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(A) $1$. Ozonolysis of $3,3-	ext{dimethylbut-1-ene}$ $(CH_3-C(CH_3)_2-CH=CH_2)$ yields $A$ and $B$.$2$. The products are $A = (CH_3)_3C-CHO$ $(2,2-

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$3-	ext{hydroxy}-4,4-	ext{dimethylpentanal}$

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(A) $1$. Ozonolysis of $3,3-	ext{dimethylbut-1-ene}$ $(CH_3-C(CH_3)_2-CH=CH_2)$ yields $A$ and $B$.$2$. The products are $A = (CH_3)_3C-CHO$ $(2,2-	ext{dimethylpropanal})$ and $B = HCHO$ (formaldehyde).$3$. The reaction $A + B \xrightarrow{dil \ NaOH / \Delta} C$ is a cross-aldol condensation.$4$. Since $HCHO$ has no $\alpha-	ext{hydrogen}$,it acts as the electrophile (acceptor),and $2,2-	ext{dimethylpropanal}$ acts as the nucleophile (donor) because it has $\alpha-	ext{hydrogens}$.$5$. The aldol product formed is $3-	ext{hydroxy}-4,4-	ext{dimethylpentanal}$.

Consider the reaction sequence and identify $'C'$.
$A + B \xrightarrow{dil \ NaOH / \Delta} C$
(The reaction sequence is: $3,3-\text{dimethylbut-1-ene} \xrightarrow[(ii) Zn/H_2O]{(i) O_3} A + B$)

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Consider the reaction sequence and identify $'C'$.$A + B \xrightarrow{dil \ NaOH / \Delta} C$(The reaction sequence is: $3,3-\text{dimethylbut-1-ene} \xrightarrow[(ii) Zn/H_2O]{(i) O_3} A + B$)