(a) In general conception, as the branching increases packing of the molecules in the crystal lattices becomes less close and hence melting point decreases accordingly.
(b) As the branching increases, surface area decreases and thus magnitude of van der Waals forces of attraction decreases and hence the boiling point decreases. 2, 2-dimethyl butane has lower surface area due to more branching and hence has lower boiling point.

The process by which simple molecules join together to form large molecules is known as polymerization.
Simple alkenes polymerize to form long chain addition polymers.
For example, ethylene gives polyethylene.

(i) Alkanes - C₆H₁₄, C₈H₁₈
(ii) Alkenes - C₄H₈
(iii) Alkynes - C₅H₈
(iv) Arenes - C₆H₆

In methane, carbon forms have single bonds with four hydrogen atoms. Since the carbon atom is attached to four other atoms, it uses sp³ hybrid orbitals to form these bonds. Hybridization of 'C' is sp³

Shape-tetrahedral having bond angle = 109.5°
In ethane, there are six C - H covalent bonds and one C-C covalent bond.
The C-H bond is the result of overlap of an sp³ hybrid orbital from carbon and s-orbital from hydrogen.
Orbital structure can be shown as

Resonance energy is the difference in energy between actual structure of compound and most stable resonating structure. The resonance energy of benzene is 150.325 J mol^-1.

For preparation of alkanes containing odd number of carbon atoms, a mixture of two alkyl halides has to be used. Since two alkyl halides can react in three different ways, therefore, a mixture of three alkanes instead of the desired alkane would be formed. For example, the Wurtz reaction between 1-bromopropane and 1-bromobutane gives a mixture of three alkanes i.e., hexane, heptane and octane as shown below:

The basic skeleton 2-methylbutane is 

Putting double bonds at various different positions and satisfying the tetracovalency of each carbon, the structures of various alkenes which give 2-methylbutane on hydrogenation are:

The hybridization state of carbon in these three compounds is:

Since s-electrons are closer to the nucleus, therefore, as the s-character of the orbital making the C-H bond increases, the electrons of C-H bond lie closer and closer to the carbon atom. In other words, the partial +ve charge on the H-atom and hence the acidic character increases as the s-character of the orbital increases. Thus, the acidic character decreases in the order: Ethyne > Benzene > Hexane.

The necessary conditions for a molecule to be aromatic are:
(i) It should have a single cyclic cloud of delocalised \(\pi\)-electrons above and below the plane of the molecule.
(ii) It should be planar. This is because complete delocalization of \(\pi\)-electrons is possible only if the ring is planar to allow cyclic overlap of p-orbitals.
(iii) It should contain Huckel number of electrons, i.e., (4n + 2) \(\pi\)-electrons where n = 0, 1, 2, 3 ................ etc.
A molecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.

Resonance and delocalisation of electrons generally lead to the stability of benzene molecule.

The dotted circle in the hybrid structure represents the six electrons which are delocalised between the six carbon atoms of the benzene ring. Therefore, presence of delocalised π-electrons in benzene makes it more stable than the hypothetical cyclohexatriene.

(i) An aldehyde with molar mass of 44 u is ethanal, CH₃CH = O
(ii) Write two moles of ethanal side by side with their oxygen atoms pointing towards each other.
CH₃CH = O      O = CHCH₃
   Ethanal            Ethanal
(ii) Remove the oxygen atoms and join them by a double bond, the structure of alkene 'A' is

As required, but-2-ene has three C-C, eight C-H \(\sigma \)-bonds and one C-C \(\pi\)-bond.

(a) Isomers of C₄H₈ having one double bond are:

(b) (i) \({ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }\overset { 2 }{ C } \equiv \overset { 1 }{ C } H\)
                   Pent-1-yne
(ii) \({ CH }_{ 3 }{ CH }_{ 2 }-C\equiv \overset { 2 }{ C } -\overset { 1 }{ C } { H }_{ 3 }\)
               Pent-2-yne
(iii) 

Chlorination of methane is a free radical reaction which occurs by the following mechanism:
   .........(i)
Termination \({ \overset .{ CH } }_{ 3 }+{ .CH }_{ 3 }\rightarrow { CH }_{ 3 }-{ CH }_{ 3 }\) .....(ii)
                                                          Ethane
\(\overset .{ C } H_{ 3 }+\overset .{ C } I\rightarrow { CH }_{ 3 }-CI\)
\(\overset { . }{ C } I+\overset { . }{ C } I\rightarrow CI-CI\)     ..(iii)
From the above mechanism, it is evident that during propagation step, \(\overset { . }{ C } H_{ 3 }\)free radicals are produced which may undergo three reactions, i.e., (i) - (iii). In the chain termination step, the two CH³ free radicals combine together to form ethane (CH₃-CH₃) molecule.