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Question: The structure of \( 3 - \) Bromoprop \( - 1 - \) ene is: (A) \( C{H_3} - C(Br) = C{H_2} \) (B) ...

The structure of 3−3 - Bromoprop −1−- 1 - ene is:
(A) CH3−C(Br)=CH2C{H_3} - C(Br) = C{H_2}
(B) CH3−CH=CH−BrC{H_3} - CH = CH - Br
(C) CH3−C(Br)=CH−BrC{H_3} - C(Br) = CH - Br
(D) Br−CH2−CH=CH2Br - C{H_2} - CH = C{H_2}

Explanation

Solution

Hint : As we know that the allyl bromide is an organic halide. It is an alkylating agent used in synthesis of polymers and other organic compounds. Physically, allyl bromide is a colourless liquid with an intense, acrid and persistent smell. We know that the preferred IUPAC name is 3−3 - Bromoprop −1−- 1 - ene.

Complete Step By Step Answer:
We know that the parent hydrocarbon is prop −1−- 1 - ene. It contains 33 carbon atoms and one double bond. BrBr is substituent present at the third carbon atom.
In this compound, according to the IUPACIUPAC rule, the double bond is given preference over the bromo group. We can write the structure of 3−3 - Bromoprop −1−- 1 - ene as Br−CH2−CH=CH2Br - C{H_2} - CH = C{H_2} .
Hence the correct option is (D) Br−CH2−CH=CH2Br - C{H_2} - CH = C{H_2} .

Note :
Before solving this kind of question we should be fully aware of their categories and their IUPAC names. IUPAC naming is a method to name chemical compounds as recommended by the International Union of Pure and Applied Chemistry.