Question bank — Types of organic reactions — addition, substitution, elimination, rearrangement

False. They are chemical opposites (reverse arrows), but a catalyst lowers the energy barrier both ways equally; it never favours one direction — only temperature and concentration shift the equilibrium.

False. A group leaves and a different group enters, so unless the two groups have identical formulas the molecular formula changes (e.g. ${\text{CH}}_3\text{Br}\to {\text{CH}}_3\text{OH}$: Br out, OH in — formula differs).

True. A $\pi$ bond (or small ring) is consumed to make two new $\sigma$ bonds, so exactly one degree of unsaturation is removed — this is the defining signature of addition.

False. A 1,2-hydride shift moves a H within the same molecule; no atom enters or leaves, so the molecular formula is identical — only connectivity/charge position changes. See Carbocation stability.

False. Both are the same principle — the most stable intermediate/product wins. Markovnikov picks the stabler carbocation in addition; Zaitsev picks the stabler alkene in elimination. One idea, two faces.

False. Both go through a free carbocation intermediate, so a nucleophile can attach ($S_{N}1$) or a base can grab a $\beta$-H (E1) from the same cation — the two pathways compete and you get a mixture.

False. Radical substitution runs on radicals (species with one unpaired electron), not nucleophile–electrophile ion pairs; initiation, propagation and termination steps trade single electrons, not electron pairs.

Error: nothing left the molecule and the $\pi$ bond was consumed. Atoms only added across the double bond → this is addition, not substitution. Check the atom budget: substitution needs something out.

Error: $S_{N}2$ is backside attack and never forms a free cation, so cation stability is irrelevant — $S_{N}2$ prefers $1^{\circ }$ (least steric block). It's $S_{N}1$/E1 that love $3^{\circ }$ because they do form the cation. See SN1 vs SN2 mechanisms.

Error: the nucleophile attacks the backside, opposite the leaving group — this Walden inversion flips the configuration like an umbrella in wind. Front-side attack is electronically forbidden.

Error: H goes to the carbon with more H's already (the less substituted one), so the $+$ charge lands on the more substituted carbon, giving the stabler cation. It's the positive charge, not the H, that seeks substitution. See Hyperconjugation.

Error: elimination splits one molecule into two — the substrate and a small expelled molecule ($\text{HX}$, ${\text{H}}_2\text{O}$). Degree of unsaturation rises by 1; the count is 1 → 2, never 3.

Error: leaving-group quality matters for substitution and elimination (something must leave). In addition nothing leaves, so leaving-group ability is irrelevant there. See Leaving groups.

The $\pi$ electrons sit outside the internuclear axis and are held loosely, so they are exposed, easy to donate, and reach out to electron-poor electrophiles — the $\sigma$ framework is buried and strong. See Nucleophiles and Electrophiles.

A 1,2-shift only proceeds if it moves the charge downhill to a more stable cation. A $3^{\circ }$ cation is already near the bottom, so there's no more-stable target to shift toward, and it stays put. See Carbocation stability.

A $\pi$ bond is weak ($\sim 264$ kJ/mol in C=C) while each new $\sigma$ bond is strong ($\sim 347$ kJ/mol); trading one weak bond for two strong ones releases energy, so addition runs downhill.

More alkyl groups on the double bond means more hyperconjugation stabilising it, lowering the transition-state energy leading to that alkene — the lower-barrier path dominates. See E1 vs E2 mechanisms.

$S_{N}1$/E1 form a free cation and anion that must be solvated to survive — protic solvents H-bond and stabilise them. $S_{N}2$'s nucleophile must stay "naked" and reactive, so aprotic solvents that don't cage it are best.

Inductive electron-donation helps, but the dominant stabiliser is hyperconjugation — more adjacent C–H bonds can overlap with the empty $p$ orbital, spreading the positive charge; a $3^{\circ }$ cation simply has more such bonds. See Inductive effect and Hyperconjugation.

No — both carbons carry the same number of H's, so the two possible cations are identical. Markovnikov is silent; the product is the same either way (e.g. $C{H}_{3}C{H}_{2}Br$).

No. Elimination needs a H on the carbon adjacent to the leaving group to form the new $\pi$ bond; with zero $\beta$-H (e.g. $(C{H}_3{)}_{3}C-C{H}_{2}X$ toward the $C{H}_2$ side wrongly, or neopentyl-type), only substitution or rearrangement can occur.

No. A 1,2-shift is intramolecular: the same molecule reorganises using its own bonding pair, so it's a rearrangement, needing no external attacker.

No. The ring temporarily loses aromaticity in the arenium intermediate, but a $\beta$-H is expelled to restore the aromatic $\pi$ system — net effect is substitution (H swapped for the electrophile), not addition. See Aromatic Electrophilic Substitution.

Judge by the net atom budget and $\pi$ count. If a $\pi$ bond is created and a molecule leaves, it's elimination; if a group swaps with net atoms conserved, substitution. Mixed-looking multistep processes are classified step-by-step, not by the overall equation alone.

Elimination (E2). The bulky base is too big to reach the crowded $3^{\circ }$ carbon for substitution, so it instead grabs an accessible $\beta$-H, and the stable alkene product wins. See E1 vs E2 mechanisms.