Identify the final product (B) in the followi | vedclass

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(B) $1$. The starting material is $1	ext{-bromo-}2	ext{-methylcyclobutane}$.$2$. Treatment with alcoholic $KOH$ (alc. $KOH$) causes dehydrohalogenat

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$OHC-CH_2-CH(CH_3)-CHO$

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(B) $1$. The starting material is $1	ext{-bromo-}2	ext{-methylcyclobutane}$.$2$. Treatment with alcoholic $KOH$ (alc. $KOH$) causes dehydrohalogenation via an $E2$ mechanism,resulting in the formation of an alkene. The major product $(A)$ is $3	ext{-methylcyclobutene}$.$3$. Ozonolysis of $3	ext{-methylcyclobutene}$ involves the cleavage of the double bond. Since it is a cyclic alkene,the ring opens to form a dicarbonyl compound.$4$. The cleavage of the $C=C$ bond in $3	ext{-methylcyclobutene}$ yields $OHC-CH_2-CH(CH_3)-CHO$ ($3$-methylbutanedial).

Reaction	Nature of reaction
$A. RBr \xrightarrow{Na}$	$1. \text{Partial reduction}$
$B. RCOO^-Na^+ \xrightarrow{NaOH/CaO, \Delta}$	$2. \text{Electrophilic addition}$
$C. R-C \equiv C-R' \xrightarrow{\text{Lindlar's catalyst } (H_2)}$	$3. \text{Wurtz reaction}$
$D. H_2C=CH_2 \xrightarrow{Br_2/CCl_4}$	$4. \text{Decarboxylation}$
	$5. \text{Decarboxylative dimerisation}$

Identify the final product $(B)$ in the following reaction sequence:
$1\text{-bromo-}2\text{-methylcyclobutane}$ $\xrightarrow{\text{alc. KOH}} A$ $\xrightarrow{\text{ozonolysis}} B$

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Identify the final product $(B)$ in the following reaction sequence:$1\text{-bromo-}2\text{-methylcyclobutane}$ $\xrightarrow{\text{alc. KOH}} A$ $\xrightarrow{\text{ozonolysis}} B$