Learning Arc 02 · Build electron, stereochemical and substitution grammar

Discover the one-step SN2 model

  • Board and NCERT Questions require mechanism, rate, energy and geometry to agree
  • no single slogan is reliable.
Foundation for this Concept22 symbols · 5 recovery lessons · 4 entry checksOpen

Start here: words and symbols used in this Concept

Open only the recovery material you need before the Mystery.

Read these words and symbols first

C*
The carbon being studied in this diagram.
The star is a temporary marker, not a charge and not part of an IUPAC name.
R
The carbon-containing part of a haloalkane.
R can represent different carbon chains, so its exact structure must be supplied.
X
A halogen atom: F, Cl, Br or I.
X is a placeholder, not a new element.
RX
A haloalkane written as its carbon part R bonded to halogen X.
Square brackets around RX later mean concentration, not a new molecule.
Nu
The nucleophile that donates one electron pair.
Nu names a role
the actual atom must still be identified.
C2
The bromoethane carbon directly bonded to Br in the supplied atom map.
C2 is a temporary atom label, not an IUPAC locant.
:
One lone pair when the colon is printed beside an atom.
A colon in ordinary prose does not carry this chemical meaning.
···
Read partly bonded: the two atoms share less than one complete bond in this proposed highest-energy arrangement.
The dots do not mean a complete ordinary single bond.
‡
Read double dagger: the mark attached to a transition-state drawing.
It does not mark a stable species that can be isolated.
>>
Much greater than in the stated comparison.
It is a qualitative ranking symbol here, not a measured numerical ratio.
SN2
Read S-N-two: S means substitution, N means nucleophilic and 2 means two reacting species in one elementary event.
The 2 does not count arrows, products or reaction steps.
energy profile
A graph of potential energy against reaction progress for one proposed pathway.
Reaction progress orders molecular arrangements
it is not a clock-time axis.
neopentyl bromide
The structure (CH3)3C-CH2-Br.
The bracketed (CH3)3 means three CH3 groups are bonded to the next C.
Br is bonded to CH2.
Its Br-bearing carbon is primary, but the adjacent three branches can block approach.
bounded model
A conclusion limited to the structures, data and fixed conditions stated in the Question.
It is not a rule for every possible substrate, reagent, solvent or temperature.
backside approach
Approach to the reacting carbon from the side opposite the leaving group.
Backside is defined by the C-X axis, not by left or right on the screen.
transition state
The proposed highest-energy arrangement along one event, marked with a double dagger.
It is not a stable intermediate and cannot be isolated.
concentration
Amount of a named reacting species per litre, written in square brackets and measured here in mol L-1.
[RX] means the concentration of RX
it is not a molecular structure or charge.
initial rate
The amount of product formed per litre per second near the start, before concentrations change greatly.
It is not the total time taken to finish.
rate constant k
The proportionality number that connects concentration to rate at fixed temperature and medium.
Changing temperature or medium can change k.
elementary event
One proposed molecular event with no smaller event written inside it.
One event can contain more than one electron-pair arrow.
molecularity
The number of reacting species in one elementary event.
It is not the number of arrows or the number of products.
activation-energy barrier
The vertical energy rise from the reactant level to a transition-state maximum.
It is an energy difference, not elapsed time.

Foundation 1

Decode SN2 before using the code

  • Read SN2 as S-N-two.
  • S means substitution.
  • N means nucleophilic.
  • 2 means two reacting species in one elementary event.
  • The 2 does not count arrows, products or steps.

Do

  • Name the substrate.
  • Name the nucleophile.
  • Count those two reacting species.

Foundation 2

Read concentration and initial rate

  • [RX] means the concentration of haloalkane RX.
  • [Nu] means the concentration of nucleophile Nu.
  • Concentration is amount per litre.
  • Initial rate is product formed per litre per second near the start.
  • In rate = k[RX][Nu], k is the fixed proportionality number for the stated temperature and medium.

Do

  • Compare two rows in which only [RX] changes.
  • Then compare two rows in which only [Nu] changes.

Foundation 3

Read the energy profile

  • Potential energy is on the vertical axis.
  • Reaction progress is on the horizontal axis.
  • Reaction progress is not time.
  • A maximum marks a transition state.
  • An internal minimum would mark an intermediate between two events.
  • The vertical rise to a maximum is the activation-energy barrier.

Do

  • Count maxima.
  • Count internal minima.
  • Do not read horizontal distance as reaction time.

Foundation 4

Read the transition-state drawing

  • Square brackets enclose the complete proposed arrangement.
  • Dotted connections show partial bonds.
  • The double dagger marks the transition state.
  • The raised sign outside the brackets is the overall charge.
  • A transition state cannot be isolated as a stable bottle of material.

Do

  • Read the partial incoming bond.
  • Read the partial leaving bond.
  • Read the overall charge separately.

Foundation 5

Decode the branched primary counterexample

  • (CH3)3C-CH2-Br is neopentyl bromide.
  • Br is bonded directly to CH2.
  • That CH2 carbon is primary.
  • The next carbon is bonded to three CH3 groups.
  • Those nearby branches can block the opposite approach line.

Do

  • Point to the Br-bearing CH2.
  • Then point to the three CH3 branches on the adjacent carbon.
Neopentyl bromide has a primary bromine-bearing carbon but severe branching on the adjacent carbon blocks backside approach.

Entry check 1

For OH− plus bromoethane, state the net made/broken bonds and total charge check.

Show the recovery
  • C-O made, C-Br broken
  • total charge -1 before and after.

Return to Concept 2.1 for the net electron/bond ledger.

Entry check 2

Classify ordinary CH3CH2Br, state the accepted local geometry at its Br-bearing carbon, and explain whether the temporary label C2 is an IUPAC locant.

Show the recovery
  • Primary
  • tetrahedral sp3 accepted HSC model
  • C2 is only an analysis label until an IUPAC parent is selected and numbered.

Return to Concepts 1.1, 1.4 and 1.6.

Primary bromoethane; one carbon group is attached to the bromine-bearing carbon.

Entry check 3

  • In OH− + CH3CH2Br, identify the substrate, partial-positive atom, available electron pair, donor, accepting atom and leaving group
  • then state both valid arrow sources and the carbon-skeleton check.
Show the recovery
  • CH3CH2Br is the substrate
  • its C bonded to Br is the partial-positive acceptor
  • an O lone pair is available
  • OH− is the donor and Br is the leaving group.
  • One arrow can start at the O lone pair and another at the C-Br bond.
  • The two substrate carbons are preserved.

Return to Concepts 1.6 and 2.1 for partial charge, substrate, lone pairs, formal charge, roles, arrow sources and the carbon ledger.

Entry check 4

What is an attraction between separate molecules or ions in the surrounding liquid called?

Show the recovery
  • An intermolecular attraction
  • it acts between separate particles, unlike a covalent bond inside one molecule.

Return to Concept 1.6's inside-versus-between particle bridge.

1 · ObjectiveOpen

What you will understand

Infer the bounded one-step backside substitution model from rate, geometry, energy and structural evidence.

Problems this Concept unlocks

  • derive rate=k[RX][Nu] from controlled data
  • draw one lawful coordinated one-event mechanism
  • distinguish transition state from intermediate
  • track labelled inversion
  • rank matched SN2 support factors and repair counter-patterns
2 · MysteryOpen

The problem to explain

Read these words and symbols first

C*
The carbon being studied in this diagram.
The star is a temporary marker, not a charge and not part of an IUPAC name.
R
The carbon-containing part of a haloalkane.
R can represent different carbon chains, so its exact structure must be supplied.
X
A halogen atom: F, Cl, Br or I.
X is a placeholder, not a new element.
RX
A haloalkane written as its carbon part R bonded to halogen X.
Square brackets around RX later mean concentration, not a new molecule.
Nu
The nucleophile that donates one electron pair.
Nu names a role
the actual atom must still be identified.
C2
The bromoethane carbon directly bonded to Br in the supplied atom map.
C2 is a temporary atom label, not an IUPAC locant.
:
One lone pair when the colon is printed beside an atom.
A colon in ordinary prose does not carry this chemical meaning.
···
Read partly bonded: the two atoms share less than one complete bond in this proposed highest-energy arrangement.
The dots do not mean a complete ordinary single bond.
‡
Read double dagger: the mark attached to a transition-state drawing.
It does not mark a stable species that can be isolated.
>>
Much greater than in the stated comparison.
It is a qualitative ranking symbol here, not a measured numerical ratio.
SN2
Read S-N-two: S means substitution, N means nucleophilic and 2 means two reacting species in one elementary event.
The 2 does not count arrows, products or reaction steps.
energy profile
A graph of potential energy against reaction progress for one proposed pathway.
Reaction progress orders molecular arrangements
it is not a clock-time axis.
neopentyl bromide
The structure (CH3)3C-CH2-Br.
The bracketed (CH3)3 means three CH3 groups are bonded to the next C.
Br is bonded to CH2.
Its Br-bearing carbon is primary, but the adjacent three branches can block approach.
bounded model
A conclusion limited to the structures, data and fixed conditions stated in the Question.
It is not a rule for every possible substrate, reagent, solvent or temperature.
transition state
The proposed highest-energy arrangement along one event, marked with a double dagger.
It is not a stable intermediate and cannot be isolated.
concentration
Amount of a named reacting species per litre, written in square brackets and measured here in mol L-1.
[RX] means the concentration of RX
it is not a molecular structure or charge.
initial rate
The amount of product formed per litre per second near the start, before concentrations change greatly.
It is not the total time taken to finish.
rate constant k
The proportionality number that connects concentration to rate at fixed temperature and medium.
Changing temperature or medium can change k.
elementary event
One proposed molecular event with no smaller event written inside it.
One event can contain more than one electron-pair arrow.
molecularity
The number of reacting species in one elementary event.
It is not the number of arrows or the number of products.
activation-energy barrier
The vertical energy rise from the reactant level to a transition-state maximum.
It is an energy difference, not elapsed time.
  • Proposal A forms a complete C-O bond while the complete C-Br bond remains.
  • This gives the reacting carbon five complete bonds.
  • Proposal B breaks C-Br completely before C-O begins to form.
  • The supplied rate depends on both reacting species.
  • The supplied energy profile has one maximum and no internal minimum.
  • The same labelled groups are tracked before and after reaction.
2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.

Which proposal can be repaired into one model that fits the rate, energy and labelled-group evidence without giving carbon five complete bonds?

Your decision

  • Predict one event or two.
  • Then check the rate evidence.
  • Check the energy profile.
  • Check the labelled groups.
3 · InvestigationOpen

Investigation

Investigation 1

Decode the viewer key

Read these words and symbols first

solid wedge
A triangular bond that points from the page towards the viewer.
It reports viewing direction, not bond strength.
hashed wedge
A striped triangular bond that points away from the viewer.
It is not a broken bond.

Do

Hold the page fixed and translate each bond into in-page, towards or away.

2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.

Check

Rotating the entire molecule is allowed only if every group rotates together.

Investigation 2

Track the same three groups

Read these words and symbols first

solid wedge
A triangular bond that points from the page towards the viewer.
It reports viewing direction, not bond strength.
hashed wedge
A striped triangular bond that points away from the viewer.
It is not a broken bond.

Do

  • Copy the atoms in A, B and C before and after.
  • Do not exchange one label for another.
2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.
The product graph preserves the exact A, B and C atom identities while hydroxyl replaces bromine. The recorded directions show each group's before-and-after position relative to the viewer.

Check

A is CH3, B is H and C is CH2CH3 in both states.

Investigation 3

Locate the open approach line

Read these words and symbols first

backside approach
Approach to the reacting carbon from the side opposite the leaving group.
Backside is defined by the C-X axis, not by left or right on the screen.

Do

  • Draw the C-Br line through carbon and extend it away from Br
  • place O on that extension.
2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.
Electron-pair mechanism

One coordinated backside substitution event

Can both electron-pair moves occur in one event without a stable five-bond carbon?

BeforeHydroxide plus labelled 2-bromobutaneBoth electron-pair moves belong to this one elementary event. The highest-energy arrangement has partial C-O and C-Br bonds; it is not a stable five-bond carbon.
Complete expanded labelled 2-bromobutane and hydroxide before the one-step event; every carbon, hydrogen, oxygen and bromine is present.CCCCBrO−HHHHHHHHHH12
  • OH approaches opposite Br; A/B/C labels track the unchanged groups.
  1. Arrow 1
    Pair starts atone hydroxide oxygen lone pairPair ends atreacting carbon opposite bromine
    Why
    Backside access aligns the entering pair opposite the bond being displaced.
    Immediate effect
    The C-O bond begins forming.
  2. Arrow 2
    Pair starts atC-Br bonding pairPair ends atbromine
    Why
    The old pair leaves carbon in the same elementary event, preventing a stable five-bond carbon.
    Immediate effect
    C-Br breaks and bromide forms.
After both arrowsLabelled 2-butanol plus bromide
Complete expanded labelled 2-butanol and bromide after the event. The same atoms remain; C-O replaces C-Br and the A/B/C tetrahedral arrangement is inverted.CCCCBr−OHHHHHHHHHH
Three compulsory checks
Atoms
The same four C, ten H, one O and one Br identities occur before and after.
Charge
Total formal charge is -1 before and -1 after.
Carbon valence
Reactant and product carbon c2 each have four full single bonds; the transition state uses two partial bonds and is not a stable five-bond carbon.

Hydroxide approaches the labelled reacting carbon opposite bromine. In one elementary event an oxygen lone pair forms C-O while the C-Br pair moves to bromine; the labelled tetrahedral arrangement inverts and no carbocation intermediate exists.

Check

The destination is carbon, not Br, and backside is not page-left by definition.

Investigation 4

Coordinate the two pair moves

Read these words and symbols first

concerted
Bond making and bond breaking belong to the same elementary event.
It does not mean a stable five-bond carbon exists.
elementary event
One proposed molecular event with no smaller event written inside it.
One event can contain more than one electron-pair arrow.

Do

Read both arrow sources and destinations, then state one event rather than two isolated stages.

Electron-pair mechanism

One coordinated backside substitution event

Can both electron-pair moves occur in one event without a stable five-bond carbon?

BeforeHydroxide plus labelled 2-bromobutaneBoth electron-pair moves belong to this one elementary event. The highest-energy arrangement has partial C-O and C-Br bonds; it is not a stable five-bond carbon.
Complete expanded labelled 2-bromobutane and hydroxide before the one-step event; every carbon, hydrogen, oxygen and bromine is present.CCCCBrO−HHHHHHHHHH12
  • OH approaches opposite Br; A/B/C labels track the unchanged groups.
  1. Arrow 1
    Pair starts atone hydroxide oxygen lone pairPair ends atreacting carbon opposite bromine
    Why
    Backside access aligns the entering pair opposite the bond being displaced.
    Immediate effect
    The C-O bond begins forming.
  2. Arrow 2
    Pair starts atC-Br bonding pairPair ends atbromine
    Why
    The old pair leaves carbon in the same elementary event, preventing a stable five-bond carbon.
    Immediate effect
    C-Br breaks and bromide forms.
After both arrowsLabelled 2-butanol plus bromide
Complete expanded labelled 2-butanol and bromide after the event. The same atoms remain; C-O replaces C-Br and the A/B/C tetrahedral arrangement is inverted.CCCCBr−OHHHHHHHHHH
Three compulsory checks
Atoms
The same four C, ten H, one O and one Br identities occur before and after.
Charge
Total formal charge is -1 before and -1 after.
Carbon valence
Reactant and product carbon c2 each have four full single bonds; the transition state uses two partial bonds and is not a stable five-bond carbon.

Hydroxide approaches the labelled reacting carbon opposite bromine. In one elementary event an oxygen lone pair forms C-O while the C-Br pair moves to bromine; the labelled tetrahedral arrangement inverts and no carbocation intermediate exists.

Check

No separate positively charged carbon species—later called a carbocation—appears in this supplied one-maximum model, and carbon is never drawn with five full bonds.

Investigation 5

Replace five complete bonds with two partial bonds

Read these words and symbols first

transition state
The proposed highest-energy arrangement along one event, marked with a double dagger.
It is not a stable intermediate and cannot be isolated.
···
Read partly bonded: the two atoms share less than one complete bond in this proposed highest-energy arrangement.
The dots do not mean a complete ordinary single bond.
‡
Read double dagger: the mark attached to a transition-state drawing.
It does not mark a stable species that can be isolated.

Do

  • Write [HO···C···Br]‡−.
  • Label each dotted connection as a partial bond.
Electron-pair mechanism

One coordinated backside substitution event

Can both electron-pair moves occur in one event without a stable five-bond carbon?

BeforeHydroxide plus labelled 2-bromobutaneBoth electron-pair moves belong to this one elementary event. The highest-energy arrangement has partial C-O and C-Br bonds; it is not a stable five-bond carbon.
Complete expanded labelled 2-bromobutane and hydroxide before the one-step event; every carbon, hydrogen, oxygen and bromine is present.CCCCBrO−HHHHHHHHHH12
  • OH approaches opposite Br; A/B/C labels track the unchanged groups.
  1. Arrow 1
    Pair starts atone hydroxide oxygen lone pairPair ends atreacting carbon opposite bromine
    Why
    Backside access aligns the entering pair opposite the bond being displaced.
    Immediate effect
    The C-O bond begins forming.
  2. Arrow 2
    Pair starts atC-Br bonding pairPair ends atbromine
    Why
    The old pair leaves carbon in the same elementary event, preventing a stable five-bond carbon.
    Immediate effect
    C-Br breaks and bromide forms.
After both arrowsLabelled 2-butanol plus bromide
Complete expanded labelled 2-butanol and bromide after the event. The same atoms remain; C-O replaces C-Br and the A/B/C tetrahedral arrangement is inverted.CCCCBr−OHHHHHHHHHH
Three compulsory checks
Atoms
The same four C, ten H, one O and one Br identities occur before and after.
Charge
Total formal charge is -1 before and -1 after.
Carbon valence
Reactant and product carbon c2 each have four full single bonds; the transition state uses two partial bonds and is not a stable five-bond carbon.

Hydroxide approaches the labelled reacting carbon opposite bromine. In one elementary event an oxygen lone pair forms C-O while the C-Br pair moves to bromine; the labelled tetrahedral arrangement inverts and no carbocation intermediate exists.

Check

  • The same drawing contains one forming partial bond and one breaking partial bond.
  • It cannot be isolated.

Investigation 6

Compare the labelled product

Read these words and symbols first

inversion
The tracked tetrahedral arrangement turns inside-out relative to the incoming/leaving axis.
Only the supplied group labels are compared here.
solid wedge
A triangular bond that points from the page towards the viewer.
It reports viewing direction, not bond strength.
hashed wedge
A striped triangular bond that points away from the viewer.
It is not a broken bond.

Do

  • Keep A, B and C identities fixed.
  • Replace Br by OH.
  • Compare the towards and away positions.
2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.
The product graph preserves the exact A, B and C atom identities while hydroxyl replaces bromine. The recorded directions show each group's before-and-after position relative to the viewer.

Check

Only the supplied group labels and directions are used.

Investigation 7

Name the surrounding liquid

Read these words and symbols first

solvent
The liquid medium surrounding the reacting particles.
A solvent label supports particle behaviour but never proves a mechanism alone.

Do

Separate reacting species from the surrounding medium, then mark full charges and partial-charge regions in a particle list.

Check

Keep temperature and medium fixed when comparing concentration effects.

Investigation 8

Decode initial-rate particles, brackets and reciprocal units

Read these words and symbols first

concentration
Amount of a named reacting species per litre, written in square brackets and measured here in mol L-1.
[RX] means the concentration of RX
it is not a molecular structure or charge.
initial rate
The amount of product formed per litre per second near the start, before concentrations change greatly.
It is not the total time taken to finish.
rate constant k
The proportionality number that connects concentration to rate at fixed temperature and medium.
Changing temperature or medium can change k.

Do

Translate each negative unit power into 'per' and name the specified species before calculating.

Check

  • Square brackets are concentration symbols, not molecular brackets or charges
  • L⁻¹ means per litre and s⁻¹ means per second.

Investigation 9

Derive one exponent at a time

Read these words and symbols first

concentration
Amount of a named reacting species per litre, written in square brackets and measured here in mol L-1.
[RX] means the concentration of RX
it is not a molecular structure or charge.
initial rate
The amount of product formed per litre per second near the start, before concentrations change greatly.
It is not the total time taken to finish.
rate constant k
The proportionality number that connects concentration to rate at fixed temperature and medium.
Changing temperature or medium can change k.

Do

Compare rows 1→2 and 1→3 separately, then predict row 4.

Check

Row 4 gives 2 × 3 = 6, independently checking the two first-power factors.

Investigation 10

Separate exponent from step count

Read these words and symbols first

reaction order
A concentration exponent inferred from controlled rate data.
It is not read from the balanced equation unless elementary-event evidence is independently supplied.

Do

Write the exponent beside each species, then add only after deriving each one.

Check

Second order here means overall concentration order two, not two reaction steps.

Investigation 11

Count species in one elementary event

Read these words and symbols first

elementary event
One proposed molecular event with no smaller event written inside it.
One event can contain more than one electron-pair arrow.
molecularity
The number of reacting species in one elementary event.
It is not the number of arrows or the number of products.

Do

Count reacting species in the event, not arrows or products.

Electron-pair mechanism

One coordinated backside substitution event

Can both electron-pair moves occur in one event without a stable five-bond carbon?

BeforeHydroxide plus labelled 2-bromobutaneBoth electron-pair moves belong to this one elementary event. The highest-energy arrangement has partial C-O and C-Br bonds; it is not a stable five-bond carbon.
Complete expanded labelled 2-bromobutane and hydroxide before the one-step event; every carbon, hydrogen, oxygen and bromine is present.CCCCBrO−HHHHHHHHHH12
  • OH approaches opposite Br; A/B/C labels track the unchanged groups.
  1. Arrow 1
    Pair starts atone hydroxide oxygen lone pairPair ends atreacting carbon opposite bromine
    Why
    Backside access aligns the entering pair opposite the bond being displaced.
    Immediate effect
    The C-O bond begins forming.
  2. Arrow 2
    Pair starts atC-Br bonding pairPair ends atbromine
    Why
    The old pair leaves carbon in the same elementary event, preventing a stable five-bond carbon.
    Immediate effect
    C-Br breaks and bromide forms.
After both arrowsLabelled 2-butanol plus bromide
Complete expanded labelled 2-butanol and bromide after the event. The same atoms remain; C-O replaces C-Br and the A/B/C tetrahedral arrangement is inverted.CCCCBr−OHHHHHHHHHH
Three compulsory checks
Atoms
The same four C, ten H, one O and one Br identities occur before and after.
Charge
Total formal charge is -1 before and -1 after.
Carbon valence
Reactant and product carbon c2 each have four full single bonds; the transition state uses two partial bonds and is not a stable five-bond carbon.

Hydroxide approaches the labelled reacting carbon opposite bromine. In one elementary event an oxygen lone pair forms C-O while the C-Br pair moves to bromine; the labelled tetrahedral arrangement inverts and no carbocation intermediate exists.

Check

Two arrows can belong to one bimolecular elementary event.

Investigation 12

Decode SN2 only after evidence

Read these words and symbols first

molecularity
The number of reacting species in one elementary event.
It is not the number of arrows or the number of products.

Do

Point from each letter/number to the already-established evidence.

Check

The 2 is not a universal step count.

Investigation 13

Read topology before labels

Read these words and symbols first

transition state
The proposed highest-energy arrangement along one event, marked with a double dagger.
It is not a stable intermediate and cannot be isolated.
activation-energy barrier
The vertical energy rise from the reactant level to a transition-state maximum.
It is an energy difference, not elapsed time.

Do

Count maxima and internal minima before naming any state.

Check

The rise from reactants to the maximum is the activation-energy barrier.

Investigation 14

Cancel units instead of memorising

Read these words and symbols first

concentration
Amount of a named reacting species per litre, written in square brackets and measured here in mol L-1.
[RX] means the concentration of RX
it is not a molecular structure or charge.
rate constant k
The proportionality number that connects concentration to rate at fixed temperature and medium.
Changing temperature or medium can change k.

Do

Cancel one mol and one L-1 factor at a time.

Check

Substituting the result back into k[A][B] returns mol L-1 s-1.

Investigation 15

Use controlled support factors

Read these words and symbols first

polar aprotic solvent
A polar medium without an O-H or N-H proton-donor bond in the solvent molecule.
It can support an available anionic donor but does not guarantee SN2.

Do

For each claim, write changed variable, held constants, direction, cause and limitation.

Methyl bromide; the reacting carbon has no alkyl group crowding the backside corridor.
Primary bromoethane; one carbon group is attached to the bromine-bearing carbon.
Secondary 2-bromopropane; two carbon groups crowd the reacting carbon.
Tertiary bromide; three carbon groups block the ordinary backside corridor.

Check

Neopentyl bromide is the counter-pattern: primary local class but poor access because the adjacent carbon is highly branched.

Investigation 16

Require independent agreement

Read these words and symbols first

backside approach
Approach to the reacting carbon from the side opposite the leaving group.
Backside is defined by the C-X axis, not by left or right on the screen.
concerted
Bond making and bond breaking belong to the same elementary event.
It does not mean a stable five-bond carbon exists.
transition state
The proposed highest-energy arrangement along one event, marked with a double dagger.
It is not a stable intermediate and cannot be isolated.
inversion
The tracked tetrahedral arrangement turns inside-out relative to the incoming/leaving axis.
Only the supplied group labels are compared here.

Do

Build the applicable evidence rows and reject a conclusion if any supplied row contradicts it.

2-bromobutane graph with reacting carbon C star, explicit tracked H and exact A, B and C atom labels around the tetrahedral carbon; Br defines the leaving side.
The product graph preserves the exact A, B and C atom identities while hydroxyl replaces bromine. The recorded directions show each group's before-and-after position relative to the viewer.
Electron-pair mechanism

One coordinated backside substitution event

Can both electron-pair moves occur in one event without a stable five-bond carbon?

BeforeHydroxide plus labelled 2-bromobutaneBoth electron-pair moves belong to this one elementary event. The highest-energy arrangement has partial C-O and C-Br bonds; it is not a stable five-bond carbon.
Complete expanded labelled 2-bromobutane and hydroxide before the one-step event; every carbon, hydrogen, oxygen and bromine is present.CCCCBrO−HHHHHHHHHH12
  • OH approaches opposite Br; A/B/C labels track the unchanged groups.
  1. Arrow 1
    Pair starts atone hydroxide oxygen lone pairPair ends atreacting carbon opposite bromine
    Why
    Backside access aligns the entering pair opposite the bond being displaced.
    Immediate effect
    The C-O bond begins forming.
  2. Arrow 2
    Pair starts atC-Br bonding pairPair ends atbromine
    Why
    The old pair leaves carbon in the same elementary event, preventing a stable five-bond carbon.
    Immediate effect
    C-Br breaks and bromide forms.
After both arrowsLabelled 2-butanol plus bromide
Complete expanded labelled 2-butanol and bromide after the event. The same atoms remain; C-O replaces C-Br and the A/B/C tetrahedral arrangement is inverted.CCCCBr−OHHHHHHHHHH
Three compulsory checks
Atoms
The same four C, ten H, one O and one Br identities occur before and after.
Charge
Total formal charge is -1 before and -1 after.
Carbon valence
Reactant and product carbon c2 each have four full single bonds; the transition state uses two partial bonds and is not a stable five-bond carbon.

Hydroxide approaches the labelled reacting carbon opposite bromine. In one elementary event an oxygen lone pair forms C-O while the C-Br pair moves to bromine; the labelled tetrahedral arrangement inverts and no carbocation intermediate exists.

Check

  • The labelled secondary case has four agreeing rows
  • the primary CH2 cyanide case uses the three observable rows and does not invent inversion.
4 · MeaningOpen

Build the meaning

Meaning 1

Decode the viewer key

  • A plain line lies approximately in the page
  • a solid wedge points towards you
  • a hashed wedge points away.

Check

Rotating the entire molecule is allowed only if every group rotates together.

Meaning 2

Track the same three groups

  • Temporary labels A, B and C identify three groups bonded to the reacting carbon.
  • The labels follow the same atoms before and after reaction.
  • They are not IUPAC locants.

Check

A is CH3, B is H and C is CH2CH3 in both states.

Meaning 3

Locate the open approach line

Backside approach means that the donating atom approaches the reacting carbon opposite Br along the O-C-Br axis.

Check

The destination is carbon, not Br, and backside is not page-left by definition.

Meaning 4

Coordinate the two pair moves

In the one-step model, O-pair-to-C bond formation and C-Br-pair-to-Br cleavage belong to the same elementary event.

Check

No separate positively charged carbon species—later called a carbocation—appears in this supplied one-maximum model, and carbon is never drawn with five full bonds.

Meaning 5

Replace five complete bonds with two partial bonds

  • At the energy maximum, C-O is partly formed and C-Br is partly broken.
  • Brackets with a double dagger mark this transition state.
  • The complete bracketed arrangement has overall charge −1.

Check

  • The same drawing contains one forming partial bond and one breaking partial bond.
  • It cannot be isolated.

Meaning 6

Compare the labelled product

  • Backside approach reverses the A/B/C tetrahedral arrangement relative to the incoming and leaving axis.
  • This tracked change is called inversion.

Check

Only the supplied group labels and directions are used.

Meaning 7

Name the surrounding liquid

  • A solvent is the liquid medium around reacting particles.
  • Attractions between the medium and full or partial charges can change how available a donor pair is, but solvent evidence alone never identifies the pathway.

Check

Keep temperature and medium fixed when comparing concentration effects.

Meaning 8

Decode initial-rate particles, brackets and reciprocal units

  • One mole means 6.022 × 10^23 specified particles. [substrate] and [nucleophile] mean amount per litre: mol L⁻¹ means mol per litre.
  • Initial rate is measured near the start
  • mol L⁻¹ s⁻¹ means mol per litre per second.

Check

  • Square brackets are concentration symbols, not molecular brackets or charges
  • L⁻¹ means per litre and s⁻¹ means per second.

Meaning 9

Derive one exponent at a time

  • When [substrate] doubles at fixed nucleophile, rate doubles
  • when [nucleophile] triples at fixed substrate, rate triples.
  • Therefore rate = k[substrate][nucleophile].

Check

Row 4 gives 2 × 3 = 6, independently checking the two first-power factors.

Meaning 10

Separate exponent from step count

  • Reaction order is the concentration exponent inferred from controlled data.
  • Each exponent is 1 here
  • overall order is 1 + 1 = 2.

Check

Second order here means overall concentration order two, not two reaction steps.

Meaning 11

Count species in one elementary event

  • An elementary event is one modelled event.
  • Molecularity counts its reacting species
  • hydroxide and the haloalkane make this event bimolecular.

Check

Two arrows can belong to one bimolecular elementary event.

Meaning 12

Decode SN2 only after evidence

S means substitution, N means nucleophilic and 2 records two reacting species in the taught elementary event.

Check

The 2 is not a universal step count.

Meaning 13

Read topology before labels

  • Potential energy is vertical
  • reaction progress, not time, is horizontal.
  • One maximum is one transition state
  • no minimum between maxima means no intermediate.

Check

The rise from reactants to the maximum is the activation-energy barrier.

Meaning 14

Cancel units instead of memorising

For rate = k[A][B], k = (mol L-1 s-1)/((mol L-1)(mol L-1)) = L mol-1 s-1.

Check

Substituting the result back into k[A][B] returns mol L-1 s-1.

Meaning 15

Use controlled support factors

  • Matched SN2 support: less backside crowding
  • a more available donor pair
  • leaving-group order I > Br > Cl >> F
  • and bounded polar-aprotic support.
  • R-F is the slow boundary because C-F is very strong and F− is a poor departing ion in this matched set.

Check

Neopentyl bromide is the counter-pattern: primary local class but poor access because the adjacent carbon is highly branched.

Meaning 16

Require independent agreement

  • Select the bounded SN2 model when the coordinated backside event, two-species first-power rate dependence and one-maximum/no-intermediate evidence agree.
  • When the reacting tetrahedral carbon has three distinguishable unchanged groups, labelled inversion must also agree
  • CH2 has two identical H atoms, so no inversion observation is available there.

Check

  • The labelled secondary case has four agreeing rows
  • the primary CH2 cyanide case uses the three observable rows and does not invent inversion.
5 · ExamplesOpen

Worked examples

Example 1 of 7

Derive the rate law from controlled rows

Use the relative-rate table.

Key evidence

  • compare 1→2
  • compare 1→3
  • predict row 4

One lawful move at a time

One move at a time

Step 1 of 3
Action
Double substrate only
Why
rate doubles
Result
substrate exponent 1

Answer

  • rate = k[substrate][nucleophile]
  • overall order 2

Independent check

Row 4 equals 6 as predicted.

Counter-pattern

Writing the law before comparing rows hides which evidence established each exponent.

6 · QuestionsOpen

Questions

Question 1 of 14

Infer each concentration exponent one variable at a time and verify the combined row.

  • Relative initial-rate rows are ([RX],[Nu],rate): (1,1,1), (2,1,2), (1,3,3), (2,3,6).
  • Which law is derived?

Evidence supplied with this Question

Controlled initial-rate evidence
RunRelative [RX]Relative [Nu]Relative initial rate
r1111
r2212
r3133
r4236

Held fixed: temperature, solvent medium, substrate identity, nucleophile identity.

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