Answer
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Hint: The reaction involved is electrophilic addition reaction .Compared to alkanes or alkynes, alkenes are more reactive. Also those alkenes which are substituted with electron releasing alkyl groups will be more reactive than unsubstituted alkenes.
Complete answer:
- As we know, bromine is a halogen. Hence in the given reaction, addition of halogen is taking place and usually it’s an electrophilic addition reaction. An electrophile (electron lover) can be defined as a molecule with a tendency to react with other molecules which contain a donatable pair of electrons.
- In alkenes there is one σ and one π bond. The π bond is a weaker one and it can break more easily and thus alkenes are more reactive towards electrophilic addition of halogens than alkanes which contain only σ bonds.
- In alkynes, one σ bond and two π bonds are present. Since there are two π bonds, alkynes are stronger than alkenes. They can react with bromine or halogens but will be less reactive than alkenes.
- In alkanes only σ bond is present and its less reactive electrophilic addition reaction. Thus the order of reactivity towards bromine is in the following order
\[Alkenes>Alkynes>Alkanes\]
- Let’s look at the options. (A) $C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}$ is an alkane and thus it will be least reactive one towards bromine. (C) $CH\equiv CH$ is an alkyne and it will be more reactive than the alkane but less reactive than the alkenes in option (B) and (D).
- (B) $C{{H}_{2}}=C{{H}_{2}}$ and (D) $C{{H}_{3}}-CH=C{{H}_{2}}$ are both alkenes. In the ethylene molecule ($C{{H}_{2}}=C{{H}_{2}}$), on bond breaking the electron density on both the carbons will be equal.
- In the propene molecule ($C{{H}_{3}}-CH=C{{H}_{2}}$), on bond breaking the electron density will be more on the substituted carbon atom and it will have a high reactivity towards the bromine atom than the ethylene molecule.
- Thus the molecule $C{{H}_{3}}-CH=C{{H}_{2}}$ will react most readily with bromine.
Therefore the answer is option (D) $C{{H}_{3}}-CH=C{{H}_{2}}$
Note: When we look into the mechanism of Electrophilic addition of bromine to alkenes, we can see that, as $B{{r}_{2}}$ molecule approaches a double bond of the alkene, the electrons in the double bond will repel electrons in bromine molecule causing polarization of the bromine bond. This will create a dipolar moment in the bromine molecule bond and heterolytic bond cleavage will occur and one of the halogens will obtain positive charge and react as an electrophile.
Complete answer:
- As we know, bromine is a halogen. Hence in the given reaction, addition of halogen is taking place and usually it’s an electrophilic addition reaction. An electrophile (electron lover) can be defined as a molecule with a tendency to react with other molecules which contain a donatable pair of electrons.
- In alkenes there is one σ and one π bond. The π bond is a weaker one and it can break more easily and thus alkenes are more reactive towards electrophilic addition of halogens than alkanes which contain only σ bonds.
- In alkynes, one σ bond and two π bonds are present. Since there are two π bonds, alkynes are stronger than alkenes. They can react with bromine or halogens but will be less reactive than alkenes.
- In alkanes only σ bond is present and its less reactive electrophilic addition reaction. Thus the order of reactivity towards bromine is in the following order
\[Alkenes>Alkynes>Alkanes\]
- Let’s look at the options. (A) $C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}$ is an alkane and thus it will be least reactive one towards bromine. (C) $CH\equiv CH$ is an alkyne and it will be more reactive than the alkane but less reactive than the alkenes in option (B) and (D).
- (B) $C{{H}_{2}}=C{{H}_{2}}$ and (D) $C{{H}_{3}}-CH=C{{H}_{2}}$ are both alkenes. In the ethylene molecule ($C{{H}_{2}}=C{{H}_{2}}$), on bond breaking the electron density on both the carbons will be equal.
- In the propene molecule ($C{{H}_{3}}-CH=C{{H}_{2}}$), on bond breaking the electron density will be more on the substituted carbon atom and it will have a high reactivity towards the bromine atom than the ethylene molecule.
- Thus the molecule $C{{H}_{3}}-CH=C{{H}_{2}}$ will react most readily with bromine.
Therefore the answer is option (D) $C{{H}_{3}}-CH=C{{H}_{2}}$
Note: When we look into the mechanism of Electrophilic addition of bromine to alkenes, we can see that, as $B{{r}_{2}}$ molecule approaches a double bond of the alkene, the electrons in the double bond will repel electrons in bromine molecule causing polarization of the bromine bond. This will create a dipolar moment in the bromine molecule bond and heterolytic bond cleavage will occur and one of the halogens will obtain positive charge and react as an electrophile.
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