The reaction CH_3-CH(C_6H_5)-CH(Br)-C_6H_5 xr | vedclass

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(B) The reaction of $1-bromo-1,2-diphenylpropane$ with $alc.KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).This reaction leads to the formati

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$1,2-$diphenylprop$-1-$ene (trans-isomer)

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(B) The reaction of $1-bromo-1,2-diphenylpropane$ with $alc.KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).This reaction leads to the formation of an alkene by removing $H$ and $Br$ from adjacent carbons.The major product is the more stable alkene,which is the trans-isomer of $1,2-diphenylprop-1-ene$.In the trans-isomer,the bulky phenyl groups are on opposite sides of the double bond,minimizing steric hindrance.Thus,the product $A$ is $1,2-diphenylprop-1-ene$ (trans-isomer).

$(I)$ Compound	$(II)$ $C-X$ Bond Length $(pm)$	$(III)$ Dipole Moment $(D)$
$(A)$ $H_3C-F$	$(i)$ $214$	$(1)$ $1.636$
$(B)$ $H_3C-Cl$	$(ii)$ $193$	$(2)$ $1.830$
$(C)$ $H_3C-Br$	$(iii)$ $178$	$(3)$ $1.847$
$(D)$ $H_3C-I$	$(iv)$ $139$	$(4)$ $1.860$

The reaction $CH_3-CH(C_6H_5)-CH(Br)-C_6H_5 \xrightarrow{alc.KOH} A$. Identify the product $A$.

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