Hydrocarbons Test - 5

Among molecules having similar mass, the strongest intermolecular hydrogen interactions exist in alcohols. Hence, methanol will have the highest boiling point among the given molecules.
The given options in the increasing order of boiling points are as follows:
Butane (intermolecular interactions are only van der Waals forces) < Difluoromethane (additional dipole-dipole interactions come into play) < Methylamine (H- bonds) < Methanol (stronger H-bonds)

The C-C chain is zigzag and not linear. Hence, statement C is incorrect.
Some examples of alkanes are methane, ethane and propane. First three members do not exhibit isomerism, but the higher members (butane and onwards) exhibit chain isomerism. Hence, statement D is incorrect.

The boiling points of alkanes decrease with the decrease in atomic mass and for isomeric alkanes it decreases with the increase in branching.
Hence, the correct decreasing order of boiling points of given alkanes is: n-pentane > 2-methylbutane > 2, 2-dimethylpropane > n-butane.

β- Elimination proceeds through the formation of a carbocation. The order of reactivity is the same as the order of stability of the intermediate carbocation.
The decreasing order of the rate of β-elimination reaction with alcoholic KOH of given alkyl halides is A > C > B.

The fully eclipsed conformation is clearly the highest in energy and least favourable, since the largest groups are interacting directly with each other. As the molecule rotates, it adopts the relatively stable gauche conformation. As it continues to rotate, it encounters less favorable eclipsed conformation, in which a methyl group eclipses a hydrogen. As rotation continues, the molecule comes to the anti-conformation, which is the most stable since the substituents are staggered and the methyl groups are as far away from each other as possible.

In the staggered conformation, the C-H bonds on the rear carbon lie between those on the front carbon with dihedral angles of 60 degrees.