CHEM 2211 1nd Edition Lecture 5Outline of Last Lecture I. Chapter two: acids and bases intro Outline of Current Lecture I. Organic acids and bases II. Predicting acid base reactions III. Factors affecting pKa of an acid Current LectureI. Organic Acids and bases Note that the high the pKa the less acidic a compound becomes. Carboxylic acids - Acidic acid o pKa = 4.76 - Formic acid o pKa = 3.75 - The functional group of carboxylic acid is the O=C-OH- The general pKa range for carboxylic acids is 3-5 AlcoholsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. COOHCH3COOHH- Methyl alcohol o pKa = 15.5 o CH3OH - Ethyl alcohol o pKa = 15.9 o CH3CH2OH- The functional group of alcohol is OH - The general pKa range for alcohols is close to 16 - Making them weaker acids than carboxylic acids Protonated alcohols - Protonationo The adding of a proton to an atom, molecule or ion - Protonated methyl alcohol o CH3O+H Ho pKa = -2.5- Protonated ethyl alcohol o CH2CH3O+HHo pKa = -2.4 - Protonated alcohols are very strong acids with low pKas Amines - Are the result of replacing one or more of the hydrogens on NH3 (ammonia) with a carbon substitute - Ex: o Methyl amine CH3NH2o pKa =40 - amines pKa are so high that they rarely act as acids II. Predicting acid base reactions An acid base reaction can be predicted using the pKa of the reactants - CH3:O:H + CH3C=:O:- OH - pKa 15.5 pKa 4.76 The lower the pKa the stronger the acid, so these pKas show that the second reactant (acetic acid) with a pKa of 4.76 is the acid in this reactionThe high pKa (15.5) in this reaction is the base, methyl alcohol - now that the acid and base are identified the reaction can be carried out - CH3:O:H + CH3C=:O:-OH CH3O+H2 + CH3C=O-O – The full reaction shows the conjugate acid and conjugate base 1. The conjugate acid is on the product side o It is the base after it accepts the H+ from the acid o In this case: CH3O+H22. The conjugate base is on the product side o It is the acid after it has given up its H+ o In this case: CH3C=O-O- It is sometimes easiest to identify the conjugate acid and bases by the charges on the product side (+) for conjugate acid (-) for conjugate base III. Factors affecting pKa 1. Electronegativity 2. Hybridization 3. Size 4. Inductive substitutes 5. Resonance Electronegativity - The greater the difference in two atoms electronegativity the more polar the bond is - The more polar the bonds the more acidic the pKa Hybridization - SP > SP2 > SP3 - Ex:o HC=CH ( triple bond) pKa 25 (SP)o H2C=CH2 (double bond) pKa 44 (SP2)o H3C-CH3 (single bond) pKa > 60 (SP3) - More bonds lower pKa Size - As size goes up acidity goes up - Atomic size - On periodic table o Moving to the right size increasing o And going down size increases - F- < Cl- <Br- < I- - HI strongest acid Inductive substituent - Replacing a hydrogen with a higher electronegative atom causes electron withdrawal - Pulling bonding electrons toward the electronegative atom stabilizes the compound o Increases acidity - EX: o CH3C=OOH pKa= 4.76 o BrCH2=OOH pKa=2.86 o ClCH2=OOH pKa=2.81 o FCH2=OOH pKa = 2.66 - The substituent with the highest electronegativity has the greatest affect on the acidity of the compound (as above) Resonance (delocalized electrons) Localized electrons - Occur when a compound loses a proton an the negative charge resides on a single atom Delocalized electrons - Occur when a compound loses a proton an the negative charge resides between two atoms, shared Best example: - Carboxylic acid (acetic acid)o CH3C=OOH o When acetic acid loses a proton its resonance structures can look like: COO-CH3COO-CH3Or But neither of these are the true structure they are resonance contributors The charge is most stable if it delocalizes between the two oxygen atoms creating a resonance hybrid - This causes the acidity of the compound too increase - And it looks like this: The lines show the charge is not just on the O- atom but between the two O
View Full Document
Unlocking...