List the four principal conformers of n-butane in order of increasing energy.

anti (most stable, $180°$) $<$ gauche ($60°$) $<$ eclipsed (methyl/H, $120°$) $<$ fully eclipsed (syn, methyl/methyl, $0°$, least stable).

Why is the chair conformation of cyclohexane the most stable?

It is free of angle strain (bond angles $\approx 111°$, near tetrahedral) and torsional strain (all adjacent bonds staggered), and it minimizes 1,3-diaxial steric interactions.

In substituted cyclohexanes, why is the equatorial position generally preferred for a substituent?

An equatorial substituent avoids 1,3-diaxial steric repulsions experienced in the axial position, lowering steric strain. Larger groups (e.g. tert-butyl) strongly favor equatorial.

What is the $A$-value of a substituent?

The $A$-value is the free-energy preference ($-\Delta G°$) for the equatorial over axial position in cyclohexane; larger $A$-value = greater equatorial preference (e.g. $\ce{CH3}\approx 1.7$, $\ce{C(CH3)3}\approx 4.9\ \mathrm{kcal/mol}$).

Summarize the rate law and molecularity of the $S_N1$ reaction.

$S_N1$ is unimolecular: $$\text{rate} = k[\text{substrate}]$$ first order overall, depending only on the substrate. The rate-determining step is ionization to a carbocation.

Describe the mechanism and stereochemical outcome of $S_N1$.

Two steps: (1) slow ionization forming a planar carbocation; (2) fast nucleophilic attack. Because the intermediate is planar (attack from both faces), $S_N1$ gives racemization, often with slight inversion preference.

Rank substrate reactivity in $S_N1$ reactions and explain.

$3° > 2° > 1° > \ce{CH3}$ — reactivity follows carbocation stability. More substituted carbocations are stabilized by hyperconjugation and $+I$, favoring the rate-determining ionization.

Summarize the rate law, molecularity, and transition state of $S_N2$.

$S_N2$ is bimolecular: $$\text{rate} = k[\text{substrate}][\text{nucleophile}]$$ second order overall. It proceeds via a single concerted step through a five-coordinate trigonal-bipyramidal transition state.

What is the stereochemical consequence of an $S_N2$ reaction?

Complete inversion of configuration (Walden inversion) — the nucleophile attacks $180°$ opposite the leaving group, flipping the carbon like an umbrella in the wind.

Rank substrate reactivity in $S_N2$ and explain.

$\ce{CH3} > 1° > 2° > 3°$ (3° essentially unreactive). Increasing substitution causes steric hindrance to backside attack, raising the transition-state energy.

🇮🇳 GATE Life Sciences · flashcards

GATE Life Sciences Structure-Reactivity Correlations and Organic Reaction Mechanisms Flashcards

50 question-and-answer cards covering Structure-Reactivity Correlations and Organic Reaction Mechanisms as it is examined in GATE Life Sciences. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

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List the four principal conformers of n-butane in order of increasing energy.
anti (most stable, $180°$) $<$ gauche ($60°$) $<$ eclipsed (methyl/H, $120°$) $<$ fully eclipsed (syn, methyl/methyl, $0°$, least stable).
Why is the chair conformation of cyclohexane the most stable?
It is free of angle strain (bond angles $\approx 111°$, near tetrahedral) and torsional strain (all adjacent bonds staggered), and it minimizes 1,3-diaxial steric interactions.
In substituted cyclohexanes, why is the equatorial position generally preferred for a substituent?
An equatorial substituent avoids 1,3-diaxial steric repulsions experienced in the axial position, lowering steric strain. Larger groups (e.g. tert-butyl) strongly favor equatorial.
What is the $A$-value of a substituent?
The $A$-value is the free-energy preference ($-\Delta G°$) for the equatorial over axial position in cyclohexane; larger $A$-value = greater equatorial preference (e.g. $\ce{CH3}\approx 1.7$, $\ce{C(CH3)3}\approx 4.9\ \mathrm{kcal/mol}$).
Summarize the rate law and molecularity of the $S_N1$ reaction.
$S_N1$ is unimolecular: $$\text{rate} = k[\text{substrate}]$$ first order overall, depending only on the substrate. The rate-determining step is ionization to a carbocation.
Describe the mechanism and stereochemical outcome of $S_N1$.
Two steps: (1) slow ionization forming a planar carbocation; (2) fast nucleophilic attack. Because the intermediate is planar (attack from both faces), $S_N1$ gives racemization, often with slight inversion preference.
Rank substrate reactivity in $S_N1$ reactions and explain.
$3° > 2° > 1° > \ce{CH3}$ — reactivity follows carbocation stability. More substituted carbocations are stabilized by hyperconjugation and $+I$, favoring the rate-determining ionization.
Summarize the rate law, molecularity, and transition state of $S_N2$.
$S_N2$ is bimolecular: $$\text{rate} = k[\text{substrate}][\text{nucleophile}]$$ second order overall. It proceeds via a single concerted step through a five-coordinate trigonal-bipyramidal transition state.
What is the stereochemical consequence of an $S_N2$ reaction?
Complete inversion of configuration (Walden inversion) — the nucleophile attacks $180°$ opposite the leaving group, flipping the carbon like an umbrella in the wind.
Rank substrate reactivity in $S_N2$ and explain.
$\ce{CH3} > 1° > 2° > 3°$ (3° essentially unreactive). Increasing substitution causes steric hindrance to backside attack, raising the transition-state energy.
Compare the solvent preferences of $S_N1$ versus $S_N2$.
$S_N1$ is favored by polar protic solvents (stabilize the carbocation and leaving group). $S_N2$ is favored by polar aprotic solvents (e.g. DMSO, DMF, acetone) which leave the nucleophile "naked" and reactive.
How does nucleophile strength affect $S_N1$ vs $S_N2$ rates?
$S_N2$ rate increases with stronger/more concentrated nucleophiles (nucleophile is in the rate-determining step). $S_N1$ rate is independent of nucleophile strength (it enters after the slow step).
Summarize the rate law and mechanism of the $E1$ reaction.
$E1$ is unimolecular: $$\text{rate} = k[\text{substrate}]$$ Two steps: slow ionization to a carbocation, then fast loss of a $\beta$-proton by a base to form the alkene. Favored by $3°$ substrates and polar protic solvents.
Summarize the rate law and mechanism of the $E2$ reaction.
$E2$ is bimolecular: $$\text{rate} = k[\text{substrate}][\text{base}]$$ A single concerted step in which a strong base removes the $\beta$-H as the leaving group departs, forming the $\pi$-bond.
What stereochemical (geometric) requirement does the $E2$ mechanism impose?
Anti-periplanar geometry: the $\beta$-H and the leaving group must be $180°$ apart (anti and coplanar) so the breaking $\sigma$-bonds overlap to form the new $\pi$-bond.
State Zaitsev's (Saytzeff) rule and Hofmann's rule for elimination orientation.
Zaitsev: elimination favors the more substituted (more stable) alkene. Hofmann: bulky bases (e.g. $\ce{t-BuOK}$) or bulky leaving groups (e.g. $\ce{-NR3+}$) favor the less substituted (least hindered) alkene.
Give a quick decision guide for $S_N1$/$S_N2$/$E1$/$E2$ based on substrate class.
Methyl/1°: $S_N2$ (E2 with strong bulky base). 2°: $S_N2$/E2 with strong nucleophiles/bases, $S_N1$/E1 in protic solvents. 3°: $S_N1$/E1 (no $S_N2$); E2 with strong base.
Define homolytic bond cleavage and the species it produces.
Homolysis splits a covalent bond so each fragment keeps one electron, producing two neutral free radicals: $$\ce{A-B -> A^{.} + B^{.}}$$ It is favored by heat, light (hv), and nonpolar solvents.
Name the three stages of a free-radical chain mechanism (e.g. halogenation of alkanes).
Initiation (homolysis generates radicals, e.g. $\ce{Cl2 ->[hv] 2Cl^{.}}$), Propagation (radical reacts to form product plus a new radical), and Termination (two radicals combine, destroying chain carriers).
Rank free-radical stability and state the governing factor.
$3° > 2° > 1° > \ce{CH3^{.}}$ — same order as carbocations, governed by hyperconjugation and $+I$ stabilization of the electron-deficient (odd-electron) carbon. Allylic/benzylic radicals are extra-stabilized by resonance.
Why is radical bromination more selective than chlorination?
Bromination is endothermic with a late, product-like transition state (Hammond postulate), so it strongly reflects radical stability and selects for the more stable $3°$ radical. Chlorination is exothermic, early TS, and far less selective.
State the orientation rules for electrophilic aromatic substitution by directing groups.
Ortho/para-directors are typically activating electron donors ($+M$/$+I$: $\ce{-OH}$, $\ce{-NH2}$, $\ce{-OR}$, alkyl); meta-directors are deactivating electron withdrawers ($-M$/$-I$: $\ce{-NO2}$, $\ce{-COOH}$, $\ce{-CN}$, $\ce{-C=O}$). Halogens are deactivating but o/p-directing.
Explain why halogens are ortho/para-directing yet deactivating in EAS.
Halogens withdraw electron density inductively ($-I$, deactivating, slowing the reaction) but donate lone-pair electron density by resonance ($+M$) that stabilizes the arenium ion at ortho/para positions, controlling orientation.
State Markovnikov's rule and the anti-Markovnikov (peroxide/Kharasch) exception.
Markovnikov: in electrophilic addition of $\ce{HX}$ to an alkene, $\ce{H}$ adds to the carbon with more hydrogens so the more stable carbocation forms (X to the more substituted carbon). With $\ce{HBr}$ + peroxides, a radical mechanism reverses orientation (anti-Markovnikov).

What this deck covers

The Structure-Reactivity Correlations and Organic Reaction Mechanisms deck follows the GATE Life Sciences Structure-Reactivity Correlations and Organic Reaction Mechanisms syllabus — 13 chapters and 11 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 3.8 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 202 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.

Structure-Reactivity Correlations and Organic Reaction Mechanisms flashcards FAQ

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