None a) Ib) IIc) IIId) IVe) V18. What is (are) the principal product(s) formed when 1 mol of ethylmagnesium bromidereacts with 1 mol of 3-(N-methylamino)cyclopentanone?CH3BrMgNH₂CCH,CH,MgBrACH2CH3+ CH₂CH,H₁COMgBrH₂CIIH₂CCH2CH3OMgBrCH2CH3H3C#וווIVdon't give Aigeneratedsolution..show workCH2CH3

Answered: Image /qna-images/answer/5641f3e2-e953-46ef-b760-a8db0500464b.jpg

Answered: a) I b) II c) III d) IV e) V 18. What…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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this was incorrect, im confused
student found a bottle of unknown origin in the basement of an "CAUTION—contains С₁H₁O.” No other A old house. The bottle was labelled information was given, but inside the bottle were some white crystals. The student took the bottle to a chemistry laboratory, where she analyzed the crystals with the help of the laboratory staff. C₁2H₁6O reacted with Br₂ in CCl4 adding one mole of Br, and forming C₁₂H₁OB2₂. When a sample of C₁₂H₂O was reacted with O3 followed by Zn/H₂O, two different samples, J and K, were obtained. Compound J had a molecular formula of C.HgO₂, and its NMR is shown below. Compound K had a molecular formula of C₂H₂O, and it could be oxidized to compound L, with a molecular formula of C₂H₂O₂. The IR spectrum of L showed a very wide, strong band centred around 3000 cm*¹. The NMR spectrum of L showed only two different absorptions: a doublet (6H) at 0.9 ppm and a septet (1H) at 3.6 ppm. There was also a singlet (1H) off the scale at 11.8 ppm. students, deduce possible…
f) 0 Baz OEt ? OEt
Just #2 please
3. Convert the Lewis structures to line-angle formulas H H a. b. с. d. H CI I I CI Н н-с Н H N: Н H С H н C I Н C -С Н Н Н C. I Н C C C н H C H Н C H Н C C C I C-C H I H :0 C Н H H H I :0 I I C-Br C C: H О Н H H
Q1. What is another name for NO
PROVISIONAL LEWIS STRUCTURES T01/S01 For each provisional structure, identify if it is the correct Lewis structure (C) or if it needs to be modified according to procedure MA or MB to obtain the correct Lewis structure. MA. Form multiple bonds to the central atom using lone pair on the terminal atom(s) until the central atom has an octet (or septet). MB. Form multiple bonds to the central atom using lone pair on the terminal atom(s) until the central atom has a formal charge of zero (0) or until two double bonds are formed. A provisional structure has the valence electrons distributed in order as (1) a bond between each pair of atoms, (2) an octet for each ligand (terminal) atom or ligand group (CH3, NH2 or OH) and (3) any remaining electrons on the central atom.
(1) Br2 CC,
Le Chateliers principle. 7 / + 3 Beg) = 2 A#= +60,000 Inc [B] Dec [D] Inc [C] €² 2 C + D (2) до Inc. I Дес Р
There are three possible Grignard reactions that lead to the given alcohol. Select the reactant needed for each of the indicated Grignard reagents. MgBr 1) ? 2) HO oi or OH OH of

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