Exam Study Notes

Synthesis with Alkynes

• If you look at these reactions we are converting three carbons in a chain to four. We are adding one carbon.
  • To extend the chain, convert the starting material into a terminal alkyne and then react with methyl chloride or methyl bromide in an SM_2 reaction.
• To do SM_2, transform your starting material into a terminal alkyne.
• The very last step looks like hydrogenation.

Reactions with Alkenes

• With an alkene, the reactions you can use include Br2 or Cl2 with light.
  • Either chlorine or bromine would halogenate the molecule.
• Any strong base can be used, like sodium hydroxide or tert-butoxide.
• To turn the starting material into an alkyne, you need two leaving groups on the molecule.
  • Halogenation of the alkene with Br2 or Cl2 adds halogens on opposite sides.
  • Stereochemistry is not important unless it's in the final product.
  • NaNH_2 is used to make the triple bond and deprotonate the alkyne.

Chain Extension with Alkynes

• To add an extra two carbons, you can do the same pathway but add ethyl bromide, ethyl chloride, or ethyl iodide.
• The same pathway is used, but you can add ethyl groups.
  • Add chlorine or bromine to make a double bond, then add halogens, make a triple bond, and add an ethyl group using ethyl chloride, bromide, or iodide.
• Instead of turning the starting material into a terminal alkyne, you can convert it into an alkyl halide.
  • This will add two carbons or more.
• You can extend the chain by adding alkyl halides.

SN2 Requirements

• For SN_2 reactions, the alkyl halide must be at a methyl or primary position.

Regioselectivity

• Br2 or Cl2 will add the halogen.
• Sodium hydroxide gives an alkene.
• HBr with hydrogen peroxide and light or heat will add the halogen to the least substituted position for an SN_2 reaction with a terminal alkyne, followed by hydrogenation.
• If both pathways end with an alkane, it doesn't matter which pathway is used.
• If the product is an alkene, the position of the double bond indicates which alkyne was added.
• If a ketone or aldehyde is formed, it indicates which pathway to use.
  • If one pathway requires more steps, try the other pathway.

Carbon-13 NMR and Mass Spectrometry

• Carbon-13 NMR range shown is from 150 to 200.
• Determine the formula of the molecule using mass spectrometry.
• The molecular ion peak (e.g., 72) represents the molecular weight of the molecule before fragmentation.
• If the molecular ion peak and the molecular ion peak + 2 are about the same height, you have chlorine. If they are in 1:3 ratio, you have bromine. A Chlorine and Bromine will not appear in the same molecule.
• Carbon-13 NMR and IR spectroscopy can provide additional information.
  • IR spectroscopy indicates the presence of a carbonyl group.
  • Carbon-13 NMR can quantify the number of carbonyl groups.

Calculating Molecular Formula

• If a carbonyl is present, subtract the mass of carbon (12) and oxygen (16), totaling 28, from the molecular weight.
• Assume the remaining mass comprises carbons and hydrogens.
• Divide the remaining mass by 12 to find the maximum number of carbons.
  • Round down to the nearest whole number.
• Multiply the number of carbons by 12 and subtract from the remaining mass to find the mass of hydrogens.
• The formula is always Carbon Hydrogen then other components.
• If subtracting leads to 1 or fewer hydrogens, decrease the number of carbons.

Example Calculation

• Given a molecular weight of 72 and a carbonyl group:
  • 72 - 28 = 44
  • 44 / 12 = 3.7 (round down to 3 carbons).
  • 3 Imes 12 = 36
  • 44 - 36 = 8 (8 hydrogens).
  • Formula: C4H8O

Mass Spectrometry Fragments

• The instructor will focus on fragments containing carbons, hydrogens, oxygens, chlorines, or bromines.
• Assume everything left after subtracting the known groups are carbons and hydrogens.

Determining Molecular Structure Using Mass Spec

• The molecular formula can be determined, then NMR spectroscopy is used to solve structure. If you get the final formula wrong at the outset, structure determination from NMR will be impossible, so these problems should be relatively straightforward.

Acylium Ion

• Cleavage of a ketone, aldehyde, or carboxylic acid results in a carbonyl with a positive charge (acylium ion).
• The oxygen can resonate, stabilizing the ion.
• The base peak is the most abundant and stable fragment.

Identifying Halides

• Alohide? Check for peaks at 2750.
• To determine the presence of halides (chlorine, bromine), look for the molecular ion peak plus two.
  • The relative height indicates the isotope.

Example 2: Mass Spec

• If the molecular ion peak is 102 and there's one carbonyl and one carboxylic acid (COOH):
  • Subtract the mass of the carboxylic acid (45) from 102 to get 57, then the carbonyl (28) from 102, instead.
    • Which ends up being the same thing since both COOH and >C=O are in the molecule.
  • Divide 57 by 12 to get 4.75 (4 carbons).
  • 4 Imes 12 = 48
  • 57 - 48 = 9 (9 hydrogens).
  • Formula: C5H{10}O_2

Example 3: Mass Spec

• If there is a peak at 122 and another peak at 124 with a one-to-one ratio:
  • This indicates the presence of bromine.
  • Subtract 79.9 (actual atomic weight) from 122, resulting in 43.
  • If there are no carbonyls OH's etc., divide 43 by 12.
  • 3 Carbons.
    • (3*12) = 36.
      • 43 -36 = 7 hydrogens
        *Formula: C3H7Br

Example 4: Flourine Mass Spec

• If there is a peak at 106 and another at 108 that is 1/3 the size, that is Flourine.
  • If there are two peaks that are relative to eachother, that means there is a chlorine.
    • The Chlorine peak is made up of Chlorine 35.

Mass Spec Masses

• Carbon mass of 12, Hydrogen mass of 1, and Oxygen mass of 16.
• Bromine (assuming select it) mass of 79.