Q.Name the following: Which of the following will exhibit geometrical isomerism?
a. CH3-CH2-C(=CH2)-CH3
b. (CH3)2C=CH2
c. CH3-C(C2H5)=C(C2H5)-CH3

Step 1. Check (a): CH3-CH2-C(=CH2)-CH3 (2-methylbut-1-ene). One carbon of the double bond is a terminal =CH2, carrying two identical hydrogens. Section 15.2.1's rule states a terminal alkene (C=CH2 unit) can never show cis-trans isomerism -- so (a) does NOT.

Step 2. Check (b): (CH3)2C=CH2 (2-methylprop-1-ene). This is again a terminal =CH2 on one carbon, and the OTHER carbon carries two identical CH3 groups (a 1,1-disubstituted pattern) -- the rule also excludes this case directly. So (b) does NOT show geometrical isomerism either. …

Checking each alkene's two double-bond carbons against the section 15.2.1 rule that geometrical isomerism require …

• Assuming any alkene that is not a simple terminal alkene automatically shows geometrical isomerism -- (b) is non-terminal in the sense of having a substituted carbon, but STILL fails because that carbon's two substituents (both CH3) are identical. …