JEE Advanced Chemistry: Some Basic Principles and Techniques (Organic Chemistry) questions with solutions

Q1. Arrange the following compounds in increasing order of their boiling points: (I) C2H5OH (ethanol) (II) C2H5Cl (chloroethane) (III) C2H5CH3 (propane) (IV) C2H5OCH3 (methyl ethyl ether / methoxyethane)

1. (III) < (II) < (I) < (IV)
2. (II) < (III) < (IV) < (I)
3. (IV) < (III) < (I) < (II)
4. (III) < (IV) < (II) < (I)

Answer: (III) < (IV) < (II) < (I)

Propane (III) has only weak London dispersion forces, giving bp -42 deg C. Methoxyethane/ether (IV) has a dipole but cannot H-bond as a donor, bp ~7 deg C. Chloroethane (II) has a larger dipole and higher MW, bp ~12 deg C. Ethanol (I) has strong O-H...O hydrogen bonding, bp ~78 deg C. Increasing order: (III) < (IV) < (II) < (I).

Q2. For the molecule with the carbon skeleton CH2=CH-CH2-CH3 type chain bearing four distinct C-H bonds labelled C1-H, C2-H, C3-H and C4-H (where C1-H and C4-H are allylic/secondary sp3 C-H, C2-H is vinylic sp2 C-H, and C3-H is a primary alkyl sp3 C-H), arrange these C-H bonds in the correct order of bond dissociation energy assuming homolytic cleavage. (The radical formed that is most stabilised corresponds to the weakest, lowest-BDE bond.)

1. C2-H > C3-H > C4-H > C1-H
2. C2-H > C3-H > C1-H > C4-H
3. C1-H > C4-H > C2-H > C3-H
4. C1-H > C4-H > C3-H > C2-H

Answer: C2-H > C3-H > C1-H > C4-H

Bond dissociation energy is highest when the radical formed is least stable. A vinylic C-H (C2-H) gives an unstable, high-energy vinyl radical, so it has the highest BDE. A primary alkyl C-H (C3-H) gives an unstabilised primary radical (next highest). The allylic positions (C1-H, C4-H) form resonance-stabilised allylic radicals, so they have the lowest BDE; the more substituted/better-stabilised allylic radical has the lowest BDE of all. This gives the order C2-H > C3-H > C1-H > C4-H.