One mole of alkene underline{X} on ozonolysis | vedclass

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Question

(A) To determine the structure of the alkene that undergoes ozonolysis,place the carbonyl oxygen atoms of the products face to face and replace the $O

Vedclass · Accepted Answer

$2$-methyl-$2$-butene

Vedclass · Answer

(A) To determine the structure of the alkene that undergoes ozonolysis,place the carbonyl oxygen atoms of the products face to face and replace the $O$ atoms with a double bond $(C=C)$.Acetaldehyde is $CH_3CHO$ and acetone is $(CH_3)_2CO$.Aligning them: $CH_3(H)C=O + O=C(CH_3)_2$.Removing the oxygen atoms and joining the carbons with a double bond gives $CH_3(H)C=C(CH_3)_2$.The structure is $CH_3-CH=C(CH_3)_2$.The longest chain contains $4$ carbon atoms with a methyl group at the $2$-position,so the $IUPAC$ name is $2$-methyl-$2$-butene.

One mole of alkene $\underline{X}$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The $IUPAC$ name of $\underline{X}$ is

Similar Questions

$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{Hg(OAc)_2/EtOH} (A) (90\%)$. The product $(A)$ of the above reaction is:

$CH_3-C(CH_3)=CH_2$ $\xrightarrow[R_2O_2]{HBr} X$ $\xrightarrow{CH_3ONa} Y$; $X$ and $Y$ are:

Write the main product of the given reactions:
$(1)$ $CH_3CH = CHCH_3 + HBr \to$
$(2)$ $CH_3-C(CH_3)=CH_2 + HBr \to$

In the reaction sequence:
$Cyclohexane$ $\xrightarrow{hv/Cl_2} (X)$ $\xrightarrow{alc. KOH/\Delta} (Y)$ $\xrightarrow[(ii) H_2O/Zn]{(i) O_3} (Z)$
$(Z)$ will be :

The major product formed in the following reaction is:

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