The absolute o value for -OCH3 group at the para position is greater than at the meta positionfor the hydrolysis of ethyl benzoate. This implies that the electronic effect of -OCH3 is governedbyA. resonance only at the para positionB. resonance and inductive at the para positionC. inductive only at the meta positionD. resonance and inductive at the meta positionЕ. В and C

Answered: Image /qna-images/answer/de3b7e91-fd6e-4c00-b335-eb86fe426571.jpg

Answered: The absolute o value for -OCH3 group at…

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter19: Eas: Electrophilic Aromatic Substitution

Section: Chapter Questions

Problem 21E

See similar textbooks

Similar questions

n ane arther. Charge repulsion: the ethylenediamine and its salts The preferred configuration for ethylenediamine are shown below. As you can see, the preferred conformation in each case depends on the protonation state of the amine groups. NH3 NH2 NH2 H. NH3 H. H. ZHN' H. H. H. H H NH3 Build the molecules and explain how the interplay between charge repulsion and hydrogen bonding explain this behavior. 1
QB. In the following compounds identify, with an arrow, the position that is most reactive towards electrophilic aromatic substitution. LOCH₁ ∞∞ da of OH O₂N- OCH,
5. Draw all resonance structures for the following molecules and indicate the minor and major contributors.
Draw the mechanism showing the Brønsted acid/base reaction between propanoic acid and methylamine. Label the acid, base, conjugate base, and conjugate acid. Show all curved arrows, lone pairs and nonzero formal charges. HO. CH;NH2 methylamine propanoic acid pKa = 4.9 Mechanism: uerm lo noits adi ol maie Using the pKa information provided/pKa tables (table appended to the end of this document), determine whether the starting reagents (propanoic acid and methyl amine) or products would be favored at equilibrium (no math needed!).
14. Phenols are stronger acids than aliphatic alcohols because of resonance stabilization by the 2??.
There are no important resonance structures for the isopropyl cation CH3C+HCH3 because there are no in the structure. There are no important resonance structures for dimethyl ether CH3OCH3 because, although there are pushable electrons on the oxygen atom, there is no . There are no important resonance structures for the 5-pentenyl cation H2C=CHCH2CH2C+H2 because the pushable electrons and the receptor are separated by two methylene groups. Thus, the electrons have no way to get to the receptor.
Draw the sigma complex for the formation of the bromonium electrophile with proper resonance forms for the meta bromination of toluene and for the para bromination of toluene. Which resonance structure places the (+) next to the methyl group? Will either one (i.e. para and meta sigma complex) yield a fourth resonance structure? Given that alkyl groups such as CH3 are electron rich (i.e. no attached electronegative atoms), which sigma complex in part (A) would be more stable?
Draw the mechanism showing the Brønsted acid/base reaction between propanoic acid and methylamine. Label the acid, base, conjugate base, and conjugate acid. Show all curved arrows, lone pairs and nonzero formal charges. einedoelM HO, CH;NH, propanoic acid pKa = 4.9 methylamine Mechanism: Using the pKa information provided/pKa tables (table appended to the end of this document), determine whether the starting reagents (propanoic acid and methyl amine) or resulting products would be favored at equilibrium (no math needed!).
NO2 would stabilize an adjacent (-) charge through resonance, making it a –R substituent What is the more acidic proton in nitrocyclohexane? explain with proper resonance structures of the conjugate base.
1.Draw the structures of all 3 resonance hybrid forms of the benzenonium intermediatefor both o and p attack for the case of activating groups an indicate which one in eachcase is the most stable form.Draw the structure of all 3 resonance hybrid forms of the benzenonium intermediatefor m attack in the case involving a deactivating group. 2.Halogens are deactivating groups but direct electrophilic attack at the o and ppositions. Why is this the case?
1. , Rank the following compounds in order from most electrophilic (1) to least electrophilic. Briefly justify your answers. HO H. OEt CH3 CH3 CH3 NCH3 N. CH3 HO, H3C CH3 H. CHF CH3 CH3 H3C
1) For each pair of compounds below, predict the hybridization of the lone pair, and specify the percentage of s and p character. Then rank the compounds in order of decreasing basicity. A :OH A 2) For each compound below, identify the atom most likely to be attacked first by a nucleophile and identify the unoccupied frontier orbital (e.g. p, *, or o*) associated with that atom. B B OEt NH₂ 3) For each compound below, identify the most reactive pair of electrons (either a bond or a lone pair) and specify the corresponding type of frontier orbital (namely n, , or σ). C B N-NH2 4) Draw a plausible arrow-pushing mechanism for each transformation below: THF 25 °C R-N=C: NH₂ D cat. H* OEt

SEE MORE QUESTIONS

Recommended textbooks for you

Organic Chemistry: A Guided Inquiry

Chemistry

ISBN:

9780618974122

Author:

Andrei Straumanis

Publisher:

Cengage Learning

Pushing Electrons

Chemistry

ISBN:

9781133951889

Author:

Weeks, Daniel P.

Publisher:

Cengage Learning

Organic Chemistry: A Guided Inquiry

Chemistry

ISBN:

9780618974122

Author:

Andrei Straumanis

Publisher:

Cengage Learning

Pushing Electrons

Chemistry

ISBN:

9781133951889

Author:

Weeks, Daniel P.

Publisher:

Cengage Learning

SEE MORE TEXTBOOKS