Dehydrohalogenation of an alkyl halide does not give an alkane. Let's understand this reaction in detail:

Dehydrohalogenation of an alkyl halide:
- In this reaction, an alkyl halide is treated with a strong base like KOH, NaOH, or NaNH2 to form an alkene.
- The halogen atom is removed along with a hydrogen atom from the adjacent carbon atoms.
- The reaction follows an E2 mechanism, where the base attacks the beta-hydrogen and eliminates the leaving group in a single step.

Example: CH3CH2Br + KOH → CH2=CH2 + KBr + H2O

Why doesn't it give an alkane?
- Alkanes are saturated hydrocarbons, meaning they have only single bonds between carbon atoms.
- Dehydrohalogenation reaction involves the formation of a double bond between the carbon atoms, thus forming an alkene.
- This reaction does not involve the addition of hydrogen atoms, which is necessary to form an alkane.

Other options:
a) Catalytic hydrogenation of alkenes:
- This reaction involves the addition of hydrogen gas to an alkene in the presence of a metal catalyst like Pt, Pd or Ni.
- The double bond is broken, and hydrogen atoms are added to the carbon atoms, forming an alkane.
- This reaction is the reverse of dehydrohalogenation reaction.

b) Wurtz reaction:
- This reaction involves the coupling of two alkyl halides in the presence of sodium metal to form a longer chain alkane.
- Example: 2CH3Cl + 2Na → CH3-CH3 + 2NaCl

c) Kolbe's electrolytic method:
- This reaction involves the electrolysis of a solution of sodium or potassium salt of a carboxylic acid to form an alkane with an even number of carbon atoms.
- Example: 2CH3COO-Na+ → CH3-CH3 + 2CO2 + 2Na+

In conclusion, dehydrohalogenation of an alkyl halide does not give an alkane because it involves the formation of a double bond between carbon atoms, whereas alkanes have only single bonds.