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(C) The reaction of an unsaturated hydrocarbon $X$ with ozone (ozonolysis) produces compound $A$. Compound $A$ gives a positive Tollen's test (forms a

Vedclass · Accepted Answer

$CH_3-CH_2-C \equiv CH$

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(C) The reaction of an unsaturated hydrocarbon $X$ with ozone (ozonolysis) produces compound $A$. Compound $A$ gives a positive Tollen's test (forms a silver mirror with ammoniacal silver nitrate),which indicates that $A$ is an aldehyde or formic acid $(HCOOH)$. Let us analyze the options: $(A)$ $CH_3-C(CH_3)=C(CH_3)-CH_3$ on ozonolysis gives $CH_3-CO-CH_3$ (acetone),which does not give Tollen's test. $(B)$ $CH_3-C(CH_3)=C(CH_2)_2$ on ozonolysis gives $CH_3-CO-CH_3$ and cyclobutanone,neither of which gives Tollen's test. $(C)$ $CH_3-CH_2-C \equiv CH$ on ozonolysis gives $CH_3-CH_2-COOH$ and $HCOOH$. Formic acid $(HCOOH)$ gives a positive Tollen's test. $(D)$ $CH_3-C \equiv C-CH_3$ on ozonolysis gives $2CH_3-COOH$ (acetic acid),which does not give Tollen's test. Therefore,the correct hydrocarbon is $CH_3-CH_2-C \equiv CH$.

An unsaturated hydrocarbon $X$ on ozonolysis gives $A$. Compound $A$ when warmed with ammoniacal silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon $X$ is ..... .

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