Organic Chemistry Reaction Mechanisms of Aliphatic Compounds Syllabus

Write the rate law for an $S_N1$ reaction.

$$\text{rate} = k[\text{substrate}]$$ It is first order overall and independent of nucleophile concentration.

What stereochemical outcome characterizes the $S_N1$ reaction?

Racemization (a mix of retention and inversion), because the planar carbocation intermediate can be attacked from either face; often slight excess of inversion is observed.

How does substrate structure affect $S_N1$ rate?

Rate: $\text{3}^{\circ} > \text{2}^{\circ} > \text{1}^{\circ} > \text{methyl}$, following carbocation stability; tertiary substrates react fastest and methyl/primary essentially do not go $S_N1$.

Why can carbocation rearrangements occur in $S_N1$ (and E1) but not in $S_N2$ (or E2)?

$S_N1$/E1 form a discrete carbocation intermediate that can rearrange via 1,2-hydride or 1,2-alkyl shifts to a more stable cation; $S_N2$/E2 are concerted with no free carbocation, so no rearrangement occurs.

What structural features make a good leaving group?

A good leaving group is a weak base that stabilizes the developing negative charge — i.e. the conjugate base of a strong acid. Larger, more polarizable, resonance/inductively stabilized groups leave better.

Rank these common leaving groups from best to worst: $\ce{I^-}$, $\ce{Br^-}$, $\ce{Cl^-}$, $\ce{F^-}$.

$\ce{I^- > Br^- > Cl^- > F^-}$. Leaving-group ability increases down the halogen group as basicity decreases and size/polarizability increases; $\ce{F^-}$ is a poor leaving group.

Why are $\ce{OH^-}$, $\ce{NH2^-}$, and $\ce{OR^-}$ poor leaving groups, and how is $\ce{OH}$ activated?

They are strong bases and thus poor leaving groups. An $\ce{-OH}$ can be activated by protonation (to leave as $\ce{H2O}$) or by converting to a tosylate/mesylate (excellent leaving groups).

How does solvent type favor $S_N1$ versus $S_N2$?

Polar protic solvents (e.g. water, alcohols) stabilize ions/carbocations and favor $S_N1$; polar aprotic solvents (e.g. DMSO, DMF, acetone) leave the nucleophile 'naked' and reactive, favoring $S_N2$.

Summarize how nucleophile strength distinguishes $S_N1$ from $S_N2$.

$S_N2$ requires a strong/good nucleophile (it participates in the rate-determining step); $S_N1$ rate is independent of nucleophile, so weak nucleophiles (often the solvent, solvolysis) suffice.

For a primary alkyl halide with a strong nucleophile, which substitution mechanism dominates and why?

$S_N2$ dominates: primary carbons allow easy backside attack, and they cannot form stable carbocations needed for $S_N1$.

For a tertiary alkyl halide with a weak nucleophile in polar protic solvent, which substitution mechanism dominates?

$S_N1$ dominates: the tertiary substrate forms a stable carbocation, the weak nucleophile/protic solvent favors ionization, and steric bulk blocks $S_N2$.

Describe the mechanism and molecularity of the E2 reaction.

A concerted, one-step bimolecular elimination: a base removes a $\beta$-hydrogen while the leaving group departs, forming a $\pi$ bond in a single transition state.

Write the rate law for an E2 reaction.

$$\text{rate} = k[\text{substrate}][\text{base}]$$ Second order overall (first order in substrate and in base).

What is the required geometry (stereochemistry) of the E2 transition state?

Anti-periplanar: the $\beta$-hydrogen and the leaving group must be on opposite sides in the same plane (dihedral angle $180^{\circ}$) so their orbitals align to form the new $\pi$ bond.

State Zaitsev's rule for elimination reactions.

Elimination predominantly gives the more substituted (more stable) alkene — the Zaitsev product — because a more substituted double bond is thermodynamically favored.

When is the Hofmann (less-substituted) alkene favored in elimination?

When a bulky, hindered base (e.g. potassium tert-butoxide) is used, or with bulky substituents; steric hindrance forces removal of the more accessible, less-hindered $\beta$-hydrogen, giving the less-substituted alkene.

Describe the mechanism and molecularity of the E1 reaction.

A stepwise, unimolecular elimination: slow ionization forms a carbocation (rate-determining), then a base removes a $\beta$-hydrogen to form the alkene in a fast second step.

Write the rate law for an E1 reaction.

$$\text{rate} = k[\text{substrate}]$$ First order overall, independent of base concentration.

Compare the conditions that favor E1 versus E2.

E2 needs a strong base and can occur on 1°/2°/3° substrates; E1 needs a weak base, favors 3°/2° substrates that form stable carbocations, and is promoted by polar protic solvents and heat.

How does temperature affect the substitution-versus-elimination competition?

Higher temperature favors elimination over substitution because elimination has a more positive $\Delta S$ (it increases the number of molecules), so the $-T\Delta S$ term makes $\Delta G$ more favorable for elimination.

How does base/nucleophile character steer a reaction toward substitution or elimination?

Strong, bulky bases favor elimination (E2); strong, small nucleophiles that are weak bases (e.g. $\ce{I^-}$, $\ce{CN^-}$) favor substitution; weak base/nucleophile with 3° substrate gives $S_N1$/E1 mixtures.

Summarize the expected pathway for methyl and primary substrates in the substitution/elimination competition.

Methyl: only $S_N2$ (no $\beta$-H for elimination on methyl). Primary: mainly $S_N2$; E2 competes only with a strong bulky base and heat. $S_N1$/E1 do not occur (unstable primary cation).

Summarize the expected pathways for tertiary substrates in the substitution/elimination competition.

Tertiary: no $S_N2$ (too hindered). With a strong base → E2; with a weak base/nucleophile in protic solvent → $S_N1$ + E1 mixture (elimination increasing with heat).

What are the three chain phases of radical halogenation of alkanes?

Initiation (homolysis of $\ce{X2}$ by heat or light to give $\ce{2X^{.}}$), Propagation (radical abstracts H, then reacts with $\ce{X2}$ to regenerate a radical), and Termination (two radicals combine).

Write the two propagation steps for the radical chlorination of methane.

$$\ce{Cl^{.} + CH4 -> HCl + CH3^{.}}$$ $$\ce{CH3^{.} + Cl2 -> CH3Cl + Cl^{.}}$$

Why is radical bromination more selective than radical chlorination?

Bromination is endothermic in the H-abstraction step, giving a late, product-like transition state (Hammond postulate) that strongly reflects radical stability, so it selectively abstracts the weakest C–H (3° > 2° > 1°). Chlorination is exothermic, early TS, and much less selective.

Rank the ease of C–H abstraction by a halogen radical based on the resulting radical stability.

$\text{allylic/benzylic} > \text{3}^{\circ} > \text{2}^{\circ} > \text{1}^{\circ} > \text{methyl}$, mirroring the stability of the carbon radical formed.

What reagent selectively performs allylic or benzylic bromination, and why?

N-bromosuccinimide (NBS), which supplies a low, steady concentration of $\ce{Br2}$. This favors radical allylic/benzylic substitution at the weak, resonance-stabilized C–H bond over ionic addition to the double bond.

Why do allylic and benzylic positions react preferentially in radical halogenation?

Abstracting an allylic or benzylic hydrogen gives a resonance-stabilized radical (delocalized over the adjacent $\pi$ system), which is much more stable and forms with a lower activation energy.

Outline the three phases of radical chain-growth polymerization and give a common initiator.

Initiation: an initiator (e.g. a peroxide such as benzoyl peroxide, or AIBN) homolyzes to radicals that add to a monomer. Propagation: the growing radical adds successive monomers across their $\ce{C=C}$ bonds. Termination: two chains combine (coupling) or disproportionate, ending growth.