An alkene X (C_6H_{12}) on ozonolysis gives a | vedclass

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(A) $1$. Ozonolysis of alkene $X$ $(C_6H_{12})$ yields acetaldehyde $(CH_3CHO)$ and ethyl methyl ketone $(CH_3COCH_2CH_3)$.$2$. By reversing the ozono

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$3-$bromo$-3-$methylpentane

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(A) $1$. Ozonolysis of alkene $X$ $(C_6H_{12})$ yields acetaldehyde $(CH_3CHO)$ and ethyl methyl ketone $(CH_3COCH_2CH_3)$.$2$. By reversing the ozonolysis process,we can determine the structure of $X$ by joining the carbonyl carbons with a double bond: $CH_3CH=C(CH_3)CH_2CH_3$.$3$. The structure of $X$ is $3-methylpent-2-ene$.$4$. When $X$ reacts with $HBr$,the reaction follows Markovnikov's rule,where the $H^+$ adds to the carbon with more hydrogen atoms and $Br^-$ adds to the more substituted carbon.$5$. The reaction is: $CH_3CH=C(CH_3)CH_2CH_3 + HBr \rightarrow CH_3CH_2-C(Br)(CH_3)CH_2CH_3$.$6$. The product formed is $3-bromo-3-methylpentane$.

An alkene $X$ $(C_6H_{12})$ on ozonolysis gives acetaldehyde and ethyl methyl ketone. What is the product formed when $X$ reacts with $HBr$?

Similar Questions

Arrange the following compounds in the decreasing order of their reactivity towards electrophilic addition of $HBr$:
$(a)$ $CH_3-C(CH_3)=CH_2$
$(b)$ $CH_3O-C(CH_3)=CH_2$
$(c)$ $CH_3OCH_2-C(CH_3)=CH_2$

What is the product formed in the given reaction?

The action of water in the presence of sulphuric acid on the following alkenes gives:
$(i)$ $CH_3-CH=C(CH_3)_2$
(ii) $CH_3-CH=CH_2$

$CH_3-CH=CH_2 + NOCl \rightarrow P$
Identify the adduct.

Markovnikov's rule guides the addition of $HBr$ to which of the following compounds?

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