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x--LiGhtNiNg-_-b0Lt
x--LiGhtNiNg-_-b0Lt asked in Science & MathematicsChemistry · 2 years ago

E2/E1/SN2/SN1 question?

I have this compound 5-Bromo-5-methyl-1-hexanol reacted with NaOH under cold conditions.

(Not sure if its names right, but its This compound, but with the OH and Br switched: http://www.chemspider.com/Chemical-Structure.96217...

I know I get a Zaitsev product from an E2 reaction getting rid off the Br. I also supposedly get another product, this time with no alcohol group...?

I have no idea what other mechanism works. I thought the Alcohol group would leave, but you would need acidic conditions to protonate it into h20 (to be a good leaving group)

Any insight? Hint: it will also have a double bond somewhere

2 Answers

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  • Anonymous
    2 years ago

    DrBob1 is right because after the Br (good LG) is kicked off, we have a carbocation forming and then the electronegative oxygen is going to attack the nucleophilic carbon or the carbocation and the proton would be kicked off from the oxygen thus having H2O and NaBr forming as side products.

  • DrBob1
    Lv 7
    2 years ago

    I don t think it would have a double bond. The second product is a cyclic ether. Note the ideal shape giving a six membered ring. Very favored!

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