Zadanie 6b, Paper 2, specimen 2025, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, specimen 2025, pytanie 6(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Halogens and halogenoalkanes are widely used as reactants in the laboratory and in industry.
Polecenie
The boiling points of the five simplest straight-chain primary amines, CH3NH2 to CH3(CH2)4NH2, and of the five simplest 1-fluoroalkanes, CH3F to CH3(CH2)4F, are plotted below against the number of carbon atoms in the molecule. An amine and a fluoroalkane with the same number of carbon atoms have similar relative formula masses.
Powiększ(i) Outline why each amine boils at a higher temperature than the fluoroalkane whose relative formula mass is closest to its own.
(ii) Explain the trend in boiling point along the series of amines.
(iii) A branched-chain fluoroalkane is compared with its straight-chain isomer. Predict which has the lower boiling point and give a reason.
(iv) Explain why replacing fluorine by iodine in a 1-halogenoalkane raises the boiling point.
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Odpowiedź z klucza
(i) hydrogen bonding «N–H···N» between amine molecules is stronger than the dipole–dipole forces between fluoroalkane molecules ✔
(ii) boiling point rises along the series because the London (dispersion) forces between the molecules become stronger ✔
each successive molecule has more electrons
OR
a larger surface area for contact with neighbouring molecules ✔
(iii) lower AND the more compact shape reduces the contact area / weakens the London (dispersion) forces ✔
(iv) iodoalkane molecules have more electrons, so the London (dispersion) forces are stronger ✔
Schemat punktowania
(i) [1]: hydrogen bonding in amines compared with dipole–dipole forces in fluoroalkanes. (ii) [2]: M1 stronger London (dispersion) forces; M2 more electrons or greater surface contact; "stronger intermolecular forces" alone does not score M1. (iii) [1]: lower with the contact-area reason. (iv) [1]: more electrons, stronger London forces; a polarity argument scores zero.
Komentarz. Amines hydrogen bond through N–H, more weakly than alcohols through O–H, but still far more strongly than the dipole–dipole forces of fluoroalkanes; the trend within one series is a London-forces argument.