Below is a sample organic chemistry packet — written from your slides,
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any day, the practice exam, or the 60-minute review on the left.
bart · organic chemistry II · day 03 of 07
day 01 · the language of organic · ≈ 60 minutes
Read a molecule the way a chemist does.
goal — look at any skeletal structure and read off its formula, charges, and hybridization without slowing down.
Before mechanisms, before reactions, before anything: you have to
read. Every chapter from here on assumes you can glance at a
line-angle drawing and instantly see the atoms that aren't drawn.
That fluency is the entire point of today.
Chemists draw three ways. Molecular (C₂H₆O) tells
you what's in the box. Condensed (CH₃CH₂OH) tells
you what's bonded to what, in order. Skeletal —
the squiggly line drawing — drops every carbon and every hydrogen
attached to a carbon, and that's the one your exam will show you
lecture 01, slide 09.
three rules, that's all
Skeletal drawings compress three facts into a line:
Every vertex and every endpoint is a carbon.
Every carbon has enough hydrogens to make four bonds. You don't draw them — you count them.
Heteroatoms (anything not C or H) are written in. Their hydrogens are drawn, because they matter.
bonds at a carbon + drawn H's + implicit H's = 4
That formula is the only thing you need. Look at a vertex, count
the lines coming off it, subtract from four. The rest are hydrogens.
Forever.
formal charge — the bookkeeping
Every atom "owns" half of each bond it's in, plus its lone-pair
electrons. Compare that to how many it should have (its
group number), and the difference is the formal charge:
FC = valence electrons − (lone-pair e⁻ + ½ bonding e⁻)
Practically: an oxygen with three bonds and one lone pair has
FC = +1 (think hydronium). A nitrogen with four bonds and no lone
pair has FC = +1 (think ammonium). These are the patterns that
show up in every mechanism this term lecture 02, slide 03.
vertex = C
Every corner is carbon. Every endpoint is carbon.
The only atoms that get a letter are the ones that aren't C —
and the only H's that get drawn are the ones bonded to those
non-C atoms.
quick checks — answer, then reveal
Q1A hexagon with alternating double bonds (benzene) is drawn
with no letters at all. How many carbons and how many hydrogens
are in the molecule?answer
6 carbons, 6 hydrogens (C₆H₆). Six
vertices = six C's. Each carbon has two bonds to neighbors (one
single, one double), so each needs one implicit H to reach four
bonds total.
Q2What's the formal charge on the oxygen in CH₃OH? On the
oxygen in CH₃O⁻?answer
0 and −1. In CH₃OH oxygen has 2 lone
pairs (4 electrons) and 2 bonds (½ × 4 = 2), totaling 6 — matches
oxygen's valence, so FC = 0. In CH₃O⁻ oxygen has 3 lone pairs (6)
and 1 bond (½ × 2 = 1), totaling 7. 6 − 7 = −1.
end of day 01 — tomorrow: arrows, the verbs of organic chemistry.
day 02 · resonance & curly arrows · ≈ 75 minutes
Arrows are how you write electron movement.
goal — push a legal arrow on demand and pick the best resonance contributor on sight.
Every reaction in this course is, underneath, electrons moving. The
curly arrow is the alphabet for that. Get
arrow-pushing right today and the next six days are just
vocabulary.
the only arrow rule
A curly arrow starts at a source of electrons (a
lone pair, a π bond, a σ bond) and ends where those
electrons go (an atom, the middle of a new bond, into a
positive center). The head is a full arrow for a pair of electrons,
half-headed (fishhook) for a single electron.
five things an arrow can't do
Most "wrong" arrows on exam pages break one of these:
do
yes
no
start point
electrons — lone pair, π bond, σ bond
an atom, or a +
end point
between two atoms or onto an atom
off into space
conserve
charge and atoms, before = after
making or losing atoms
octet
row-2 atoms stay ≤ 8 electrons
five bonds on carbon
resonance σ bonds
don't break σ bonds for resonance — only π and lone pairs move
resonance — different drawings, same molecule
Resonance structures aren't different molecules in equilibrium —
they're different pictures of one molecule whose electrons
are delocalized. The real molecule is the weighted average. The
best contributor is the one that:
Has the most full octets.
Has the fewest formal charges.
Puts negative charge on the most electronegative atom (and positive on the least).
arrows = electrons
From electrons. To electrophile.
Read every arrow as a sentence: "this lone pair becomes that
bond," "this π bond becomes that lone pair." If you can't speak
it, you can't draw it.
quick checks — answer, then reveal
Q1For the acetate ion (CH₃CO₂⁻), the negative charge is shown
on one oxygen. What does the other resonance structure look
like, and which contributes more?answer
Both contribute equally. Push the
C=O π bond onto the doubly-bonded oxygen and push the
single-bonded oxygen's lone pair into a new C=O π bond. You get
the same structure with the charge on the other oxygen. They're
equivalent — the real ion has both C–O bonds identical, with
charge spread across both oxygens.
Q2Between two resonance forms of a cation — one with the + on
carbon, one with the + on oxygen — which contributes more?answer
The one with + on oxygen. Oxygen is
more electronegative and holds the positive charge poorly — but
the structure where it carries + has a full octet on
every atom (it gained a bond). The carbocation form has only 6
electrons on the carbon. Octet beats electronegativity here.
end of day 02 — tomorrow: arrows in their first real fight (SN1 vs SN2).
day 03 · nucleophilic substitution · ≈ 90 minutes
SN1 vs. SN2, decided on sight.
goal — look at any substrate and call the mechanism without guessing.
Yesterday you learned to push arrows. Today you use them. Nucleophilic
substitution is two reactions wearing one name, and almost every exam
question is really asking which one is running. Get
that call right and the product, the rate, and the stereochemistry all
fall out for free.
Start with the kinetics, because the rate law is the cleanest tell.
SN2 is one concerted step — the nucleophile and the substrate meet in
the rate-determining step, so both appear in the rate law:
rateSN2 = k [substrate] [nucleophile]
SN1 goes through a carbocation. The slow step is the substrate losing
its leaving group by itself; the nucleophile arrives
afterward, off the clock. So only the substrate shows up:
rateSN1 = k [substrate]
That single difference — second order vs. first order — is why your
slides call SN2 bimolecular and SN1 unimolecularlecture 06, slide 14. Everything else today
is a consequence of it.
the backside attack
In SN2 the nucleophile attacks the carbon directly opposite
the leaving group. The three other groups flip through — like an
umbrella in the wind — and the product has inverted
configuration. This is why it's stereospecific, and why a crowded
carbon kills it: there's no room at the back door.
SN2 — Nu⁻ enters opposite the leaving group; the substrate inverts.
the four tells
You almost never compute a rate on an exam. You read four features
off the structure and let them vote.
factor
favors SN2
favors SN1
substrate
methyl / 1° — open back side
3° — stable carbocation
nucleophile
strong (OH⁻, CN⁻, RO⁻)
weak (H₂O, ROH)
solvent
polar aprotic (DMSO, acetone)
polar protic (water, alcohols)
leaving group
good in both — I⁻ > Br⁻ > Cl⁻ > F⁻
The substrate vote is the loudest. 2° substrates are
the genuine toss-ups — that's when the nucleophile and solvent break
the tie your notes · pg. 7.
SN2 = 2 things
Two molecules in the slow step. Two-faced result.
Bimolecular rate law, backside attack, inverted product —
all "two." SN1 stands alone: one molecule, one carbocation, a
racemic mix.
quick checks — answer, then reveal
Q1Which is most likely to react by SN1: 2-bromobutane,
tert-butyl bromide, or bromomethane?answer
tert-Butyl bromide. It's a 3°
substrate, so losing Br⁻ gives a stabilized 3° carbocation — the SN1
slow step. Bromomethane is the SN2 extreme (wide-open back side);
2-bromobutane is the 2° toss-up that depends on conditions.
Q2You run a 2° substrate with NaCN in DMSO. Which mechanism
wins, and what happens to the stereocenter?answer
SN2. CN⁻ is a strong nucleophile and
DMSO is polar aprotic — both push SN2, overruling the ambiguous 2°
substrate. The stereocenter inverts via backside
attack.
Q3Why does an SN1 reaction on a single enantiomer give a
racemic mixture?answer
The carbocation is flat. The sp²
intermediate is planar, so the nucleophile attacks either face with
roughly equal probability — you get both enantiomers, ≈50/50.
end of day 03 — tomorrow: stereochemistry, where today's inversion becomes a tool you reach for.
day 04 · stereochemistry · ≈ 85 minutes
R, S, and the molecules that look identical in a mirror.
goal — assign R/S on any stereocenter in under 30 seconds and tell enantiomers from diastereomers on sight.
Yesterday's SN2 inverted a stereocenter. To use that, you
need to read one. A stereocenter is any carbon
bonded to four different groups — and switching any two of them
gives a molecule that is not superimposable on the
original. That mirror-image pair is called
enantiomers.
CIP priorities, the only rule
To label a stereocenter R or S:
Rank the four groups by atomic number (higher = higher priority). Ties go to the next atom out.
Put the lowest priority (group 4) pointing away from you.
Trace 1 → 2 → 3 around the remaining three.
Clockwise = R, counterclockwise = S.
If group 4 is already pointing toward you, do the trace anyway and
flip your answer. Faster than redrawing lecture 11, slide 22.
R = rectus (right) · S = sinister (left)
enantiomers vs. diastereomers vs. meso
type
relationship
physical properties
enantiomers
mirror images, non-superimposable
identical — except rotation of polarized light (and biology)
diastereomers
stereoisomers that are not mirror images
different mp, bp, solubility, NMR — different compounds
meso
two+ stereocenters but an internal mirror plane
achiral overall — superimposable on its own mirror image
A molecule with n stereocenters has at most 2ⁿ
stereoisomers — fewer if there's a meso form pulling two of them
together.
CIP = chase the heavy atom
Higher atomic number wins. Tied? Walk one bond further out.
A double bond counts the atom twice: a C=O carbon
"sees" (O, O, C) for priority purposes. That's the trick most
students miss.
quick checks — answer, then reveal
Q1You take (R)-2-bromobutane and swap two groups on the
stereocenter. Is the product still (R), or is it (S)?answer
(S). Any single swap of two groups on
a stereocenter inverts its configuration. Two swaps returns you to
the original.
Q22,3-dibromobutane has two stereocenters. How many distinct
stereoisomers does it have?answer
Three. Theoretical max is 2² = 4:
(R,R), (S,S), (R,S), (S,R). But (R,S) and (S,R) are the same
molecule — they have an internal mirror plane (meso). So you get
the (R,R)/(S,S) pair of enantiomers, plus one meso compound.
end of day 04 — tomorrow: directing effects, where one substituent decides where the next reaction lands.
day 05 · electrophilic aromatic substitution · ≈ 80 minutes
One group on the ring decides where the next one goes.
goal — predict the major product of any EAS on a mono-substituted benzene.
A benzene ring with one substituent reacting under EAS (nitration,
halogenation, Friedel–Crafts) has three open positions:
ortho (next to), meta (two away),
and para (across from). You don't get a
statistical mix — the existing group routes the incoming
electrophile to specific spots and either speeds up or slows down
the whole reaction.
two categories, three behaviors
group on ring
where
how fast
lone-pair donor (–NH₂, –OH, –OR)
ortho / para
strongly activating — faster than benzene
alkyl (–CH₃, –R)
ortho / para
weakly activating
halogens (–F, –Cl, –Br, –I)
ortho / para
deactivating — slower than benzene
EWGs (–NO₂, –C(=O)R, –SO₃H, –CN)
meta
strongly deactivating
why o/p vs. m
Every EAS goes through a positively-charged intermediate called the
sigma complex (or arenium ion). When you draw its
resonance forms, the + sits on the ring carbons that are
ortho and para to where the electrophile attacked.
If the existing group is a donor, it can stabilize
the + when the + lands on a carbon it's bonded to (or next to it) —
which only happens when the electrophile lands ortho or para.
EWGs do the opposite: they make the + worse, so
the electrophile avoids ortho/para and ends up meta by default
lecture 18, slide 11.
Does the substituent have a lone pair to donate into the ring?
If yes — o/p directing (and usually activating; halogens are
the exception, deactivating by induction but still directing by
resonance). If no — meta.
quick checks — answer, then reveal
Q1You nitrate toluene (PhCH₃). What's the major product, and
where does the NO₂ go?answer
Para-nitrotoluene (with some ortho).
–CH₃ is weakly activating and o/p directing. The para isomer
dominates as a single product because ortho is split across two
equivalent positions and also suffers steric clash with the
methyl.
Q2You try to brominate nitrobenzene. Where does the Br end
up — and is the reaction fast or slow?answer
Meta to the nitro, and very slow.
–NO₂ is a strong EWG: it destabilizes the sigma complex at ortho
and para, so the electrophile lands meta. It also pulls so much
electron density out of the ring that the rate is roughly 10⁻⁷
that of benzene — you usually need harsher conditions to get any
reaction at all.
end of day 05 — tomorrow: multi-step synthesis, where today's tools become moves in a longer game.
day 06 · multi-step synthesis · ≈ 90 minutes
Plan synthesis by walking backward.
goal — given a target, propose a 2–4 step synthesis using only the reactions you know.
A synthesis question doesn't ask you to be clever — it asks you to
be organized. The trick is to refuse to think forward. Start from
the target, ask what could have made the last
bond, then ask what made that molecule. You're doing
retrosynthesis, and the symbol for it is the open
arrow ⇒.
the two operations
Retrosynthesis only does two things:
Disconnect a bond in the target. The bond you
break should be one you actually know how to form. The
two pieces become your "synthons" — imaginary fragments — which
you then map to real reagents.
FGI (functional group interconversion). Change
one group into another using a reaction you know. Use this when
you can't disconnect directly but can convert the target into
something disconnectable.
target ⇒ precursor ⇒ precursor ⇒ starting material
worked example
Target: 2-phenylpropan-2-ol from benzene and any
other organic starting materials.
Step back once. A tertiary alcohol with two methyls and one phenyl
can come from a ketone + a Grignard: acetophenone
(PhCOCH₃) + CH₃MgBr gives exactly this product after workup. So:
PhC(OH)(CH₃)₂ ⇒ PhC(=O)CH₃ + CH₃MgBr
Now step back again. Acetophenone comes from benzene by
Friedel–Crafts acylation with CH₃COCl / AlCl₃.
That closes the loop:
PhC(=O)CH₃ ⇒ PhH + CH₃COCl / AlCl₃
Forward synthesis (the answer you write down):
Benzene + CH₃COCl, AlCl₃ → acetophenone.
Acetophenone + CH₃MgBr, then H₃O⁺ workup → 2-phenylpropan-2-ol.
work backward
Stare at the target. Find the youngest bond.
The "youngest" bond is the one you know a reaction for —
usually C–C bonds formed by Grignards, aldol, or EAS.
Disconnect there. Repeat.
quick checks — answer, then reveal
Q1How do you make cyclohexyl methyl ether (C₆H₁₁–O–CH₃) from
cyclohexanol and any other simple reagents?answer
Williamson ether synthesis.
Deprotonate cyclohexanol with NaH to give the alkoxide, then add
CH₃I — an SN2 on methyl (the open back side) gives the ether.
Don't reverse the partners: making the methyl alkoxide and adding
it to cyclohexyl iodide would try SN2 on a 2° center and
elimination wins.
Q2Target: 1-phenyl-1-propanol. Starting materials: benzaldehyde
and anything organic. One step or two?answer
One step. Benzaldehyde (PhCHO) +
ethylmagnesium bromide (CH₃CH₂MgBr), then H₃O⁺ workup, gives the
secondary alcohol PhCH(OH)CH₂CH₃ directly. A Grignard adding to
an aldehyde always makes a secondary alcohol after workup.
end of day 06 — tomorrow: a decision tree for any exam question.
day 07 · full review · ≈ 90 minutes
One decision tree that handles any exam question.
goal — read any orgo II question and know the first move within five seconds.
Six days in and you have all the pieces. Today is about
routing: knowing, on sight, which day's tools a problem
belongs to. The tree below is the one you'll run mentally during
the exam.
the master decision tree
every question is one of three types; every step is one arrow.
top five traps
trap
the fix
seen in
treating SN1 / E1 separately
they share the carbocation — they always compete
day 03
forgetting to flip when LP4 is toward you
trace anyway, then flip R↔S
day 04
calling halogens activating
halogens are o/p directors but deactivators
day 05
Grignard on the wrong partner
Grignards add to aldehydes / ketones; never to acidic H's (alcohols, acids)
day 06
missing a meso compound
2ⁿ stereoisomers at most — check for an internal mirror
day 04
quick checks — answer, then reveal
Q1You're given a 3° alkyl bromide in ethanol and asked for
the product. No nucleophile is added. What runs?answer
SN1 and E1 in competition. A 3°
substrate in a weak nucleophile / polar protic solvent (ethanol)
ionizes to a carbocation. From there ethanol can attack (SN1,
giving an ether) or a β-hydrogen can leave (E1, giving an
alkene). Higher temperature pushes toward E1.
Q2Synthesize 4-bromonitrobenzene from benzene. Two steps —
but which order?answer
Brominate first, then nitrate. –Br
is o/p directing, so nitration on bromobenzene goes mostly para.
If you nitrated first, the –NO₂ would direct bromine to
meta, giving the wrong isomer. Order matters — always
ask what the existing group directs before adding the
next.
end of day 07 — next: the practice exam, then the one-hour review sheet.
goal — work each question untimed first, then redo timed. The answer key explains the why, because that's what transfers.
Each question samples a different day. Together they're the spread
your real exam will draw from. Cover the answer, commit to a
response out loud, then reveal.
Q1Draw the major resonance contributor of the acetate ion
(CH₃CO₂⁻). Say briefly why it's better than the alternatives.show key
Both C–O⁻ structures contribute equally.
Acetate has two equivalent resonance forms with the negative on
either oxygen — both have full octets on every atom and place
the charge on the most electronegative atom. The real ion has
identical C–O bond lengths (between single and double) and half a
negative charge on each oxygen.
Q2(R)-2-iodobutane is treated with NaOH in DMSO. Predict the
product and the stereochemistry.show key
(S)-butan-2-ol. NaOH is a strong
nucleophile; DMSO is polar aprotic. On a 2° substrate this combo
runs SN2 cleanly. Backside attack inverts the stereocenter, so
(R)-iodide → (S)-alcohol. (Priorities of I and OH happen to rank
the same way relative to the other two groups, so the descriptor
flips with the geometry.)
Q3You brominate anisole (PhOCH₃) with Br₂ / FeBr₃. What's
the major product, and is the reaction faster or slower than
bromination of benzene?show key
para-Bromoanisole (with some ortho), much
faster than benzene. –OCH₃ is a strong activator (lone
pair on oxygen donates into the ring) and o/p director. Para
dominates over ortho because of less steric clash with the
methoxy. The activation is strong enough that mild conditions
will already over-brominate — you'd dial things back if you only
want mono.
Q4Assign R or S to the stereocenter in (S)-alanine,
NH₂–CH(CH₃)–COOH.show key
(S) — verify with CIP. Priorities:
–NH₂ (N) > –COOH (C bonded to O,O,O via double-bond
duplication) > –CH₃ > –H. Place H pointing back; the
remaining 1 → 2 → 3 trace goes counterclockwise = S. If H pointed
toward you, the trace would appear clockwise — flip it.
Q5Propose a synthesis of 1-phenyl-1-butanol
(PhCH(OH)CH₂CH₂CH₃) from benzene and any other organic
reagents.show key
Two steps via Friedel–Crafts + Grignard.
Retrosynthesis: a 2° alcohol with phenyl on the same carbon comes
from PhCHO + a Grignard. PhCHO comes from benzene by the
Gattermann–Koch reaction (CO, HCl, AlCl₃). So write forward:
(1) Benzene + CO, HCl, AlCl₃ → benzaldehyde.
(2) Benzaldehyde + CH₃CH₂CH₂MgBr, then H₃O⁺ workup → 1-phenyl-1-butanol.
Q6Without computing rates, rank 1-bromobutane,
2-bromobutane, and 2-bromo-2-methylbutane by the rate of SN2
with NaOH in acetone.show key
1-bromobutane > 2-bromobutane >>
2-bromo-2-methylbutane. SN2 wants an open back side: 1°
beats 2° easily, and 3° is essentially zero — there's no room
for backside attack. Under these conditions the 3° substrate
largely sits; under different conditions (polar protic, weaker
nu) it would run SN1/E1 instead.
end of practice exam — read the 60-minute review sheet the day of.
⏱ 60-minute review · the one hour before the exam
The one-hour-before sheet.
goal — read this once, slowly, and walk in calm. No new ideas — only the ones you already know, condensed.
This sheet exists because the last hour before an exam is for
confidence, not learning. Each day collapses to two or three lines.
If a line surprises you, that's the topic to skim — but don't open
the full lesson now. Trust the work you've already done.
the seven, in one breath each
Day 01 — notation. Every vertex is C. Each C needs
enough H's to make 4 bonds. Heteroatoms are written in; their H's
are drawn. Formal charge = valence − (lone-pair e⁻ + ½ bonding e⁻).
Day 02 — arrows & resonance. Arrows start at
electrons, end at the electrophile. Best resonance contributor:
full octets first, fewest charges second, negative on the most
electronegative atom third.
Day 03 — SN1 / SN2. SN2: 1° substrate, strong nu,
polar aprotic, inversion. SN1: 3° substrate, weak nu, polar protic,
racemization. 2° is the toss-up — nu and solvent decide.
Day 04 — stereochemistry. Rank by atomic number,
put LP4 away, trace 1→2→3. Clockwise = R. Max 2ⁿ stereoisomers;
meso compounds eat a pair. Single swap = inversion.
Day 06 — synthesis. Work backward with ⇒.
Disconnect at a bond you know how to form. FGI when you
have to. Grignard on aldehyde = 2° alcohol; on ketone = 3° alcohol;
on ester = 3° alcohol (adds twice).
Day 07 — routing. Three question types: predict,
push, propose. When stuck, find the most electron-rich and most
electron-poor atoms — an arrow goes from one to the other.
the five mistakes you almost made last time
SN2 on a 3° substrate. Doesn't run. Re-route.
Calling halogens activators. They direct o/p but deactivate the ring.
Forgetting the H-toward-you flip when assigning R/S.
Friedel–Crafts on a ring deactivated by –NO₂. Won't react. Use a different sequence.
Drawing a five-bond carbon while pushing arrows. Stop. Carbon is four bonds, always.
walk in calm
You already know this material.
The exam is asking you to label things you already
know. Read each question twice. Underline the substrate.
Identify what's running. Then write.
that's the packet. go do well.
✦ and that's one packet of many
every packet ships with all of this.
The plan above is one example. Yours is built end to end from the materials
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◷
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Your syllabus, sequenced into one day for every day you have until exam day — not a pile of flashcards.
✎
Daily lessons
Dense, readable, and citing your actual notes — like the days you just read.
◇
Schematics & mnemonics
The hard ideas drawn out, plus memory hooks that survive exam-day nerves.
★
A full practice exam
Written to your course's level, with a worked answer key — not generic filler.
⏱
A 60-minute review guide
The one-hour-before-the-exam sheet: every must-know, nothing else.
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