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Organic Chem P1 Question!

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What is the major organic product formed from the reaction of (CH3)3CBr with concentrated ethanolic solution of KOH?

A. (CH3)3COH

B. (CH3)2CCH2

C. (CH3)2CO

D. (CH3)2CHO

I thought the answer was A because of the nucleophilic substitution reaction of tertiary halogenoalkane by SN1 mechanism. 

However, the markscheme says it's B

 

Can anyone explain why the answer is B?

Thanks in advance :)

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Seems to be an elimination reaction, which I'm pretty sure is not on the current IB syllabus.  Following bits are my own thoughts, which aren't directly related to the IB syllabus, so feel free to skim through it.  

I think the reaction is an E1 reaction, and it's a good thing you can recognize that it should be unimolecular due to it being a tertiary halogenoalkane.  KOH is a strong base, so I think what happens is that the acidic carbocation formed ends up donating a proton to OH, and therein forms the double bond.  Solvent is ethanol, which is polar protic, which generally favours E2 over SN2 AFAIK, so I'm not too sure how this helps determine whether it's E1 or SN1.  Apparently the ethanol isn't polar enough to facilitate the SN1 reaction.  There isn't really enough information in the question to really determine whether or not it's SN1 or E1 - it's also affected by temperature, with higher temperatures favouring elimination reactions too.  

Is this from an old past paper?  If so, then don't worry too much about it - just focus on substitution reactions (SN1/SN2) and you should be fine.  

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53 minutes ago, SC2Player said:

Seems to be an elimination reaction, which I'm pretty sure is not on the current IB syllabus.  Following bits are my own thoughts, which aren't directly related to the IB syllabus, so feel free to skim through it.  

I think the reaction is an E1 reaction, and it's a good thing you can recognize that it should be unimolecular due to it being a tertiary halogenoalkane.  KOH is a strong base, so I think what happens is that the acidic carbocation formed ends up donating a proton to OH, and therein forms the double bond.  Solvent is ethanol, which is polar protic, which generally favours E2 over SN2 AFAIK, so I'm not too sure how this helps determine whether it's E1 or SN1.  Apparently the ethanol isn't polar enough to facilitate the SN1 reaction.  There isn't really enough information in the question to really determine whether or not it's SN1 or E1 - it's also affected by temperature, with higher temperatures favouring elimination reactions too.  

Is this from an old past paper?  If so, then don't worry too much about it - just focus on substitution reactions (SN1/SN2) and you should be fine.  

yes, it's from 2014, the old syllabus. It's great to know that it's not from the current syllabus.

Thanks for your reply!

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