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I did this quiz and I forgot the hydride shift.

With the hydride shift, would these be the right answer?

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    $\begingroup$ Huh? Check your images again--they both have 9 carbons, but you're adding an EtOH (2 carbons) to an 8-carbon molecule. I think once you make the numbers add up, you'll see the right answer. $\endgroup$
    – chipbuster
    Commented Nov 1, 2014 at 23:27
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    $\begingroup$ Instead of the H, we should see a CH3 group. $\endgroup$
    – yolo123
    Commented Nov 2, 2014 at 2:05
  • $\begingroup$ Excellent! Now if you draw the molecules with CH3 instead of hydrogens, what happens? $\endgroup$
    – chipbuster
    Commented Nov 2, 2014 at 2:22
  • $\begingroup$ We get the right amount of carbons? :P $\endgroup$
    – yolo123
    Commented Nov 2, 2014 at 20:49

1 Answer 1

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When you solvolyze the starting bromide a secondary carbocation would be generated. Since a secondary carbocation is not particularly stable, the molecule will explore other pathways such as hydrogen shift, alkyl shift, etc., in order to produce a more stable carbocation. The figure below, explores these options for your starting bromide. Realize that although I've drawn a discreet secondary carbocation that then undergoes rearrangement, in reality cation generation and alkyl or hydrogen shift may occur in a more or less concerted fashion.

enter image description here

In this case there are 3 options (actually there are 4 options, but the fourth one would produce a primary carbocation, so I left it out - can you find it?), a, b and c. Pathways a and c produce secondary carbocations; however pathway b, involving hydrogen shift, produces a more stable tertiary carbocation. Capture of the tertiary carbocation from pathway c by solvent should produce the expected product.

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  • $\begingroup$ Very clear. Thanks ron. I guess I just wrongly wrote H instead of CH3 in my attempt to correct my answer. $\endgroup$
    – yolo123
    Commented Nov 2, 2014 at 20:53

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