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    Hello..
    I am so confused with this question:
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    Start with the basics, adding ammonia to bromoethane...

    The lone pair on the N attacks the delta positive C next to the Br and ethylamine forms: NH2CH2CH3; that should look a bit funny, but bare with me. Simple SN2, so far.

    Start again with methylamine instead of ammonia and there is still a lone pair on the N, so SN2 once again andCH3NHCH2CH3.

    But, there is still a lone pair on the N, so we can go again: CH3N(CH2CH3)2 forms.

    BUT, there is still a lone pair, so off we go again: CH3N+(CH2CH3)3
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    In addition to the previous post by Pigster, each subsequent substitution reaction occurs quicker than the previous one. This is due to the fact that alkyl substituents are electron donating. When the nitrogen lone pair is donated to the carbon to kick out the nitrogen becomes positively charged. The alkyl substituents can now stabilise this positive charge through hyper-conjugation, which makes the nitrogen more nucleophilic. Due to this, you would probably not be able to stop this reaction before the quaternary amine.
 
 
 
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