The product (A) of the above reaction is: | vedclass

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(D) The given reactant is $CH_3O-CH(CO_2H)-CH_2-CH_2-CH=CH-CH_2-CH_3$. Ozonolysis ($O_3$ followed by $(CH_3)_2S$) cleaves the $C=C$ double bond to for

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$CH_3O-CH(CO_2H)-CH_2-CH_2-CH_2-CHO$

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(D) The given reactant is $CH_3O-CH(CO_2H)-CH_2-CH_2-CH=CH-CH_2-CH_3$. Ozonolysis ($O_3$ followed by $(CH_3)_2S$) cleaves the $C=C$ double bond to form two carbonyl compounds. The cleavage results in: $CH_3O-CH(CO_2H)-CH_2-CH_2-CHO \ (A)$ and $CH_3-CH_2-CHO \ (B)$. Compound $(A)$ is optically active due to the chiral center at the carbon atom attached to the $-OCH_3$ and $-CO_2H$ groups. Therefore,the correct option is $(D)$.

The product $(A)$ of the above reaction is:

Similar Questions

Statement-$I$ :- Ozonolysis reaction is used to locate the position of a double bond in an alkene.
Statement-$II$ :- When $but-1-ene$ and $but-2-ene$ are subjected to ozonolysis separately,both yield the same product.
In the light of the above statements,select the correct option.

The order of increasing reactivity towards $HCl$ of the following compounds will be:
$(1)$ $CH_2 = CH_2$
$(2)$ $(CH_3)_2C = CH_2$
$(3)$ $CH_3CH = CHCH_3$

Propene on hydroboration followed by oxidation with alkaline peroxide forms.

Consider the following statements: $A$ hydrocarbon of molecular formula $C_5H_{10}$ is a
$I.$ monosubstituted alkene
$II.$ disubstituted alkene
$III.$ trisubstituted alkene
Which of the following statement$(s)$ is(are) correct?

Allyl chloride on dehydrochlorination gives:

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