In: Chemistry
A. While both NaBH4 and LiAlH4 can both be used to reduce ketones and aldehydes, the procedure that you followed in lab would not result in the fluorenol product if NaBH4 were simply replaced with LiAlH4.
i) Explain why.
ii) What change to the procedure would have to be made to use LiAlH4 tosuccessfully reduce fluorenone to fluorenol?
B. NaBH4 is not capable of reducing esters since estersare less reactive than ketones. Explain structurally why this is the case.
C. In a lab, NaBH4 is kept in a dessicator to avoid prolonged contact with water in the atmosphere. This is done since waterwill react with NaBH4. Write the balanced equation for the decomposition of NaBH4 by water.
D. It is found that a bottle of NaBH4 in a lab contains 70% NaBH4 by mass as a significant portion has reacted with atmospheric water as described in question 3 above. What is the minimum that a student should weigh out to fully reduce 3.18 g of fluorenone?
B)Very general mechanism of reduction of carbonyl compounds by metal hydrides (LiAlH4, NaBH4) is shown here.
It’s a simple nucleophilic addition of hydride ion (H?) at carbonyl carbon. Two factors a)Nucleophilicity and availability of Hydride ion and b) Electrophilicity of carbonyl C
1)Ketones/Aldehyde having carbonyl C substituted with either alkyl group(R) or H atom. An alkyl group have just +Inductive effect on carbonyl carbon and thereby the stabilization of partial +ve charge(+?) charge on carbonyl C.
2)In esters this alkyl substituent is replace by –OR i.e.alkoxy group. It has small –I but larger +R effect which stabilizes partial +ve charge (+?) on carbonyl carbon and make it less electropositive.
3)LiAlH4 and NaBH4 gives rise to H- (hydride ion) and +LiAlH3 and +NaBH3 species. LiALH3 is stable species as AlH3 is better Lewis base i.e. it donates its lone pair to Li+ easily whereas BH3 is less good at doing so H- is much associated with Na and hence less potential nucleophile. So NaBH4 provides relatively less Hydride ions than LiALH4. Moreover NaBH4 can be used thus in water but AlLiH4 being reactive have to be used in non-protic solvent like ether.
Esters being having less electropositive carbonyl carbon get attacked by reactive hydride donar LiAlH4 and less reactive NaBH4 do not reduce esters.
But ketone/aldehydes are having much electrophilic C and get attacked by both AlH4 and NaBH4.
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C) Balanced equation for the decomposition of NaBH4 by water in slightly acidic medium.
NaBH4 + 4H2O + 2H+ ---------- > NaOH + B(OH)3 + 5H2(g)
B(OH)3 is H3BO3 i.e. Boric acid.
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D)
From reaction mechanism its clear that,
1 mole Fluorenone react with 1 mole NaBH4 to give 1 mole Fuluorenol.
180 g Flurenone react with 38 g of NaBH4 to give 182 g Flurenol.
Hence 180 g Flurenone = 38 g NaBH4
Then, 3.18 G Flurenone = say ‘A’ g NaBH4
A = 38x3.18 /180 = 0.6713 g
A = 0.6713 g NaBH4 will be required for complete reduction of 3.18 g Flurenone.
But purity of NaBH4 is just 70%
i.e. 100 g NaBH4 sample = 70 g NaBH4 actually
Hence say ‘B’ g NaBH4 sample = 0.6713 g NaBH4(which actually theoretically required)
B =100x0.6713 / 70 = 0.959 g
Hence 0.9590 g of NaBH4 will suffice actually 0.6713 g NaBH4 required for complete reduction of 3.18 g Fluoreneone.
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A)In case of Methanol as solvent if we try using LiALH4 then
Methanol (H3C-OH) has active H (H in -O-H) and LiALH4 react with this and forms H2 gas and LIALH4 get consumed in the reaction completely. Because Methanol being used as solvent must be excess in the reaction.As no LiALH4 remained in solution no reduction possible. For LIALH4 use solvent must be without active (protic H) H like ethers.
But in ether flurenone will have its C=O less polarized i.e. with less electrophilic Carbonyl C and i think fluorenol is less miscible with ether solvent.
NaBH4 on the other hand do not react with such active H compounds and can be used as reductant in solvent like methanol.
(This reaction will be violent and explosive as H2 gas formation is instantaneous). I hope you are not asked by your instructure to follow it actually to check. Please don't try to attemp this reaction i.e. with LiALH4.