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BIHAR CET — free MCQ practice

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What is the IUPAC name of the compound $\mathrm{CH_2=CH-CH_2-CH_3}$?Read the answer and worked solution →According to Markovnikov's rule, the addition of HBr to propene ($\mathrm{CH_3-CH=CH_2}$) gives predominantly:Read the answer and worked solution →When propene reacts with HBr in the presence of organic peroxides, the major product is:Read the answer and worked solution →Reductive ozonolysis (O$_3$, then Zn/H$_2$O) of $2$-methylbut-$2$-ene gives:Read the answer and worked solution →

Worked practice questions

What is the IUPAC name of the compound $\mathrm{CH_2=CH-CH_2-CH_3}$?

  1. But-1-ene
  2. But-2-ene
  3. Butyne
  4. Butane
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Correct answer: A. But-1-ene

A four-carbon chain with a double bond starting at C1 is but-1-ene. Butyne would have a triple bond but-2-ene has the double bond between C2 and C3.

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According to Markovnikov's rule, the addition of HBr to propene ($\mathrm{CH_3-CH=CH_2}$) gives predominantly:

  1. propane
  2. $1{,}2$-dibromopropane
  3. $2$-bromopropane
  4. $1$-bromopropane
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Correct answer: C. $2$-bromopropane

**Markovnikov's rule:** the H of HX attaches to the carbon of the double bond bearing more hydrogens, leaving X on the more substituted carbon. Here Br goes to C2 → 2-bromopropane.

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When propene reacts with HBr in the presence of organic peroxides, the major product is:

  1. propan-$1$-ol
  2. propane
  3. $2$-bromopropane
  4. $1$-bromopropane
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Correct answer: D. $1$-bromopropane

Peroxides switch the mechanism from ionic to free-radical (peroxide / Kharasch effect). The radical adds anti-Markovnikov so Br lands on the *less* substituted carbon — giving 1-bromopropane. The trap is option B (Markovnikov product) — that's the answer without peroxides. The peroxide effect is specific to HBr HCl and HI do not show it.

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