Q.1. Explain why Grignard's reagents should be prepared under anhydrous conditions.

Q.2. Predict the order of reactivity of four isomeric bromobutanes in S_N1 reaction.

Q.3. Which would undergo S_N2 reaction faster in the following pair and why?

    CH₃−CH₂−Br and CH₃−CH₂−I

Q.4. Write the structure of an isomer of compound C₄H₉Br which is most reactive towards S_N1 reaction.

Q.5. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?

Q.6. Out of and  which is more reactive towards S_N1 reaction and why?

Q.7. Which would undergo S_N2 reaction faster in the following pair and why?

Q.8. Which would undergo S_N1 reaction faster in the following pair and why?

Q.9. Identify the chiral molecule in the following pair:

Q.10. Which halogen compound in each of the following pairs will react faster in S_N2 reaction:

    (i) CH₃Br or CH₃I

    (ii) (CH₃)₃CCl or CH₃Cl

Q.11. What happens when CH₃−Br is treated with KCN?

Q.12. What happens when ethyl chloride is treated with aqueous KOH?

Q.13. Why is (±)-butan-2-ol is optically inactive?

Q.14. Which compound in the following pair undergoes faster S_N1 reaction?

Q.15. How may methyl bromide be preferentially converted to methyl isocyanide?

 

Answers

Ans.1. Grignard reagents react with water to form alkanes.

   

So, they must be prepared under anhydrous conditions.

Ans.2.  

   

Ans.3. Since I⁻ is a better leaving group than Br⁻, thus, CH₃CH₂I undergoes S_N2 reaction faster than CH₃CH₂Br.

Ans.4.

   

Ans.5. Benzyl chloride gets easily hydrolysed by aq. NaOH due to formation of stable benzyl carbocation. But due to partial double bond character of C−Cl bond in chlorobenzene, it does not hydrolyse.

Ans.6. The S_N1 reaction proceeds through carbocation formation thus, the compound which forms more stable carbocation will be more reactive.

   

As, 2° carbocation is more stable than l° carbocation thus, 2-chlorobutane is more reactive towards S_N1 reaction.

Ans.7. CH₃−CH₂−Br would undergo S_N2 reaction faster due to formation of less sterically hindered transition state.

Ans.8.   will undergo S_N1 reaction faster due to the formation of stable carbocation.

Ans.9.   is a chiral molecule.

Ans.10. (i) CH₃I will give faster S_N2 reaction.

(ii) CH₃Cl will give faster S_N2 reaction.

Ans.11. CH₃CN is formed by nucleophilic substitution reaction.

    CH₃Br + KCN → CH₃CN + KBr

Ans.12. When ethyl chloride is treated with aqueous KOH, ethanol is formed,

    CH₃CH₂Cl + KOH(aq) → CH₃CH₂OH + KCl

Ans.13. (±)-Butan-2-ol is optically inactive because it exists in two enantiomeric forms which are non-superimposable mirror images of each other. Both the isomers are present in equal amounts therefore, it does not rotate the plane of polarized light and is optically inactive.

   

Ans.14. Tertiary halide  reacts faster than the secondary halide because of the greater stability of tert-carbocation.

Ans.15. AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as main product.

   

 

 

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