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8. If a resonance structure must have a negative formal charge, that charge is most stable on an electronegative atom. Draw all relevant resonance structures of the molecule below with curved arrows only using pattern 1. Circle the most stable structure(s). Explain the choice of most stable structure with 1-3 complete sentences.

Draw a more stable resonance structure for the following molecule. Use a curved arrow to show how to transform the original structure to the new one and please specify charges.

For each example, specify whether the two structures are resonance contributors to the same resonance hybrid. CH2 HC a) HC=CH2 HC-CH2 HC-CH2 ö: ö: b) H3C- H3C- NH2 NH2 c) H3C-NH-CH2 H3C-NH=CH2

Answer all parts (A, B, & C)

Curved arrows are used to illustrate the flow of electrons. Using the provided resonance structures, draw the curved electron- pushing arrows to show the interconversion between resonance hybrid contributors. Be sure to account for all bond- breaking and bond-making steps. 0:0 11 Select to Add Arrows ISO H 'Н ■ Select to Add Arrows U H H

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structure. For any correct answers, select the "No problems" box. Items (5 images) (Drag and drop into the appropriate area below) For each answer above that was identified as incorrect, select a reason why the curved arrow(s) did not generate a valid resonance ONH2 NH2 H,C CH2 H.C HC CH2 Incorroct Aspect

Hi, I found the answers for the first three subparts (A, B, C) but couldn't find the answers for the last three. Any help would be appreciated.

Can you show me im trying to determine which portions are sp2 and sp3 so I can determine where ressonance occurs. Like label them so I can understand kind of the way I did, and then the Oxygen part I dont understand either. For instance I labeled a few but I want each C point and then the oxygen bc after doing so im going to use it to determine delocalized lone pairs and localized lone pairs.

Please answer fast I give you upvote.

Draw curved arrows to illustrate the bond-making and bond-breaking processes. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be created. D M 10 + [1] H₂C י H EXP. CONT. L 1 CH₂ HBr: d H₂C CH3 H C N O S CI Br I P F

Click on those molecules below which have a dipole moment. If there are none, submit your answer without clicking. HC CH3 || CH₂ Н. H3C CH3 'H ه CH₂ CI

Answer A please explain answer

PQ-2. What is the equivalent structure for this molecule using a bond-line formula? CH2CH3 CH3 (CH3)2CH(CH2)3CHBRCHCH3 (A) (B) Br Br (C) (D) Br Br

Draw the Lewis structure that corresponds to each of the following Newman projections. H H CH3 PH CH3 (a) CH3 H CI CI H (b) CI F F F CF3 (c) F

What is the correct answer?

Q8: Draw the resonance structures for the following molecule. Show the curved arrows (how you derive each resonance structure). Circle the major resonance contributor. од

My answer is in pencil and was marked wrong, can someone explain why it’s wrong and what the correct version would be ?

For 1 and 2, use curved arrows to illustrate the potential overall electron movements or bond changes, and identify the type of reaction by examining the overall chemical transformation. Show A-H bonds as needed.

3. Use curved arrow notation to derive a second reasonable resonance structure for each structure shown below. Show your curved arrows and all formal charges. (a) H I H-CH 11 H I H (b) :O: || H₂C CH3 H (c) use your answer from #2b for formal charges

Would you help me, please?

how do you find the dipole moment?

Draw the conjugate base of C5F5H. Include lone pairs of electrons on all atoms. Add non-zero formal charges where appropriate.

Make a continuous model for C4H10 by using 4 black 4 hole carbon atoms, 10 white one hole hydrogen atoms, and 13 pink bonds. Then write in wedge dash notation C4H10 and add in missing hydrogen atoms.

The bond angles around all of the carbons in 2-methylbutane are the same because all ot the carbons in this molecule have the same "ABE" classification. Can you help me with part A and part B?