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

Define geometrical (cis–trans) isomerism and state the condition required.

Stereoisomerism arising from restricted rotation about a double bond (or ring), where substituents differ in spatial arrangement. Requirement: each doubly bonded carbon must bear two different groups.

How is $E/Z$ nomenclature assigned for alkene geometrical isomers?

Assign CIP priority to the two groups on each $\ce{sp^2}$ carbon. If the higher-priority groups are on the same side it is $Z$ (zusammen); on opposite sides it is $E$ (entgegen).

Why does cis-but-2-ene have a higher heat of hydrogenation (less stable) than trans-but-2-ene?

In the cis isomer the two methyl groups are on the same side, producing steric strain (van der Waals repulsion). The trans isomer minimizes this strain, making it thermodynamically more stable.

Can $E/Z$ and cis/trans assignments differ for the same alkene? Explain.

Yes. cis/trans refers to identical or reference substituents' relative positions, while $E/Z$ uses CIP priorities. An alkene labeled cis by one convention may be $E$ if the higher-priority groups happen to lie trans.

What are conformers (conformational isomers)?

Different spatial arrangements of a molecule interconvertible by rotation about single ($\sigma$) bonds, without breaking bonds. They are not separable isomers under normal conditions because the rotational barrier is low.

For ethane, name the two key conformers and give the magnitude of the rotational (torsional) barrier.

Staggered (most stable, dihedral $60°$) and eclipsed (least stable, dihedral $0°$). The energy barrier is about $12.5\ \mathrm{kJ/mol}$ ($\approx 3\ \mathrm{kcal/mol}$), due to torsional strain.

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).