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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
Complete the following:
\[\ce{2 - butyne ->[Lindlar Catalyst]}\]
\[\begin{array}{cc}
\phantom{......................................................}\ce{CH3}\phantom{............}\ce{Br}\phantom{}\\
\phantom{....................................................}|\phantom{.............}/\phantom{}\\
\ce{A + Mg ->[Dry][ether] CH3-Mg-Br + B ->[i. Dry ether][ii. H-OH/H^+] CH3-C-OH + Mg}\\
\phantom{.....................................................}|\phantom{.............}\backslash\phantom{}\\
\phantom{......................................................}\ce{CH3}\phantom{..........}\ce{OH}\phantom{}
\end{array}\]
Thus, 'B' in the above reaction is:
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\ce{CH3}\phantom{....}\\
|\phantom{.......}\\
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|\phantom{.......}\\
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\end{array}\]?
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