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Solved Examples on Organic Compounds Containing Halogens

Question 1: Compound (A),C₈H₉Br

, gives white precipitate when warmed with alcoholic

AgNO₃.

Oxidation of (A)

gives an acid (B), C₈H₆O₄

.(B) easily forms anhydride on heating. Identify the

compound (A). (IIT-JEE 2013)

Answer: C

Solution: It is clear from the molecular formula and also from the given options that the compound contains benzene ring. As it gives white ppt on heating with alcoholic AgNO₃, the Br atom must be linked to a sp3 hybridised carbon i.e. it is not linked directly with benzene ring. Further the formation of anhydride indicates that the two carboxylic groups and therefore two alkyl groups are present at adjacent positions. So, the only possible structure is C.

Hence, the correct option is C.

Question 2: The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amine and an alkyne. The bromoalkane and alkyne respectively are (IIT-JEE 2010)

a. BrCH₂CH₂CH₂CH₂CH₃ and CH₃CH₂C≡CH
b. BrCH₂CH₂CH₃ andCH₃CH₂CH₂C≡CH
c. BrCH₂CH₂CH₂CH₂CH₃ and CH₃C≡CH
d. BrCH₂CH₂CH₂CH₃ andCH₃CH₂C≡CH

Answer:d

Solution: 3-octyne contains a triple bond at carbon number 3. This indicates that both the reacting molecules must contain four carbon atoms. The only option in which both the molecules have four carbon atoms is d. Reactions involved are
CH₃CH₂C≡CH+NaNH₂ →CH₃CH₂C≡CNa
CH₃CH₂C≡CNa +BrCH₂CH₂CH₂CH₃ →CH₃CH₂C≡CCH₂CH₂CH₃

Question 3: The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is: (IIT-JEE 2011)

Answer:5

Solution: Dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH results in beta elimination. The given molecules contain three c atoms at beta position to Br atom. Hence, three products are formed.

Out of these three products , A and B exist in two isomeric form each i.e. E and Z forms. So the total number of possible products is 5.
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