Consider the following equimolar pure liquid samples: 1. n-propyl chloride 2. isopropyl chloride 3. 1-chlorobutane 4. 2-chloro-2-methylpropane 5. 1-chloropentane If they are arranged in strictly incre

Question

Consider the following equimolar pure liquid samples:
1. n-propyl chloride
2. isopropyl chloride
3. 1-chlorobutane
4. 2-chloro-2-methylpropane
5. 1-chloropentane
If they are arranged in strictly increasing order of their normal boiling points, what is the assigned index number of the specific compound that sits exactly in the middle (the median) of the resulting series?

Prashant Kotian · Accepted Answer

Correct answer: 4. First, group by molecular mass. Boiling point increases with mass: . compounds: Isopropyl chloride (branched) < n-propyl chloride (straight). compounds: 2-chloro-2-methylpropane (highly branched) < 1-chlorobutane (straight). compound: 1-chloropentane (highest mass). The strictly increasing overall sequence is: isopropyl chloride < n-propyl chloride < 2-chloro-2-methylpropane < 1-chlorobutane < 1-chloropentane. The median (3rd out of 5) is 2-chloro-2-methylpropane, which corresponds to index number 4.

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