Explain why metal hydride reduction gives an endo alcohol as the major product in one reaction and an exo alcohol as the major product in the other reaction. [1] LIAIH4 [2) H20 [1] LIAIH, OH OH endo OH group exo OH group
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- #B: Methyl acetate has methoxy, -OCH3 as the remaining of the alcohol part in the ester. Isopropyl acetate has isopropoxy, -OCH(CH3)2 as the remaining of the alcohol part in the ester. -OCH(CH3)2 is more electron donating than the methoxy, -OCH3 group due to the presence of two electron-donating -CH3 group in the former. Hence saponification reaction of Isopropyl acetate is much slower than methyl acetate. Hence the rate of saponification of methyl acetate, CH3CCO2CH3 is 50 times greater than that for isopropyl acetate.Draw the organic products formed when allylic alcohol A is treated with each reagent.a.H2 + Pd-C b.mCPBA c. PCC d.CrO3, H2SO4, H2O e.(CH3)3COOH, Ti[OCH(CH3)2]4, (+)-DET f. (CH3)3COOH, Ti[OCH(CH3)2]4, (−)-DET g. [1] PBr3; [2] LiAlH4; [3] H2O h.HCrO4−–Amberlyst A-26 resinDraw the structure corresponding to each name.a. 3,3-dimethylpentanoic acidb. 4-chloro-3-phenylheptanoic acidc. (R)-2-chloropropanoic acidd. m-hydroxybenzoic acide. potassium acetatef. sodium α-bromobutyrateg. 2,2-dichloropentanedioic acidh. 4-isopropyl-2-methyloctanedioic acidi. 3,3-dimethylpentanenitrile j. 4,5-diethyl-2-isopropylnonanenitrile
- Draw the structure corresponding to each name.a. 3,3-dimethylpentanoic acidb. 4-chloro-3-phenylheptanoic acidc. (R)-2-chloropropanoic acidd. m-hydroxybenzoic acide. potassium acetatef. sodium a-bromobutyrateg. 2, 2-dichloropentanedioic acidh. 4-isopropyl-2-methyloctanedioic acidConsider the reaction below to answer the following questions.( the blurred chemical is FeBr3 above the arrow) a. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds___________(faster or slower) than benzene. d. Draw the structure of product DExplain why pentane-2,4-dione forms two different alkylation products (Aor B) when the number of equivalents of base is increased from one totwo.
- Describe the preparation (E and F only)Draw the products formed when phenol(C6H5OH) is treated with each reagent. Give an explanation. d. (CH3CH2)2CHCOCl, AlCl3 j. product in (d), then NH2NH2, – OHKMnO4, warm, conc'd reacts with hept-1-ene to yield __________. CO2, hex-1-ene CO2, hexanoic acid Formic acid, pentanoic acid Ethanoic acid, pentanal Formic acid, hexanone
- Number 1 (a to f). Kindly give it's new iupac name, and common name. Number 2 (a to d). Kind draw it. Number 3 (a and b). Kindly write the formula and gibe the common name for the sulfur containing organic products of the following reactions.Draw the organic products formed when attached allylic alcohol A is treated with following reagent. (CH3)3COOH, Ti[OCH(CH3)2]4, (+)-DETDraw the correct organic product of the following oxidation reaction: