Exam2an09NEW

Exam2an09NEW - Some Answers to Examination #2. 1. Watch...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Some Answers to Examination #2. 1. Watch out! There are many many right answers to this one. What comes below is only one possibility for each. H 3 C CH 2 CH 3 OH O H 3 C 1. EtLi 2. H 2 O OH H 3 C 2. H 2 O KMnO 4 Ph = H 3 C H O 1. PhLi H 3 C H O H 3 C H OH CrO 3 pyridine EtOH 1. PBr 3 2. Li EtLi 1. Br 2 /FeBr 3 2. Li PhLi H 2 C CHCH 3 CH 3 CHCH 3 CH 3 O NH 2 NH 2 KOH heat CHCH 3 CH 3 HO KMnO 4 2. H 2 O 1. PhLi CHCH 3 CH 3 HO CHCH 3 CH 3 O CrO 3 pyridine CHCH 3 CH 3 O H H
Background image of page 1

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

View Full DocumentRight Arrow Icon
2. Trivial E2: Br (CH 3 ) 3 CO 50 °C 1 H E2 At high temp there is a Cope: 1 2 3 4 5 6 Make 1-6, Break 3-4 = 3. Dimethylamine, a secondary amine, will form an enamine, with two hydrogens in the proper NMR region. H 3 C N H H 3 C H 3 C O H 3 C H 3 O + H 2 O H 3 C OH H 3 C + H 3 C OH H 3 C N(CH 3 ) 2 H CH 3 H 2 C N(CH 3 ) 2 H + CH 3 H 2 C N(CH 3 ) 2 ! = 5.0 ppm + H 3 C OH H 3 C N(CH 3 ) 2 H 3 C N H H 3 C H 3 C OH 2 H 3 C N(CH 3 ) 2 + H 3 O +
Background image of page 2
Methylamine, a primary amine, will form an imine, with no hydrogens at about delta 5.0.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

Exam2an09NEW - Some Answers to Examination #2. 1. Watch...

This preview shows document pages 1 - 4. Sign up to view the full document.

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