Final_Exam_Key 30CL Fall 2009

Final_Exam_Key 30CL Fall 2009 - 1.a. The first reaction is a

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1.a. The first reaction is a Friedel-Crafts-acylation which affords predominantly the para-substituted product. Reaction 2 is an aldol condensation that yields an -unsaturated ketone. The last reaction leads to a mixture of two oximes (anti and syn). The product does not form a cycle (dihydroisoxazole) if there is no base catalyst present. OCH3 CH3COCl Compound X X CHO [OH - ] (NH3OH)Cl KOH (1:1) OCH3 O CH3 H 3 CO O NO 2 H 3 CO N NO 2 OH ( A ) Reaction 1 ( B ) Reaction 2 ( C ) Reaction 3 ( D ) b. Compound X in reaction has to be a Lewis acid like AlCl 3 or BF 3 . This compound acts as a catalyst to generate the acylium ion needed for the Friedel-Crafts acylation. Cl O + AlCl4 - + AlCl3 O Despite acting as a catalyst, the amount of compound X needed is more than one equivalent because it remains coordinated to the acylation product and cannot act as a catalyst anymore afterwards. OCH 3 O H 3 C AlCl 3 c. The solvent for reaction 1 has to be moderately polar, but not be a strong Lewis base which is able to coordinate to the Lewis acid catalyst. Often times, solvents like dichloromethane or carbon disulfide are used in these reactions since they are unreactive, but are able to dissolve the reagents reasonably well and are also weak Lewis bases. d. The p-nitrobenzaldehyde should be used in excess here to prevent the self- condensation of the ketone B , which is slower than the reaction between the ketone and the aldehyde, but still would occur if there is an excess of the ketone present and the reaction is not properly monitored. In addition, many aldehydes are contaminated by carboxylic acids due to the oxidation in air. This is particularly true for liquid aldehydes.
Background image of page 1

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

View Full DocumentRight Arrow Icon
B and p-nitrobenzaldehyde are the reactants here while KOH is the catalyst. n B = 5.25 g/(150.17 g/mol)= 0.0350 mol n aldehyde = 5.82 g/(151.12 g/mol)= 0.0384 mol n KOH = 0.1 g/(56.11 g/mol)= 0.00178 mol (~5 mol%) Since the ketone and the aldehyde react in a 1:1 ratio, the ketone is the limiting reagent here. Thus, 0.035 mol of the product should be formed. n C = 7.93 g/(283.28 g/mol)= 0.0280 mol The yield for the second reaction is 80% (=0.028 mol/0.035 mol * 100%). f. Under the given conditions (polar stationary phase, weakly polar mobile phase), compound D would exhibit the lowest R f -value, compound A would have the highest R f -value. Compound D possesses a hydroxy function that interacts strongly with the polar stationary phase. Compound A is an ether which has a fairly moderate polarity compound to compound B and compound C , which much larger dipole moments. g. Extra Credit: The base peak of m/z=135 in mass spectrum of compound B indicates the presence of a fragment C 8 H 7 O 2 radical cation which is obtain after the loss of a methyl group. The cation is highly stabilized by the methoxy group which is an electron- donating group. H
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/10/2010 for the course CHEM 30CL 142189201 taught by Professor Bacher during the Spring '97 term at UCLA.

Page1 / 12

Final_Exam_Key 30CL Fall 2009 - 1.a. The first reaction is a

This preview shows document pages 1 - 3. Sign up to view the full document.

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