Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 14, Problem 14.45AP
Interpretation Introduction
Interpretation:
The structure of compound A is to be proposed with the help of given information.
Concept introduction:
Ozonolysis is the oxidative cleavage of the double bond, where
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An unknown hydrocarbon A with the formula C6H10 reacts with 1 molar equivalent of H2 over a palladium catalyst to give B C6H12 (Rxn 1). Hydrocarbon A also reacts with OsO4 to give the glycol C (Rxn 2). A gives 5-oxohexanal on ozonolysis (Rxn 3).
Draw the structures of A, B, and C. Give the reactions.
An optically active monoterpene (compound A) with molecular formula C10H18O undergoes catalytic hydrogenation to form an optically inactive compound with molecular formula C10H20O (compound B). When compound B is heated with acid, followed by reaction with O3 and then with dimethyl sulfide, one of the products obtained is 4-methylcyclohexanone. Give possible structures for compounds A and B.
Hydrocarbon A has the formula C9H12 and absorbs 3 equivalents of H2 to yield B,
C9H18, when hydrogenated over a Pd/C catalyst. On treatment of A with aqueous
H2SO4 in the presence of mercury(II), two isomeric ketones, C and D, are produced .
Oxidation of A with KMnO4 gives a mixture of acetic acid (CH3COOH) and the
tricarboxylic acid E. Propose structures for compounds A-D, and write the reactions.
Chapter 14 Solutions
Organic Chemistry
Ch. 14 - Prob. 14.1PCh. 14 - Prob. 14.2PCh. 14 - Prob. 14.3PCh. 14 - Prob. 14.4PCh. 14 - Prob. 14.5PCh. 14 - Prob. 14.6PCh. 14 - Prob. 14.7PCh. 14 - Prob. 14.8PCh. 14 - Prob. 14.9PCh. 14 - Prob. 14.10P
Ch. 14 - Prob. 14.11PCh. 14 - Prob. 14.12PCh. 14 - Prob. 14.13PCh. 14 - Prob. 14.14PCh. 14 - Prob. 14.15PCh. 14 - Prob. 14.16PCh. 14 - Prob. 14.17PCh. 14 - Prob. 14.18PCh. 14 - Prob. 14.19PCh. 14 - Prob. 14.20PCh. 14 - Prob. 14.21PCh. 14 - Prob. 14.22PCh. 14 - Prob. 14.23PCh. 14 - Prob. 14.24PCh. 14 - Prob. 14.25PCh. 14 - Prob. 14.26APCh. 14 - Prob. 14.27APCh. 14 - Prob. 14.28APCh. 14 - Prob. 14.29APCh. 14 - Prob. 14.30APCh. 14 - Prob. 14.31APCh. 14 - Prob. 14.32APCh. 14 - Prob. 14.33APCh. 14 - Prob. 14.34APCh. 14 - Prob. 14.35APCh. 14 - Prob. 14.36APCh. 14 - Prob. 14.37APCh. 14 - Prob. 14.38APCh. 14 - Prob. 14.39APCh. 14 - Prob. 14.40APCh. 14 - Prob. 14.41APCh. 14 - Prob. 14.42APCh. 14 - Prob. 14.43APCh. 14 - Prob. 14.44APCh. 14 - Prob. 14.45APCh. 14 - Prob. 14.46APCh. 14 - Prob. 14.47AP
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- Compound A (C7H11Br) is treated with magnesium in ether to give B (C7H11MgBr) which reacts violently with D2O to give 1-methylcyclohexene with a deuterium atom on the methyl group (C). Reaction of B with acetone (Ch3COCH3) followed by hydrolysis gives D (C10H18O). Heating D with concentrated H2SO4 gives E (C10H16), which reacts with 2 equivalents of Br2 to give F (C10H16Br4). E undergoes hydrogenation with excess H2 and Pt catalyst to give 2-methylpropylcyclohexane. Determine the structures of compounds A through F, and show your reasoning throughout.arrow_forwardCompound P (C2H4) which is an alkene undergoes reaction with HCl to produce compound Q (C2H5Cl). Reaction of compound Q with benzene in the presence of AlCl3 as catalyst produces compound R. Then, nitration of compound R in the presence ofnH2SO4 produces two compounds, S and T. But when compound R is reacted with a hot acidified solution of alkaline KMnO4 gives compound U. Deduce the structure of compounds P to U.arrow_forwardAn organic compound A which has a characteristic odour is treated with 50% NaOH to give B (C7H8O)and C which is a sodium salt of an organic acid . Oxidation of B gives back A. Heating C with soda lime yields an aromatic hydrocarbon D . Deduce the structures of A,B,C and Darrow_forward
- Compound A (C7H11Br) is treated with magnesium in ether to give B (C7H11MgBr), which reacts violently with D2O to give 1-methylcyclohexene with a deuterium atom on the methyl group (C). Reaction of B with acetone (CH3COCH3) followed by hydrolysis gives D (C10H18O). Heating D with concentrated H2 SO4 gives E (C10 H16), which decolorizes two equivalents of Br2 to give F (C10H16 Br4). E undergoes hydrogenation with excess H2 and a Pt catalyst to give isobutylcyclohexane. Determine the structures of compounds A through F, and show your reasoning throughout.arrow_forward4. Compound A has the formula C 8H 8. It reacts rapidly with KMnO 4 to give CO 2 and a carboxylic acid, B (C 7H 6O 2), but reacts with only 1 molar equivalent of H 2 on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4, equivalents of H 2 are taken up and hydrocarbon C (C 8H 16) is produced. What are the structures of A, B, and C.arrow_forwardA hydrocarbon (X), with the molecular formula: C8H14 is reduced in presence of sodium and liquid ammonia to give the only product (Y) with the molecular formula: C8H16. Compounds X and Y both resulting 2,5-dimethylhexane when treated with hydrogen and platinum catalyst (H2/Pt). As a result of the oxidative cleavage of compound Y (by using KMnO4 / H2SO4), a single carboxylic acid derivative with C4H8O2 molecular formula is formed. Again, as a result of the reaction of Y with perbenzoic acid, the chiral compound C8H14O is observed, but the reaction of compound Y with bromine gives the achiral C8H14Br2 as the product.arrow_forward
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