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In my textbook (NCERT 12th class Chemistry, Part 2, India), $\ce{H2}$ with $\ce{Pd}$ catalyst is the only method mentioned.

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    $\begingroup$ Yes they certainly can, although I don't know why anyone would want to use LiAlH4 when you can just use NaBH4. $\endgroup$ – orthocresol Sep 23 '16 at 13:16
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Yes, $\ce{LiAlH4}$ can be used for the reduction of an aldehyde to an alcohol. Compared to $\ce{NaBH4}$ it is a stronger reducing agent due to the electronegativity differences (Pauling electronegativity of $\ce{Al}$ is 1.61 & $\ce{B}$ is 2.04) making the metal-H bond more polar in case of $\ce{LiAlH4}$. Therefore, it reacts very violently with water and other acidic group containing compounds, and careful addition of $\ce{LiAlH4}$ is required.

The mechanism is rather straightforward, with alkoxide formation and protonation steps to get the final product.

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In the lithium aluminum hydride reduction, water is usually added in a second step, whereas in the sodium borohydride reduction the methanol solvent system achieves this protonation immediately.

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It really depends on the molecule itself. If you're talking about a molecule where the aldehyde is virtually the only functional group present, then yes $\ce{LiAlH4}$ will be very efficient. However, in the presence of other sensitive functional groups such as nitriles which can also be reduced with $\ce{LiAlH4}$, using milder reagents such as $\ce{NaBH4}$ makes more sense since nitriles cannot be reduced by it.

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I would suggest you to kindly refer page no. 360 of Chemistry Part II NCERT of class 12.

The paragraph having title Reduction reads:

Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride $\ce{NaBH4}$ or lithium aluminium hydride $\ce{LiAlH4}$ as well as by catalytic hydrogenation.

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  • $\begingroup$ Don't use LATEX for italiacizing. Use asterisk sign instead. $\endgroup$ – Nilay Ghosh Feb 10 '17 at 11:10

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