Consider the following E_1/S_N1 reaction. The | vedclass

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(D) The reaction proceeds via the formation of a carbocation intermediate. $1$. The loss of the leaving group $Br^-$ generates a tertiary carbocation

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$1, 2, 3$ and $4$

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(D) The reaction proceeds via the formation of a carbocation intermediate. $1$. The loss of the leaving group $Br^-$ generates a tertiary carbocation at the $C_1$ position. $2$. $S_N1$ pathway: The nucleophile $(H_2O)$ attacks the carbocation from both the top and bottom faces,leading to the formation of two diastereomers $(1)$ and $(2)$,which are the cis and trans isomers of $4-$methylcyclohexan$-1-$ol. $3$. $E_1$ pathway: The base (solvent) abstracts a proton from the adjacent carbons to form alkenes. - Abstraction of a proton from the $CH_3$ group attached to $C_1$ leads to the exocyclic alkene $(3)$. - Abstraction of a proton from $C_2$ leads to the endocyclic alkene $(4)$. Thus,all four products $(1)$,$(2)$,$(3)$,and $(4)$ are formed.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

Consider the following $E_1/S_N1$ reaction. The missing product$(s)$ is(are):

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