destroyer ochem q 162 w/question

[pidbull]

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http://www.chemexper.com/cheminfo/servlet/com.actelion.research.chem.MEAServlet?url=%2Fcheminfo%2Fservlet%2Forg.dbcreator.MainServlet%3Faction%3DSendFieldAction%26tableName%3Dmol%26fieldName%3Dvalue%26uniqueIDValue%3D3004867&&suppressChiralText=false&maxWidth=250&maxHeight=120 + KOH (ethanol, reflux)

I was thinking A = SN1 = Br good Lv group, K from the KOH bonds to it, OH binds in its place. But it is actually elimination. Can someone explain the problem. Thanks, appreciate any help.

Also wanted to ask: if u had a strong bulky base and a tertiary halide - what happens?

 

[tranv117]

You have to remember that OH- is a very strong base (conjugate acid H2O is a very weak acid), so elimination would occur rather then nucleophilic substitution.

With a strong bulky base and tertiary halide, the major product would be the least substituted alkene. It is harder for the base to come in and pull of an internal H so it rather pulls off a H from the terminal ends.

 

[pidbull]

You have to remember that OH- is a very strong base (conjugate acid H2O is a very weak acid), so elimination would occur rather then nucleophilic substitution.

With a strong bulky base and tertiary halide, the major product would be the least substituted alkene. It is harder for the base to come in and pull of an internal H so it rather pulls off a H from the terminal ends.

ahh I get the first part, thanks.
One thing I want to clear up tho is that if you have a big bulky base, then you always get least substituted alkene? (destroyer says primary and secondary halide u get least substituted alkene, and you have said tertiary as well, so all forms of halides with big bulky base = elimination at least substituted correct?)

 

[keliao]

not all halides with tertiary will go to eliminaton. I (WeakBase-better leaving group) F(strong base)

only strong bases SB goes to elimination E2, now you have to keep in mind if you have a hindered/ or unhindered base:

hindered SB = E2 at secondary and tertiary (to the most subt-Saytzeff)
unhindered SB = E2 at primary, secondary, and tertiary ( least subt-Hoffman's)