Which hydrogen compound (E) is easily replaceable during bromination reaction in presence of light?
$ \underset{\delta }{\mathop{C{{H}_{3}}}}\,-\underset{\gamma }{\mathop{C{{H}_{2}}}}\,-\underset{{}}{\mathop{\underset{\beta }{\mathop{CH}}\,}}\,=\underset{\alpha }{\mathop{C{{H}_{2}}}}\,. $
(A) $ \beta -Hydrogen $
(B) $ \gamma -Hydrogen $
(C) $ \delta -Hydrogen $
(D) $ \alpha -Hydrogen $
Answer
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Hint: We know that When alkane is made to react with halogen in the presence of light, it results in the formation of alkyl halide i.e. halogenation and when this is made to react with the lithium and copper iodide , it results in the formation of lithium alkyl cuprate which on further reaction with alkyl halide undergoes Corey-house synthesis reaction.
Complete answer:
Alkanes are the simplest hydrocarbon and are saturated as it doesn’t consist of any double bond or single in it. In the above given statement, is an alkane as there is no double and triple bond present in it. Since, the given alkene consists of three carbon atoms, so it is propane. Free radical substitution reaction is the one that involves the radicals. Reaction of toluene with chlorine in presence of light is one of the examples of free radical substitution reaction. Hydrogen atoms from side chain get replaced by chlorine
The reaction between methylbenzene and chlorine in the presence of light – typically sunlight is a free radical substitution reaction. The organic product is (chloromethyl) benzene. One of the atoms in the methyl group is replaced by a chlorine atom, so this is a substitution reaction. And all the three hydrogens in the methyl group can in turn be replaced by chlorine atoms.
$ \gamma -Hydrogen $ is easily replaceable during bromination reaction in presence of light , because Allylic substitution is being preferred.
Therefore, the correct answer is option B.
Note:
Remember that Corey-house synthesis is the method of increasing the carbon chain and results in the formation of higher alkanes i.e. in this method, higher alkanes are formed from the alkyl lithium curates.
Complete answer:
Alkanes are the simplest hydrocarbon and are saturated as it doesn’t consist of any double bond or single in it. In the above given statement, is an alkane as there is no double and triple bond present in it. Since, the given alkene consists of three carbon atoms, so it is propane. Free radical substitution reaction is the one that involves the radicals. Reaction of toluene with chlorine in presence of light is one of the examples of free radical substitution reaction. Hydrogen atoms from side chain get replaced by chlorine
The reaction between methylbenzene and chlorine in the presence of light – typically sunlight is a free radical substitution reaction. The organic product is (chloromethyl) benzene. One of the atoms in the methyl group is replaced by a chlorine atom, so this is a substitution reaction. And all the three hydrogens in the methyl group can in turn be replaced by chlorine atoms.
$ \gamma -Hydrogen $ is easily replaceable during bromination reaction in presence of light , because Allylic substitution is being preferred.
Therefore, the correct answer is option B.
Note:
Remember that Corey-house synthesis is the method of increasing the carbon chain and results in the formation of higher alkanes i.e. in this method, higher alkanes are formed from the alkyl lithium curates.
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