The mechanism of Sn1 reaction is given as:
$$R - X \rightarrow R^\oplus X^\ominus \rightarrow R^\oplus \| X^\ominus \xrightarrow{Y^\ominus} R - Y + X^\ominus$$
Ion pair    Solvent separated ion pair
A student writes general characteristics based on the given mechanism as:
(a) The reaction is favoured by weak nucleophiles.
(b) $$R^\oplus$$ would be easily formed if the substituents are bulky
(c) The reaction is accompanied by racemization
(d) The reaction is favoured by non-polar solvents.
Which observations are correct?

For an $$\mathrm{S_N1}$$ (unimolecular nucleophilic substitution) process, the mechanism proceeds in two distinct kinetic steps. First, the substrate $$R\!-\!X$$ ionises to form a carbocation $$R^{\oplus}$$ and a leaving group $$X^{\ominus}$$. Second, the nucleophile $$Y^{\ominus}$$ attacks the planar carbocation. Symbolically, the elementary events are

$$R - X \;\longrightarrow\; R^{\oplus} + X^{\ominus}$$

followed by

$$R^{\oplus} + Y^{\ominus} \;\longrightarrow\; R - Y.$$

The experiment-based rate law for an $$\mathrm{S_N1}$$ reaction is stated first:

$$\text{rate} \;=\; k\,[R\!-\!X]$$

because the slow step is the formation of the carbocation. This rate law contains no concentration term for the nucleophile, so increasing or decreasing nucleophile strength hardly alters the rate.

Now we judge each student observation one by one using the above principles.

(a) “The reaction is favoured by weak nucleophiles.” Since the nucleophile does not appear in the rate law, even very weak nucleophiles react readily after the carbocation is formed. Therefore the statement is true.

(b) “$$R^{\oplus}$$ would be easily formed if the substituents are bulky.” Bulky alkyl substituents mean more alkyl groups attached to the positively charged carbon. More alkyl groups stabilise a carbocation by the inductive effect and by hyperconjugation. The order of carbocation stability is $$\text{tertiary} > \text{secondary} > \text{primary} > \text{methyl}.$$ Hence a bulky (tertiary) substrate ionises faster, so this statement is also true.

(c) “The reaction is accompanied by racemization.” After the leaving group departs, the carbocation centre is trigonal planar. The nucleophile $$Y^{\ominus}$$ can attack from either side with almost equal probability, giving both inversion and retention products. The net result is racemisation (usually partial but often substantial). Thus statement (c) is true.

(d) “The reaction is favoured by non-polar solvents.” Formation of a carbocation and a free anion creates highly polar species. Polar protic solvents (e.g. water, alcohols) stabilise these ions through solvation and therefore accelerate the reaction. Non-polar solvents cannot stabilise ions efficiently, so the reaction rate drops. Thus statement (d) is false.

Collecting the results, (a), (b) and (c) are correct, while (d) is incorrect.

Hence, the correct answer is Option C.