Among the following compounds, how many can exhibit geometric (cis-trans) isomerism? (i) HO-C(CH3)=CH-C(F)(CH3)2 (ii) CH3-C(CH3)=C(F)(CH3) (iii) CH3-N(=O)-OH (iv) H-N-C(CH3)=CH-CH(CH3)2 (v) CH3-C(=O)Cl (vi) (CH3)2C=CH-CH3 (vii) A cyclohexane ring bearing two different substituents on the t

Question

Among the following compounds, how many can exhibit geometric (cis-trans) isomerism? (i) HO-C(CH3)=CH-C(F)(CH3)2 (ii) CH3-C(CH3)=C(F)(CH3) (iii) CH3-N(=O)-OH (iv) H-N-C(CH3)=CH-CH(CH3)2 (v) CH3-C(=O)Cl (vi) (CH3)2C=CH-CH3 (vii) A cyclohexane ring bearing two different substituents on the two carbons of an endocyclic C=C bond (viii) CH3-C(CH3)=C(CH2CH3)2 (ix) CH3-C(Cl)=C(CH3)2

Accepted Answer

4. (i) C(CH3)(OH)=CH-: left C bears OH and CH3 (different); right C bears H and C(F)(CH3)2 (different) -> YES. (ii) C(CH3)(CH3)=: two identical CH3 on left C -> NO. (iii) CH3-N(=O)-OH: N=O with N carrying CH3 and OH (different) -> shows syn/anti isomerism -> YES. (iv) C(CH3)=CH-: left C bears CH3 and N-group (different); right C bears H and CH(CH3)2 (different) -> YES. (v) C=O: one side is only O -> NO. (vi) C(CH3)2=: two CH3 on one carbon -> NO. (vii) Two different substituents on each double-bond carbon in the ring -> YES. (viii) C=C(CH2CH3)2: right C has two identical ethyl groups -> NO. (ix) C=C(CH3)2: right C has two CH3 -> NO. Count: (i), (iii), (iv), (vii) = 4.

Solution

Related JEE Advanced Chemistry questions