Page No 307: - Chapter 10 - Haloalkanes & Halo Arenes - Intext Solutions - Ncert Solutions class 12 - Chemistry

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Question 10.7:

Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism? Explain your Answer.

(i)

(ii)

(iii)

Answer:

(i)

2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an S_N2 mechanism.

(ii)

2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an S_N2 mechanism.

(iii)

Both the alkyl halides are primary. However, the substituent −CH₃ is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by S_N2 mechanism.

Question 10.8:

In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction?

(i)

(ii)

Answer:

(i)

S_N1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of S_N1 reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e. 2−chloro-2-methylpropane, undergoes faster S_N1 reaction than (II) i.e., 3-chloropentane.

(ii)

The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster S_N1 reaction than (II), 1-chlorohexane.

Question 10.9:

Identify A, B, C, D, E, R and R¹ in the following:

Answer:

Since D of D₂O gets attached to the carbon atom to which MgBr is attached, C is

Therefore, the compound R − Br is

When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R¹−X, is

Therefore, compound D is

And, compound E is