Var Posted April 19, 2009 Share Posted April 19, 2009 Does LiAlH4 react with formaldehyde? If so, what's the product? My book focuses on LiAlH4 reacting with other things but doesn't mention formaldehyde. Link to comment Share on other sites More sharing options...
crazybrain68 Posted April 20, 2009 Share Posted April 20, 2009 Yes. It should reduce it to methanol. Link to comment Share on other sites More sharing options...
hellomister Posted April 21, 2009 Share Posted April 21, 2009 LiAlH4 reduces aldehydes and ketones to their corresponding alcohols. Link to comment Share on other sites More sharing options...
crazybrain68 Posted April 23, 2009 Share Posted April 23, 2009 Don't forget carboxylic acids Link to comment Share on other sites More sharing options...
vedmecum Posted April 25, 2009 Share Posted April 25, 2009 Addition of a hydride anion to an aldehyde or ketone would produce an alkoxide anion, which on protonation should yield the corresponding alcohol. Aldehydes would give 1º-alcohols (as shown) and ketones would give 2º-alcohols. RCH=O + H:(–) RCH2O(–) + H3O(–) RCH2OH practical source of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH4). so LiAlH4 reduces all aldehydes and ketones . Link to comment Share on other sites More sharing options...
John Cuthber Posted April 26, 2009 Share Posted April 26, 2009 Don't forget carboxylic acids I think you can forget them in this case, they get deprotonated to the carboxylate ion which is negatively charged and so it's not attacked by the hydride. Esters, IIRC are reduced by LiAlH4 Link to comment Share on other sites More sharing options...
UC Posted April 26, 2009 Share Posted April 26, 2009 I think you can forget them in this case, they get deprotonated to the carboxylate ion which is negatively charged and so it's not attacked by the hydride.Esters, IIRC are reduced by LiAlH4 Carboxylic acids are reduced to the alcohols as well, but one equivalent of hydride will be wasted deprotonating the free acid. For example, http://www.orgsyn.org/orgsyn/prep.asp?prep=cv8p0434 Link to comment Share on other sites More sharing options...
vedmecum Posted April 27, 2009 Share Posted April 27, 2009 The carbon atom of a carboxyl group is in a relatively high oxidation state. Reduction to a 1º-alcohol takes place rapidly on treatment with the powerful metal hydride reagent, lithium aluminum hydride, as shown by the following equation. One third of the hydride is lost as hydrogen gas, and the initial product consists of metal salts which must be hydrolyzed to generate the alcohol. These reductions take place by the addition of hydride to the carbonyl carbon, in the same manner for aldehydes and ketones. The resulting salt of a carbonyl hydrate then breaks down to an aldehyde that undergoes further reduction. 4 RCO2H + 3 LiAlH4 ether__> 4 H2 + 4 RCH2OM + metal oxides H20__> 4 RCH2OH + metal hydroxides. Link to comment Share on other sites More sharing options...
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