A finding that opened a route to B-blockers was the discovery that B-blocking activ- ity is retained if an oxygen atom is interposed between the aromatic ring and the side chain. To see this difference, compare the structures of labetalol propranolol. Thus, alkylation of phenoxide ions can be used as a way to introduce this side chain. The first of this new class of drugs was propranolol. and OH ОН 1-Napthol (B-Naphthol) Propranolol

A finding that opened a route to B-blockers was the discovery that B-blocking activ- ity is retained if an oxygen atom is interposed between the aromatic ring and the side chain. To see this difference, compare the structures of labetalol propranolol. Thus, alkylation of phenoxide ions can be used as a way to introduce this side chain. The first of this new class of drugs was propranolol. and OH ОН 1-Napthol (B-Naphthol) Propranolol

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter21: Benzene And The Concept Of Aromaticity

Section: Chapter Questions

Problem 21.58P

See similar textbooks

Similar questions

Using resonance structures, explain why a nitroso group (– NO) is an ortho, para director that deactivates a benzene ring toward electrophilic attack.
a) Put these three common types of carbonyl compound in order of decreasing reactivity ester amide acid chloride b) For the least reactive, show the interconversion to its other resonance form: How does this electron delocalisation make it stable? c) For the most reactive, draw the mechanism of its undergoing hydrolysis (reaction with H2O): Why makes this type of carbonyl so reactive to nucleophiles?
The question is: "Draw the curved arrow mechanism for the reaction between pentan-2-one and (CH3)3O– in t-butanol to form an enolate. Draw all electrons and charges on both resonance structures. Then answer the question about the reaction." I got the initial arrows correct, but am not entirely sure what the carbanion intermediate would look like and then what the curved arrows would be to convert it to its final oxanion form
Given the information pictured, The final step of the reaction sequence uses PhLi. Provide a reason and possible mechanism by which this final elimination and hydrolysis reaction would occur.
Like acid chlorides, acid anhydrides react with alcohols in the presence of pyridine to form esters. For the following reaction : (a) Draw a circle around the nucleophile ; (b) Draw a square around the electrophile; (c) Put a star next to the base; (d) Draw the structure of the tetrahedral intermediate in the box below. Then draw a reasonable 3-step arrow pushing mechanism for the reaction.
In paragraph/essay form:Provide an explanation (in detail) on how base catalysts speed up nucleophilic reactions of carbonyl compounds.
Chemistry Please help explain this textbook question: Although N, N -dimethylaniline is extremely reactive toward electrophilic aromatic substitution and is readily substituted by weak electrophiles, such as diazonium and nitronium ions, this reactivity is greatly diminished by the introduction of an alkyl substituent in an ortho position.
Carbodiimides are excellent coupling agents to use to allow an amide or ester coupling to occur under milder conditions. DCC (dicyclohexylcarbodiimide) is extremely common, but the byproduct is insoluble and can contaminate products, whereas other coupling agents have different byproducts with different products. DIC (diisopropylcarbodiimide) and its byproduct are soluble and are easily removed from a solid-phase synthesis. Draw a full mechanism for the amide coupling shown below using DIC as a coupling agent. OH CH3NH₂ -N=C=N- base HINT: The first step of the mechanism uses the organic base to deprotonate something... lohexane as When doing solid-phase peptide synthesis, it is vitally important to get the amino acids in the correct order. However, all amino acids have an amine handle and a carboxylic acid handle, so it's easy to couple an amino acid with itself or in the wrong sequence. How can a chemist ensure that they make only the sequence Gly-Ala-Arg, without ending up with the…
BS21008 Organic Chemistry for the Life Sciences Assignment Week 7 Bearing in mind Huckel's rule, determine whether each of the heterocycles involved in the following acid-base equilibria is aromatic or not (show your working). Use your findings to explain which of the nitrogen atoms in the indazole ring is more basic, i.e. which would be the product of indazole protonation, compound A or compound B? + H* + H* N: - H* - H* H CH3 CH3 CH3 A 1-methyl-1H-indazole
Complete the reaction wheels below. For the starred reactions, draw the full reaction mechanism. Once complete, compare and contrast the common reactions shared by acyl halides and acid anhydrides.
As shown below, electrophilic aromatic substitution on N,N-dimethylaniline is faster than on the unsubstituted aniline.In other words, the aromatic ring in N,N-dimethylaniline is more activated than the ring is in aniline itself. If the ring is methylated at the 2 and 6 positions, however, then the N,N-dimethyl-substituted compound reacts more slowly in electrophilic aromatic substitution than the unsubstituted compound. Explain both of these results. Hint: It does not have to do with the number of H atoms on the ring.
Amidation Reactions Part A Learning Goal: To identify characteristics common to all amidation reactions, and to apply these generalities in drawing specific amidation reactants or products. Part B Amidation reactions are condensation reactions. In a condensation reaction, two molecules join with a new covalent bond, and this results from the removal of a small molecule such as a water molecule. The reactants in amidation are a carboxylic acid and an amine, and the products are an amide and a water molecule. The carboxylic acid reactant can be of any type. The amine can be simple ammonia (NH₂). a primary amine. or a secondary amine. It cannot be a tertiary amine because tertiary amines lack the N-H Aspartame, an amide, is the sweetener commonly used in diet cola. COOH CH2 < 33 of 38 Review | Constants | Periodic Table CH2 O H₂N CH C NH-CH-COOCH3 7 H 2D EXP. CONT. aspartame Draw the carboxylic acid that is a reactant in the production of aspartame. Draw the molecule on the canvas by…