Draw the skeletal structure of the final major organic product at the end of the second reaction above.
http://www.webassign.net/ksorgchem2/221-13-003.jpg
http://www.webassign.net/ksorgchem2/221-13-003.jpg
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I'm assuming you've learned about the SN1, SN2, E1, and E2 reactions from your class. This is a pretty cool and fun reaction! :D
Okay, so you know this is a two step reaction. Looking at the reactant, you can tell OH is a poor leaving group and PBr3 is a good electrophile. So based on this first reaction, you want to kick off the OH and replace it with a better leaving group. The most important reaction of PBr3 is with alcohols, where it replaces an OH group with a bromine atom to produce an alkyl bromide. The mechanism (shown for a primary alcohol) involves initial activation of the alcohol oxygen by the electrophilic phosphorus...meaning you'll form a good leaving group, which is what you want to get after the first step. In the second step, you see that (NaOC(CH3)3) is a good base because it is very sterically hindered so it cannot act as a nucleophile. In addition, the 3 methyl groups in the base are electron-donating to an area where there is already a negative charge, which is the Br- (your good leaving group). Because of this, you know that the second step is a E2 reaction! I think you can get the answer.
Okay, so you know this is a two step reaction. Looking at the reactant, you can tell OH is a poor leaving group and PBr3 is a good electrophile. So based on this first reaction, you want to kick off the OH and replace it with a better leaving group. The most important reaction of PBr3 is with alcohols, where it replaces an OH group with a bromine atom to produce an alkyl bromide. The mechanism (shown for a primary alcohol) involves initial activation of the alcohol oxygen by the electrophilic phosphorus...meaning you'll form a good leaving group, which is what you want to get after the first step. In the second step, you see that (NaOC(CH3)3) is a good base because it is very sterically hindered so it cannot act as a nucleophile. In addition, the 3 methyl groups in the base are electron-donating to an area where there is already a negative charge, which is the Br- (your good leaving group). Because of this, you know that the second step is a E2 reaction! I think you can get the answer.