula for the volume of a cylinder, which is pi r squared pH. So the volume of the poor is going to be pi times R squared. So at 17.5 squared times 30 and this is going through this, our volume of 28,849 remember Nanometers cubes. We have to keep track of our unit, sir. So you've got Nanometers Cube now, right? So this is the volume, and we have 5000 repeats. What does that really mean? Well, we're going to work out how many malls 5000 repeats is, So we're gonna have 5000, and we're gonna just divided by aggregator is constant. So six, um, times 10 to 23. And this gives us not, um, scripts. US 5/6 times tend to be minus 20. I'll leave in this form now, so make it easier to use in future equations. But you can get a rough idea of what we're looking at here. So we have five of six times sentiments. 20 months, right? Perfect. We have our amount. We have our volume. To get the density or we have to do is divide one by the other. So we're going to work with our There were 5000 over six times five of six times 10 to the minus 20 like so. And we're going to divide this by our area. So we're going to divide this by 28,849 nanometers cubes and note that we're still using nanometers cubed. We have to deal with that. If we wanted to go from from nanometers to, say, centimeters, that would be a factor of seven sense for seven but its cubes There's gonna be a factor of 10 to 21. And furthermore, we are also, uh, we're also working with moles. We're going to end up in mill malls, but anyway, onwards. So we're also going to multiply our end results by 10 to 21 to bring us out of nanometers and into centimeters cubed. So all of this gives us 289. And if we do not 0.289 and if we convert it into minimal, most polite by thousands, we get 289 Miller malls. So it's all so this is our overall concentration. And as you can see, it is more than sufficient to form a gel. We're looking for at least 50. We're getting 287 so we have we have plenty, right? That's part a. Now we're gonna be looking at part B. Alright. For part B, we're looking at the diffusion of um, a couple of molecules. First, we're looking at X M V p. A protein, and then we're looking at a tagged M V P. But also it has important attached foot. And if we look at the diagrams and book, we see that without important, it does not pass through a membrane and nuclear membrane. But with important, it does no. Is the fusion of this important bound tagged protein fast enough to account for the efficient flow of materials and were given some more data? Let's just write down the information we have. We have the diffusion ordinates D, which is not 0.1 micrometers cubed per second. We also have the equation for diffusion, which is helpful, and we want to see how long it takes to move through a poor, uh, that is 30 nanometers. So we have X equals 30 nanometers, Correct. So the equation is going to be when we rearrange it, we're going to get t equal to X squared over two D. So we have t is equal to 30 nanometers squared, 30 squared, and we'll keep in mind. We're looking at nanometer strip, um, so divided by D, which is not quite one micrometers square per second. And so we have nanometers micrometers here. We want to We want to change this. Um, So what I'm going to do is I'm going to more supply by Tend to be minus three on this hot here. There we go. So that this will now be in nanometers squared per second. So we have nanometers squared over nanometer square per second, which is going to give us what we need. And this gives us not point. Not, not for five seconds. So will change out to, uh, 4.5 milliseconds. So again, the most important bit for part B is to make sure you change your not 0.1, uh, micrometers in 29.1 times 10 to 9 to three nanometers to keep everything working with units. But this gives us 4.4 point five milliseconds, and if he wants to make a comment on that, that seems reasonable. That's very fast. Um, and so we would expect that to match in fever because in Viva, we will be looking at a rate of about five milliseconds to import a protein through a nuclear pore


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