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NCERT Exemplar · Q6

Q.Which of the following will not show geometrical isomerism?

(i) CClF=CDH\overset{F}{\underset{Cl}{C}}=\overset{H}{\underset{D}{C}}
(ii) CClF=CClF\overset{F}{\underset{Cl}{C}}=\overset{F}{\underset{Cl}{C}}
(iii) CC2H5H3C=CCH3C2H5\overset{H_3C}{\underset{C_2H_5}{C}}=\overset{C_2H_5}{\underset{CH_3}{C}}
(iv) CCH3CH3=CC2H5CH3\overset{CH_3}{\underset{CH_3}{C}}=\overset{CH_3}{\underset{C_2H_5}{C}}
Assam AhsecMCQ· 1mImportance★★★★★est
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Geometrical isomerism requires each carbon of the double bond to carry two different groups. Option (iv) has two identical methyl groups on one carbon, so it cannot show geometrical isomerism.

Geometrical isomerism arises from the restricted rotation about a carbon–carbon double bond: the π bond locks the two carbons in a plane, so groups cannot rotate past one another, giving distinct cis and trans (or E/Z) arrangements. The essential requirement is that each carbon of the double bond is bonded to two different groups; if either carbon carries two identical groups, the two arrangements are superimposable and no isomerism results.

(i) CClF=CDH\overset{F}{\underset{Cl}{C}}=\overset{H}{\underset{D}{C}} — the left carbon carries F and Cl (different), the right carbon H and D (deuterium is a distinct isotope of hydrogen). All four groups are different, so cis and trans forms exist. Shows geometrical isomerism.

(ii) CClF=CClF\overset{F}{\underset{Cl}{C}}=\overset{F}{\underset{Cl}{C}} — each carbon carries F and Cl (two different groups). The two fluorines can be on the same side (cis) or opposite sides (trans). Shows geometrical isomerism. …

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