Alkanes

Alkyl halide reacts with sodium in presence of dry ether forms alkane is known as Wurtz reaction.

The general formula of a cycloalkane is .

Step 1: Understanding Bond Fission.
Chemical reactions involve the breaking of existing bonds and the formation of new bonds. The breaking of a covalent bond, known as bond fission, can occur in two ways:

• Homolytic Fission (Homolysis): The shared pair of electrons is distributed equally between the two separating atoms. This results in the formation of neutral species with an unpaired electron, called free radicals. These reactions are typically initiated by heat, light, or peroxides and are called homopolar reactions.
  Example:
  
• Heterolytic Fission (Heterolysis): The shared pair of electrons is transferred completely to one of the atoms. This results in the formation of a positively charged ion (cation) and a negatively charged ion (anion). This type of fission occurs in polar bonds and is often facilitated by polar solvents. Reactions involving these charged intermediates are called heteropolar reactions or ionic reactions.
  Example: (if B is more electronegative)

Step 2: Identifying Intermediates in Heteropolar Reactions.
As defined above, heteropolar reactions proceed through heterolytic bond fission. This process generates charged intermediates. Depending on which atom takes the electron pair, either a cation or an anion (or both, in different steps of a mechanism) can be formed as intermediates. For example, in electrophilic addition to an alkene, a carbocation is formed. In nucleophilic substitution, a carbocation (S 1) or a transition state with developing charges is formed. In reactions involving carbanions, an anion is the intermediate. Therefore, the intermediates in heteropolar reactions are charged species, which can be either cations or anions.
Step 3: Final Answer.
The intermediates in heteropolar reactions are both anions and cations.

Step 1: Analyzing Statement I.
Kjeldahl's method is a classical method for the quantitative determination of nitrogen in organic compounds. However, it has limitations. The method is not applicable to compounds where the nitrogen atom is part of a heterocyclic ring (like pyridine), or in nitro (-NO ) and azo (-N=N-) groups. This is because the nitrogen in these compounds is very stable and is not completely converted to ammonium sulfate under the conditions of the digestion step. Therefore, the statement that nitrogen in pyridine cannot be estimated by Kjeldahl's method is true.
Step 2: Analyzing Statement II.
This statement claims that the nitrogen in pyridine *does* change to ammonium sulfate when heated with concentrated H SO . This contradicts the reason why Kjeldahl's method fails for pyridine. The nitrogen atom in the pyridine ring is part of an aromatic system, which makes it very resistant to the acid digestion. It does not get converted to ammonium sulfate. Therefore, Statement II is false.
Step 3: Conclusion.
Statement I is true because Kjeldahl's method has limitations and does not work for heterocyclic compounds like pyridine. Statement II is false because the reason the method fails is precisely that the nitrogen is *not* converted to ammonium sulfate.
Step 4: Final Answer.
Statement I is true but Statement II is false.