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Chapter 15: Q19. (page 570)

Question: Draw the structure of the four allylic halides formed when 3-methylcyclohexene undergoes allylic halogenation with NBS + hν.

Short Answer

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Halogenation

Treatment of compound with the halogen yield substituted product. It involves the presence of light.

Formation of products

The reaction of methylcyclohexene with NBS in the presence of light. The reaction is shown below:

Formation of products

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Most popular questions from this chapter

Question:Draw all the monochlorination products that are formed from (S,S)-1,2-dimethylcyclopropane.

Question. Devise a synthesis of OHC(CH2)4CHO from cyclohexane using any required organic or inorganic reagents.

Question: What reagents are needed to convert 1-ethylcyclohexene into (a) 1-bromo-2-ethylcyclohexane;

(b) 1-bromo-1-ethylcyclohexane; (c) 1,2-dibromo-1-ethylcyclohexane?

Question: Five isomeric alkanes (A–E) having the molecular formula C₆H₁₄ are each treated with Cl₂ + hv to give alkyl halides having molecular formula C₆H₁₃Cl . A yields five constitutional isomers. B yields four constitutional isomers. C yields two constitutional isomers. D yields three constitutional isomers, two of which possess stereo-genic centres. E yields three constitutional isomers, only one of which possesses a stereo-genic centre. Identify the structures of A–E.

Question:Treatment of a hydrocarbon A (molecular formula C₉H₁₈) with Br₂ in the presence of light forms alkyl halides B and C, both having molecular formula C9H17Br. Reaction of either B or C with KOC (CH₃)₃ forms compound D ( C₉H₁₆) as the major product. Ozonolysis of D forms cyclohexanone and acetone. Identify the structures of A–D.