Consider the reaction sequence shown below:Br | vedclass

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(A) Step $1$: Formation of $A$.The reaction of $Br-CH_2-CHO$ with excess $EtOH$ in the presence of dry $HCl$ gas is an acetal formation reaction. The

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$Br-CH_2-CH(OEt)_2$ and $CH_2=C(OEt)_2$

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(A) Step $1$: Formation of $A$.The reaction of $Br-CH_2-CHO$ with excess $EtOH$ in the presence of dry $HCl$ gas is an acetal formation reaction. The aldehyde group $(-CHO)$ is converted into an acetal group $(-CH(OEt)_2)$. Thus,$A$ is $Br-CH_2-CH(OEt)_2$.Step $2$: Formation of $B$.The product $A$ $(Br-CH_2-CH(OEt)_2)$ is treated with potassium tert-butoxide $(^tBuO^-K^+)$,which is a strong,bulky base. This promotes an $E2$ elimination reaction. The base abstracts a proton from the $\alpha$-carbon (the carbon attached to the bromine atom),leading to the elimination of $HBr$ and the formation of a double bond. The resulting product $B$ is $CH_2=C(OEt)_2$ (a ketene acetal).Therefore,the major products are $A = Br-CH_2-CH(OEt)_2$ and $B = CH_2=C(OEt)_2$.

Consider the reaction sequence shown below:
$Br-CH_2-CHO$ $\xrightarrow[\text{dry } HCl \text{ gas}]{\text{EtOH (excess)}} A$ $\xrightarrow{^tBuO^-K^+} B$
[Where $Et \Rightarrow -C_2H_5$,$^tBu \Rightarrow (CH_3)_3C^-$]
Identify the major products $A$ and $B$ formed respectively.

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Consider the reaction sequence shown below:$Br-CH_2-CHO$ $\xrightarrow[\text{dry } HCl \text{ gas}]{\text{EtOH (excess)}} A$ $\xrightarrow{^tBuO^-K^+} B$[Where $Et \Rightarrow -C_2H_5$,$^tBu \Rightarrow (CH_3)_3C^-$]Identify the major products $A$ and $B$ formed respectively.