Question
Question: Which of the following reactions do not present in the major product of the given Birch reductions? ...
Which of the following reactions do not present in the major product of the given Birch reductions?
(A) (i), (iii), (iv)
(B) (iv), (vi), (vii)
(C) (iv), (v), (vi)
(D) (i), (ii), (v), (vii)
Solution
Hint : The Birch Reduction is the process in which benzene (and its aromatic relatives) is converted to 1,4− cyclohexadiene using sodium (or lithium) as a reducing agent in liquid ammonia as solvent (boiling point: −33∘C ) in the presence of an alcohol such as ethanol, methanol or t-butanol. Here, Na acts as a reducing agent in the presence of liquid NH3 .
Complete Step By Step Answer:
Reaction (i) satisfies the Birch reduction as there is a formation of 1,4− cyclohexadiene ring from a benzene ring.
Reaction (ii) also satisfies the Birch reduction as benzene is simply converted to 1,4− cyclohexadiene.
Reaction (iii) also satisfies the Birch reduction as Toluene (an aromatic compound) is converted to 1− methyl −1,4− cyclohexadiene.
Reaction (iv) does not satisfy the Birch reduction. In Benzoic acid, COOH is an electron withdrawing group. On treatment with Na in the presence of NH3 , it results in the protonation of carbon which has the electron withdrawing group, like COOH , where the electron is transferred to oxygen (which is more stable) due to regioselectivity. However, in birch reduction, only the phenyl group needs to be protonated and not the functional groups.
Reactive (v) satisfies the birch reduction. Here, naphthalene gets converted into 1,4− dihydronaphthalene.
Reactive (vi) does not satisfy the birch reduction. Here, anthracene only loses one double bond and forms conjugate to become 9,10− dihydroanthracene.
In reaction (vii), neither the reactants nor the products have a benzene ring or any other aromatic compound. It does not qualify for birch reduction.
Hence, the correct option is B.
Note :
You should be clear that birch reduction occurs only with Na/NH3 . It should not be confused with NaNH2/NH3 . Also, you might sometimes replace C−H with anyne, NH2 which is again incorrect.