Step 1: Understanding the Concept
Sn with HCl gives nascent hydrogen, which is a strong reducing agent for the nitro group.
Step 2: Reaction
\(\text{R-NO}_{2}+6[\text{H}]\xrightarrow{\text{Sn/HCl}}\text{R-NH}_{2}+2\text{H}_{2}\text{O}\)
Nitrogen stays bonded to the same single alkyl group.
Step 3: Conclusion
The product is a primary amine, not a secondary, tertiary or quaternary one.
Final Answer:
Nitroalkanes on reduction with Sn/HCl give primary amines, option (A).
\[ \boxed{\text{(A) }\text{R-NH}_{2}} \]