Question

Please help answer P1-P4 for this Perkin Reaction Lab

Perkin Reaction Organic REActivity PERKIN REACTION (ALDOL CONDENSATION) The Reaction Scheme: Formation of trans-cinnamic acid The following reaction is named the Perkin Reaction. It is a very specific type of Aldol condensation that forms a variety of cinnamic acid derivatives. In this lab, you will perform a Perkin reaction using benzaldehyde and potassium carbonate in acetic anhydride to form trans- cinnamic acid (1). Trans-cinnamic acid is a natural compound found in the oil of cinnamon and is often used as a flavor component of cinnamon. O K₂CO₃ O O CyH60 Mol. Wt.: 106.12 bp: 178°C C4H603 Mol. Wt.: 102.09 bp: 140°C C9H2O2 Mol. Wt.: 148.16 mp: 135°C Paired Experiment Formation of trans-cinnamic acid via the Perkin condensation involves a 60 minute reflux. Ample time is available during the reaction to stop and think more critically about what's happening at the molecular level. Isolation and characterization of the products will be required. Set-up: A. Set up sand bath on a hot plate into which seal round bottomed flask can fit. B. Obtain the following items in no particular order: 50 mL round bottom flask (RBF) 5 mL acetic anhydride 3g benzaldehyde 3.8g K2CO3 1 magnetic stir bar Reflux condenser 2 condenser lines Heating plate Sand bath Thermometer C. Clamp your 50 mL round bottomed flask to the monkey bars. D. Add the stir bar to the flask.

Answer 1

P1.

Step 1: K is a metal whereas C and O is nonmetals. So the compound is a ionic one. So potassium (K) will completely donate its 1 valence electron to the carbonate group and form K+ ion.

Step 2: Count the total valence elctron of the covalent part, i.e. carbonate (covalent since carbonate is made of two nonmetals- C and O). Here total valence electron = 4 (for 1 carbon) + 18 (6 for each oxygen) + 2 (donated by 2 K)

= 24

Step 3: Determine the central atom to which all other atoms will be attached. Rule of thumb for this is that the atom which is least in number (mostly 1 atom is present.) in the molecule. Here, C is the central atom.

Step 4: Place two elctrons in between the central carbon atom and all of the 3 oxygen atom to form three single bonds. 6 electrons are used up to form 3 C-O bonds.

Step 5: Then place all other remaining elctrons (18) to the oxygen atoms as lone pairs.

Step 6: The central atom will not have an octet of electrons here (C will have 6 electrons upto now). So to fill up octet one of the oxygen will donate and the equally share (as is the rule in covalent bonds) one pair of elctron to form a C-O double bond (The C-O bond directed upwards in the picture below.).

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P2.

Do (2 P2. I LH-0- > Hilor Bicarbonate formed during the generation of nucleophile in the very 1st step of the reaction mechanism (see P4) This bicarbonate may once again acte as a base to produce another nucleophile and itself converting to carbonic acid .. C ma CA bon ab07 n VC He og Heorh CH₂ → Hojet CQ (9) H, O

P3. The evolution of bubbles of CO2 (the foaming of the solution), indicates the formation of carbonic acid and the consequent decomposition of it in the reaction mixture.

P4.

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