Why does LiAlH4 reduce esters, amides, or carboxylic acids, while NaBH4 cannot reduce them?
Carboxylic acids and esters are much less reactive to reduction than are ketones and aldehydes and sodium-borohydride, NaBH4 (aq) is too weak a reducing agent for them. NaBH4 is preferred for aldehydes and ketones because it does not react violently with H2O, the way LiAlH4 does and can be used as an aqueous solution, whereas the LiAlH4 must be delivered in an anhydrous solution of diethyl-ether, Et2O, and then neutralized by water and acid to isolate the product/s. But, ultimately, LiAlH4 can be used for all of these reactions. As for the "why"... Refer to the graphic and consider the following: On the right, you have aldehydes, the easiest to reduce. The electronegativity of the oxygen atom is disproportionally strong and is drawing electron density upward, leaving a partial positive at the central carbon and a partial negative at the oxygen. As you move left in the diagram, to the ketone, that partial polarization decreases with the addition of another R group, a second electron density releasing alkyl group, making reaction slightly more difficult, (but still manageable by the NaBH4 solution). Another step left and we now have a second oxygen atom. At this point, the reactivity is too weak to occur without a stronger reducing agent. The short answer is that NaBH4 is simply too weak a reducing agent for carboxylic acids and esters and the two will simply mingle in solution with little to no reaction for months.
Is carboxylic acid a kind of ester?
No, carboxylic acid is NOT a kind of ester. But, you can make ester using carboxylic acid. Carboxylic acid + alcohol = Ester + water Eg: Ethanoic acid + Propanol = Propil etanoate + Water CH3COOH + C3H7OH = CH3COOCH3H7 + H2O Ester formula: RCOOR' R => represents the hydrocarbon part of the carboxylic acid. In the example given above, R = CH3COO R' => represents the hydrocarbon part of the alcohol. In the example given above, R' = C3H7 This shows that carboxylic acid is NOT a kind of ester. Carboxylic acid is a homologue series. (Its members: Methanoic acid, ethanoic acid, propanoic acid, butanoic acid, etc) Ester is a different homologue series. (It has many members, among them: Metil etanoate, etil etanoate, propil etanoate, metil metanoate, etil metanoate, butil pentanoate, and etc) Quote: "But I wonder the structure of acid fits that of ester and would like to know what make two so different." Carboxylic acid has the caboxylic functional group (COOH), whereas ester has the carboxylate bond as the functional group (COO). P.S. I learnt chemistry in Malay (I'm Malaysian!). So, forgive me if I misspelt any of the chemical terms.
Structure of two esters
Put the ester functional group in the middle: we know that we will have 1 carbon (the acid carbon) double bonded to an oxygen atom and single bonded to another O that is then bonded to another C (the alcohol carbon). We have 1 C left: it can go on either the first or second carbon, making two esters (methyl acetate or ethyl formate). Hard to explain in words - see the articles for the diagrams.
How to name esters?
Hello! I don't understand how to name esters.. =/ for example they give you: CH3(CH2)2COO(CH2)4CH3 ok. one with an answer: CH3(CH2)4COOCH2CH3 is ethyl hexanoate. did they say it's ethyl cuz there's two carbons on the right side of the oxygens? and then there's 6 carbons on the left side, so it's hexanoate? if so, then why is CH3COOH named ethyl ethanoate? please help me.. thanks!!! <3
I need to separate an ester from its reactants (an alcohol and a carboxylic acid) after reflux using Na2CO3?
Na2CO3 will form sodium carboxylate with the carboxylic acid.filter the products.only the alcohol and ester left.add some solid Na to it.alocohol will react with it forming sodium alkanoate or whatever.filter it.and u get ester.
Help: What ester that can be derived from '2-propanol' and 'propanoic acid'?
The names of esters are made up of two parts. The second part is derived from the name of the acid: by changing the -ic of the acid to -ate. So in an ester made from propanoic acid will end in propanoate. The first part of the name comes from the alcohol. Because you have a secondary alcohol - the OH group is attached to C2 in the propanol chain, and it is necessary to specify in the name, where the C-O bond occurs. So, in your case, where propan-2-ol was used, where the -C-O- will be from the middle carbon in the propyl group, the ester will be named prop-2-yl propanoate ...........O[H + HO]O=C - CH2 - CH3 propanoic acid the water removed is shown [] .....H....l....H ......l....l.....l H - C - C - C - H = 2-propanol ......l....l.....l .....H...H....H ...........O-O=C - CH2 - CH3 .....H....l....H ......l....l.....l H - C - C - C - H = prop - 2- yl- propanoate. + H2O ......l....l.....l .....H...H....H I hope this structure copies OK - not the easiest to do here.