In Exp. D (the practicum) for CHM 375, one of the neutral compounds is compound
ID: 524241 • Letter: I
Question
In Exp. D (the practicum) for CHM 375, one of the neutral compounds is compound X biphenyl (mp 68-70 degree C), and one of the acids is compound Y, 4-chlorobonzoic add (mp 240 - 243 degree C). Explain why x is easily lost by sublimation and Y can dry on a watch glass for a week with negligible loss. label each molecule below as either chiral, meso (achiral overall with chiral centers), or achiral (i.e., achiral overall with no chiral centers). Propose a structure for a hydrocarbon that shows a parent ion M^+ = 84 in its mass spectrum. Answer: Consider a ketone with M+ = 86 and abundant fragments m/z = 71 and 43. Draw the parent ion and the fragments. M^+ m/z = 71 m/z = 43Explanation / Answer
Ans 4. The compound X biphenyl has a low melting point i.e 68 - 70o C whereas the compund Y has a higher melting point.
The lower melting point of X suggests that it needs less energy for breakage of its intermolecular bonds, which it can gain from the surroundings. Hence it gains this energy from the environment and the breakage of intermolecular bonds cause the molecules to escape and hence sublime when left in open.
The compound Y has higher melting point , and it cannot gain enough energy from the surroundings, so that its bonds are broken and the compound can sublime.