Question

For the compound having molecular formula \(C_4H_9Br\), write:
(a) the isomer which is most reactive towards S_N1 displacement.
(b) the isomer which, on reacting with Na metal in dry ether, gives 2,5-dimethylhexane.

Hint:

For \(S_N1\), choose the halide giving the most stable carbocation. For Wurtz reaction, split the product into two identical alkyl groups.

Answer 1

Step 1: List the four isomers of $C_4H_9Br$.
(i) 1-Bromobutane: $CH_3CH_2CH_2CH_2Br$ (primary). (ii) 2-Bromobutane: $CH_3CHBrCH_2CH_3$ (secondary). (iii) 2-Bromo-2-methylpropane: $(CH_3)_3CBr$ (tertiary, tert-butyl bromide). (iv) 1-Bromo-2-methylpropane: $(CH_3)_2CHCH_2Br$ (primary, isobutyl bromide).
Step 2: Part (a): Most reactive isomer towards SN1.
SN1 reactivity depends on carbocation stability. Tertiary carbocations (3 alkyl groups) are most stable (maximum hyperconjugation and inductive donation). $(CH_3)_3CBr$ forms a stable tertiary carbocation: \[ (CH_3)_3CBr \rightarrow (CH_3)_3C^+ + Br^- \] So 2-bromo-2-methylpropane (tert-butyl bromide) is most reactive in SN1.
Step 3: Recall the Wurtz reaction.
Wurtz reaction: \[ 2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX \] Two identical alkyl halide molecules couple to give a symmetrical alkane.
Step 4: Part (b): Work backward from 2,5-dimethylhexane.
2,5-Dimethylhexane = $(CH_3)_2CHCH_2CH_2CH(CH_3)_2$. It has 8 carbons. In Wurtz reaction, two 4-carbon fragments couple. Each fragment = $(CH_3)_2CHCH_2-$ (isobutyl group). The corresponding halide = 1-bromo-2-methylpropane: $(CH_3)_2CHCH_2Br$.
Step 5: Verify the Wurtz reaction gives 2,5-dimethylhexane.
\[ 2(CH_3)_2CHCH_2Br + 2Na \rightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr \] Product $(CH_3)_2CHCH_2CH_2CH(CH_3)_2$ is 2,5-dimethylhexane. Correct!
Step 6: State both answers.
(a) Most reactive towards SN1: $(CH_3)_3CBr$ (2-bromo-2-methylpropane, tert-butyl bromide). (b) Gives 2,5-dimethylhexane via Wurtz: $(CH_3)_2CHCH_2Br$ (1-bromo-2-methylpropane, isobutyl bromide). \[ \boxed{(a)\;(CH_3)_3CBr \quad (b)\;(CH_3)_2CHCH_2Br} \]