Q.An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.

The key concept here is using the properties of the Wurtz reaction and the regioselectivity of free radical halogenation to deduce the structure of an alkane and its derivative.

1. The Wurtz reaction of a primary alkyl halide ($R-X$) yielding a single alkane $\text{C}_8\text{H}_{18}$ implies that the alkane is symmetrical ($R-R$) and formed from two identical primary alkyl groups ($R$). Since the alkane has 8 carbons, $R$ must be a $\text{C}_4\text{H}_9$ primary alkyl group. The two possible primary $\text{C}_4\text{H}_9$ groups are $n$-butyl ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2-$) and isobutyl ($(\text{CH}_3)_2\text{CHCH}_2-$).
2. If $R$ is $n$-butyl, the alkane is $n$-octane ($\text{CH}_3(\text{CH}_2)_6\text{CH}_3$). If $R$ is isobutyl, the alkane is 2,5-dimethylhexane ($\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_3$). …

The problem describes an alkane (C8H18) obtained through a specific synthesis (Wurtz reaction of a primary alkyl halide) and then characterised by its monobromination product. We need to deduce the structures of both the alkane and the tertiary bromide.

Here's the conceptual breakdown:

• Wurtz Reaction: This reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a larger alkane. If a single primary alkyl halide ($R-X$) is used, the resulting alkane will be symmetrical, formed by joining two identical alkyl groups ($R-R$). The general reaction is $2R-X + 2Na \rightarrow R-R + 2NaX$.
• Alkane Formula C8H18: This is the molecular formula for an alkane. Since it's formed by coupling two identical alkyl groups ($R$), each $R$ group must contain 4 carbon atoms ($C_4H_9$).
• Primary Alkyl Halide: The starting material is specified as a primary alkyl halide. This means the carbon atom bonded to the halogen (bromine, in this case) must be attached to only one other carbon atom.
• Monobromination: This is a free radical substitution reaction where one hydrogen atom in the alkane is replaced by a bromine atom. The reactivity of different types of hydrogens in free radical bromination follows a specific order: tertiary ($3^\circ$) > secondary ($2^\circ$) > primary ($1^\circ$). This means tertiary hydrogens are the most reactive.
• "Single isomer of a tertiary bromide": This is the crucial piece of information. It implies that upon monobromination, the alkane produces only one unique structural isomer that is a tertiary bromide. This can happen if there is only one type of tertiary hydrogen in the alkane, or if all tertiary hydrogens are chemically equivalent due to molecular symmetry, leading to the same product regardless of which one is substituted. Given the higher reactivity of tertiary hydrogens, this product is expected to be the major one.

Let's work through the problem step-by-step:

1. Determine the possible structures of the primary alkyl group (R) and the resulting symmetrical alkane.

   • The alkane is C8H18, formed from $R-R$. Therefore, the alkyl group $R$ must be $C_4H_9$.
   • We need to identify primary alkyl groups with 4 carbon atoms:
     • n-butyl group: $CH_3CH_2CH_2CH_2-$ (The carbon attached to the rest of the molecule is primary).
     • Isobutyl group: $(CH_3)_2CHCH_2-$ (The carbon attached to the rest of the molecule is primary).
     • Other C4 alkyl groups (sec-butyl, tert-butyl) are secondary or tertiary, respectively, and thus cannot be the starting primary alkyl halide.
   • Now, let's form the C8H18 alkane ($R-R$) from these primary alkyl groups:
     • If $R$ is n-butyl, the alkane is $CH_3CH_2CH_2CH_2-CH_2CH_2CH_2CH_3$, which is n-octane.
     • If $R$ is isobutyl, the alkane is $(CH_3)_2CHCH_2-CH_2CH(CH_3)_2$, which is $2,5$-dimethylhexane.

2. Evaluate these alkanes based on the monobromination condition.

   • Case 1: Alkane is n-octane.

     • Structure: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$
     • n-octane contains only primary hydrogens (on C1 and C8) and secondary hydrogens (on C2-C7). It does not have any tertiary hydrogens.
     • Therefore, n-octane cannot yield a tertiary bromide upon monobromination. This rules out n-octane.

   • Case 2: Alkane is $2,5$-dimethylhexane.

     • Structure:

         CH3   CH3
         |     |
       CH3-CH-CH2-CH2-CH-CH3
       

     • Let's identify the types of hydrogens in $2,5$-dimethylhexane:
       • Primary hydrogens: Present on the six methyl ($CH_3$) groups. Due to the molecule's symmetry, all 12 primary hydrogens are chemically equivalent. …