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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5$\begingroup$ There could be an intramolecular attack on the chlorine bearing carbon once the hydride ion attacks the carbonyl carbon and the oxygen gets a whole negative charge. This will lead to the formation of methyl substituted oxolane. But this route will be followed if the halide part isn't reduced before the aldehyde part. In terms of reactivity, LiAlH4 will most likely attack the aldo group before the carbon bearing chlorine as the carbonyl carbon is more electrophilic. $\endgroup$– Raviraj BhosaleJun 11 at 4:47
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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.
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$\begingroup$ I took the opportunity to add 2-methyl part to THF. You can rall back if you don't like my editing. $\endgroup$ Jun 11 at 7:12
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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.
References:
- E. C. Ashby, T. N. Pham, and A. Amrollah-Madjdabadi, "Concerning the mechanism of reaction of lithium aluminum hydride with alkyl halides," J. Org. Chem. 1991, 56(4), 1596–1603 (DOI: https://doi.org/10.1021/jo00004a047).
answered Jun 11 at 7:00