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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
- In the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+. Assuming there is 2.00 L of solution, answer the following: The Ka of the left & right solution is? The pH of the left & right solution is? The acid on the left & right is what kind of acid?arrow_forwardNonearrow_forwardNonearrow_forward
- Nonearrow_forwardWhat spectral features allow you to differentiate the product from the starting material? Use four separate paragraphs for each set of comparisons. You should have one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of functional group changes.arrow_forwardQuestion 6 What is the major product of the following Diels-Alder reaction? ? Aldy by day of A. H о B. C. D. E. OB OD Oc OE OAarrow_forward
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