Chemistry: The Science in Context (Fourth Edition)
4th Edition
ISBN: 9780393124187
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 10, Problem 10.89QP
Interpretation Introduction
Interpretation: The reason for concave meniscus of water and convex meniscus of mercury in a glass tube is to be explained.
Concept introduction: The shape of meniscus depends upon the cohesive and adhesive forces.
To determine: The reason for concave meniscus of water and convex meniscus of mercury in a glass tube.
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Rate = k [I]1.7303[S2O82-]0.8502,
Based on your rate, write down a mechanism consistent with your results and indicate which step is the rate determining step.
36. Give the major product(s) of each of the following reactions. Aqueous work-up steps (when necessary) have
been omitted.
a. CH3CH=CHCH3
b.
CH3CH2CH2CCH3
H,PO₂, H₂O, A
(Hint: See Section 2-2.)
1. LIAIH. (CH,CH,),O
2. H', H₂O
H
NaBH, CH,CH₂OH
d.
Br
LIAIH. (CH,CH,)₂O
f.
CH3
NaBH, CH,CH,OH
(CH3)2CH
H
NaBH, CH,CH₂OH
H
Predict the major products of this reaction:
+ H
excess
NaOH
Δ
?
Note that the second reactant is used in excess, that is, there is much more of the second reactant than the first.
If there won't be any products, just check the box under the drawing area instead.
Chapter 10 Solutions
Chemistry: The Science in Context (Fourth Edition)
Ch. 10.1 - Prob. 1PECh. 10.2 - Prob. 2PECh. 10.3 - Prob. 3PECh. 10.4 - Prob. 4PECh. 10.4 - Prob. 5PECh. 10.5 - Prob. 6PECh. 10.6 - Prob. 7PECh. 10 - Prob. 10.1VPCh. 10 - Prob. 10.2VPCh. 10 - Prob. 10.3VP
Ch. 10 - Prob. 10.4VPCh. 10 - Prob. 10.5VPCh. 10 - Prob. 10.6VPCh. 10 - Prob. 10.8VPCh. 10 - Prob. 10.15QPCh. 10 - Prob. 10.16QPCh. 10 - Prob. 10.17QPCh. 10 - Prob. 10.18QPCh. 10 - Prob. 10.19QPCh. 10 - Prob. 10.20QPCh. 10 - Prob. 10.23QPCh. 10 - Prob. 10.24QPCh. 10 - Prob. 10.25QPCh. 10 - Prob. 10.26QPCh. 10 - Prob. 10.27QPCh. 10 - Prob. 10.28QPCh. 10 - Prob. 10.29QPCh. 10 - Prob. 10.30QPCh. 10 - Prob. 10.31QPCh. 10 - Prob. 10.32QPCh. 10 - Prob. 10.33QPCh. 10 - Prob. 10.34QPCh. 10 - Prob. 10.35QPCh. 10 - Prob. 10.36QPCh. 10 - Prob. 10.37QPCh. 10 - Prob. 10.38QPCh. 10 - Prob. 10.39QPCh. 10 - Prob. 10.40QPCh. 10 - Prob. 10.41QPCh. 10 - Prob. 10.42QPCh. 10 - Prob. 10.43QPCh. 10 - Prob. 10.44QPCh. 10 - Prob. 10.45QPCh. 10 - Prob. 10.46QPCh. 10 - Prob. 10.47QPCh. 10 - Prob. 10.48QPCh. 10 - Prob. 10.49QPCh. 10 - Prob. 10.50QPCh. 10 - Prob. 10.51QPCh. 10 - Prob. 10.52QPCh. 10 - Prob. 10.53QPCh. 10 - Prob. 10.54QPCh. 10 - Prob. 10.55QPCh. 10 - Prob. 10.56QPCh. 10 - Prob. 10.57QPCh. 10 - Prob. 10.58QPCh. 10 - Prob. 10.59QPCh. 10 - Prob. 10.60QPCh. 10 - Prob. 10.61QPCh. 10 - Prob. 10.62QPCh. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Prob. 10.66QPCh. 10 - Prob. 10.67QPCh. 10 - Prob. 10.68QPCh. 10 - Prob. 10.69QPCh. 10 - Prob. 10.70QPCh. 10 - Prob. 10.71QPCh. 10 - Prob. 10.72QPCh. 10 - Prob. 10.73QPCh. 10 - Prob. 10.74QPCh. 10 - Prob. 10.75QPCh. 10 - Prob. 10.76QPCh. 10 - Prob. 10.77QPCh. 10 - Prob. 10.78QPCh. 10 - Prob. 10.79QPCh. 10 - Prob. 10.80QPCh. 10 - Prob. 10.81QPCh. 10 - Prob. 10.82QPCh. 10 - Prob. 10.83QPCh. 10 - Prob. 10.84QPCh. 10 - Prob. 10.85QPCh. 10 - Prob. 10.86QPCh. 10 - Prob. 10.87QPCh. 10 - Prob. 10.88QPCh. 10 - Prob. 10.89QPCh. 10 - Prob. 10.90QPCh. 10 - Prob. 10.91QPCh. 10 - Prob. 10.92QPCh. 10 - Prob. 10.93QPCh. 10 - Prob. 10.94QPCh. 10 - Prob. 10.95QPCh. 10 - Prob. 10.96QPCh. 10 - Prob. 10.97QPCh. 10 - Prob. 10.98QPCh. 10 - Prob. 10.99APCh. 10 - Prob. 10.100APCh. 10 - Prob. 10.101APCh. 10 - Prob. 10.102APCh. 10 - Prob. 10.103APCh. 10 - Prob. 10.104APCh. 10 - Prob. 10.105APCh. 10 - Prob. 10.106APCh. 10 - Prob. 10.107APCh. 10 - Prob. 10.108APCh. 10 - Prob. 10.109APCh. 10 - Prob. 10.110APCh. 10 - Prob. 10.111APCh. 10 - Prob. 10.112APCh. 10 - Prob. 10.113APCh. 10 - Prob. 10.114APCh. 10 - Prob. 10.115APCh. 10 - Prob. 10.116APCh. 10 - Prob. 10.117APCh. 10 - Prob. 10.118APCh. 10 - Prob. 10.119APCh. 10 - Prob. 10.120APCh. 10 - Prob. 10.121AP
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- An organic chemistry Teaching Assistant (TA) suggested in your last discussion section that there is only one major organic product of the following reaction and that this reaction builds a ring. If the TA is right, draw the product in the drawing area below. If the TA is wrong, just check the box below the drawing area. 1. NaOMe CH3O N. OCH3 ? 2. H3O+arrow_forwardComplete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. + More... ☐ ☐ : ☐ + G 1. NaOMe Click and drag to start drawing a structure. 2. H +arrow_forward6. Ammonia reacts with nitrogen monoxide and oxygen to form nitrogen and water vapor. If the rate of consumption of NO is 4.5 mollitermin) (a) Find the rate of reaction (b) Find the rate of formations of N; and HO (c) Find the rate of consumption of NH, and O 4NH: 4NO 0:4: +60arrow_forward
- 34. Give the expected major product of each of the following reactions. Conc. HI a. CH3CH2CH2OH b. (CH3)2CHCH2CH2OH Conc. HBr H Conc. HI C. OH Conc.HCI d. (CH3CH2)3COHarrow_forward42. Which of the following halogenated compounds can be used successfully to prepare a Grignard reagent for alcohol synthesis by subsequent reaction with an aldehyde or ketone? Which ones cannot and why? H3C CH3 a. Br H OH b. Cl C. I H H d. Cl e. H OCH3 Br Harrow_forwardFor each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. ? Will the first MgBr product that forms in this reaction create a new CC bond? olo ? OH جمله O Yes Ⓒ No MgCl ? Will the first product that forms in this reaction create a new CC bond? Click and drag to start drawing a structure. Yes No X ☐ : ☐ टे PHarrow_forward
- Assign all the carbonsarrow_forward9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00arrow_forwardAssign all the carbonsarrow_forward
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