Question

The following boiling points correspond to the substances listed. Match the boiling points to the substances by considering the relative strengths of their intermolecular forces. b.p. (C): 162, 88.5, 28, 36, 64.5, 78.3, 82.5, 140, 205, 290; substance: CH4, CH3CHOHCH3, C6H5CH2OH (has a benzene ring), CH3CH3, C5H9OH (cyclic), (CH3)2CHCH2CH3, CH3OH, HOCH2CHOHCH2OH, CH3(CH2)3CH3, CH3CH2OH.

Answer 1

Answer:

CH₄                                - 162 ⁸C

CH₃CH₃                          -88.5 ⁸C

(CH₃)₂ CHCH₂CH₃         28 ⁸C

CH₃3(CH2)₃CH₃             36 ⁸C

CH₃OH                           64.5 ⁸C

CH₃CH₂OH                     78.3 ⁸C

CH₃CHOHCH₃                82.5 ⁸C

C₅H₉OH                         140 ⁸C

C₆H₅CH₂OH                   205 ⁸C

HOCH₂CHOHCH₂OH    290 ⁸C

Explanation:

To answer this question we need first to understand that for organic compounds:

a. Non polar compounds have lower boiling points than polar ones of similar structure and molecular weight.

b. Boiling points increase with molecular weight.  In alkane compounds if we compare  isomers, the  straight chain isomer will have a higher boiling point than the branched one (s) because of London dispersion intermolecular forces.

a. The introduction of hydroxyl groups increase the intermolecular forces and hence the boiling points because the electronegative oxygen, and, more importantly the presence  of hydrogen bonds.

Considering the observations above, we can match the boiling points as follows:

CH₄                                - 162 ⁸C

CH₃CH₃                          -88.5 ⁸C

(CH₃)₂ CHCH₂CH₃         28 ⁸C

CH₃3(CH2)₃CH₃             36 ⁸C

CH₃OH                           64.5 ⁸C

CH₃CH₂OH                     78.3 ⁸C

CH₃CHOHCH₃                82.5 ⁸C

C₅H₉OH                         140 ⁸C

C₆H₅CH₂OH                   205 ⁸C

HOCH₂CHOHCH₂OH    290 ⁸C

Note: There was a mistake in the symbols used for the  162 and 88.5 values which are negative and correspond to the common gases methane and ethane