Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb
enzeneb. 1-methyl-3-decylbenzenec. 1-methyl-4-decylbenzened. None of these
2 answers:
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
Answer:
c. 1-methyl-4-decylbenzene
Explanation:
Hello,
On the attached document you will find the major product for the stated chemical reaction even thought the listed product are possible, nonetheless, by cause of the steric hindrance, the most probable and abundant product is the shown one, 1-methyl-4-decylbenzene, as it has more space for the decyl to become part of the ring. Such reaction is a typical Friedel-Crafts alkylation of an aromatic compound whereas aluminium chloride is used as the catalyst to attach the alkyl chloride to the aromatic ring.
Best regards.
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