The maximum number of bromine atoms present i | vedclass

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(D) Step $1$: Dehydrohalogenation using $Alc. KOH/\Delta$ (E2 elimination). The starting material has two $-Br$ groups. Elimination of two moles of $H

Vedclass · Accepted Answer

$8$

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(D) Step $1$: Dehydrohalogenation using $Alc. KOH/\Delta$ (E2 elimination). The starting material has two $-Br$ groups. Elimination of two moles of $HBr$ results in the formation of two double bonds: one terminal vinyl group $(-CH=CH_2)$ and one exocyclic double bond ($-C(CH_3)=CH_2$ is not formed,rather the double bond forms between the ring and the side chain). Wait,looking at the structure,the elimination creates a vinyl group and an exocyclic double bond. The alkyne remains unaffected.Step $2$: Electrophilic addition of excess $Br_2/CCl_4$. The molecule now contains one terminal alkene,one exocyclic alkene,and one internal alkyne. - The terminal alkene adds $1$ mole of $Br_2$ ($2$ $Br$ atoms).- The exocyclic alkene adds $1$ mole of $Br_2$ ($2$ $Br$ atoms).- The internal alkyne adds $2$ moles of $Br_2$ ($4$ $Br$ atoms).Total $Br$ atoms added = $2 + 2 + 4 = 8$ atoms.Including the $2$ $Br$ atoms already present in the starting material is incorrect as the question asks for the number of bromine atoms in the final product $P$. Counting the structure of $P$ in the solution image: there are $2$ $Br$ on the terminal side chain,$2$ $Br$ on the ring,$4$ $Br$ on the alkyne side chain. Total = $8$ atoms.

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

The maximum number of bromine atoms present in the final product $P$ upon complete bromination is

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