Learning Arc 03 · Choose elimination and carbon-construction routes

Choose substitution or elimination

  • The same haloalkane can have more than one structurally possible product route.
  • A dependable answer connects the substrate, electron-pair destination, complete conditions and product bond changes.
  • This prevents one-word reagent guessing.
Foundation for this Concept23 symbols · 0 recovery lessons · 3 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

aqueous, aq
Water-rich reaction medium.
The abbreviation aq means aqueous.
Aqueous is a condition label, not proof of one detailed mechanism.
alcoholic, alc
KOH is used in ethanol, or in another alcohol solvent named by the question.
The abbreviation alc means alcoholic.
Alcoholic names the surrounding liquid
it does not identify the only reacting oxygen species.
heat, Δ
The reaction mixture is kept at an increased temperature.
The symbol Δ above an arrow means heat is supplied.
Heat supports a pathway comparison
it does not create a missing beta hydrogen.
favours
The stated evidence makes one route the main expected route in this Board comparison.
Favours does not mean that every competing pathway is completely absent.
reaction pathway
A connected account of how reactant bonds and electron pairs become product bonds and electron pairs.
A condition word alone does not prove the detailed timing of a pathway.
nucleophile role
An electron-pair donor performs the nucleophile role when its pair forms a bond to carbon.
Name the donating atom and the carbon destination.
base role
An electron-pair donor performs the base role when its pair forms a bond to hydrogen.
The arrow starts at a lone pair and ends at the selected hydrogen.
electron-pair destination
The atom or bond region that receives the electron pair at the arrow head.
The destination, not the reagent name alone, identifies the local role.
substitution
One atom or group bonded to carbon is replaced by another atom or group.
The carbon skeleton remains unchanged in the bounded examples here.
beta elimination
A beta hydrogen and the halogen leave adjacent carbons while the carbon-carbon bond becomes double.
The substrate must already contain an eligible beta hydrogen.
competing pathways
The same starting material can have more than one structurally possible reaction route.
The exact product ratio requires evidence beyond a single condition word.
KOH
KOH is potassium hydroxide.
In this simplified solution model it separates into K+ and OH−.
The formula KOH names the reagent
solvent and temperature must be read separately.
K+
K+ is a potassium ion with charge plus one.
K+ does not change a bond in the teaching frames here.
OH−
OH− is a hydroxide ion.
Its oxygen has a lone pair that can be donated.
The destination of the donated pair decides the local role.
Br−
Br− is a bromide ion.
It is formed when the C-Br bonding pair ends on bromine.
Br− has charge minus one.
ion
An ion is an atom or group of bonded atoms with an overall electric charge.
A plus or minus sign states total charge
it is not an electron-pair arrow.
spectator ion
A spectator ion is present for charge balance but does not change bonds in the reaction step being tracked.
K+ is the spectator ion in the teaching frames here.
lone pair
A lone pair is a pair of outer electrons located on one atom and not already forming a bond.
A curved electron-flow arrow starts at a lone pair or a bond that owns the moving pair.
condition strip
The reagent, solvent and temperature written together around a reaction arrow.
Read every field
do not infer a missing field.
bond ledger
A short list of bonds made, bonds broken and bond orders changed.
Only list changes supported by the reactant and product structures.
structural gate
A check that the starting structure contains every bond required by a proposed route.
Beta elimination fails this gate when no beta C-H bond exists.
product family
A group of products with the same key functional feature, such as an alcohol or an alkene.
A family name does not replace the exact product structure.
teaching electron-flow frame
A simplified structure-to-structure account that conserves atoms, charge, valence and electron pairs.
It accounts for the net bond changes but does not by itself prove whether all moves occur at exactly the same instant.

Entry check 1

For 2-bromopropane, name the substitution bond ledger.

Show the recovery
  • C-O is made.
  • C-Br is broken.
  • The carbon skeleton remains.

Return to Arc 2 Concept 2.6 for the complete substitution evidence comparison.

Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.

Entry check 2

For 2-bromopropane, identify Cα, one Cβ and one Hβ.

Show the recovery
  • C2 is Cα.
  • Either terminal carbon is Cβ.
  • A hydrogen on that terminal carbon is Hβ.

Return to Concept 3.1 for alpha-beta mapping and hidden-hydrogen restoration.

Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.

Entry check 3

State what a solvent medium describes.

Show the recovery

It describes the surrounding liquid environment in which the reacting particles are present.

Return to Arc 2 Concept 2.2 for the solvent-medium bridge.

1 · ObjectiveOpen

What you will understand

Use visible bond changes, electron-pair destination, substrate structure, solvent and temperature to choose the main expected Board route without treating one condition as proof.

Problems this Concept unlocks

  • decode aqueous, alcoholic and heat conditions
  • distinguish nucleophile and base roles from electron-pair destination
  • compare substitution and elimination bond ledgers
  • apply the beta-hydrogen gate before a condition rule
  • write a complete Maharashtra HSC condition comparison
  • repair absolute pathway claims
2 · MysteryOpen

The problem to explain

Read these words and symbols first

KOH
KOH is potassium hydroxide.
In this simplified solution model it separates into K+ and OH−.
The formula KOH names the reagent
solvent and temperature must be read separately.
aqueous, aq
Water-rich reaction medium.
The abbreviation aq means aqueous.
Aqueous is a condition label, not proof of one detailed mechanism.
alcoholic, alc
KOH is used in ethanol, or in another alcohol solvent named by the question.
The abbreviation alc means alcoholic.
Alcoholic names the surrounding liquid
it does not identify the only reacting oxygen species.
heat, Δ
The reaction mixture is kept at an increased temperature.
The symbol Δ above an arrow means heat is supplied.
Heat supports a pathway comparison
it does not create a missing beta hydrogen.
reaction pathway
A connected account of how reactant bonds and electron pairs become product bonds and electron pairs.
A condition word alone does not prove the detailed timing of a pathway.
nucleophile role
An electron-pair donor performs the nucleophile role when its pair forms a bond to carbon.
Name the donating atom and the carbon destination.
base role
An electron-pair donor performs the base role when its pair forms a bond to hydrogen.
The arrow starts at a lone pair and ends at the selected hydrogen.
substitution
One atom or group bonded to carbon is replaced by another atom or group.
The carbon skeleton remains unchanged in the bounded examples here.
beta elimination
A beta hydrogen and the halogen leave adjacent carbons while the carbon-carbon bond becomes double.
The substrate must already contain an eligible beta hydrogen.
competing pathways
The same starting material can have more than one structurally possible reaction route.
The exact product ratio requires evidence beyond a single condition word.
bond ledger
A short list of bonds made, bonds broken and bond orders changed.
Only list changes supported by the reactant and product structures.
condition strip
The reagent, solvent and temperature written together around a reaction arrow.
Read every field
do not infer a missing field.
  • Both rows begin with 2-bromopropane and KOH.
  • The first row uses aqueous KOH and forms propan-2-ol.
  • The second row uses alcoholic KOH with heat and forms propene.
Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.
Fully expanded propene. Every hydrogen is shown. C1=C2 is double and C2-C3 is single.
  • The products have different bonds.
  • Which bond changes, electron-pair destination and conditions support each product?

Your decision

  • Write both product bond ledgers.
  • Find where the oxygen pair must go in each route.
  • Then read every condition word.
3 · InvestigationOpen

Investigation

Investigation 1

Compare the bond changes before naming the role

Read these words and symbols first

nucleophile role
An electron-pair donor performs the nucleophile role when its pair forms a bond to carbon.
Name the donating atom and the carbon destination.
base role
An electron-pair donor performs the base role when its pair forms a bond to hydrogen.
The arrow starts at a lone pair and ends at the selected hydrogen.
electron-pair destination
The atom or bond region that receives the electron pair at the arrow head.
The destination, not the reagent name alone, identifies the local role.
substitution
One atom or group bonded to carbon is replaced by another atom or group.
The carbon skeleton remains unchanged in the bounded examples here.
beta elimination
A beta hydrogen and the halogen leave adjacent carbons while the carbon-carbon bond becomes double.
The substrate must already contain an eligible beta hydrogen.
lone pair
A lone pair is a pair of outer electrons located on one atom and not already forming a bond.
A curved electron-flow arrow starts at a lone pair or a bond that owns the moving pair.
teaching electron-flow frame
A simplified structure-to-structure account that conserves atoms, charge, valence and electron pairs.
It accounts for the net bond changes but does not by itself prove whether all moves occur at exactly the same instant.

Do

  • Compare each displayed reactant with its product.
  • Write the new bond in each route.
  • Mark the atom at the head of the oxygen arrow.
Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Hydroxide ion. Oxygen is bonded to one hydrogen, carries formal charge minus one and has three lone pairs.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.
Fully expanded propene. Every hydrogen is shown. C1=C2 is double and C2-C3 is single.
Electron-pair mechanism

Oxygen donor used as a nucleophile

Where does the oxygen lone pair go in the substitution route?

Before2-bromopropane, K+ and hydroxideThis teaching frame accounts for C-O formation and C-Br cleavage. It does not prove whether the two electron-pair moves occur simultaneously or in separate steps.
Complete expanded 2-bromopropane, hydroxide and potassium ion. C2 is bonded to C1, C3, H4 and Br. Hydroxide oxygen has three lone pairs and charge minus one. K+ has charge plus one and changes no bond.CCCBrHHHHHHHO−HK+12
  1. Arrow 1
    Pair starts atoxygen lone pairPair ends atC2 bonded to Br
    Why
    The oxygen pair is the supplied donor and C2 is the carbon destination in substitution.
    Immediate effect
    A C2-O bond forms.
  2. Arrow 2
    Pair starts atC2-Br bonding pairPair ends atBr
    Why
    C2 cannot keep C-Br after gaining C-O without exceeding ordinary valence.
    Immediate effect
    C2-Br breaks and Br− forms.
After both arrowspropan-2-ol, K+ and bromide
Complete propan-2-ol, bromide and potassium ion. C2-O has replaced C2-Br. Every carbon-hydrogen bond and O-H remain. Br− and K+ balance each other. K+ changes no bond.CCCBr−HHHHHHHOHK+
Three compulsory checks
Atoms
The same three C, eight H, one O, one Br and one K atom identities appear before and after.
Charge
Before, K+ and OH− give total charge zero. After, K+ and Br− give total charge zero.
Carbon valence
C2 loses C-Br as it gains C-O and remains at four bond-order units.

The oxygen lone pair goes to C2. C-O forms as the C2-Br pair goes to bromine. K+ remains a spectator ion. Atom identities, total charge and carbon valence are conserved.

Electron-pair mechanism

Oxygen donor used as a base

Where does the oxygen lone pair go in the beta-elimination route?

Before2-bromopropane, K+ and a representative oxygen-containing baseThe three arrows form one teaching electron-pair accounting frame. This frame accounts for the net bond changes. It does not prove that every substrate and condition follows one universal detailed mechanism.
Complete 2-bromopropane, representative oxygen-containing base and potassium ion. C2 is alpha because it bears Br. C1 is a beta carbon and H1 is the selected beta hydrogen. K+ has charge plus one and changes no bond.CCCBrHHHHHHHO−HK+123
  1. Arrow 1
    Pair starts atoxygen lone pairPair ends atbeta hydrogen H1
    Why
    The oxygen donor performs the base role because its pair is directed to hydrogen.
    Immediate effect
    An O-H1 bond forms.
  2. Arrow 2
    Pair starts atC1-H1 bonding pairPair ends atC1-C2 bond region
    Why
    The pair released from C1-H1 supplies the second pair between C1 and C2.
    Immediate effect
    C1-C2 changes from single to double.
  3. Arrow 3
    Pair starts atC2-Br bonding pairPair ends atBr
    Why
    C2 must lose C-Br while C1=C2 forms so ordinary carbon valence is retained.
    Immediate effect
    C2-Br breaks and Br− forms.
After both arrowspropene, water, K+ and bromide
Complete propene, water, bromide and potassium ion. C1=C2 is double. H1 is now bonded to oxygen. C2-Br is absent. Br− and K+ balance each other. K+ changes no bond.CCCBr−HHHHHHHOHK+
Three compulsory checks
Atoms
The same three C, eight H, one O, one Br and one K atom identities appear before and after.
Charge
Before, K+ and the oxygen-containing anion give total charge zero. After, K+ and Br− give total charge zero.
Carbon valence
C1 loses C1-H1 as C1-C2 becomes double. C2 loses C2-Br as C1-C2 becomes double. Both carbons remain at four bond-order units.

The oxygen lone pair goes to beta hydrogen H1. The C1-H1 pair forms the second bond of C1=C2 while the C2-Br pair goes to bromine. K+ remains a spectator ion. Propene, water, K+ and bromide conserve atoms, charge and carbon valence.

Check

  • One electron-pair arrow has one source.
  • One electron-pair arrow has one destination.
  • The same oxygen-containing donor can perform different roles in different routes.

Investigation 2

Decode the complete condition strip

Read these words and symbols first

KOH
KOH is potassium hydroxide.
In this simplified solution model it separates into K+ and OH−.
The formula KOH names the reagent
solvent and temperature must be read separately.
aqueous, aq
Water-rich reaction medium.
The abbreviation aq means aqueous.
Aqueous is a condition label, not proof of one detailed mechanism.
alcoholic, alc
KOH is used in ethanol, or in another alcohol solvent named by the question.
The abbreviation alc means alcoholic.
Alcoholic names the surrounding liquid
it does not identify the only reacting oxygen species.
heat, Δ
The reaction mixture is kept at an increased temperature.
The symbol Δ above an arrow means heat is supplied.
Heat supports a pathway comparison
it does not create a missing beta hydrogen.
favours
The stated evidence makes one route the main expected route in this Board comparison.
Favours does not mean that every competing pathway is completely absent.
condition strip
The reagent, solvent and temperature written together around a reaction arrow.
Read every field
do not infer a missing field.
substitution
One atom or group bonded to carbon is replaced by another atom or group.
The carbon skeleton remains unchanged in the bounded examples here.
beta elimination
A beta hydrogen and the halogen leave adjacent carbons while the carbon-carbon bond becomes double.
The substrate must already contain an eligible beta hydrogen.

Do

  • Copy aq, alc and Δ from the displayed condition strips.
  • Write the everyday meaning of each printed symbol.
  • Do not choose a product yet.
Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.
Fully expanded propene. Every hydrogen is shown. C1=C2 is double and C2-C3 is single.

Check

  • A condition strip supports a main expected route only after the substrate passes the structural gate.
  • Favours does not mean only product.

Investigation 3

Run the complete comparison checklist

Read these words and symbols first

aqueous, aq
Water-rich reaction medium.
The abbreviation aq means aqueous.
Aqueous is a condition label, not proof of one detailed mechanism.
alcoholic, alc
KOH is used in ethanol, or in another alcohol solvent named by the question.
The abbreviation alc means alcoholic.
Alcoholic names the surrounding liquid
it does not identify the only reacting oxygen species.
heat, Δ
The reaction mixture is kept at an increased temperature.
The symbol Δ above an arrow means heat is supplied.
Heat supports a pathway comparison
it does not create a missing beta hydrogen.
favours
The stated evidence makes one route the main expected route in this Board comparison.
Favours does not mean that every competing pathway is completely absent.
reaction pathway
A connected account of how reactant bonds and electron pairs become product bonds and electron pairs.
A condition word alone does not prove the detailed timing of a pathway.
nucleophile role
An electron-pair donor performs the nucleophile role when its pair forms a bond to carbon.
Name the donating atom and the carbon destination.
base role
An electron-pair donor performs the base role when its pair forms a bond to hydrogen.
The arrow starts at a lone pair and ends at the selected hydrogen.
electron-pair destination
The atom or bond region that receives the electron pair at the arrow head.
The destination, not the reagent name alone, identifies the local role.
substitution
One atom or group bonded to carbon is replaced by another atom or group.
The carbon skeleton remains unchanged in the bounded examples here.
beta elimination
A beta hydrogen and the halogen leave adjacent carbons while the carbon-carbon bond becomes double.
The substrate must already contain an eligible beta hydrogen.
competing pathways
The same starting material can have more than one structurally possible reaction route.
The exact product ratio requires evidence beyond a single condition word.
KOH
KOH is potassium hydroxide.
In this simplified solution model it separates into K+ and OH−.
The formula KOH names the reagent
solvent and temperature must be read separately.
K+
K+ is a potassium ion with charge plus one.
K+ does not change a bond in the teaching frames here.
OH−
OH− is a hydroxide ion.
Its oxygen has a lone pair that can be donated.
The destination of the donated pair decides the local role.
Br−
Br− is a bromide ion.
It is formed when the C-Br bonding pair ends on bromine.
Br− has charge minus one.
ion
An ion is an atom or group of bonded atoms with an overall electric charge.
A plus or minus sign states total charge
it is not an electron-pair arrow.
spectator ion
A spectator ion is present for charge balance but does not change bonds in the reaction step being tracked.
K+ is the spectator ion in the teaching frames here.
lone pair
A lone pair is a pair of outer electrons located on one atom and not already forming a bond.
A curved electron-flow arrow starts at a lone pair or a bond that owns the moving pair.
condition strip
The reagent, solvent and temperature written together around a reaction arrow.
Read every field
do not infer a missing field.
bond ledger
A short list of bonds made, bonds broken and bond orders changed.
Only list changes supported by the reactant and product structures.
structural gate
A check that the starting structure contains every bond required by a proposed route.
Beta elimination fails this gate when no beta C-H bond exists.
product family
A group of products with the same key functional feature, such as an alcohol or an alkene.
A family name does not replace the exact product structure.
teaching electron-flow frame
A simplified structure-to-structure account that conserves atoms, charge, valence and electron pairs.
It accounts for the net bond changes but does not by itself prove whether all moves occur at exactly the same instant.

Do

  • For each displayed row, list the starting bonds, possible product bonds, arrow destination, solvent and heat.
  • Leave the final route choice blank until every field is recorded.
Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.
Fully expanded propene. Every hydrogen is shown. C1=C2 is double and C2-C3 is single.
Fully expanded 1-bromo-2,2-dimethylpropane. Every hydrogen is shown. Br is bonded to C1. The only beta carbon C2 has four carbon neighbours and no hydrogen.
Electron-pair mechanism

Oxygen donor used as a nucleophile

Where does the oxygen lone pair go in the substitution route?

Before2-bromopropane, K+ and hydroxideThis teaching frame accounts for C-O formation and C-Br cleavage. It does not prove whether the two electron-pair moves occur simultaneously or in separate steps.
Complete expanded 2-bromopropane, hydroxide and potassium ion. C2 is bonded to C1, C3, H4 and Br. Hydroxide oxygen has three lone pairs and charge minus one. K+ has charge plus one and changes no bond.CCCBrHHHHHHHO−HK+12
  1. Arrow 1
    Pair starts atoxygen lone pairPair ends atC2 bonded to Br
    Why
    The oxygen pair is the supplied donor and C2 is the carbon destination in substitution.
    Immediate effect
    A C2-O bond forms.
  2. Arrow 2
    Pair starts atC2-Br bonding pairPair ends atBr
    Why
    C2 cannot keep C-Br after gaining C-O without exceeding ordinary valence.
    Immediate effect
    C2-Br breaks and Br− forms.
After both arrowspropan-2-ol, K+ and bromide
Complete propan-2-ol, bromide and potassium ion. C2-O has replaced C2-Br. Every carbon-hydrogen bond and O-H remain. Br− and K+ balance each other. K+ changes no bond.CCCBr−HHHHHHHOHK+
Three compulsory checks
Atoms
The same three C, eight H, one O, one Br and one K atom identities appear before and after.
Charge
Before, K+ and OH− give total charge zero. After, K+ and Br− give total charge zero.
Carbon valence
C2 loses C-Br as it gains C-O and remains at four bond-order units.

The oxygen lone pair goes to C2. C-O forms as the C2-Br pair goes to bromine. K+ remains a spectator ion. Atom identities, total charge and carbon valence are conserved.

Electron-pair mechanism

Oxygen donor used as a base

Where does the oxygen lone pair go in the beta-elimination route?

Before2-bromopropane, K+ and a representative oxygen-containing baseThe three arrows form one teaching electron-pair accounting frame. This frame accounts for the net bond changes. It does not prove that every substrate and condition follows one universal detailed mechanism.
Complete 2-bromopropane, representative oxygen-containing base and potassium ion. C2 is alpha because it bears Br. C1 is a beta carbon and H1 is the selected beta hydrogen. K+ has charge plus one and changes no bond.CCCBrHHHHHHHO−HK+123
  1. Arrow 1
    Pair starts atoxygen lone pairPair ends atbeta hydrogen H1
    Why
    The oxygen donor performs the base role because its pair is directed to hydrogen.
    Immediate effect
    An O-H1 bond forms.
  2. Arrow 2
    Pair starts atC1-H1 bonding pairPair ends atC1-C2 bond region
    Why
    The pair released from C1-H1 supplies the second pair between C1 and C2.
    Immediate effect
    C1-C2 changes from single to double.
  3. Arrow 3
    Pair starts atC2-Br bonding pairPair ends atBr
    Why
    C2 must lose C-Br while C1=C2 forms so ordinary carbon valence is retained.
    Immediate effect
    C2-Br breaks and Br− forms.
After both arrowspropene, water, K+ and bromide
Complete propene, water, bromide and potassium ion. C1=C2 is double. H1 is now bonded to oxygen. C2-Br is absent. Br− and K+ balance each other. K+ changes no bond.CCCBr−HHHHHHHOHK+
Three compulsory checks
Atoms
The same three C, eight H, one O, one Br and one K atom identities appear before and after.
Charge
Before, K+ and the oxygen-containing anion give total charge zero. After, K+ and Br− give total charge zero.
Carbon valence
C1 loses C1-H1 as C1-C2 becomes double. C2 loses C2-Br as C1-C2 becomes double. Both carbons remain at four bond-order units.

The oxygen lone pair goes to beta hydrogen H1. The C1-H1 pair forms the second bond of C1=C2 while the C2-Br pair goes to bromine. K+ remains a spectator ion. Propene, water, K+ and bromide conserve atoms, charge and carbon valence.

Check

  • No single condition word proves one universal detailed mechanism.
  • No single condition word proves an exact product ratio.
  • No single condition word proves that every competing route is absent.
4 · MeaningOpen

Build the meaning

Meaning 1

Compare the bond changes before naming the role

  • Substitution makes C2-O.
  • The oxygen pair must end at C2.
  • That is the nucleophile role.
  • Beta elimination makes O-Hβ.
  • The oxygen pair must end at Hβ.
  • That is the base role.

Check

  • One electron-pair arrow has one source.
  • One electron-pair arrow has one destination.
  • The same oxygen-containing donor can perform different roles in different routes.

Meaning 2

Decode the complete condition strip

  • Aqueous KOH is potassium hydroxide in a water-rich medium.
  • Alcoholic KOH is potassium hydroxide in ethanol, or in another named alcohol solvent.
  • Δ means heat is supplied.
  • For this Board comparison, aqueous KOH commonly favours substitution.
  • For this Board comparison, alcoholic KOH with heat commonly favours beta elimination.

Check

  • A condition strip supports a main expected route only after the substrate passes the structural gate.
  • Favours does not mean only product.

Meaning 3

Run the complete comparison checklist

  • First read the substrate bond map.
  • Next generate every structurally possible product.
  • Then identify the electron-pair destination.
  • Then read solvent and heat.
  • Finally state the main expected route and its limit.

Check

  • No single condition word proves one universal detailed mechanism.
  • No single condition word proves an exact product ratio.
  • No single condition word proves that every competing route is absent.
5 · ExamplesOpen

Worked examples

Example 1 of 6

Aqueous route

Predict the main expected Board route for 2-bromopropane with aqueous KOH.

Fully expanded 2-bromopropane. Every hydrogen is shown. C1-C2-C3 is the carbon chain. Br is bonded directly to C2.
Fully expanded propan-2-ol. Every hydrogen is shown. The C1-C2-C3 skeleton remains. O-H is bonded through O to C2.
Electron-pair mechanism

Oxygen donor used as a nucleophile

Where does the oxygen lone pair go in the substitution route?

Before2-bromopropane, K+ and hydroxideThis teaching frame accounts for C-O formation and C-Br cleavage. It does not prove whether the two electron-pair moves occur simultaneously or in separate steps.
Complete expanded 2-bromopropane, hydroxide and potassium ion. C2 is bonded to C1, C3, H4 and Br. Hydroxide oxygen has three lone pairs and charge minus one. K+ has charge plus one and changes no bond.CCCBrHHHHHHHO−HK+12
  1. Arrow 1
    Pair starts atoxygen lone pairPair ends atC2 bonded to Br
    Why
    The oxygen pair is the supplied donor and C2 is the carbon destination in substitution.
    Immediate effect
    A C2-O bond forms.
  2. Arrow 2
    Pair starts atC2-Br bonding pairPair ends atBr
    Why
    C2 cannot keep C-Br after gaining C-O without exceeding ordinary valence.
    Immediate effect
    C2-Br breaks and Br− forms.
After both arrowspropan-2-ol, K+ and bromide
Complete propan-2-ol, bromide and potassium ion. C2-O has replaced C2-Br. Every carbon-hydrogen bond and O-H remain. Br− and K+ balance each other. K+ changes no bond.CCCBr−HHHHHHHOHK+
Three compulsory checks
Atoms
The same three C, eight H, one O, one Br and one K atom identities appear before and after.
Charge
Before, K+ and OH− give total charge zero. After, K+ and Br− give total charge zero.
Carbon valence
C2 loses C-Br as it gains C-O and remains at four bond-order units.

The oxygen lone pair goes to C2. C-O forms as the C2-Br pair goes to bromine. K+ remains a spectator ion. Atom identities, total charge and carbon valence are conserved.

Key evidence

  • structurally possible alcohol
  • water-rich alkali

One lawful move at a time

One move at a time

Step 1 of 4
Action
generate alcohol
Why
The displayed C-O product is structurally possible.
Result
Propan-2-ol is a possible product.

Answer

  • Substitution is favoured.
  • Propan-2-ol is the main expected product.

Independent check

  • C-O replaces C-Br.
  • The three-carbon skeleton remains.

Counter-pattern

  • Aqueous supports the answer.
  • Aqueous alone does not prove SN1 or SN2.
6 · QuestionsOpen

Questions

Question 1 of 14

Translate aq into a water-rich medium and keep the label separate from pathway proof.

What does KOH(aq) tell you before any product is chosen?

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Your selection is not checked or saved.

Select one answer.
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