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प्रश्न
An alkyl halide with molecular formula C6H13Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and Y gave four compounds CH3COCH3, CH3CHO, CH3CH, CHO and (CH3)2 CHCHO. Find the alkyl halide.
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उत्तर
1. C6H13Br is 3 – Bromo – 4 methylpentanc.
\[\begin{array}{cc}\ce{Br}\\
|\\\ce{CH3 - CH - C - CH2 - CH3}\\
|\phantom{......}|\phantom{.......}\\
\ce{CH3}\phantom{...}\ce{H}\phantom{.......}
\end{array}\]
2. 3 – Bromo -4 methylpentane on dehydrogenation give two isomers X and Y as follows:
Therefore C6H13 Br is 3 – Bromo – 4 – methy ipentane.
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संबंधित प्रश्न
Write the reaction to convert propene to n-propyl alcohol.
What is the action of following on but-2-ene?
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\ce{CH3}\phantom{....}\\
|\phantom{.......}\\
\ce{CH3-C-CH = CH2}\\
|\phantom{.......}\\
\ce{H3C}\phantom{.........}
\end{array}\]?
What is the action of following on but-2-ene?
dil alkaline KMnO4
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Read the following reaction and answer the questions given below.
\[\begin{array}{cc}
\phantom{...............................}\ce{CH3}\\
\phantom{............................}|\\
\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide] CH3 - CH - CH2Br}\\
|\phantom{....................................}\\
\ce{CH3}\phantom{.................................}\\
\end{array}\]
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What is the action of following on but-2-ene?
dil alkaline \[\ce{KMnO4}\]
