Interpretation:
The reagent used to accomplish the given transformation has to be identified.
Concept Introduction:
Elimination reactions are the one in which the groups are lost and the saturated bonds are converted to unsaturated bonds. Usually the substitution reaction compete with elimination reaction.
In elimination reaction, the beta proton is removed together with the leaving group to form a double bond.
E2 reaction proceeds through a single step without any formation of intermediates. The base abstracts a proton form the substrate and the loss of leaving group also happens resulting in the formation of double bond. E2 stands for bimolecular elimination.
The proton can be abstracted by the base in two different ways leading to a regiochemical outcome. In the given reaction below, two
The two alkenes that are formed can be given different names. If the alkene is more substituted means it is known as Zaitsev product, and if the alkene formed is less substituted means it is known as Hofmann product. Generally more substituted product is the major product (Zaitsev product).
If the reaction is performed with a sterically hindered base, the major product will be less substituted alkene (Hofmann product). The regiochemical outcome can be decided by choosing the base. Some of the sterically hindered bases are,
Addition reactions are the one in which the groups are added to an unsaturated double bond. The double bond is destroyed because the groups are added across the double bond. If the considered alkene is a symmetrical one, then there is no difference in the product obtained but if the alkene is an unsymmetrical one, then regiochemistry decides the products that are obtained.
When two similar groups are added across a double bond, regiochemistry is irrelevant. When adding two different groups across the double bond of an unsymmetrical alkene, the regiochemistry becomes relevant. In simple words, it can be said that, when two different groups are added across a double bond, then regiochemistry becomes relevant.
Markovnikov Addition
In the vinylic bond, if the bulky group gets substituted in the carbon atom that is more substituted means, it is known as Markovnikov’s addition.
anti-Markovnikov Addition
In the vinylic bond, if the bulky group gets substituted in the carbon atom that is less substituted means, it is known as anti-Markovnikov’s addition.
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Chapter 11 Solutions
Organic Chemistry As a Second Language: First Semester Topics
- An organic chemistry Teaching Assistant (TA) suggested in your last discussion section that there is only one major organic product of the following reaction and that this reaction builds a ring. If the TA is right, draw the product in the drawing area below. If the TA is wrong, just check the box below the drawing area. NaOH ?arrow_forwardA student suggests that the molecule on the right can be made from a single molecule that doesn't have a ring. If the student is correct, draw the starting material below, otherwise, check the box under the drawing area. Click and drag to start drawing a structure. : ☐ + NaOH टेarrow_forwardRate = k [I]1.7303[S2O82-]0.8502, Based on your rate, write down a mechanism consistent with your results and indicate which step is the rate determining step.arrow_forward
- 36. Give the major product(s) of each of the following reactions. Aqueous work-up steps (when necessary) have been omitted. a. CH3CH=CHCH3 b. CH3CH2CH2CCH3 H,PO₂, H₂O, A (Hint: See Section 2-2.) 1. LIAIH. (CH,CH,),O 2. H', H₂O H NaBH, CH,CH₂OH d. Br LIAIH. (CH,CH,)₂O f. CH3 NaBH, CH,CH,OH (CH3)2CH H NaBH, CH,CH₂OH Harrow_forwardPredict the major products of this reaction: + H excess NaOH Δ ? Note that the second reactant is used in excess, that is, there is much more of the second reactant than the first. If there won't be any products, just check the box under the drawing area instead.arrow_forwardAn organic chemistry Teaching Assistant (TA) suggested in your last discussion section that there is only one major organic product of the following reaction and that this reaction builds a ring. If the TA is right, draw the product in the drawing area below. If the TA is wrong, just check the box below the drawing area. 1. NaOMe CH3O N. OCH3 ? 2. H3O+arrow_forward
- Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. + More... ☐ ☐ : ☐ + G 1. NaOMe Click and drag to start drawing a structure. 2. H +arrow_forward6. Ammonia reacts with nitrogen monoxide and oxygen to form nitrogen and water vapor. If the rate of consumption of NO is 4.5 mollitermin) (a) Find the rate of reaction (b) Find the rate of formations of N; and HO (c) Find the rate of consumption of NH, and O 4NH: 4NO 0:4: +60arrow_forward34. Give the expected major product of each of the following reactions. Conc. HI a. CH3CH2CH2OH b. (CH3)2CHCH2CH2OH Conc. HBr H Conc. HI C. OH Conc.HCI d. (CH3CH2)3COHarrow_forward
- 42. Which of the following halogenated compounds can be used successfully to prepare a Grignard reagent for alcohol synthesis by subsequent reaction with an aldehyde or ketone? Which ones cannot and why? H3C CH3 a. Br H OH b. Cl C. I H H d. Cl e. H OCH3 Br Harrow_forwardFor each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. ? Will the first MgBr product that forms in this reaction create a new CC bond? olo ? OH جمله O Yes Ⓒ No MgCl ? Will the first product that forms in this reaction create a new CC bond? Click and drag to start drawing a structure. Yes No X ☐ : ☐ टे PHarrow_forwardAssign all the protonsarrow_forward
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