What would be the major product between 4-chloropentanal and $\ce{LiAlH4}$?
After the reduction of the aldehyde, would the hydroxyl group formed at $\ce{C-1}$ attack $\ce{C-4}$ causing cyclization to occur?
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Sign up to join this communityWhat would be the major product between 4-chloropentanal and $\ce{LiAlH4}$?
After the reduction of the aldehyde, would the hydroxyl group formed at $\ce{C-1}$ attack $\ce{C-4}$ causing cyclization to occur?
It will depend on strict control of the equivalents of reductant, your work-up conditions and your control of temperature during the workup.
The intial product of the reduction is the aluminium alkoxide. If you use a proton source such as $\ce{HCl (aq)}$ or $\ce{NH4Cl (aq)}$ to quench the alkoxide and keep the reaction mixture cold then you will get the 4-chloropentan-1-ol product. If you use the commonly used $\ce{NaOH (aq)}$ workup then you may get the 2-methyltetrahydrofuran.
It is worth noting that 4-chlorobutan-1-ol is commercially available in large quantities.
The reduction of alkyl halides to the corrosponding alkane by $\ce{LiAlH4}$ is well known reaction:
$$\ce{R-X + LiAlH4 -> R-H + AlH3 + LiX}$$
This reaction has been studied in some details (For example, Ref.1). Accordingly, the major product from the reaction of 4-chloropentanal and $\ce{LiAlH4}$ would be pentanol after simultaneous carbonyl and halide reduction. However, if $\ce{LiAlH4}$ is replaced by the milder $\ce{NaBH4}$, the major product you would obtain is 2-methyltetrahydrofuran (2-Me-THF). The reduction of carbonyl group would result in 4-chloropentoxide intermediate, which would readily undergo intramolecular $\mathrm{S_N}$2 reaction to give the final product.
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