CHM 219 Module 4 TEST 2025/2026 ACTUAL COMPREHENSIVE
REAL EXAM QUESTIONS AND VERIFIED ACCURATE
SOLUTION ( CORRECT DETAILED & ELABORATED) |
GET IT 100% ACCURATE!! 2025 TEST!!
Two Main Types of Nucleophilic Substitution REactions -
ANSWER-Sn1 and
Sn2
Sn2 - ANSWER--single step (concerted)
-nucleophile attacks from the backside of the C-L bond.
Nucleophile uses a lone pair of electrons to begin making a
bond to the carbon. Simultaneously the bond between the C
and the L begins to break. "Backside attack"
-Transition state: nucleophile and leaving group are both partly
attached to substrate.
-L departs from the substrate with the pair of electrons from the
C-L bond and nucleophile is fully bonded to the C
-Inversion of the tetrahedral geometry occurs.
*increasing the concentration of either the nucleophile or
substrate will increase the rate of the reaction
*2 is used becuae nucleophile and substrate are involved in the
only step (key step)
, *methyl and primary alkyl halide react this way more easily
FActors that influence the mechanistic path... of Sn2 -
ANSWER--Strength of the nucleophile - SN2 depends on a
strong Nucleophile, the more negative, the stronger.
-Size of nucleophile - larger are more efficient
-Electronegativity of the Nucleophilic atom - the more
electronegative, the less nucleophilic
-Solvent chosen - they are polar and require polar solvents ;
best solvents are polar aprotic solvents
Sn1 - ANSWER--two steps
-Step 1 - Leaving group leaves first. C-L bond breaks and
produces 2 ionic products: carbocation intermediate and a
leaving group anion (this step is energy intensive and very
slow) known as the rate determining step (RDS) -Step 2 -
nucleophile attacks the carbocation intermediate, using a lone
pair to form a covalent bond.
*no inversion of tetrahedral geometry
*produces 50:50 enantiomers.
*secondary and tertiary typically react this way
REAL EXAM QUESTIONS AND VERIFIED ACCURATE
SOLUTION ( CORRECT DETAILED & ELABORATED) |
GET IT 100% ACCURATE!! 2025 TEST!!
Two Main Types of Nucleophilic Substitution REactions -
ANSWER-Sn1 and
Sn2
Sn2 - ANSWER--single step (concerted)
-nucleophile attacks from the backside of the C-L bond.
Nucleophile uses a lone pair of electrons to begin making a
bond to the carbon. Simultaneously the bond between the C
and the L begins to break. "Backside attack"
-Transition state: nucleophile and leaving group are both partly
attached to substrate.
-L departs from the substrate with the pair of electrons from the
C-L bond and nucleophile is fully bonded to the C
-Inversion of the tetrahedral geometry occurs.
*increasing the concentration of either the nucleophile or
substrate will increase the rate of the reaction
*2 is used becuae nucleophile and substrate are involved in the
only step (key step)
, *methyl and primary alkyl halide react this way more easily
FActors that influence the mechanistic path... of Sn2 -
ANSWER--Strength of the nucleophile - SN2 depends on a
strong Nucleophile, the more negative, the stronger.
-Size of nucleophile - larger are more efficient
-Electronegativity of the Nucleophilic atom - the more
electronegative, the less nucleophilic
-Solvent chosen - they are polar and require polar solvents ;
best solvents are polar aprotic solvents
Sn1 - ANSWER--two steps
-Step 1 - Leaving group leaves first. C-L bond breaks and
produces 2 ionic products: carbocation intermediate and a
leaving group anion (this step is energy intensive and very
slow) known as the rate determining step (RDS) -Step 2 -
nucleophile attacks the carbocation intermediate, using a lone
pair to form a covalent bond.
*no inversion of tetrahedral geometry
*produces 50:50 enantiomers.
*secondary and tertiary typically react this way
