How many stereoisomeric tetrabromides will be | vedclass

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(C) The reaction of $1,4-	ext{pentadiene}$ with $2$ equivalents of $Br_2$ involves the electrophilic addition of bromine to both double bonds.Each ad

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$4$

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(C) The reaction of $1,4-	ext{pentadiene}$ with $2$ equivalents of $Br_2$ involves the electrophilic addition of bromine to both double bonds.Each addition of $Br_2$ creates two chiral centers.Since the molecule is $1,4-	ext{pentadiene}$,the two double bonds are separated by a $CH_2$ group,making the two chiral centers formed at each double bond independent of each other.Each double bond addition produces a pair of enantiomers ($d$ and $l$ forms).Let the chiral centers formed at the first double bond be $C_1$ and $C_2$,and at the second double bond be $C_3$ and $C_4$.Since the molecule is not symmetric in a way that would create a meso compound (the chiral centers are not in the same environment relative to the whole molecule),we have $2$ chiral centers from the first addition and $2$ from the second,leading to $2^2 	imes 2^2 = 16$ possible stereoisomers if all centers were independent. However,the addition of $Br_2$ to an alkene is anti-addition.For $1,4-	ext{pentadiene}$,the product is $1,2,4,5-	ext{tetrabromopentane}$.This molecule has $2$ chiral centers at $C_2$ and $C_4$.Thus,the number of stereoisomers is $2^n = 2^2 = 4$.

Column-$(I)$	Column-$(II)$
$(A)$ $C_2H_5-C(CH_3)(OH)-H$	$(p)$ $Cl-C(Br)(I)-H$
$(B)$ $Br-C(Cl)(I)-H$	$(q)$ $OH-C(CH_3)(C_2H_5)-H$
	$(r)$ $H_5C_2-C(OH)(CH_3)-H$
	$(s)$ $Cl-C(I)(Br)-H$

How many stereoisomeric tetrabromides will be formed in the following reaction?
$1,4-\text{pentadiene} + 2Br_2 \xrightarrow{CCl_4} \text{Product}$

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Column-$(I)$ Column-$(II)$
$(A)$ $C_2H_5-C(CH_3)(OH)-H$ $(p)$ $Cl-C(Br)(I)-H$
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How many stereoisomeric tetrabromides will be formed in the following reaction?$1,4-\text{pentadiene} + 2Br_2 \xrightarrow{CCl_4} \text{Product}$