Q.
Geometrical isomerism is shown by
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a
CH2 = C(Br) I
b
CH3CH = C(Br)I
c
(CH3)2 C = C(Br)I
d
CH3 CH = CCl2
answer is B.
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Detailed Solution
Conclusion:→ In order to exhibit geometrical isomerism, double bonded carbon should possess two different groups.→ The C1 carbon two groups and C2 carbons two groups may (or) may not be same.→ Either C1 carbon two groups (or) C2 carbon two groups are identical then geometrical isomerism is not possibel.→ If all the four groups are different [a,b,c,d], then they donot exhibit CIS, trans isomerism but they exhibit geometrical isomerism i.e., E, Z-isomerism