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How can we get 1-methylcyclohexanol from cyclohexanone , by reduction using the reagent LiALH4 in H2O/H+ , I got cyclohexanol , then what should I do?
##"LiAlH"_4## does not ever add a methyl group; it is a hydride donor, so naturally you would get cyclohexanol. (You can alternatively tell because there is no carbon in the compound.)
Also, it is typically NOT used on ketones or aldehydes, because those are reactive enough to be reduced by ##"NaBH"_4##, which acts similarly, but is safer to use because it reacts less violently and is more controllable.
LITHIUM ALUMINUM HYDRIDE ON CYCLOHEXANONE
Noting that the lithium cation does not participate and simply acts as a counterion, we can focus on the (tetrahedral) ##"AlH"_4^(-)##.
Due to the empty ##p## orbital on aluminum, we can expect that the mechanism for ##"LiAlH"_4## goes as follows:
The hydride is donated straight off of the aluminum hydride, giving an oxyanion intermediate. Then, the intermediate can form an alcohol once you add a bit of acid for a reaction workup/"finishing-up".
You still technically have ##"LiAlH"_4## in there; it just has to equilibrate from what's drawn above.
And that's why you got cyclohexanol.
CYCLOHEXANONE TO 1-METHYLCYCLOHEXANOL
To instead get 1-methylcyclohexanol, one option you should have been taught is a Grignard reagent.
What you could use here is methyl magnesium bromide.
(You can make it by simply reacting methyl bromide with magnesium solid in dissolved in anhydrous diethyl ether.)
This mechanism then follows:
The Grignard reagent acts like an anionic nucleophile; in this case, a methyl anion, which is quite good---it's because the pKa of methane is about ##50##, and so its tendency to donate electrons is quite high.
The diethyl ether has to be anhydrous so that the Grignard reagent doesn't accidentally grab a proton from water and deactivate itself.
The acid workup then finishes up the reaction.
The resultant water can then easily donate a proton to the methyl anion, and the Grignard reagent becomes methane, while the magnesium bromide becomes a side product.
Technically, any remaining acid would ultimately give you some ##"Mg"^(2+)##, ##"Br"^(-)##, and the conjugate base of the acid you used as the final-final side products.
From this, actually, you can see how the deactivation occurs; methane is a terrible nucleophile.