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: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter19: Eas: Electrophilic Aromatic Substitution

Section: Chapter Questions

Problem 35CTQ

See similar textbooks

Related questions

Q: Draw the skeletal structures of: phenyl benzoate, benanilide, benzoic acid, and 2-napthaldehyde and…

A: In this question, we will draw the structure of the given compounds and then identify their carbonyl…

Q: Draw the structure of another product that could form in the reaction mixture?

A: When benzophenone reacts with 2- propanol in presence of light and alcohol it will form benzopinacol…

Q: Explain why aldehydes and ketones react with a weak acid such as hydrogen cyanide but do not react…

A: HCl and H2SO4, being strong acids, releases anions in the solution which are very weak conjugate…

Q: Draw the structural formula of the principal organic product formed when the unsaturated δ-ketoester…

A: Diels–Alder reaction:The reaction between a conjugated diene and dienophile (substituted alkene) to…

Q: 5. Draw a structure for the following names. 4-Chloro-2-methylphenol 2.6-dimethoxystyrene…

Q: 3B. Will phenyl magnesium bromide react with formic acid to form an ester? If so, write the…

A: The question is regarding reaction of substrate formic acid with an organometallic reagent phenyl…

Q: Predict the product or products from the hydroformylation of cyclohexene.

A: Cyclohexene is an alkene having cyclic 6-membered ring structure. Here we have to find out the…

Q: Draw the structure of the organic product formed when the following compounds undergo the three-step…

Q: b) Prepare ethanoic acid from ethene

A: First convert ethene (alkene) to alcohol. then alcohol is oxidized to give carboxylic acid.

Q: Acetal derivatives of aldehydes and ketones are prepared by an acid-catalyzed dehydration reaction…

A: Acetal is a derivative of aldehyde or ketone that is formed by acid-catalyzed dehydration reaction…

Concept explainers

Designing a Synthesis

Organic synthesis is a branch of chemical synthesis that involves the planned construction of organic compounds/molecules from smaller units that possess functional groups. Friedrich Wöhler was the first scientist who designed the synthesis of an organic compound, urea from inorganic starting materials silver cyanate with ammonium chloride via a simple, one-step synthesis.

Question

Show how propanolol can be synthesized from 1-naphthol, epichlorohydrin
, and isopropylamine.

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

Expert Solution

Step 1

When isopropylamine is treated with epichlorohydrin an amino compound with an epoxide ring is formed.
1-napthol is treated with the epoxide. It gives a protonated alcohol.
By the transfer of proton we get the final product.

Step by step

Solved in 2 steps with 1 images

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Know classification/mechanism of inhibition for: phenolics, halogens, oxidizers, heavy metals, surfactants (quaternary ammonium halide, soaps, detergents), alcohols, alkylating agents. Be able to give an example for each class of chemical.
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.
Please help me with these two I am very confused and I want to study.
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
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.
Draw the reaction scheme. Use one arrow direct from starting materials to products.
Why is this wrong?
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…
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.
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
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…