13.8 Postlab Questions 1. Three of the liquids, butanone, 1-butanol and pentane

Question

13.8 Postlab Questions 1. Three of the liquids, butanone, 1-butanol and pentane had nearly the same molecu- lar weights, but significantly different AT values. Explain the difference in AT values of these substances based on their intermolecular forces. 2. Which of the alcohols studied has the strongest intermolecular forces of attraction? The weakest intermolecular forces? Use principles of intermolecular forces to ex- plain these results. 3. Which of the non-alcohols studied has the stronger intermolecular forces of attrac- tion? The weaker intermolecular forces? Use principles of intermolecular forces to explain these results. 4. Use Excel to plot a graph of values of the four alcohols versus their respective molec- ular weights. Plot molecular weight (u) on the horizontal axis and on the vertical axis.

Explanation / Answer

a)

butanone --> has C=O group

butanol --> has OH group

pentane --> has CH2 group

note that even ghough the changes are approx 12-15 g/mol, their interacitons are way different

C=O is polar, OH is hydrogen bonded, CH2 is london dispersion force

these intermoleclar forces vary greatly between each other. therefore, different vlaues are shown

Q2

typically --< the shortest alcohol will have much more attraction between dipole-dipole + hydrogne bonding

the least --< will be the heaviest alcohols, i.e. hexanol and heptanol are xamples.

this is mainly because of the strenght of the dipole in methanol, CH3-OH is 50% polarized whereas CH3CH2CH2CH2CH2OH is so small dipole compared.

Q3

typically, ketone will be larger than all other alkanes, this is mainly because of the C=O polar formed

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