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(a)
Interpretation:
Propylene does not undergo free radical
Concept introduction:
In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up ) is used to represents the movement of a single electron in a homolysis process.
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Answer to Problem 26.37P
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. The mechanism and products for these two competing steps are given below:
Explanation of Solution
The mechanism for the propagation and its corresponding product for a given reaction is shown below:
So in the propagation step, secondary radical is formed as a product.
The mechanism for the hydrogen atom abstraction and its corresponding product in a given reaction is shown below:
So in the hydrogen atom abstraction step, allylic radical is formed as a product.
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. The mechanism and products for these two competing steps are drawn.
(b)
Interpretation:
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is to be explained which step produces a more stable product.
Concept introduction:
In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up polymer and this process is called a chain-growth polymerization or addition polymerization. Finally, in terminating step, the growth of polymer is stopped by two ways: combination or disproportionation. In combination step, two growing polymer radicals combine to form the uncharged product, polymer. The termination occurs when the one radical abstracts a hydrogen atom from another radical is called disproportionation. The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or homolysis. So in homolysis generally radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow () is used to represents the movement of a single electron in a homolysis process. The weakest bond possesses the smallest
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Answer to Problem 26.37P
Due to resonance stabilization, an allylic radical is more stable than the secondary radical.
Explanation of Solution
The products of both these steps are:
The stability order of the radical is
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is explained which step produce a more stable product on the basis of the stability of radicals.
(c)
Interpretation:
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is to be explained why propylene is poor reactive in free radical polymerization.
Concept introduction:
In any free radical polymerization reaction generally, three steps are involved viz. initiation, propagation, and termination. In initiation step, free radicals are generated from either monomer or from another compound such as hydrogen peroxide or benzoyl peroxide called a radical initiator. In propagation, step radicals react with the monomer to grow the chain to build up polymer and this process is called a chain-growth polymerization or addition polymerization. In a such case for propagation step less stable radical must be formed. Finally, in terminating step, the growth of polymer is stopped by two ways: combination or disproportionation. In combination step, two growing polymer radicals combine to form the uncharged product, polymer. The termination occurs when the one radical abstracts a hydrogen atom from another radical is called disproportionation. The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or homolysis. So in homolysis generally radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow () is used to represents the movement of a single electron in a homolysis process. The weakest bond possesses the smallest bond dissociation energy. The stability order for radicals is
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Answer to Problem 26.37P
In free radical polymerization reaction less stable radical must be formed to propagate the chain. But in given case more stable, allylic carbocation is formed which affects the propagation step results in rate of polymerisation is very low.
Explanation of Solution
The products of the both steps in free radical polymerization of propylene are given below:
Due to the formation of more stable, allylic radical in hydrogen atom abstraction step it will interfere with the formation of polymer.
In free radical polymerization reaction less stable radical must be formed to propagate the chain. But in given case more stable, allylic carbocation is formed which affects the propagation step results in rate of polymerisation is very low.
Propylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction. It is explained why propylene is poor reactive in free radical polymerization.
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Chapter 26 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- 12. Choose the best diene and dienophile pair that would react the fastest. CN CN CO₂Et -CO₂Et .CO₂Et H3CO CO₂Et A B C D E Farrow_forward(6 pts - 2 pts each part) Although we focused our discussion on hydrogen light emission, all elements have distinctive emission spectra. Sodium (Na) is famous for its spectrum being dominated by two yellow emission lines at 589.0 and 589.6 nm, respectively. These lines result from electrons relaxing to the 3s subshell. a. What is the photon energy (in J) for one of these emission lines? Show your work. b. To what electronic transition in hydrogen is this photon energy closest to? Justify your answer-you shouldn't need to do numerical calculations. c. Consider the 3s subshell energy for Na - use 0 eV as the reference point for n=∞. What is the energy of the subshell that the electron relaxes from? Choose the same emission line that you did for part (a) and show your work.arrow_forwardNonearrow_forward
- (9 Pts) In one of the two Rare Earth element rows of the periodic table, identify an exception to the general ionization energy (IE) trend. For the two elements involved, answer the following questions. Be sure to cite sources for all physical data that you use. a. (2 pts) Identify the two elements and write their electronic configurations. b. (2 pts) Based on their configurations, propose a reason for the IE trend exception. c. (5 pts) Calculate effective nuclear charges for the last electron in each element and the Allred-Rochow electronegativity values for the two elements. Can any of these values explain the IE trend exception? Explain how (not) - include a description of how IE relates to electronegativity.arrow_forwardPlease explain thoroughly and provide steps to draw.arrow_forwardAs you can see in the picture, the instrument uses a Xe source. Given that the instrument is capable of measuring from 200-800nm, if Xe was not used, what other source(s) could be used? Refer to figure 7-3. How many monochrometers does this instrument have? Why? Trace the light as it goes from the Xenon lamp all the way to the circle just slightly to the right and a little bit down from S4. What do you think that circle is? In class we talked about many types of these, which kind do you think this one is for a fluorimeter? Why? Explain. What is/are some strategy(ies) that this instrument has for dealing with noise that you see present in the optics diagram? Why does a fluorescence cuvette have to be clear on four sides?arrow_forward
- Provide steps and thoroughly solve.arrow_forwardNonearrow_forwardDevise a synthesis to prepare 4-tert-butyl-2-nitrotoluene from toluene. Complete the following reaction scheme. Part 1 of 4 Step 1 Step 2 A B Draw the structure for compound B, 4-tert-butyl-2-nitrotoluene. Click and drag to start drawing a structure. 'O Х ப:arrow_forward
- What is N hybridized? sp3 or sp2? whyarrow_forwardDate Unknown o Hydrated Salt Lab Sec. Name Trial I Trial 2 1. Mass of fired crucible and lid (g) 2. Mass of fired crucible, lid, and hydrated sah (g) 3. Instructor's approval of flame and apparatus 4. Mass of crucible, lid, and anhydrous salt Ist mass measurement (g) 2nd mass measurement (g) 3rd mass measurement (g). Desk No. Trial 3 48.833 46.808 213.692 51.507 9.359 46,615 50.296 48.211 45.351 50.142 48.146 45.1911 50.103 48.132 45.186 5. Final mass of crucible, lid, and anhydrous salt (g) 50.180 4.13 45.243 Calculations 1. Mass of hydrated salt (g) 2. Mass of anhydrous salt (g) 2.674 2.491 2.9239 1.3479 1.2959 1.5519 3. Mass of water lost (g) 1.32791969 1.322g 4. Percent by mass of volatile water in hydrated salt (%) 49.6% 48% 216.9% 5. Average percent HO in bydrated salt (%H,O) 5. Standard deviation of %H,O Relative standard deviation of %H,O in hydrated salt (RSD) how calculations on next page. 48.17% Data Analysis, B Data Analysis, C Data Analysis, D Experiment 5 89arrow_forwardConsidering the irregular electronic configurations we discussed for certain transitionmetals, think about the possibility of silicon (Si) having a [Ne]3s 2 3p 2 configuration vs.[Ne]3s 1 3p 3. Discuss the pros and cons of both configurations. Which one does Si actuallyadopt and why?arrow_forward
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