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Question: One mole of \[1,2-\] Dibromopropane on treatment with X moles of \[NaN{{H}_{2}}\]followed by treatme...

One mole of 1,2−1,2- Dibromopropane on treatment with X moles of NaNH2NaN{{H}_{2}}followed by treatment with ethyl bromide gives a pentyne. The value of X is:
A.One.
B.Two.
C.Three.
D.Four.

Explanation

Solution

We know that in organic chemistry, hydrocarbons are an important topic. The hydrocarbons are majorly classified as three groups. There are alkane, alkene and alkyne. The alkane means carbon-carbon single bond. The alkene has a carbon-carbon double bond. The alkyne means carbon-carbon having triple bonds in the molecule.

Complete answer: We also need to know that Markownikoff's rule is one of the important rules for hydrogen bromide addition. In this law some rules for addition of hydrogen and halogen in alkene. According to this halogen that means the most electronegative atom in the reagent going to addition in the carbon having least hydrogen atom in the alkene or alkyne molecule. The hydrogen in the reagent is going to attack the carbon having more number of hydrogen atoms in the alkene or alkyne. According to Markownikoff’s rule, hydrogen in hydrogen bromide to add in carbon atoms having more number of hydrogen accounts in alkene. Bromide is going to add in carbon having less number of hydrogen accounts in alkene. We have to know that the functional group of the molecule is very important for the naming time. If one molecule having more than one functional group means some priority list is also provided from IUPAC. If the number of substituent and functional group is dependent on the minimum number will come from the left side or right of the terminal carbon atom in the chain.
CH3CH(Br)CH2(Br)→2NaNH2CH3C≡CH→NaNH2CH3−C≡C−Na+C{{H}_{3}}CH(Br)C{{H}_{2}}\left( Br \right)\xrightarrow{2NaN{{H}_{2}}}C{{H}_{3}}C\equiv CH\xrightarrow{NaN{{H}_{2}}}C{{H}_{3}}-C\equiv {{C}^{-}}N{{a}^{+}}
⇒X=3.\Rightarrow X=3.
Therefore, the correct answer is option C.

Note:
Remember that the conversion of one type of hydrocarbon to another hydrocarbon by oxidation and reduction. The oxidation of alkane to give alkene. The oxidation of alkene to give alkyne. The reduction of alkyne to give alkene. The reduction of alkene to give alkane. It has some general formulas.