Recall the condition for tautomerism

Tautomerism, especially keto-enol tautomerism, is shown by compounds having:

• a carbonyl group $\left(\ce{C=O}\right)$, and
• at least one $\alpha$-hydrogen on the carbon adjacent to the carbonyl carbon.

Then the equilibrium is: $\ce{R-CO-CH2-R' <=> R-C(OH)=CH-R'}$

Check each option

A: Lactic acid

Lactic acid is $\ce{CH3-CH(OH)-COOH}$. It contains a carboxylic acid group, not a simple aldehyde/ketone system suitable for normal keto-enol tautomerism in this context. Hence, it is not typically considered to exhibit tautomerism here.

B: 2-Pentanone

Structure: $\ce{CH3-CO-CH2-CH2-CH3}$ It is a ketone and has $\alpha$-hydrogens on both sides of the carbonyl group. Therefore, it shows keto-enol tautomerism.

Example: $\ce{CH3-CO-CH2-CH2-CH3 <=> CH3-C(OH)=CH-CH2-CH3}$ So, this compound exhibits tautomerism.

C: Phenol

Phenol mainly exists as $\ce{C6H5OH}$. Although special valence tautomeric forms may be discussed in advanced contexts, it is not the standard answer for simple tautomerism questions of this type. So, not correct here.

D: 2-Butene

Structure: $\ce{CH3-CH=CH-CH3}$ This is an alkene and does not have the required carbonyl-$\alpha$-hydrogen arrangement. Hence, it does not show keto-enol tautomerism.

Conclusion

The compound that exhibits tautomerism is: $\boxed{\text{2-Pentanone}}$

Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

These match.