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BIHAR CET Haloalkanes and Haloarenes — practice questions

Explore 57 free Haloalkanes and Haloarenes questions in BIHAR CET Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Haloalkanes and Haloarenes questions and exercises with answers

Showing questions 1–30 of 57.

Hydrocarbons are organic compounds composed exclusively of:Read the answer and worked solution →In the presence of organic peroxides, addition of HBr to propene gives predominantly:Read the answer and worked solution →Wurtz reaction is used to prepare:Read the answer and worked solution →The C–X bond enthalpies follow the order:Read the answer and worked solution →Among alkyl halides, the order of reactivity towards nucleophilic substitution (for the same R group) is:Read the answer and worked solution →The order of reactivity of alcohols with a given haloacid (HCl, HBr, HI) is:Read the answer and worked solution →The reaction of alkyl chloride/bromide with NaI in dry acetone to form alkyl iodide is called:Read the answer and worked solution →Among methyl chloride, methyl bromide and methyl iodide, the order of boiling points is:Read the answer and worked solution →Grignard reagent (RMgX) is prepared by:Read the answer and worked solution →A molecule that is non-superimposable on its mirror image is called:Read the answer and worked solution →Stereoisomers that are non-superimposable mirror images of each other are called:Read the answer and worked solution →A racemic mixture is:Read the answer and worked solution →Haloalkane has general formula:Read the answer and worked solution →Haloarene example:Read the answer and worked solution →Classification of haloalkane by C bearing halogen: 1° if X is on:Read the answer and worked solution →IUPAC name of CH₃CH₂CH₂Cl:Read the answer and worked solution →Polarity of C-X bond order (most polar first):Read the answer and worked solution →Reactivity order for SN2 reactions (1°, 2°, 3°):Read the answer and worked solution →Finkelstein reaction:Read the answer and worked solution →Grignard reagent: R-X + Mg in dry ether gives:Read the answer and worked solution →Reaction of Grignard with formaldehyde HCHO gives:Read the answer and worked solution →Reactivity of haloalkanes in SN2 by leaving group:Read the answer and worked solution →For SN1 reaction, rate is independent of:Read the answer and worked solution →Reaction of (CH₃)₃C-Br with H₂O gives:Read the answer and worked solution →Saytzeff product from 2-bromo-3-methylbutane + base:Read the answer and worked solution →NEET · Chemistry · SN1 or SN2Read the answer and worked solution →Fix the wrong word on halide reactionsRead the answer and worked solution →Sort each clue into SN1 or SN2Read the answer and worked solution →Match each named reactionRead the answer and worked solution →Sort the SN1 and SN2 traitsRead the answer and worked solution →

Worked practice questions

Hydrocarbons are organic compounds composed exclusively of:

  1. Carbon and oxygen
  2. Carbon and hydrogen
  3. Carbon and nitrogen
  4. Hydrogen and nitrogen
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Correct answer: B. Carbon and hydrogen

By definition, hydrocarbons contain **only** carbon and hydrogen atoms. Adding oxygen, nitrogen, halogens, etc. produces other functional-group families (alcohols, amines, halides). Three main classes: - **Alkanes**: only single C-C bonds, $C_nH_{2n+2}$. - **Alkenes**: contain at least one C=C double bond, $C_nH_{2n}$. - **Alkynes**: contain at least one C≡C triple bond, $C_nH_{2n-2}$. Aromatic hydrocarbons (e.g. benzene) are a special case.

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In the presence of organic peroxides, addition of HBr to propene gives predominantly:

  1. 1-bromopropane (anti-Markovnikov)
  2. propan-1-ol
  3. 2-bromopropane (Markovnikov)
  4. propane
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Correct answer: A. 1-bromopropane (anti-Markovnikov)

Kharasch peroxide effect: HBr in the presence of organic peroxides (e.g. benzoyl peroxide) adds **anti-Markovnikov** via a free-radical mechanism. Br ends up on the **less** substituted carbon. Product: 1-bromopropane. Why: the radical chain involves a Br• radical adding to the alkene first. Br• prefers the less-substituted carbon, forming the more stable carbon radical. Then HBr donates H• to complete the addition. Notably, the peroxide effect is **specific to HBr**. HCl and HI don't show it.

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Wurtz reaction is used to prepare:

  1. Alkenes from alkyl halides
  2. Symmetric alkanes from alkyl halides
  3. Alcohols from alkanes
  4. Carboxylic acids from alkanes
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Correct answer: B. Symmetric alkanes from alkyl halides

**Wurtz reaction**: $2 R X + 2 Na \xrightarrow{\text{dry ether}} R{-}R + 2 NaX$. Two alkyl halide molecules combine, with sodium metal as the reducing agent, to give a higher symmetric alkane with double the carbon chain. Example: $\mathrm{CH_3 I + CH_3 I \xrightarrow{2Na} CH_3 CH_3}$ (ethane). Limitation: doesn't work well for asymmetric alkanes — mixing two different RX gives a statistical mixture of R-R, R'-R' and R-R', hard to separate. So Wurtz is best for symmetric products.

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More Chemistry topics

Ideal and Non-Ideal SolutionsIdeal SolutionsNon-Ideal SolutionsReactions in Aqueous SolutionElectrochemistryElectrochemical SeriesCoordination CompoundsSolutions