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I'm studying the synthesis of alcohols. One of the methods given is "Reduction with metallic hydrides like $\text{LiAlH}_4$". Under this title the following reaction has been listed [1, p.635]:

$$\text{CH}_{3}-\text{CH=CH}-\text{CHO}+\text{LiAlH}_{4}\rightarrow \text{CH}_{3}-\text{CH=CH}-\text{CH}_{2}\text{OH}$$

Also on [1, p.298], the following proposition is made which seems to contradict with obtained results of the above reaction.

In certain scenarios, these reagents may reduce double bonds in conjugation with $\text{C=O}$ as well as reduce the $\text{C=O}$ bond.

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It would be great if someone could clarify exactly under what cases and conditions would the double bond reduce as well. Thanks.

Reference:

  1. Shahi, R. Essential Organic Chemistry for JEE Main & Advanced; Arihant Prakashan: New Delhi. ISBN: 978-93-13192-10-7.
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    $\begingroup$ Conjugate reduction with NaBH4 in protic medium is an option because protonation of the resultant enolate will occur and the ketone will reduce to cyclopentanol. Conjugate reduction with LiAlH4 in aprotic solvent leaves a non-reducible enolate. If protic workup protons the enolate faster than excess LiAlH4 is destroyed, then cyclopentanol can form. Compare with: chemistry.stackexchange.com/questions/87138/… $\endgroup$
    – user55119
    Commented Dec 24, 2020 at 18:32

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