How The Atoms Are Arranged Within A Molecule

Empirical formulas represent the ratio of atoms in a formula. It is expressed in the simplest, small, whole number ratio. The molecular formula can either be the empirical formula, or a multiple of the empirical

The aldehyde group in pentanal is polar and because none of the Hydrogen atoms are directly bonded to oxygen so no Hydrogen bonding is possible. From the other hand 1-pentanol is an alcohol with five carbon atoms with the molecular formula C5H12O, also 1-Pentanol is both Hydrogen donor and accepter which causes an increase in the boiling point. (Smith, 2001) The figure below shows the similarities in the two compounds Pentanal and 1-Pentanol that they both have the same number of carbon atoms with the same arrangement. The difference is on that end carbon atom where two Hydrogen atoms are removed to make the aldehyde group.

n-Pentane and 1-Butanol have similar molecular weights, as well as fairly linear structures. This leads to them both having roughly similar London forces. However the 1-Butanol is packing a Hydroxyl group that the n-Pentane lacks, so when 1-Butanol is found experimentally to have a much lower ?T than n-Pentane, it is because it has much higher IMF of attraction due to Hydrogen Bonding compared to the n-Pentane. It also benefits from dipole-dipole forces due to its polarity as a molecule.

2. Using the atom and bond library below, construct the following molecule. It may be

3. In Molecular Series I (Straight-Chain Alcohols): As you go through the group from methanol  ethanol  1-propanol  1-butanol:

The term chiral can be used to describe an organic molecule if it is consists of a carbon atom which is attached to four different groups around it.

These are just a few of the possible Addition that can occur across the double bond

Molecule- a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical

- The C-H bonds in this structure are shown at 1444 and 1368cm-1. These two bands indicate the two different types of C-H bends that occur on the molecule. One is that of the alkene and the other is that of the several alkanes on the molecule.

These are Carvone, Limonene, and Thymol. Each of these molecules has an enantiomer where its mirror image has the same molecular formula yet when arranged differently they produce something different. To begin let us look at Limonene. Limonene has the molecular formula C10H16.

Figure 4: Reaction 4 [1-bromopentane + K+ -OC(CH3)3 (Potassium tert-butoxide)] and its (theorized) major and minor products are shown. The major product was 1-t-butoxypentane and the minor was 1-pentene (in consecutive order). Note that 1-pentene increases in reaction 4 relative to reaction 3. This is due to steric hindrance (bulky tert-butoxide) which decreases the SN2 product in reaction 4 relative to reaction 3.

Never add the substitutent group to a terminal carbon atom to avoid repeating structures. • When n = 5, The methyl group (-CH3) cannot be added to the terminal carbon atoms, C1 and C5. Otherwise, it would just duplicate the straight chain isomer drawn in step 1. There are only three positions available for the methyl group (-CH3) to add to, C2, C3 and C4.

The results from the NMR of 1-propanol showed 3 different prominent peaks with the peak at 2.2 cm-1 being the acetone. Because 1-bromopropane has three non-equivalent hydrogens it was found to represent this set of NMR data. The other product, 2-bromopropane only had 2 different types of hydrogens and would have only had 2 peaks. Further analysis of the structure of 1-bromopropane showed that the hydrogens closest the bromine group were an indication of peak A in the graph. Because of the electronegativity of the bromine, this peak was located further downfield. There were 2 neighboring hydrogens so using the n+1 rule gave the 3 peaks. Going down peak B showed the next carbon which had 5 neighboring hydrogens thus giving 6 peaks. Finally, the carbon furthest away from the bromine was found at peak C. It had 2 neighboring hydrogens and provided 3 peaks.

This experiment was done to identify one ketone with Thin Layer Chromatography and one using nuclear magnetic resonance spectrometry. When using Thin Layer Chromatography, I was able to determine my unknown by observing my slides and seeing that the spots of my unknown are most like those of 2-heptanone in height and size. Furthermore, I calculated my Rf values and determined that my Rf value of Unknown K (.5) was most near my Rf value of 2-heptanone (.51). Finally, comparing the derivative melting points of these two substances confirms that 2-heptanone is my unknown. The derivative melting point of 2-heptanone is 89 while my unknown was 87.

Polycyclic Aromatic Hydrocarbons (PAHs) consist of two or more benzene-ringed compounds made of only carbon and hydrogen. Compounds which are made of two rings are known as di- aromatics whereas those made of three benzene rings are known as tri- aromatics and so forth. As the number of rings increase, molecular weights of these compounds also increase and the compounds tend towards higher thermodynamic stability. Most common examples of PAHs are naphthalene, phenanthrene, pyrene etc.