Exercises — Hydrogen bonding — intermolecular, intramolecular; consequences (boiling points, water density)

Recall the rule: an H-bond needs an H covalently bonded to F, O, or N (the donor) meeting a lone pair on an F, O, or N (the acceptor). Both partners are checked.

• (a) CH₄ + H₂O → NO. Methane's H is bonded to carbon (not F/O/N), so CH₄ cannot donate. It has no lone pairs either, so it cannot accept. No H-bond. (Only weak Van der Waals forces.)
• (b) NH₃ + NH₃ → YES. N–H is a valid donor, and the neighbouring N carries a lone pair (acceptor). $\text{N–H}\cdots \text{N}$.
• (c) HCl + HCl → NO (not strong). Cl is electronegative but large and diffuse, so its charge/lone pair is spread out and cannot approach the proton closely. Only dipole–dipole, not a proper H-bond.
• (d) CH₃OH + H₂O → YES. The O–H of methanol donates to water's O lone pair (and vice-versa). $\text{O–H}\cdots \text{O}$.

• F–H is the covalent bond. Because F holds the shared electrons, this whole F–H unit is the donor (it "donates" the polarised H).
• The N on the right supplies a lone pair and is the acceptor.
• The $\cdots$ (dotted) link is the hydrogen bond itself — the electrostatic attraction between the exposed H proton and N's lone pair.

The three heavier hydrides follow a near-linear rise with molar mass. Extending that same line back to the very light ${\text{H}}_2\text{O}$ predicts a boiling point below that of ${\text{H}}_2\text{S}$ — the standard textbook extrapolation lands around $\approx -80{}^{\circ }\text{C}$.

The gap: $\Delta =(+100)-(-80)=180{}^{\circ }\text{C}.$ This roughly $180{}^{\circ }\text{C}$ jump is the extra energy needed to break water's hydrogen-bond network — energy that plain London forces alone (which the extrapolation assumes) would never require. See Boiling point trends of hydrides.

See the figure below for the "kink" this creates.

Count "hands" (donors + usable acceptors) per molecule:

A molecule can only make as many H-bonds as its scarcer resource allows. NH₃ has three H's but only one lone pair → bottlenecked at ~1 network bond. HF has three lone pairs but only one H → bottlenecked at ~1 donation, forming zig-zag chains (~2 bonds/molecule). Water has a matched 2-and-2 → a full 3-D network (~4 bonds/molecule).

$\boxed{{\text{NH}}_3<\text{HF}<{\text{H}}_2\text{O}}$ Real b.p.: $-33{}^{\circ }\text{C}<19.5{}^{\circ }\text{C}<100{}^{\circ }\text{C}$ ✓

Same formula ⇒ same molar mass ⇒ same London forces. So the difference is purely geometric H-bonding.

• ortho: –OH and –NO₂ sit on adjacent carbons, exactly the right distance to close a 6-membered intramolecular ring $\text{O–H}\cdots \text{O=N}$. The molecule satisfies its own H-bond internally, so it has little left over to bond to neighbours → weak intermolecular attraction → lower b.p. → steam-volatile.
• para: the two groups are on opposite ends, too far to reach each other, so each –OH bonds to other molecules (intermolecular) → strong network → higher b.p.

$\Delta T=279-214=65{}^{\circ }\text{C}$

Steam-volatile isomer: ortho. See the geometry:

Volume of $1\text{g}$: $V_{\text{water}}=\frac{1}{1.000}=1.000{\text{cm}}^3,V_{\text{ice}}=\frac{1}{0.917}=1.0905{\text{cm}}^3.$ Percentage increase: $\frac{V_{\text{ice}}-V_{\text{water}}}{V_{\text{water}}}\times 100\%=\frac{1.0905-1.000}{1.000}\times 100\%\approx 9.05\%.$

Physical origin: on freezing, each water molecule locks into 4 rigid H-bonds in an open hexagonal tetrahedral lattice (Water — anomalous properties). That lattice has large hexagonal holes, so the same mass spreads over ~9% more volume → lower density → ice floats.

(a) In a non-polar solvent, two acetic acid molecules join head-to-head via two H-bonds ($\text{O–H}\cdots \text{O}$ on each side) into a cyclic dimer. Two molecules travelling as one particle → measured particle mass looks doubled.

(b) Let $\alpha$ = fraction associated into dimers. Starting with 1 mole of monomer, $\alpha$ moles combine into $\alpha /2$ moles of dimer, leaving $(1-\alpha )$ free. Total particles: $n=(1-\alpha )+\frac{\alpha }{2}=1-\frac{\alpha }{2}.$ Apparent molar mass scales inversely with particle count: $M_{\text{app}}=\frac{M_{\text{true}}}{1-\alpha /2}.$ Plug in $M_{\text{app}}=120$, $M_{\text{true}}=60$: $120=\frac{60}{1-\alpha /2}\Rightarrow 1-\frac{\alpha }{2}=0.5\Rightarrow \alpha =1.$ So 100% association — every molecule is dimerised.

(c) The carboxyl group has both a donor (O–H) and an acceptor (C=O) positioned so two molecules form a symmetric, strain-free 8-membered double-H-bonded ring. Two bonds, mutually reinforcing, make the dimer robust enough for full association.

• A–T pairs: $30\times 2=60$ H-bonds.
• G–C pairs: $70\times 3=210$ H-bonds.
• Total: $60+210=270$ H-bonds (DNA structure).

Melting: the G–C-rich region is harder to separate. Each G–C pair holds with 3 H-bonds vs 2 for A–T, so G–C-rich DNA needs a higher temperature to unzip. (This is why "GC content" predicts DNA melting temperature.)

Coulomb ($1/r^2$) argument. Attractive electrostatic force scales as $F\propto 1/r^2$, so it blows up at small $r$. In a normal polar molecule the ${\delta }^{+}$ end still has inner electrons shielding its nucleus, keeping the acceptor at a distance. In $\text{X–H}$, hydrogen has no inner electrons — once X pulls away its only electron, the bare proton is exposed. A lone pair can approach much closer (smaller $r$) → the $1/r^2$ term is far larger → much stronger attraction.

Orbital argument. There's a small charge-transfer/covalent contribution: the acceptor's lone pair donates slightly into the X–H antibonding orbital. This partial covalency adds extra binding a pure dipole–dipole interaction lacks.

Counterfactual. If H had inner-shell electrons, they would shield the proton and prevent the close approach — the acceptor couldn't get near, $r$ would stay large, $1/r^2$ would stay small, and the "hydrogen bond" would collapse to an ordinary weak dipole–dipole force. H-bonding exists precisely because hydrogen is the one atom with a shieldless nucleus.

Two opposing effects fight over volume as temperature drops toward $0{}^{\circ }\text{C}$:

1. Normal thermal contraction (dominates above 4 °C): cooling slows molecules, they crowd closer, volume ↓, density ↑ — the ordinary behaviour.
2. Open-cluster formation (dominates below 4 °C): as $0{}^{\circ }\text{C}$ nears, molecules start snapping into open, ice-like tetrahedral clusters with hexagonal holes, volume ↑, density ↓.

Above $4{}^{\circ }\text{C}$ effect 1 wins (density rising as it cools). Below $4{}^{\circ }\text{C}$ effect 2 wins (density falling again). The crossover — maximum density — sits at $4{}^{\circ }\text{C}$, the tug-of-war balance point. See Water — anomalous properties.

Biology: the $4{}^{\circ }\text{C}$ densest water sinks to the lake bottom; colder $0{}^{\circ }\text{C}$ water and ice stay on top and insulate the water below. Lakes freeze top-down, so fish survive winter in the liquid layer beneath the ice.

Number of H-bonds a water molecule makes in ice ::: 4 (two as donor via its O–H, two as acceptor via its lone pairs) Why o-nitrophenol boils lower than p-nitrophenol ::: ortho forms an intramolecular ring, so it clings less to neighbours (less intermolecular H-bonding) Volume increase when water freezes ::: about 9%, from the open hexagonal lattice holes Degree of association of acetic acid at $M_{\text{app}}=120$ ::: $\alpha =1$ (fully dimerised)