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Organic Chemistry Reaction Mechanisms of Aliphatic Compounds Flashcards
50 question-and-answer cards covering Reaction Mechanisms of Aliphatic Compounds as it is examined in Organic Chemistry. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Reaction Mechanisms of Aliphatic Compounds deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
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.
What this deck covers
The Reaction Mechanisms of Aliphatic Compounds deck follows the Organic Chemistry Reaction Mechanisms of Aliphatic Compounds syllabus — 4 chapters and 14 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 12.5 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 181 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Reaction Mechanisms of Aliphatic Compounds flashcards FAQ
How many Reaction Mechanisms of Aliphatic Compounds flashcards are in this Organic Chemistry deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these Organic Chemistry flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Reaction Mechanisms of Aliphatic Compounds cards cover?
They follow the Organic Chemistry Reaction Mechanisms of Aliphatic Compounds syllabus — 4 chapters and 14 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.