Interpret the two reactions

The question is about two elimination/substitution-type reactions in haloalkanes/haloarenes, and the options ask how the rate changes with:

• concentration of base, and
• nature of base ($OH^-$ vs $RO^-$).

Such questions typically distinguish between:

• Reaction (1): a unimolecular process, i.e. $E1$/$S_N1$-type, where the rate-determining step is ionization of the substrate.
• Reaction (2): a bimolecular elimination, i.e. $E2$, where the base participates in the rate-determining step.

Rate law for reaction (1)

For an $E1$-type reaction, $\text{Rate} = k[\text{substrate}]$

because the slow step is formation of a carbocation by departure of the leaving group.

Therefore, changing the concentration of base does not affect the rate of reaction (1).

So, statement A is true.

Rate law for reaction (2)

For an $E2$-type reaction, $\text{Rate} = k[\text{substrate}][\text{base}]$

Hence the rate does depend on base concentration.

Therefore:

• Statement B (“Changing the concentration of base will have no effect on reaction (2)”) is false.

Effect of changing base from $OH^-$ to $RO^-$ in reaction (2)

In an $E2$ reaction, the base is directly involved in the transition state. Changing from $OH^-$ to alkoxide $RO^-$ changes basicity/steric effects and thus generally changes the rate.

Therefore statement C is false.

Check statement D

Statement D says doubling base concentration doubles the rate of both reactions.

• For reaction (1): no effect.
• For reaction (2): rate doubles.

Hence D is false.

Conclusion

The only true statement is: $\boxed{A}$

Comparison with stored answer

Stored correct answer = A.

My derived answer also = A. So they agree.