Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Biacetyl has a molecular formula of C\(_4\)H\(_6\)O\(_2\). From its molecular formula, we can determine that it contains two carbonyl groups (C=O). The Lewis structure of biacetyl is: H H | H | | | | C-C=O C=C=O | | | H Acetoin has a molecular formula of C\(_4\)H\(_8\)O\(_2\). It contains one carbonyl group (C=O) and one hydroxyl group (OH). The Lewis structure of acetoin is: H H H H | | | | C-C=O C-C-O-H | | | H H H

For both biacetyl and acetoin, we can determine the hybridization of the carbon atoms by counting the electron domains around each carbon atom. In biacetyl, the carbon atoms in the middle (C\(_1\) and C\(_2\)) are \(\mathrm{sp^2}\) hybridized, while the carbon atoms at the ends (C\(_3\) and C\(_4\)) are \(\mathrm{sp}\) hybridized. In acetoin, the first and fourth carbon atoms (C\(_1\) and C\(_4\)) are \(\mathrm{sp^3}\) hybridized, the second carbon atom (C\(_2\)) is \(\mathrm{sp^2}\) hybridized, and the third carbon atom (C\(_3\)) is also \(\mathrm{sp^2}\) hybridized. For the C-C-O bond angles, we can predict the following: - Biacetyl: C\(_1\)-C\(_2\)-O will be 120° (due to \(\mathrm{sp^2}\) hybridization) and C\(_2\)-C\(_1\)-O will be 180° (due to \(\mathrm{sp}\) hybridization). - Acetoin: C\(_1\)-C\(_2\)-O will be 120° (due to \(\mathrm{sp^2}\) hybridization) and C\(_3\)-C\(_4\)-O will be about 109.5° (due to \(\mathrm{sp^3}\) hybridization).

In biacetyl, the carbon atoms C\(_1\) and C\(_2\) are \(\mathrm{sp^2}\) hybridized, which means that their atoms and bonding electron pairs are arranged in a trigonal planar shape. Additionally, the carbonyl groups (C=O) also have a planar arrangement. Therefore, all four carbon atoms and two oxygen atoms in biacetyl lie in the same plane.

We can count the number of sigma (single bond) and pi (double and triple bonds) bonds in both biacetyl and acetoin by analyzing their Lewis structures: - Biacetyl: 7 \(\sigma\) bonds (5 C-C and 2 C-H) and 3 \(\pi\) bonds (1 C=C and 2 C=O) - Acetoin: 9 \(\sigma\) bonds (6 C-C, 2 C-H, and 1 C-O) and 1 \(\pi\) bond (1 C=O) In conclusion, biacetyl and acetoin have different bond angles and hybridizations due to the presence of carbonyl and hydroxyl groups. Biacetyl has a planar arrangement of atoms, while acetoin has both planar and tetrahedral sections. The number of sigma and pi bonds in these compounds differs significantly, with biacetyl having more pi bonds and acetoin having more sigma bonds.