(A) (C_7H_{14}) xrightarrow[Zn]{O_3} (B) + (C | vedclass

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(C) Compound $(A)$ is $(CH_3)_3C-CH=CH-CH_3$ ($4,4$-dimethylpent-$2$-ene). Upon ozonolysis,it yields $(CH_3)_3C-CHO$ (pivalaldehyde) and $CH_3CHO$ (ac

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$(CH_3)_3C-CH=CH-CH_3$

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(C) Compound $(A)$ is $(CH_3)_3C-CH=CH-CH_3$ ($4,4$-dimethylpent-$2$-ene). Upon ozonolysis,it yields $(CH_3)_3C-CHO$ (pivalaldehyde) and $CH_3CHO$ (acetaldehyde). $(CH_3)_3C-CH=CH-CH_3 \xrightarrow[Zn]{O_3} (CH_3)_3C-CHO + CH_3CHO$ Pivalaldehyde $(B)$ lacks $\alpha$-hydrogens and thus undergoes the Cannizzaro reaction. Additionally,$(A)$ exhibits geometrical isomerism because the groups attached to each carbon of the double bond are different ($H$ and $t$-butyl on one,$H$ and methyl on the other).

$(A) (C_7H_{14}) \xrightarrow[Zn]{O_3} (B) + (C)$
Compound $(A)$ exists as geometrical isomers and $(B)$ gives Cannizzaro reaction. $(A)$ will be:

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$(A) (C_7H_{14}) \xrightarrow[Zn]{O_3} (B) + (C)$ Compound $(A)$ exists as geometrical isomers and $(B)$ gives Cannizzaro reaction. $(A)$ will be: