1. Find the hydrocarbon with molecular mass $72u$

For an alkane, general formula is: $C_nH_{2n+2}$ Its molecular mass is: $12n + (2n+2) = 14n+2$ Given: $14n+2=72$ $14n=70$ $n=5$ So the hydrocarbon is: $C_5H_{12}$

2. Write the isomers of $C_5H_{12}$

The structural isomers are:

• $n$-pentane
• isopentane ($2$-methylbutane)
• neopentane ($2,2$-dimethylpropane)

The question asks for the branched chain isomer that gives only one isomer of monosubstituted alkyl halide.

3. Check each branched isomer

(i) Isopentane: $2$-methylbutane

Structure: $CH_3-CH(CH_3)-CH_2-CH_3$ Here, hydrogen atoms are not all equivalent. There are different types of carbon atoms, so substitution of one H by halogen can occur at different positions.

Thus, isopentane gives more than one monosubstituted alkyl halide.

(ii) Neopentane: $2,2$-dimethylpropane

Structure: $C(CH_3)_4$ All four methyl groups are equivalent due to symmetry. Therefore, all $12$ hydrogens are equivalent.

So replacing any one H by halogen gives the same single product.

Hence, neopentane gives only one monosubstituted alkyl halide.

4. Evaluate options

• A: Tertiary butyl chloride $o$ not a hydrocarbon, already a haloalkane, so not relevant.
• B: Neopentane $o$ correct.
• C: Isohexane $o$ formula is not $C_5H_{12}$.
• D: Neohexane $o$ formula is not $C_5H_{12}$.

5. Final answer

The required branched chain isomer is: $\boxed{\text{Neopentane}}$