The question involves a reaction between p-anisaldehyde and formaldehyde in the presence of concentrated sodium hydroxide. This is an example of the Cannizzaro reaction, which typically occurs with aldehydes that do not contain an alpha hydrogen.
- In the Cannizzaro reaction, one molecule of the aldehyde is oxidized to an acid (or its salt), while another molecule is reduced to an alcohol. This reaction usually requires the presence of a strong base like sodium hydroxide.
- p-Anisaldehyde is C_6H_4(OCH_3)CHO, and it does not have any alpha hydrogen, making it a suitable candidate for the Cannizzaro reaction.
- Concurrently, formaldehyde, or HCHO, also lacks an alpha hydrogen. In this reaction setup, formaldehyde will also undergo a similar disproportionation.
- The result of the Cannizzaro reaction is that p-anisaldehyde is reduced to p-Methoxy benzyl alcohol (C_6H_4(OCH_3)CH_2OH), while formaldehyde gets oxidized to formic acid, which reacts with NaOH to give sodium formate.
Thus, the correct answer is p-Methoxy benzyl alcohol. This product results from the reduction part of the Cannizzaro reaction affecting p-anisaldehyde. The other options are not feasible products from this reaction because they don't align with the typical outcomes of a Cannizzaro reaction under these conditions.