Question

\[1,2 - \]dibromopropane on treatment with \[X\]moles of \[NaN{H_2}\]followed by treatment with \[{C_2}{H_5}Br\] gives a pentyne. The value of \[X\] is:
A. \[1\]
B. \[2\]
C. \[3\]
D. \[4\]

Answer 1

Hint: Hydrocarbons are well-known in organic chemistry. The hydrocarbons are divided into three categories: alkane, alkene, and alkyne. The term alkane refers to a single carbon-carbon bond. A carbon-carbon double bond exists in the alkene. The term alkyne refers to a carbon-carbon molecule with triple bonds.

Complete Step by Step Solution:
We should also be aware that Markovnikov's rule is an important rule for hydrogen bromide addition. There are certain rules for adding hydrogen and halogen to alkenes in this law. According to this definition, the most electronegative atom in the reagent will be added to the carbon in the alkene or alkyne molecule with the least hydrogen atom. The hydrogen in the reagent will attack the carbon in the alkene or alkyne that has the most hydrogen atoms.
According to Markovnikov's rule, hydrogen in hydrogen bromide should be added to carbon atoms in alkenes with a higher number of hydrogen atoms. Bromide will add carbon to an alkene with a reduced number of hydrogen atoms.
We should write the reaction as follows:

From the above reaction, there are three moles of \[NaN{H_2}\] in the reaction.
As a result, the correct answer is option C \[3\].

Note: Remember that oxidation and reduction are used to convert one form of hydrocarbon to another. Alkane is oxidised to produce alkene. Alkene is oxidised to produce alkyne. Alkyne is reduced to an alkene in this process. The process of converting alkene to alkane. There are some general formulas in it.