(Solved): Please help me with the pre lab pages. Pre-lab notebook preparation (sections I-III): Prepare as usu ...

Please help me with the pre lab pages. Pre-lab notebook preparation (sections I-III): Prepare as usual, including a balanced generic chemical equation similar to that shown above (what's shown above is not fully balanced), and a table containing structures of possible aldehydes, ketones, and aldol condensation products (do not just copy the table below). You don't need to look up and report any data for these possible reactants and products for your pre-lab notebook entry. Limiting reagent \& theoretical yield calculations: AFTER you have identified your unknown you can write in your lab notebook a correct balanced chemical equation, calculations for moles of reactants/reagents used, and calculations for limiting reagent and theoretical yield. Treat this as a practice problem for your lab exam - you should write this in your lab notebook, but you should NOT turn in those pages. Background: The aldol condensation reaction is an important carbon-carbon bond forming reaction in organic chemistry. Besides being utilized routinely in industrial and research applications, the aldol condensation is involved in various biological processes such as glycolysis, gluconeogenesis and photosynthesis. In this experiment you will do the following: 1) Combine \( \geq 2 \) molar equivalents of an unknown aldehyde with an unknown ketone in the presence of NaOH to prepare the corresponding unknown aldol condensation product. 2) You will determine the identity of the product (and, indirectly, the identities of the aldehyde and ketone starting materials) by comparing the observed melting point ( mp ) range of your product with the melting points of the potential products listed in Table 1, and also by analyzing your product's spectral data (to be posted on Bb ). Table 1. Melting Points ( \( { }^{\circ} \mathrm{C} \) ) of Possible Aldol Condensation Products (product structures not shown) Ketone Aldehyde ( \( \geq 2 \) molar equiv.)