Kevin B. answered • 10/29/22
Enthusiastic Orgo Professor Can Help You Unlock Organic Chemistry
This is a fairly straightforward reaction sequence. Hopefully, before asking here, you have gone to the appropriate chapter in your textbook (or lecture notes), which are very likely to steer you in the correct direction. (Learning how to answer your own questions with the resources you have available to you is an invaluable skill in any class, but perhaps particularly organic chemistry.)
But if it's not clear, I'll try to break it down here. The reactant (1-bromo-2-methylpropane) is also "isobutyl bromide", or as a condensed formula, (CH3)2CH-CH2-Br. The first reaction (Li, ether) replaces the -Br with -Li to make an alkyllithium reagent (in this case, isobutyllithium, or (CH3)2CH-CH2-Li). Alkyllithium reagents are extremely strong nucleophiles (as well as extremely strong bases -- this reaction would normally done under rigorously anhydrous conditions, as even traces of water will react with and deactivate the alkyllithium reagent formed, or prevent it from being formed in the first place). This is because the C-Li bond is very polarized toward carbon. C is not particularly electronegative, but it is much more electronegative than Li, so the covalent bond is a very /polar/ covalent bond, with a large partial negative charge on carbon (and a corresponding large partial positive on Li, though that's relatively unimportant). Since a negative on C is not very stable, this carbon is "desperate" to react with something, and use those electrons to make a nicely shared covalent bond, at which point the C can breathe a big sigh of relief. As a result, it will grab a hydrogen ion (H+) from nearly anything (that is, it is an extremely strong base), or donate its electrons to nearly any electrophile, pushing any "pushable" pair of electrons (like the pi bond of a C=O carbonyl) out of the way in the process (that is, it is a very good nucleophile).
In the second reaction, 2-butanone is added. The extremely nucleophilic isobutyllithium does a nucleophilic addition to the carbonyl (C=O) of the 2-butanone, pushing the C=O pi bond up to form a new lone pair on the oxygen atom. The result is a new C-C bond between the -CH2 end of the isobutyl group (where the lithium was attached) and the carbon which was originally the carbonyl of 2-butanone. The lithium is kinda bonded to the oxygen at this point, but the Li-O bond is considerably ionic, rather than covalent, so it's generally acceptable to think of this as an alkoxide with a lithium counterion. It's hard to convey the structure with text, but the C-O carbon of the alkoxide has three C-C bonds -- one to -CH3 and one the -CH2CH3 (the two groups which were originally bonded to the C=O in the 2-butanone reactant), and a new bond to the isobutyl group: -CH2-CH(CH3)2,
The last step (H3O+) simply puts a proton onto the alkoxide oxygen -- or, alternatively, breaks the O-Li bond and replaces it with O-H. It works out to be the same thing. This does not change the carbon structure at all, so the C-OH of the alcohol product is connected to the same three carbon groups -- one -CH3, a -CH2CH3, and a -CH2CH(CH3)2.
Giving this an IUPAC name results in 3,5-dimethyl-3-hexanol. The longest carbon chain has six carbons, and you need to count from the end closest to the alcohol FG, which means starting to count from the end of the ethyl group. That results in a six-carbon chain, with the -OH on the 3rd carbon (3-hexanol) and methyl substituents on C3 and C5.
