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During free radical bromination of isobutane, an alkene is produced as by product via disproportionation of the intermediate alkyl free radical. What is this alkene?
- a)1-butene
- b)2-butene
- c)2-methyl propene
- d)Both ‘b’ and ‘c’
Correct answer is option 'C'. Can you explain this answer?
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During free radical bromination of isobutane, an alkene is produced as...
As shown in the above mechanism 1-butene and 2-butene cannot be formed by this free radical. release of 1 hydrogen radical gives 2-methyl propene as disproportionated product.
Hence, the correct option is C.
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During free radical bromination of isobutane, an alkene is produced as...
The alkene produced as a byproduct during the free radical bromination of isobutane is 2-methyl propene (option c). Let's understand the process in detail:
1. Free Radical Bromination:
During free radical bromination, isobutane (C4H10) reacts with a bromine radical (Br•) to form a primary alkyl radical (C4H9•). This reaction is initiated by the presence of light or heat.
C4H10 + Br• → C4H9• + HBr
2. Disproportionation of the Intermediate Alkyl Free Radical:
The intermediate alkyl free radical (C4H9•) can undergo different reactions, including abstraction of a hydrogen atom from another isobutane molecule or rearrangement. However, in this case, it undergoes disproportionation to form a new alkene.
C4H9• → Alkene + Alkane
3. Formation of 2-Methyl Propene:
The disproportionation of the C4H9• radical leads to the formation of 2-methyl propene. This alkene is obtained when the alkyl radical undergoes a β-scission, resulting in the formation of a double bond.
C4H9• → 2-methyl propene + Alkane
Hence, during the free radical bromination of isobutane, 2-methyl propene is produced as a byproduct through the disproportionation of the intermediate alkyl free radical.
To summarize:
- Isobutane reacts with a bromine radical to form a primary alkyl radical.
- The primary alkyl radical can undergo various reactions, but in this case, it undergoes disproportionation.
- The disproportionation of the alkyl radical leads to the formation of 2-methyl propene as a byproduct.
1. Free Radical Bromination:
During free radical bromination, isobutane (C4H10) reacts with a bromine radical (Br•) to form a primary alkyl radical (C4H9•). This reaction is initiated by the presence of light or heat.
C4H10 + Br• → C4H9• + HBr
2. Disproportionation of the Intermediate Alkyl Free Radical:
The intermediate alkyl free radical (C4H9•) can undergo different reactions, including abstraction of a hydrogen atom from another isobutane molecule or rearrangement. However, in this case, it undergoes disproportionation to form a new alkene.
C4H9• → Alkene + Alkane
3. Formation of 2-Methyl Propene:
The disproportionation of the C4H9• radical leads to the formation of 2-methyl propene. This alkene is obtained when the alkyl radical undergoes a β-scission, resulting in the formation of a double bond.
C4H9• → 2-methyl propene + Alkane
Hence, during the free radical bromination of isobutane, 2-methyl propene is produced as a byproduct through the disproportionation of the intermediate alkyl free radical.
To summarize:
- Isobutane reacts with a bromine radical to form a primary alkyl radical.
- The primary alkyl radical can undergo various reactions, but in this case, it undergoes disproportionation.
- The disproportionation of the alkyl radical leads to the formation of 2-methyl propene as a byproduct.
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During free radical bromination of isobutane, an alkene is produced as by product via disproportionation of the intermediate alkyl free radical. What is this alkene?a)1-buteneb)2-butenec)2-methyl propened)Both ‘b’ and ‘c’Correct answer is option 'C'. Can you explain this answer?
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