Question 25 of 26advanced🔍 AnalyzeLong Answer5 marks

Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Correct Answer

Structural Formula of Compound (a):

Compound (a) is isobutyl bromide (1-bromo-2-methylpropane).

CH₃
 |

CH₃—CH—CH₂—Br

Reasoning:

Since (a) is a primary alkyl halide with formula C₄H₉Br, it can be either n-butyl bromide or isobutyl bromide. The key clue is that compound (d), formed by reaction of (a) with sodium, is different from the compound formed when n-butyl bromide reacts with sodium. This confirms (a) cannot be n-butyl bromide.

Reaction Equations:

1. Reaction of (a) with alcoholic KOH:

(CH₃)₂CHCH₂Br + KOH (alcoholic) → (CH₃)₂C=CH₂ + KBr + H₂O

Isobutyl bromide (a) undergoes elimination reaction to form 2-methylprop-1-ene (b).

2. Reaction of (b) with HBr:

(CH₃)₂C=CH₂ + HBr → (CH₃)₃CBr

The alkene (b) undergoes electrophilic addition following Markovnikov's rule to form tert-butyl bromide (c), which is an isomer of (a).

3. Reaction of (a) with sodium metal (Wurtz reaction):

2(CH₃)₂CHCH₂Br + 2Na → (CH₃)₂CHCH₂CH₂CH(CH₃)₂ + 2NaBr

This forms 2,5-dimethylhexane (d), C₈H₁₈.

4. For comparison - n-butyl bromide with sodium:

2CH₃CH₂CH₂CH₂Br + 2Na → CH₃(CH₂)₆CH₃ + 2NaBr

This forms n-octane, which is different from compound (d).

Exercise: EXERCISES | Q: 6.21 | (Page 33)
For More Understanding

Explanation

The question tests understanding of elimination reactions, addition reactions, and Wurtz reaction. The key to solving this problem is recognizing that: (i) alcoholic KOH causes elimination to form alkenes, (ii) HBr adds to alkenes following Markovnikov's rule, and (iii) the Wurtz reaction couples two alkyl groups. The critical clue is that compound (d) differs from the product of n-butyl bromide with sodium, confirming (a) must be isobutyl bromide, not n-butyl bromide.

Solution Steps

  1. Step 1: Identify that (a) must be a primary alkyl halide - either n-butyl bromide or isobutyl bromide.

  2. Step 2: Apply elimination reaction with alcoholic KOH to form alkene (b).

  3. Step 3: Apply Markovnikov addition of HBr to alkene (b) to form isomer (c).

  4. Step 4: Use the Wurtz reaction condition to distinguish between the two possible structures - since (d) differs from n-octane, (a) must be isobutyl bromide.

  5. Step 5: Write all balanced chemical equations for the reactions.

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