Consider the following five compounds.
a. CH3CH2CH2CH2CH3
b. OH
'
CH3CH2CH2CH2
c. CH3CH2CH2CH2CH2CH3
d. O
''
CH3CH2CH2CH3
e. CH3
'
CH3CCH3
'
CH3
The boiling points of these five compounds are 9.5 C, 36 C, 69 C, 76 C, and 117 C. Which compound boils at 36 C? Explain.
a. CH3CH2CH2CH2CH3
b. ----->OH
--------->'
CH3CH2CH2CH2
c. CH3CH2CH2CH2CH2CH3
d. ----->O
-------->''
CH3CH2CH2CH3
e. CH3
-->'
CH3CCH3
-->'
-->CH3
I tried to have the bond/double bond connect correctly by using arrows, but that didn't work. I hope you can figure it out.
I cant, but I can offer advice. Arrange the molecules in order of intermolecular bond strength. Look for hydrogen bonding, etc. That will allow you to put the molecules in order of bp.
I can't figure it out either; however, Bob Pursley gave good advice.
First, the lowest b.p. will be those compounds without hyrogen bonding and that means a, c, or e.
The one with the most branching(e) will be the lowest (9.5). a and c are both straight chains; the longest will have the higher b.p.; therefore, a (the 5-C chain) will have b.p. of 36 and c (the 6-C chain) will have b.p. of 69. The alcohol and ketone will have the highest b.p. By the way, I see a 5 valent C atom on b and a 6 valent C atom on d. That's a no-no. I assume b is 2-butanol and I assume d is methyl ethyl ketone but I can't be sure. My tables list a, c, and e b.p. the same as your post; my tables don't list methyl ethyl ketone or 2-butanol at the other b.p. in your post so I may not have them interpreted correctly.
1 answer