Which of the following alkenes exhibits geometrical (cis–trans) isomerism?
APent-$1$-ene
BBut-$2$-ene
CPropene
D$2$-methylpropene
Answer & Solution
Correct answer: B. But-$2$-ene
**Test.** For cis–trans isomerism each sp²-carbon of the C=C bond must carry two *different* groups. If either carbon has two of the same substituent, the two configurations are identical.
- **A. 2-methylpropene** $(CH_3)_2C=CH_2$ — left carbon has two CH₃ groups → no.
- **B. Propene** $CH_3CH=CH_2$ — the CH₂ end has two H → no.
- **C. But-2-ene** $CH_3CH=CHCH_3$ — each sp²-C carries one CH₃ and one H → **yes**, gives cis (Z) and trans (E).
- **D. Pent-1-ene** $CH_2=CHCH_2CH_2CH_3$ — terminal C has two H → no.
The rule of thumb: terminal alkenes never show cis–trans; internal alkenes do *only if* the two substituents on each end differ.
Related questions
Pent-$2$-ene is treated with hot, acidic KMnO$_4$. What are the products?Reductive ozonolysis (O$_3$, then Zn/H$_2$O) of $2$-methylbut-$2$-ene gives:When propene reacts with HBr in the presence of organic peroxides, the major product is:According to Markovnikov's rule, the addition of HBr to propene ($\mathrm{CH_3-CH=CH_2}$) What is the IUPAC name of the compound $\mathrm{CH_2=CH-CH_2-CH_3}$?