'A' Level Chemistry Problem Analysis: Analyzing Lithium Aluminium Hydride (LiAlH4)

(i) What is the difference between using lithium aluminium hydride LiAlH4, versus using sodium borohydride NaBH4, as a reducing agent?


(ii) LiAlH4 must be used in dry ether, followed by solvating the product of the reduction in a protic solvent (eg. via hydrolysis with water). Explain why these must be done in two separate steps.


(iii) LiAlH4 can be used to reduce almost all organic compounds, except two notable exceptions. One of them is nitrobenzene, for which the recommended reduction pathway involves Sn/Zn/Fe and concentrated HCl, or alternatively hydrogen gas with nickel catalyst. Using LiAlH4 would result in undesirable side-reactions, which are beyond the interest of the H2 syllabus. What is the other class of organic compounds, for which you cannot use LiAlH4 to reduce? Explain why LiAlH4 fails to reduce this class of compounds.


Thought process:


(i) Because Al is one electron shell larger than B (the Li+ and Na+ are counterbalancing spectator ions), hence the electron density of LiAlH4 is more polarizable, consequently it is a more ready/effective source of hydride anion nucleophiles (and therefore a stronger reducing agent) as compared to NaBH4. NaBH4 can only reduce aldehydes and ketones, while LiAlH4 can be used to reduce almost all organic compounds (other than the two in iii), including carboxylic acids, acyl chlorides, esters, nitriles and amides.


(ii) Firstly, understand why a protic solvent is REQUIRED after the reduction, then we look at why AFTER the reduction. As the hydride anion nucleophiles attack the partial positively charged carbonyl carbon, the pi electron density / pi-bond shifts up to become a lone pair on carbonyl oxygen. To stabilize this negative formal charged oxygen (forming an alkoxide anion), we need to give it a source of protons. Usually water is used for this, hence certain school lecture notes represent this as "LiAlH4 (in dry ether) followed by hydrolysis", since hydrolysis simply means the reaction of a species with water; what happens here involves a proton transfer from water to the alkoxide anion, to stabilize it.


Why add water only AFTER the reduction? Obviously, if you attempt to use wet or aqueous LiAlH4 (as opposed to in DRY ether), the protons H+ from water would combine with the hydride anions H- from LiAlH4, and you would lose your precious hydride anions (your ammunition for reduction!) in the form of fast escaping hydrogen gas!


(iii) C=C alkenes cannot be reduced by LiAlH4, since the double C=C bonds are electron rich and nucleophilic, and the H- hydride anions from LiAlH4 are also electron rich and nucleophilic. Hence they repel each other. Even if the repulsion hadn't taken place (eg. say you have sufficiently high activation energies to overcome the repulsion), no reaction would occur because in Organic Chemistry, reactions happen because electrons always naturally flow from electron-rich to electron-poor sites, forming dative covalent bonds as nucleophiles attack electrophiles (note that so-called 'electrophilic attacks' is nevertheless still a nucleophile attacking (ie. forming a dative covalent bond with) an electrophile, yet some folks resort to using the term 'electrophilic attack' simply because the electrophile is added to the substrate nucleophile; thus suggesting incorrectly that the substrate has been 'attacked').


Incidentally, unsaturated aromatic compounds such as benzene, cannot be reduced by LiAlH4 for the same reason it cannot reduce C=C unsaturated alkenes. The benzene ring is electron-rich, and will have no reaction with hydride anion nucleophiles. To reduce the benzene ring into cyclohexane, you will need to use a hydrogen and nickel catalyst, at 200 deg C and 30 atm. Such extreme conditions are needed to overcome the strong resonance stabilization energy of benzene.


A typical H2 exam question related to this is found in the H2 Specimen paper, "Explain why even though the benzene ring is unsaturated, it does not usually undergo addition reactions". The answer, of course, is : Benzene is resonance stabilized by the delocalization of pi electrons within the 6-carbon aromatic ring. Electrophilic addition that would destroy the exceptional stability of the aromatic ring is thus resisted. The official SEAB Mark Scheme iterates it as : "The delocalised electrons give increased stability / the resonance energy stabilises benzene".



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The above content is contributed by Mr Heng, owner and 'A' Level Chemistry tutor at Bedok Funland JC. He also goes by the handle UltimaOnline on various online popular homework forums.


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