Question:

The isomeric deuterated bromide with molecular formula $C _4 H _8 DBr$ having two chiral carbon atoms is

Updated On: Mar 21, 2025
  • 2 - Bromo - 1 - deuterobutane
  • 2 - Bromo - 2 - deuterobutane
  • 2 - Bromo - 3 - deuterobutane
  • 2 - Bromo-1 - deutero - 2 - methylpropane
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The Correct Option is C

Approach Solution - 1

The correct answer is (C) : 2 - Bromo - 3 - deuterobutane

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Approach Solution -2

The compound described has two chiral centers, which means that two carbon atoms are each attached to four different groups. The structure consists of a four-carbon chain with a deuterium atom and a bromine atom located on separate carbons. Based on the given information, the correct name for the compound is 2-Bromo-3-deuterobutane, where: The bromine is attached to the second carbon in the chain, and The deuterium (D) is attached to the third carbon.
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Concepts Used:

Preparation of Haloalkanes

  1. Preparation by Alcohols: By using alcohol, we can very easily prepare Haloalkane as when R-OH is reacted with a suitable reagent it will form R-X. The following reagents will help in the formation of the reactions:
    1. Halogen Acids (HX)
    2. Thionyl chloride (SOCl2)
    3. Phosphorous halides (PX5 or PX3)
  2. Preparation by Free Radical Halogenation: The formation of alkyl bromides and alkyl chloride is very much possible with the help of free radical halogenation, but as these radicals are highly reactive and non-specific in nature, they can result in the formation of a mixture of products. Though, it's not a preferred method for the preparation of haloalkanes. For example, free radical chlorination can result in the formation of a number of haloalkanes, which in turn makes it difficult to set apart a single product.
  3. Preparation by Alkenes: An electrophilic addition reaction can be used to transform an alkene into a haloalkane as alkene will react with HX to form R-X.

Read More: Haloalkanes and Haloarenes