Identify the products A and B formed in the following reaction : CH 3 — CH 2 — CH = CH — CH 3 + HCl → A + B

The HCl gets added to the double bond of the given alkene in accordance with Markovnikov's rule to give two products, as shown below: A is 3 -Chloropentane while B is 2 -Chloropentane.

Arrange the following in order of decreasing S N 1 reactivity: ( CH 3 ) 3 CCl , CH 3 Cl , ( CH 3 ) 2 CHCl , CH 3 CH 2 Cl

In S N 1 mechanism the compound produces carbocation by removal of Cl − . Therefore, higher the stability of the carbocation intermediate, more will be the reactivity. Greater the number of alkyl group, more will be the + I effect, the greater is the stability of carbocation. Therefore, the is in the order, 3 ° > 2 ° > 1 ° Hence, the decreasing order of S N 1 reactivity is ( CH 3 ) 3 CCl 3 ° > ( CH 3 ) 2 CHCl 2 ° > CH 3 CH 2 Cl 1 ° > CH 3 Cl 1 ° Ethyl has a greater + I effect than the methyl group.

How do the bond lengths of C - X bonds ( X = F , Cl , Br , I ) in haloalkanes vary and why?

The C - X bond strength decreases with an increase of the size of the halogen ( X ), because the size of the p -orbital of the halogen increases as well. Thus, the p -orbital becomes more diffused and the extent of overlap with the orbital of the carbon decreases. As a result, the C - X bond is weakened and elongated. Therefore, the increasing order of the C - X bond length is as given below: C - F < C - Cl < C - Br < C - I

Write the reaction with conditions for conversion of 2 -Methylpropene into 1 -Bromo- 2 -methylpropane.

In the presence of peroxides, anti-Markovnikov's addition of hydrogen bromide to 2 -Methylpropene gives 1 -Bromo- 2 -methylpropane. The reaction can be written as given below:

Write the products of the following reactions: C 2 H 5 I + KNO 2 → ? C 2 H 5 I + AgCN → ?

Ethyl iodide on heating with alcoholic potassium nitrite forms ethyl nitrite. This is a nucleophilic substitution reaction.The reaction can be written as given below: C 2 H 5 I + KNO 2 → C 2 H 5 - ONO + KI Ethyl iodide on heating with alcoholic silver cyanide gives ethyl isocyanide. Silver cyanide is a covalent molecule, hence, nitrogen atom donates an electron pair to form isocyanide. C 2 H 5 I + AgCN → C 2 H 5 - NC + AgI

Write the products of the following reaction. Which one is the major product and why?

Haloalkanes when heated with alcoholic KOH undergoes dehydrohalogenation reaction. In dehydrohalogenation reaction, according to Saytzeff rule the major product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Thus, But- 2 -ene is the major product and But- 1 -ene is the minor product.

(a) Explain why thionyl chloride ( SOCl 2 ) method is preferred for preparing alkyl chlorides from alcohols. (b) For isomeric haloalkanes, the boiling point decreases with branching of chain. Why?

SOCl 2 method is preferred than other methods because both the products of the reaction ( SO 2 and HCl ) are gases and can easily escape leaving behind pure alkyl halide. Therefore, the alkyl halide can be purified easily. R - OH + SOCl 2 → R - Cl + SO 2 + HCl For isomeric haloalkanes, the boiling point decreases with branching of chain. This is because branching of the chain makes the molecule more compact and therefore, decreases the surface area. Due to decrease in surface area, the magnitude of Van der Waals forces of attraction decreases and consequently, the boiling points of the branched chain compound is less than those of the straight chain compounds.

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