Question bank — Reaction mechanisms — curved-arrow notation, bond formation - breaking (heterolysis vs homolysis)

False — a curved arrow only ever shows electrons moving; the atoms are dragged along by the electron flow but the arrow never represents an atom. See Resonance and arrow pushing.

False — which atom keeps the pair depends on electronegativity. C–MgBr heterolyzes to a carbanion (C⁻) because Mg is electropositive and lets carbon take both electrons.

False — each atom keeps exactly one electron, so both fragments are neutral radicals; no atom gains or loses a net electron, hence no charge.

False — a full arrow moves an electron pair (2 e⁻), a fishhook moves one electron. Using the wrong one changes the whole reaction type (ionic vs radical).

True — net charge is always conserved; if a bond splits to give +1 and −1, they sum to 0, matching a neutral reactant.

True — the shared pair splits into two single electrons, and each electron needs its own half-arrow pointing to its atom.

False — a positive charge is electron-poor; there are no electrons there to move. Tails start only on lone pairs or bonds. See Nucleophiles and Electrophiles.

False — non-polar bonds (equal electronegativity, e.g. Cl–Cl) break homolytically, because neither atom is greedy enough to grab both electrons and no charge could be stabilized.

False — they are the same act of electron motion seen from opposite ends: an arrowhead landing between two atoms is a new bond; a tail leaving a bond is a breaking bond.

The tail must sit on the C–Br bond's electrons, not on the Br atom, and it must point onto Br. "Leaving" is modeled as the bond pair moving to the more electronegative atom.

A full arrow moves a pair, which would give ions, not radicals. Homolysis needs two fishhook arrows, one electron to each Cl.

The π electrons attack the δ⁺ hydrogen (the electron-poor site), not Br. A separate arrow then breaks H–Br, sending that pair onto Br. See Electrophilic addition to alkenes.

Period-2 atoms cannot exceed the octet; carbon can hold only 8 electrons. Either a bond must break as the new one forms, or the arrow is illegal.

Charge isn't conserved — a neutral start (0) cannot give net +1. The electrons that left carbon must land on some atom, producing the matching anion.

Nucleophilic attack moves a pair of electrons, so it needs a full arrow. Fishhooks are reserved for single-electron (radical) steps.

Because the arrow is the moving electrons — you cannot move electrons from a place that has none, so the source must be a lone pair or a bond.

The large electronegativity gap means one atom strongly wants both electrons, so an uneven split producing stabilizable ions is energetically preferred. See Inductive effect and electronegativity.

High energy input can split a bond, but with no way to stabilize separated charge, the system avoids ions and splits evenly into neutral radicals. See Bond dissociation energy.

Each atom "owned" one electron in the shared pair; keeping both gives an extra electron (−1), losing your share leaves you one short (+1).

A more stable cation lowers the energy cost of the uneven split, so the bond heterolyzes more easily. See Carbocations — stability and structure.

A π bond is a region rich in electrons (a nucleophile), and any bond or lone pair with available electrons is a legal source for a full arrow.

It depends on conditions: polar/protic solvents and ion-stabilizing groups push toward heterolysis, while heat/UV in the gas phase pushes toward homolysis — polarity alone doesn't decide.

No — a double-barbed arrow always moves a pair; two of them would move four electrons. Radicals require single-barbed fishhooks.

Zero on each — the identical atoms split the pair evenly, so both fragments are neutral radicals with one unpaired electron.

Carbon keeps both electrons (giving a carbanion, C⁻) because magnesium is electropositive and readily becomes Mg⁺; this is the "heterolysis ≠ always carbocation" edge case.

Exactly −1 — net charge is conserved in every step, so the products' charges must also sum to −1.

Yes — electrons always flow from the rich site (nucleophile) to the poor site (electrophile); the arrowhead landing between two atoms creates the new shared pair. See Nucleophiles and Electrophiles.