March 23, 2023, 10:38:54 AM
Forum Rules: Read This Before Posting

Topic: Preparation of a halogeonalkane  (Read 2853 times)

0 Members and 1 Guest are viewing this topic.

Offline fatroro

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Preparation of a halogeonalkane
« on: October 15, 2008, 08:50:18 AM »
I have some questions about this preparation. :-\

What is the underlying principle of the preparation of 2-chloro-2-methylpropane from 2-methylproan-2-ol and hydrochloric acid?

Why should we use the simple distillation to obtain the product?

Why is the SN1 substitution favoured in this reaction?

thz for answering =]
« Last Edit: October 15, 2008, 09:13:12 AM by fatroro »

Offline T-rex

  • New Member
  • **
  • Posts: 7
  • Mole Snacks: +0/-0
  • Gender: Male
Re: Preparation of a halogeonalkane
« Reply #1 on: October 19, 2008, 04:15:05 PM »
1.  Electrons from the H on the HCl get dumped on to the chlorine. That hydrogen gets grabbed by the alcohol. This forms a leaving group (water), and a tertiary carbocation (Sn1). The Nucleophilic Chlorine attacks the carbocation (electrophilic) from either orientation of the molecule not just the backside (Sn1). Leaving you with a final product that is a racemic mixture.

2. Simple distillation is only used when products in a mixture have very high boiling point differences. 
2-chloro-2-methylpropane = 51 C
2-methylproan-2-ol = 83 C

Simple distillation to separate is feasible.

3. answered above.

Sponsored Links