Chem 51LB Week Summer Session 1 Kevin Chen

Grades Curve within the sections Curve within the class What does this mean for you? – If your lab reports are ~90%, but class average is also ~90%, you are not getting an A – If your lab reports are ~50%, but class average is also ~50%, you are not going to fail! Drop the lowest lab report grade (exclude prelab)

Research Opportunity Nowick lab is looking for undergraduate researchers Pros – Gain research experience – Learn new lab techniques – Get name on publications – Apply for graduate school and/or job after graduation Cons – Nothing… maybe less social life and Pokemon Go Meet the graduate students (Yang and Mike), 3PM today, Natural Science 1, 3402B.

Practicals 110 min exam 40 min wet lab 70 min data analysis, etc. This is the single MOST IMPORTANT thing for your grade… so STUDY!!!

Scared yet?

You shouldn’t be…

Scared yet? You shouldn’t be… as long as you study…

Wet-Lab To test student’s lab proficiency and evaluate his or her technique (“under stress”) 4 possible choices: – Recrystallization – Acid/base extraction – Thin layer chromatography – Melting point Assume there is very limited amount of experimental detail What does this mean for you? – Mentally work through all the steps without looking at instructions

Recrystallization 1 of 2 ways for isolating desire compounds from undesired impurities (other being extraction) Often require two solvents, a good and a bad one – Good solvent (dissolves material) – Bad solvent (crashes out material) Patience is a virtue! DO NOT ADD TOO MUCH OF THE GOOD SOLVENT!

Recrystallization procedure Heat solvents Add small amount of hot, good solvent – Goal is to dissolve everything Add small amount of bad solvent until you see a little bit of precipitate – Goal is to reduce solubility of the solvent mixture Add very small amount of hot good solvent – Goal is to dissolve everything again Now… the solubility is just right and all you have to do is wait and crystallization should happen Percent recovery is important. Usually cannot get 100% recovery (if you do… solid is probably not dry! So the extra weight is all solvent)

Extraction Needs two immiscible solvents – Water (usually bottom-layer) – Organic solvent (usually top-layer, less dense!) EXCEPTION: dichloromethane aka methylene chloride aka DCM

Acid/base extraction Use solubility to separate compound mixtures Change pH to change protonation state – Low pH protonates (add HCl) – High pH deprotonates (add NaOH) At low pH – Acid and neutral does nothing – Base protonates  + charged ions  soluble in water At high pH – Base and neutral does nothing – Acid deprotonates  - charged ions  soluble in water

Add NaOH Add water Add organic Add HCl Acid, base, and neutral extraction Note: organic solvent is the top layer (grey) Note 2: the box color represent the pH of the aqueous layer Add HCl Add organic

IMPORTANT SAFETY ANNOUNCEMENT Pressure can build up when one shakes a full separatory funnel IT CAN EXPLODE!!! VENT OFTEN!!!

Thin Layer Chromatography Separation of organic compounds based on polarity Two phases – Stationary phase (solid)  usually silica, which is a silicon based material that is polar – Mobile phase (liquid)  usually a solvent mixture of two organic solvent, one polar and one “relatively” nonpolar

Rf = distance to spot/ distance to top (unitless!) Comparing Rf can give you an idea of relative polarity of each compounds Which compound A, B, or C is the most polar? Why?

Mobile phase matters A LOT Polarity differences between compounds A, B, and C is the consistent (B > C > A)

TLC common problems Streaking – Spot too much compound – Properties of the compounds (no solution) No spot – Spot too little compound – Properties of the compounds (no solution) Side ways lanes – Incorrect placement of TLC

Melting point A common (pretty accurate) way to characterize an organic compound Melting point range of a pure sample is abrupt and tight Melting point range of an impure sample is depreciated and wider Mix melting point “introduce” impurity… unless it is the SAME compound

Mix melting point example #1 Given a pure sample of an unknown compound (either A or B) Given pure sample of A (100˚C) and B (200˚C) Mix melting point expampl – Mix unknown + A  100˚C – Mix unknown + B  150˚C What is the unknown? Why?

Mix melting point example #1 Given a pure sample of an unknown compound (either A or B) Given pure sample of A (100˚C) and B (200˚C) Mix melting point example – Mix unknown + A  100˚C – Mix unknown + B  150˚C What is the unknown? Why? A, melting point unchanged, it is not B because the melting point is higher than A but lower than B (which also makes sense because that sample is not just pure compound B)

Dry lab questions Unknown compound identification (requires knowledge of NMR and IR) Mechanisms Data analysis based on hypothetical wet lab technique Stoichiometry (math…) Basically, anything that you have done in this quarter is fair game!

Unknown compound identification Given molecular formula, proton NMR, carbon NMR, and IR MUST know the “degree of unsaturation” rule, because it will not be provided!!! DOU = (2C N – X – H)/2 – C = carbon – N = nitrogen – X = halogen – Note: oxygen and sulfur does not factor into DOU

Degree of unsaturation DOU = (2C N – X – H)/2 – 1 ring or double bond Example: benzene, C 6 H 6 DOU = (2 x – X – H)/2 = 8/2 = 4 – 4 ring or double bonds In benzene… 1 ring and 3 double bonds

Another DOU example Acetylnitrile (C 2 H 3 N) DOU = (2C N – X – H)/2 – (2 x – 0 - 3)/ 2 = 4/2 = 2 – 2 rings or double bonds – Triple bond = “2 double bonds”

NMR The #1 structural determination technique Chemical shift, splitting pattern and integration – Different magnetic environment means different chemical shift – Splitting pattern comes from (n+1 rule) and leads to singlet, doublet, triplet, etc – Integration means # of protons

Chemical shift Note: Memorize!!! DownfieldUpfield

Splitting pattern N+1 rule: N = number of protons on the carbon(s) ADJACENT to the proton you pick Ha  3H, n = 0  n+1 = 1 (singlet) Hb  2H, n = 2  n+1 = 3 (triplet) Hc  2H, *n = = 5  n+1 = 6 (sextet) Hd  3H, n = 2  n+1 = 3 (triplet)

Disubstituted benzene!

1,4 disubstituted benzene Ha = 2H, doublet Hb = 2H, doublet Key: very clean NMR spectrum!

1,3-disubstituted benzene Ha = 1H, singlet Hb = 1H, doublet Hd = 1H, doublet – Note: Hd ~ Hb Hc = 1H, triplet OR doublet of doublet Key: there is a singlet! (Ha)

1,2-disubstituted benzene Ha = 1H, doublet Hd = 1H, doublet – Note = Ha ~ Hd Hb = 1H, triplet or doublet of doublet Hc = 1H, triplet or doublet of doublet Key: very messy NMR spectrum!

Carbon NMR Works exactly the same as proton NMR Only two things things – Chemical shift – “integration” Count different type of carbon! – No splitting pattern!!! 1 type of carbon 4 types of carbon ??? types of carbon

Mechanisms Experiment #2: Nucleophilic substitution (SN1 versus SN2) Experiment #3: Bromination of alkene Experiment # Dehydration of alcohol, alkylbromide (E1 versus E2) There are only so many…

Stoichiometry and math Bring a calculator (calculator on phone is not ALLOWED!) Percent yield, theoretical yield, percent recovery, Rf, density and etc… Be able to convert between reagent(s) and product(s) when given… – Mass – Moles – Volume (always need… density!) – Equivalence