9 - Chem 10172, Spring 2008 PROBLEM SET 9 Due Tuesday,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 10172, Spring 2008 PROBLEM SET 9 Due Tuesday, April 1 1 (1) (e) Bromide is a better leaving group than chloride. Thus for I , (i) the rate will increase, but since the carbocation formed is identical to that formed from the chloride (ii) and (iii) are unchanged. For II , (i) the rate will increase since both S N 1 and S N 2 reactions will increase in rate, but since the sensitivity of S N 1 reactions to the leaving group is greater, more of the reaction will take place by S N 1. Thus (ii) more alcohol will be formed, and (iii) the products (especially the azide) will be less inverted (the alcohol will be essentially racemic under all conditions since it is formed almost exclusively by S N 1). (f) The less polar solvent will strongly slow S N 1 and will have only a modest effect on S N 2 (it may increase the reactivity of the azide ion slightly since it will be less hydrogen bonded). So, for I : (i) rate will be slower; (ii) more azide and less alcohol will be formed (the concentration of H 2 O has decreased, and the N 3 has become more nucleophilic), and (iii) products still racemic. For II : (i) rate will be not much changed (S
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

9 - Chem 10172, Spring 2008 PROBLEM SET 9 Due Tuesday,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online