http://www.mnstate.edu/jasperse/Chem350/Practice%20Tests/Test3-350-PracticeV2.pdf
2006-12-14 07:04:07 · 2 answers · asked by lovely_antionette 1 in Science & Mathematics ➔ Chemistry
This is methoxymercuration in the first step, followed by removal of the mercury in the second step. The first compound is (CH3)2C=CHCH2CH3, 2-methyl-2-pentene. This reacts with mercuric acetate, Hg(C2H3O2)2, in methanol, CH3OH, to put (C2H3O2)Hg- on the C-3 carbon and -OCH3 on the C-2 carbon. This is classical Markownikoff addition, addition across a C=C bond, with the AcOHg+ and CH3O-. In the second reaction, NaBH4 replaces AcOHg- with H-. The product is 2-methoxy-2-methylpentane.
2006-12-14 07:23:06 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
This is an alkoxy mercuration-demercuration reaction.
-If the oxymercuration reaction is carried out in the presence of alcohols, the alcohol will open the mercurinium ion formed in one step.
- Final product after demercuration, is an ether with Markovnikov orientation (nucleophile is alchol it attacks anti to more substitued carbon).
12. 2-methyl-3-methoxypentane
2006-12-14 08:10:11 · answer #2 · answered by Peter B 3 · 0⤊ 0⤋