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Question: An alkyl halide with molecular formula \({C_6}{H_{12}}Br\) on dehydrohalogenation gives two isomeric...

An alkyl halide with molecular formula C6H12Br{C_6}{H_{12}}Br on dehydrohalogenation gives two isomeric alkenes X and Y with molecular formula C6H12{C_6}{H_{12}} on reductive ozonolysis X and Y gives compounds CH3COCH3,CH3CHO,CH3CH2CHOC{H_3}COC{H_3},C{H_3}CHO,C{H_3}C{H_2}CHO and (CH3)2CHCHO{\left( {C{H_3}} \right)_2}CHCHO the alkyl halide is-
(A) 4-bromo-2-methylpentane.
(B) 3-bromo-2-methylpentane
(C) 2-bromo-2,3 dimethylbutane.
(D) 2,2-dimethyl-1-bromobutane.

Explanation

Solution

Hint: The reaction follows β\beta - dehydrohalogenation which is characterized as an elimination reaction. Let’s consider the structure of ethyl bromide CH3−CH2−BrC{H_3} - C{H_2} - Br. Now, we know that β\beta - dehydrohalogenation means hydrogen atom will be eliminated from the β\beta position.

Complete step by step solution:

After the reaction with alcoholic KOH on heating gives-
(Br attached to the carbon atom will be named as the α\alpha while the other atom will be named as β\beta .
⇒CH2−CH2−Br→−HBrAlc.KOH/ΔCH2=CH2\Rightarrow C{H_2} - C{H_2} - Br\xrightarrow[{ - HBr}]{{Alc.KOH/\Delta }}C{H_2} = C{H_2}
Now, the second step is given as ozonolysis. Whenever you react any alkene with ozone in the presence of dichloromethane followed by zinc hydrolysis which means reductive hydrolysis, it makes an ozonide complex.
Let’s first consider 3-bromo-2-methylpentane-
(we will have two beta positions and one alpha position. This alpha position will be the one carbon to which bromide is attached and the beta carbons are the carbons present on the right and left side of the alpha carbon. This means that there will be two compounds formation)
⇒CH3−CH(CH3)−CH(Br)−CH2CH3→−HBralc.KOHCH3−C(CH3)=CH−CH2CH3\Rightarrow C{H_3} - CH(C{H_3}) - CH(Br) - C{H_2}C{H_3}\xrightarrow[{ - HBr}]{{alc.KOH}}C{H_3} - C(C{H_3}) = CH - C{H_2}C{H_3}
Let this product be product 1.
If we eliminate hydrogen from the beta position present on the right side-
⇒CH3−CH(CH3)−CH(Br)−CH2CH3→−HBralc.KOHCH3C(CH3)−CH=CH−CH3\Rightarrow C{H_3} - CH(C{H_3}) - CH(Br) - C{H_2}C{H_3}\xrightarrow[{ - HBr}]{{alc.KOH}}C{H_3}C(C{H_3}) - CH = CH - C{H_3}
Let this product be our product number second.
Now, when we make both the products go through ozonolysis, we get-
The first product after the ozonolysis will be will be-
⇒(CH3)2C=O+OHC−CH2CH3\Rightarrow {\left( {C{H_3}} \right)_2}C = O + OHC - C{H_2}C{H_3}
The second product after the ozonolysis will be will be-
⇒(CH3)2CH−CHO+CH3CHO\Rightarrow {\left( {C{H_3}} \right)_2}CH - CHO + C{H_3}CHO

This clearly gives us the answer that option B will be the correct option.

Note: Ozonolysis is an organic chemical reaction to determine the position of a double carbon-carbon bond in unsaturated compounds. The ozone compound contributes to the production of ozonide and the ozonide produces a combination of aldehydes , ketones or carboxylic acids after the process of hydrogenation or treatment with acid.