17 - Why cannot phenylacetaldehyde be prepared by LiAlH4...

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Unformatted text preview: Why cannot phenylacetaldehyde be prepared by LiAlH4 reduction of methyl phenylacetate? Does the C=O group of phenylacetaldehyde react slower with LiAlH4? A. yes B. no Why? phenylacetaldehyde methyl phenylacetate Which C=O group is more stabilized by resonance? O H A B Which C=O group is more stabilized by resonance? O H A G E B R What is the structure of the first intermediate? fast O H OCH3 This intermediate is very unstable, why? slow H H Al H H How could aldehydes be prepared by LiAlH4 reduction of esters? O H fast slow How could aldehydes be prepared by LiAlH4 reduction of esters? O H slow fast Why does diisobutylaluminum hydride reduce esters to aldehydes but lithium aluminum does not? Which reducing agent is the stronger hydride (H:) donor? Which reducing agent is the stronger acid? O H 1. H-Al(i-Bu)2 2. H2O Al H H H Al H H A B Why does diisobutylaluminum hydride reduce esters to aldehydes but lithium aluminum does not? Which carbonyl group is the stronger base? O H A 1. H-Al(i-Bu)2 2. H2O Al H H H Al H H B Why does diisobutylaluminum hydride reduce esters to aldehydes but lithium aluminum does not? Why? Which carbonyl group is the stronger base? A 1. H-Al(i-Bu)2 2. H2O slow fast fast fast B fast more favorable 11 12 Cory Coe Roohi Maini Ravi Patel Linshuai Qin Tanya Turkewitz Xiao Jing Liu Martin Weiss Order of the Curved Arrow If you can use the curved arrow formalism to show reasonable bond making and bond breaking that could occur in the formation of the product from the reactants then you will become a charter member of the "Order of the Curved Arrow". Membership in the "Order of the Curved Arrow" is high achievement in Organic Chemistry. All inductees will be awarded a mug as in recognition of their achievement. What is the major product of the following reaction? fast 1. MgBr O OH slow 2. H3O A OH B OH O OCH3 C D What is the mechanism of this reaction? What is the structure of the first intermediate? O N 1. MgBr NH 2. H3O A NH C B NH N C D What is the mechanism of this reaction? What is the structure of the first intermediate? O -2 1. MgBr N 2. H3O H3O MgBr very, very slow C N slow C N How does this reaction occur? O 1. MgBr O 2. H3O H3O H H C N C N Can imines be prepared from C=O groups? What is the mechanism? O Propose a good syntheses of the following ketone from benzene and compounds containing four carbon atoms or less? O Which of the following reaction sequences would prepare the following ketone? MgBr 1. O 2. H3O MgBr 1. N C 1 H O 2 2. H3O 3. PCC O 1. BrMg C O CH3 2. H3O 3 F = 1 + 2 +3 A = 1 B = 2 C = 3 D = 1 + 2 E = 2 + 3 Propose a good syntheses of the following ketone from benzene and compounds containing four carbon atoms or less? MgBr 1. O 2. H3O MgBr 1. N C 1 H O 2 1. O 2. H3O 3. PCC 2. H3O 3. PCC Br2 FeBr3 Mg Br BrMg A = 1 B = 2 C = 3 D = 1 + 3 E = 2 + 3 F = 1 + 2 How could benzyl cyanide be prepared from benzene and carbon compounds containing four carbon atoms or less? MgBr 1. O 2. H3O N C benzyl cyanide Can benzyl cyanide be prepared from benzaldehyde and cyanide ion? A. yes B. no MgBr 1. O 2. H3O N C Can benzyl cyanide be prepared from benzyl bromide and cyanide ion? A. yes B. no MgBr 1. O 2. H3O N C How can benzyl bromide be prepared from benzene and carbon compounds containing four carbon atoms or less? MgBr 1. O 2. H3O N C ...
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This note was uploaded on 03/27/2012 for the course CHE 322 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.

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