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CHE 8B Midterm 2 Study Guide

by: Ellyn Kwan

CHE 8B Midterm 2 Study Guide CHE 8B

Marketplace > University of California - Davis > Chemistry > CHE 8B > CHE 8B Midterm 2 Study Guide
Ellyn Kwan

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Chapters 11, 12, and 16
Organic Chemistry-Brief
Study Guide
Organic Chemistry 8B
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This 4 page Study Guide was uploaded by Ellyn Kwan on Wednesday February 24, 2016. The Study Guide belongs to CHE 8B at University of California - Davis taught by Eskandari in Winter 2016. Since its upload, it has received 130 views. For similar materials see Organic Chemistry-Brief in Chemistry at University of California - Davis.

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Date Created: 02/24/16
CHE 8B Midterm 2 Study Guide: Chapters 11, 12, 16 A. Nomenclature Carboxylic Acids “-oic Acid” (i.e. butanoic acid) Acyl Chlorides “-noyl” chloride (i.e. ethanoyl chloride) Esters “-oate” (i.e. propanoate) Amides “amide” added to the (i.e. butanamide) Aldehydes “-al” (i.e. methanal) Ketones “-one” (i.e. cyclohexanone) Nitriles “-nitrile” (i.e ethanenitrile) B. Relative Reactivites: Carboxylic Acids * The WEAKER the BASE, the more reactive/ THE BETTER THE LEAVING GROUP. • Stronger the acid, weaker the base: Acid strength: HCl (strong) > ROH, H2O > NH3 (weak) * (the more EN, the more acidic) • Base strength: Cl (weak) < OR, OH < NH2 (strong) * The weak base forms weaker bonds, so it is easier to break off and leave • Nucleophilic substitution will happen if and only if the “floating group” (nucleophile) is a STRONGER BASE than the group attached to the acyl group *The stronger base will replace the good leaving/weaker base group by kicking it off. * If the “floating group” (nucleophile) is a WEAK BASE, nucleophilic substitution will NOT happen because the base on the major compound is too strong and will not want to leave. C. Reactions of Carboxylic Acid and its Derivatives: OXYGEN GETS PROTONATED FIRST IF THERE IS A STRONG ACID INVOLVED IN THE REACTION Acyl Chloride Charged nucleophile: carbon gains electrons; forms ester Neutral nucleophile: BASE deprotonates nucleophile; forms an ester. Esters Hydrolysis: water + strong acid; forms carboxylic acid + alcohol Trans-esterification: ester + alcohol; forms ester + alcohol Under BASIC conditions (use OH): forms oxygen on carbonyl carbon + ROH Carboxylic Acids Reactivities: OH > NH2 > O- Esterification: Carboxylic Acid + Alcohol + acid catalyst; forms ester Carboxylic Acid + Amine; forms oxygen on carbonyl carbon + NH3R Amides * Amides are VERY UNREACTIVE compounds Acid Catalyzed Hydrolysis: (use strong acid and water); forms carboxylic acid + ammonium (NH4) Nitriles (use HCl and water); forms carboxylic acid Acid Anhydride Anhydride + water : splits the middle oxygen and forms 2 compounds each with an OH group. Anhydride + ROH: one compound is OH, the other is RO. D. Ketones and Aldehydes • Aldehyde is more reactive than Ketone because… i. The H group is more electron withdrawing than ketone’s CH3 group. ii. Steric effect: Hydrogen is smaller than a CH3 group, so the carbonyl carbon is more accessible to a nucleophile. • Relative Reactivites: o Formaldehyde (2 H groups) > Aldehyde (1 H group) > Ketone • Ketones and Aldehydes go through Nucleophilic Addition Reactions because they are too basic to be replaced by another group. Reactions: a) MgBr, good nucleophiles/strong base * GRIGNARD REAGENTS CANNOT BE PREPARED FROM COMPOUNDS THAT CONTAIN ACIDIC GROUPS (OH, NH2, SH, CCH, COOH) Aldehyde/Ketone + Grignard Reagents b) Reactions 1) Formaldehyde: (use R-MgBr then add H3O+) = primary alcohol 2) Aldehyde: forms secondary alcohol 3) Ketone: forms tertiary alcohol 4) Carbon dioxide: forms carboxylic acid Step 1: Add CN to the carbonyl carbon Step 2: Add H3O+ to make OH. Acid Catalyzed Hydrolysis (HCl and water): Use the CN product then add HCl Aldehyde/Ketone + CN and water to form a Carboxylic Acid from the CN carbon. Catalytic Addition (adding H2): Use the CN product then add 2 equivalents of H2; the carbon and nitrogen in CN get double protonated. Actually adding H-BH3 followed by H3O+ This is a REDUCTION REACTION Aldehyde/Ketone + NaBH4 (adding a H- because H3O+ is added and protonates to ion) form OH group. Actually adding H-AlH3 followed by H3O+ Esters: use (2) H-AlH3 and (1) H3O+; + LiAlH4 (adding a H- ion) forms primary alcohol Carboxylic Acid: use (2) H-AlH3 followed by H3O+; forms primary alcohol *Use (2) H-AlH3 because the intermediate formed is aldehyde, and it is unstable/reactive. C=O turns into a CH2 group. Amide + LiAlH4 *Oxygen will leave because it is a better leaving group than NH2; the carbon will gain the hydrogen. Adding (1) Alcohol + (1) strong acid = HEMIACETAL (OH and OR) Aldehyde/Ketone + Alcohol Adding (2) Alcohol + (2) strong acid = ACETAL ( (2) OR groups) E. Carbohydrates • Monosaccharides can be categorized as… a. Aldehydes (glucose) = aldoses; aldohexose b. Ketones (fructose) = ketoses; ketohexose • D and L Notation nd a. OH group on RIGHT of 2 ndrbon from bottom: D-sugar b. OH group on LEFT of 2 carbon from bottom: L-sugar D and L sugars are ENANTIOMERS of each other • Epimers: Diastereomers that differ in configuration at only 1 carbon center (i.e C-2 epimers and C-3 epimers) • Converting Fischer Projections into Cyclic Hemiacetals a. OH group on RIGHT in Fischer = DOWN on cyclic b. OH group on LEFT in Fischer = UP on cyclic Anomers = 2 sugars differeing in configuration at the anomeric carbon. *If OH group points DOWN on cyclic @ anomeric carbon, then hemiacetal is ALPHA. *If OH group points UP on cyclic @ anomeric carbon, then hemiacetal is BETA. GLUCOSE IS THE MOST STABLE ALDOHEXOSE BECAUSE.. a. Glucose can have 2 chair conformations: alpha and beta. i. Alpha: The OH group is pointing DOWN at the anomeric position; AXIAL. AXIAL POSITION IS VERY UNSTABLE. ii. Beta: The OH group is pointing UP at the anomeric position; EQUITORIAL. EQUITORIAL POSITION IS VERY STABLE because there is less steric strain. b. In BETA GLUCOSE, all of the OH groups are in the equatorial position, which makes the compound extremely stable, and it will predominate in aqueous solutions. • Forming Glycosides (cyclic hemi-acetal + alcohol = 2 acetals) a. Glycosidic bond = the bond between the anomeric carbon and the O in the OR group that forms at the anomeric carbon. b. 2 acetals form because there are 2 conformers: alpha and beta * In the mechanism, the alcohol approaches the chair from the top and the bottom, leading to OR pointing UP and DOWN at the anomeric carbon.


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