orgo destroyer #178

[akimhaneul]

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In this question, the starting compound 2-butanol. Then it is treated with 1) SOCL2, 2) (CH3)3CO- K+ (CH3)3COH, 3) CH3COOOH, and 4) CH3OH, H+. The second step is E2 reaction. But it seems that this step leads to a creation of double bond at the less subsituted carbon, creating 1-butene instead of 2-butene.

I thought the more subsituted means greater stability? Why is this happening here?

 

[iphone2020]

You creating epoxied and under acidic condition it will open epoxied on more substituent side!!!!


Sent from my iPhone using SDN mobile

 

[DAT Destroyer]

In this question, the starting compound 2-butanol. Then it is treated with 1) SOCL2, 2) (CH3)3CO- K+ (CH3)3COH, 3) CH3COOOH, and 4) CH3OH, H+. The second step is E2 reaction. But it seems that this step leads to a creation of double bond at the less subsituted carbon, creating 1-butene instead of 2-butene.

I thought the more subsituted means greater stability? Why is this happening here?

When an E2 reaction occurs, we can get the more stable alkene when it is the more substituted ( Zaitsev product ) only when a suitable strong base such as CH3ONa , NaOH, C2H5ONa, NaNH2, KNH2 is employed. HOWEVER,,,,,,when a larger, more sterically hindered base is employed , such as LDA, DBU, or (CH3)3COK as in this problem, we get the HOFMANN or what some call the anti-Zaitsev product. This is a VERY important concept that you must understand. Large sterically hindered bases react FASTEST with the easiest accessible proton. This gives the less substituted product as the major isomer. For more details, if this is nebulous, consult the text books written by Leroy Wade, or David Klein.

Hope this helps.

Dr. Romano