CHEM 233 Study tips for Midterm 1


Section 1

  • typically multiple choice
  • “Indicate the hybridization of the carbon labeled a” -> in a line bond structure, count the invisible Hydrogen atoms. If the carbon is part of a single bond ring, it’s likely SP3 hybridized. If there’s a double involved, it’s probably sp2 especially it’s connected one ring to another ring.
  • If asked to identify the hybridization of a nitrogen, oxygen, sulfur etc, always check for resonance!
    • Draw out the substituent group -> is there a double bond in there? If your atom e.g. Nitrogen has an unbonded lone pair and there is a double bond one single bond away, then think resonance. So instead of sp3 like it looks like at first glance, it’s actually sp2.
    • If the N is not involved in resonance, the electron pair could be in sp2
    • If the N is involved in resonance, then it’s in the P orbital ** This is a little confusing. but remember ELECTRONS IN RESONANCE ARE ALWAYS IN P ORBITAL
  • Resonance patterns
    • = – (neg)
    • =-+
    • +- :
    • : = EN atom
    • conjugated ring = – = – =
  • Chiral molecule
    • Is the mirror image of this image nonsuperimposable?
    • For some, think about whether the plane of the  paper could be a plane of symmetry. If there’s planar ring, everything else is planar sp2, the only group that is not sp3 is a CH3 (the hydrogen would be same on both side.
    • Also look for whether there is a chiral centre (four different atoms attached)
    • meso compound – compound with a chiral centre and an internal plane of symmetry. Lacks enantiomers
    • Stereochemistry R or S – the lowest priority group must be in the back
    • It is achiral if symmetrical, attached to two of the same gropu

There’s always an equilibrium question

  • choose the option with
    • greater s chracter
    • larger group in equatorial positino
    • more resonance structure ( having resonance is better than no resonance)
    • sometimes you’re going to have use your memorized pKas
    • closer inducers
    • more electronegative inducers
  • beware of the resonance structures. you might see 4 structures but actually it’s resonance

There’s always identity if this is oxidation or reduction type of question

  • double bond to oxygen is same oxidation state as carbon with a single bond to oxygen 1 and single bond to oxygen 2
  • gaining C-H bond is reduction
  • secondary alcohol means the carbon that the alcohol is connected to 2 different carbons
  • Number the corresponding carbons. if that carbon wasn’t in the earlier structure, then don’t give it any attention because it doesn’t matter to you
  • Don’t let the weird looking shapes trick you
  • oxidizing agent is required to convert between different oxidation levels but is not required to convert between 2 molecules of the same oxidation level
  • each bond to an atom more electronegative than C -> +1
  • each bond to hydrogen -> -1

There’s always a question about identical, constitutional isomer, enantiomer, distereomer, different compound

  • beware the line bond. remember that every – represents ch3. don’t be fooled into thinking that they are hydrogens.
  • If there is a sp2 tail, make sure that it will make a mirror image. it could be a diastereomer


  • Planar
  • Cyclic
  • no interruption by sp3
  • Huckel’s number 4n +2
  • B is a strange one. It is able to participate in resonance
  • empty p orbital can participate in aromaticity e.g. sp2,  ( +)

A question on true, false, cannot be determined

  • use cannot be determined sparingly .  If you have a ton of cannot be determined, you’re probably wrong. Sorry buddy.
  • Examples of questions
    • the pKa of a strong acid has a lower numerical value compared to pKa of weaker acid TRUE
    • mixtures of chiral molecules are always optically active ( false. if it’s a 1:1 mix of enantiomers, then inactive)
    • molecules that contain polar bonds must have a dipole moment ( false!)
    • FALSE water have a relatively high boiling point because the OH bond in water. Truth: the OH bond is not important
    • the GC base pair is stronger than A-T because GC has triple bond while AT is double bond
    • to be a mesocompound, you have to have a stereogenic center.
    • Not all nucleophiles have a negative charge

Draw bonding or the structure given __

  • Draw the pi bonding framework for a compound that has resonance structure -> draw the p orbitals of all the atoms involved and link them together
  • When you’re drawing the major resonance strcutures -> the most significant resonance structure can be the one with a positive charge on an electronegative atom. It’s more important to have a full octet
  • Draw the conjugate base -> remove a hydrogen and add an electron pair -> and probably a negative charge sign. Do it in the right location! Tertiary is better than secondary or primary.
  • When drawing the conjugate acid, add a hydrogen and maybe a positive charge

the tricky 4 mark questions where you have explain why

  • Draw out the conjugate bases
  • Key point: Remember ARIO. Choose the conjugate base that most effectively stabilizes negative charge -> weaker base that corresponds to the stronger acid
  • How do bile salts help disperse fats into small droplets?
    • The bile salt has a large nonpolar segment to a polar end group.
    • To minimize energetically unfavourable interactions between the non polar segments and water, the non polar segments aggregate around fats to form a micelle.
    • The polar ends are exposed to water.
    • And this is how Fats can be solubilized into small droplets
  • Does the compound with the alkyl or the methyl group have a greater pka?
    • alkyl
    • When the compound is deprotonated, the conjugate base must stabilize a negative charge
    • In the alkyl atom, the negative charge resides in a sp orbital
    • in the methyl atom, the negative charge resides in a sp3 orbitall
    • the sp orbital has greater s character ( 50% s character) so electrons in the sp orbital are closer to the positively charged nucleus and are more stabilized compared to electrons in the sp3 orbital.
  • When comparing acid strength, always compare the structure of the conjugate base (whether they are negatively charged or not)
    • identify which lone pair of electrons would accept a proton in an acid base reaction, and look for ways to stabilize that lone pair
  • in acid base reaction, the most acidic proton in a species is deprotonated first no matter the strength of the base used
  • When asked to explain why acids and base differ in their strength, use
    • atom size
    • electronegativity
    • hybridization
    • induction ( electron withdrawal and donation) -> For halides, induction matter more than resonance. F, Cl, Br, I
    • resonance. For O and N, resonance matters more than Ind
  • Electronegativity
    • Sample answer: a has a more stable conjugate base because the negative charge is distributed over an oxygen and nitrogen atom. In the conjugate base of B, the negative charge is distributed over a oxygen and a carbon atom. Nitrogen is more electronegative than carbon so nitrogen can better stabilize the negative charge. Since stronger acids have more stable conjugate bases, A is the stronger acid
    • A is the stronger acid because it has the more stable conjugate base. The conjugate base of A stabilizes a negative charge over a nitrogen and oxygen atom via resonance. The conjugate base of B stabilizes a negative charge over two nitrogen atoms. Oxygen is more electronegative than nitrogen and so stabilizes a negative charge better. Therefore the conjugate base of A is more stable and since the stronger acid has the stronger base, A is the stronger acid.
  • Atom size
    • A has  more stable conjugate base because the negative charge is distributed over a oxygen and sulfur atom via resonance. In the conjugate base of B, the negative charge is distributed over two oxygen atoms. Sulfur is a larger atom than oxygen and stabilizes a negative charge better.
    • The larger atom volume has the charge more “spread out” than a smaller atom.
    • since stronger acids have more stable conjugate bases, A is the stronger acid.
  •  aromaticity
    • The conjugate base of A displays aromaticity because it is cyclic, continuous conjugation, and the number of electrons it possess fulfills Huckel’s rule.
    • The conjugate base of B is resonance stabilized but does not display aromaticity because the structure is not cyclic.
    • Aromatic structure in a conjugate base means that the conjugate base is more stable than a conjugate base that does not display aromaticity. A more stable conjugate base means that its corresponding acid is a stronger acid with a lower pKa.
  • Electronegativity & inductive effect
    • The -Cl substitutent is more electronegative than the -Br substituent and so causes a greater inductive effect to generate a partial positive charge that stabilizes the conjugate base. So the compound with -Cl substitutent will have a more stable conjugate base and so the acid with -Cl substitutent will be the stronger acid
  • Atom size (same column)
    • S atom has a larger atomic raidus than oxygen atom. The negative charge is spread over a larger volume fo space so more stable.
  • resonance
    • There are more possible resonance contributors if the (-) is between two = EN rather than next to just one =EN
  • More basic – > more willing to share electrons, generally opposite of the trends for acidity
    • electron-withdrawing groups reduce the basicity if neutral compound. because the lone pair becomes more stable and is less reactive and that’s bad for basicity
  • Solubility?
    • Cycloheptrienyl is much more soluble in water than typical alkyl halides because ionization creates an aromatic compound.

Draw the product of the reaction

  • If there’s an oxygen with a negative charge, consider make a new cyclic ring!

Rank boiling point

  • Hydrogen bonding means that boiling point is relatively high
  • Some dipole dipole is better than nothing.
  • Molecular weight – the heavier, the higher the boiling point
  • If the molecular weight is the same, go for the straighter structure over the branched
  • Hydrogen bond = attractive force felt between H bond in a covalent bond to O,N , F and the lone pair of electrons on a O,N F. The H must be attached to an electronegative atom. A H attached to a carbon does not count. The O N F possessing the lone pair does not need to be attached to a hydrogen though
  • Intramolecular H bond is allow -> think about how that can affect acidity
  • Protic = H bond donors.
  • The more opportunities to H bond, the better. e.g. if there is OH on both ends > OH on just one end

Draw the lewis structure

  • if it indicate S or R, be sure to use your wedges and dashes
  • There might be only three major contributing resonance structures. Don’t go crazy trying to draw too many

Given the structure of a compound, draw its conjugate acid and its conjugate base.

  • If you’re given something with NH2, draw NH3+ for its conjugate acid. N is more basic because it is less electronegative
  • If you’re given something with OH, draw O- for its conjugate base. -> O stabilizes the negative charge better, is more acidic
  • indicate the wedge or dash

Amino Acid

  • You need to have your pKa values memorized to complete this quickly. Some pKa values are provided though.
  • Say the pKa of a carboxylic acid functional group on this particular amino acid is 2.1, at a pH greater than 2.1 the functional group will be in its basic form (remove a hydrogen and indicate negative charge) and at pH less than 2.1 the functional group is plain ol’ COOH
  • If the pKa of NH3+ functional group is 10.7, at pH greater than 10.7 = NH2. At pH less than 10.7, NH3+
  • typically the pH must be more than 1 pH unit different from your pKa for you to see a change in the structure of the amino acid


Apply your chemistry knowledge to the real world

  • How is mayonnaise formed?
    • A phospholipid has a long hydrophobic chain and a polar head that interacts with H2o and CH3COOH.
    • The hydrophobic chains want to minimize contact with polar substances and so the hydrophobic chain aggregate around fats  to form spherical micelles.

Functional groups

  • What distinguishes an ether from an acetal?
    • Ether  R-o-R
    • Acetal R-O R-O R H
  • Functional groups
    • acetal: OR OR R H
    • ketal OR OR R R
    • hemiacetal OR OH R H
    • hemiketal OR OH R R
    • lactone  = cyclic structure containing ester O = C- O
    • imine  R-N1=(R2 and R3)
    • carbonyl groups include esters, aldehydes, ketones

random other notes

  • For LDA to be a suitabl solvent, the conjugate base of your target acid must be more stable than LDA. LDA has a negative charge on its nitrogen atom
  • Typically the proton attached to a C=O bond is removed because its removal generates a resonance stabilized conjugate base.
  • Lewis acids are electron pair acceptors in chemical reactino
  • If the difference in pKa is 8 units, then the ratio between the two sides is 10^ 8
  • Bronsted lowry – > protons
  • lewis acids and base -> electrons, includes bronsted
  • polar bonds can cancel out to give net dipole of zero for molecule
  • boron trifluoride -> the atoms attached are 120 degree in plane
  • acid base reactions are fast
  • good solvent?
    • strong enough base to deprotonate
  • lower pKa than typical alcohol because negative charge is delocalized -> lots of resonance contributors
    • resonance is typically a bigger factor for acidity than inductive effects
  • enolate
  • resonance
    • spread over more atoms is better
  • electronegativity -> more electronegative inducer, pulls electronegative density towards it, make more stable
  • Acidity and basicity]
    • Keq = Ka[H20]
    • Keq = [A-][H30+]/[HA][H20)
  • resonance structures all represent the same structure (reality is a hybrid)
  • ethylammonium chloride would be more soluble in water than in hexane because it is a salt that will ionize in water

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