Organic Chemistry - Nucleophilic Substitution & Elimination Mechanisms Flashcards

Question 1

$S_N2$ Mechanism & Kinetics

Accepted Answer

A concerted, single-step bimolecular substitution reaction where the nucleophile attacks as the leaving group departs. Rate law: $	ext{Rate} = k[	ext{substrate}][	ext{nucleophile}]$.

Question 2

$S_N2$ Stereochemistry (Walden Inversion)

Accepted Answer

Proceeds with $100\%$ inversion of stereochemical configuration at the electrophilic carbon due to back-side attack of the nucleophile opposite the leaving group.

Question 3

Substrate Reactivity Trends in $S_N2$

Accepted Answer

Methyl > $1^\circ$ > $2^\circ$ >> $3^\circ$ (unreactive). Driven by steric hindrance around the electrophilic carbon; unhindered substrates react significantly faster.

Question 4

$S_N1$ Mechanism & Kinetics

Accepted Answer

A step-wise unimolecular substitution. Step 1 (rate-determining): Loss of leaving group to form a carbocation intermediate. Step 2: Nucleophilic attack. Rate law: $	ext{Rate} = k[	ext{substrate}]$.

Question 5

$S_N1$ Stereochemical Outcome

Accepted Answer

Results in racemization (a mixture of inversion and retention) because the nucleophile can attack either face of the planar, $sp^2$-hybridized carbocation intermediate.

Question 6

Substrate Reactivity Trends in $S_N1$

Accepted Answer

$3^\circ$ > $2^\circ$ >> $1^\circ$, methyl (unreactive). Driven by carbocation stability via hyperconjugation and inductive electron donation.

Question 7

$E2$ Mechanism & Kinetics

Accepted Answer

A concerted, single-step bimolecular elimination where a strong base abstracts a $\beta$-hydrogen while the leaving group departs to form a $\pi$ bond. Rate law: $	ext{Rate} = k[	ext{substrate}][	ext{base}]$.

Question 8

$E2$ Anti-Periplanar Requirement

Accepted Answer

Requires an anti-periplanar arrangement (dihedral angle $	heta = 180^\circ$) between the $\beta$-$	ext{H}$ and the leaving group for optimal orbital overlap during double bond formation.

Question 9

$E1$ Mechanism & Kinetics

Accepted Answer

A step-wise unimolecular elimination. Step 1 (rate-determining): Dissociation of leaving group to form a carbocation. Step 2: Deprotonation by a weak base. Rate law: $	ext{Rate} = k[	ext{substrate}]$.

Question 10

Zaitsev's Rule

Accepted Answer

Elimination reactions preferentially yield the most substituted and thermodynamically stable alkene by abstracting a proton from the $\beta$-carbon with fewer hydrogen atoms.

Question 11

Hofmann's Rule

Accepted Answer

In $E2$ eliminations, using a sterically hindered base (e.g., $	ext{KO}t	ext{Bu}$) leads to the least substituted, less hindered alkene as the major product.

Question 12

Polar Protic Solvents

Accepted Answer

Solvents containing $	ext{O-H}$ or $	ext{N-H}$ bonds (e.g., $	ext{H}_2	ext{O}$, $	ext{MeOH}$, $	ext{EtOH}$). They solvate both cations and anions via hydrogen bonding, lowering nucleophilicity and favoring $S_N1$/$E1$ pathways.

Question 13

Polar Aprotic Solvents

Accepted Answer

Solvents without acidic protons (e.g., $	ext{DMSO}$, $	ext{DMF}$, acetone). They solvate cations well but leave nucleophiles bare/unsolvated and highly reactive, favoring $S_N2$ pathways.

Question 14

Nucleophilicity vs. Basicity

Accepted Answer

Nucleophilicity is a kinetic property measuring how fast an electron pair attacks an electrophilic carbon atom. Basicity is a thermodynamic property measuring affinity for a proton ($	ext{H}^+$).

Question 15

Leaving Group Ability

Accepted Answer

Good leaving groups are weak, stable bases (conjugate bases of strong acids). Relative order: $	ext{I}^- > 	ext{Br}^- > 	ext{Cl}^- \gg 	ext{F}^-$. Sulfonate esters ($	ext{TsO}^-$, $	ext{TfO}^-$, $	ext{MsO}^-$) are also excellent leaving groups.

Question 16

Halide Nucleophilicity in Protic vs. Aprotic Solvents

Accepted Answer

In polar protic solvents, nucleophilicity increases down the group ($	ext{I}^- > 	ext{Br}^- > 	ext{Cl}^- > 	ext{F}^-$) due to polarizability. In polar aprotic solvents, nucleophilicity parallels basicity ($	ext{F}^- > 	ext{Cl}^- > 	ext{Br}^- > 	ext{I}^-$).

Question 17

Carbocation Rearrangements ($S_N1$ & $E1$)

Accepted Answer

Because $S_N1$ and $E1$ proceed through carbocation intermediates, 1,2-hydride or 1,2-alkyl shifts occur whenever a secondary carbocation can rearrange into a more stable tertiary or resonance-stabilized position.

Question 18

Predicting Pathway: Primary ($1^\circ$) Substrates

Accepted Answer

Favors $S_N2$ with unhindered strong or weak nucleophiles. Shifts to $E2$ when reacted with a strong, sterically hindered base (e.g., $	ext{KO}t	ext{Bu}$). $S_N1$ and $E1$ pathways do not occur.

Question 19

Predicting Pathway: Tertiary ($3^\circ$) Substrates

Accepted Answer

Cannot undergo $S_N2$ due to steric hindrance. Undergoes rapid $E2$ with strong bases, and a mixture of $S_N1$ and $E1$ under solvolysis conditions with weak nucleophiles/bases.

Question 20

$E2$ Regioselectivity on Cyclohexane Rings

Accepted Answer

Requires both the leaving group and the adjacent $\beta$-hydrogen to occupy trans-diaxial positions (axial-axial) in the chair conformation to satisfy the anti-periplanar requirement.

Organic Chemistry - Nucleophilic Substitution & Elimination Mechanisms

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