3Quiz A- Basic Organic Chem. Principles How many bonds do the following atoms form in a covalent compound?a) Ob) Cc) Sd) Ne) Hf) P2. Write down the formula of the functional group in each of the following compounds.a) amineb) amidec) alcohold) carboxylic acide) ketonef) aldehydeg) ester3. Give the name for each of the following molecules.a) CHCHe)b) CH3CH2OHf)c) C2H6g)d)
4Quiz A- Basic Organic Chem. Principles Which of the following molecules might be unsaturated and which are definitelyunsaturated?a) C3H8b) C4H8c) C3H4d) CH3CHCHCH35 a) Name the following molecules.i) CH3CH2CH2CH2CH2CH2CH2CH2OHii) CH3CH2CH2CH2CH2OHiii) CH3CH2OHb) Identify the 2 types of intermolecular bonding present in each of the above molecules.c) Which of the three molecules do you expect to have the highest boiling pointtemperature? Explain your answer.
5Quiz A- Basic Organic Chem. Principles 6. Which type of reactions do you expect the following molecules to undergo?a) CH2CH2b) CHCHc) CH3CH2CH – CH3Cl7. Classify the type of reaction occurring in each of the following cases.a)b)c)8. Write down all the possible isomers for the compounds which have the followingformulae.a) C3H8Oa) C4H8
6Quiz A- Basic Organic Chem. Principles 9. Identify the functional groups present in the following molecule.
7Proteins Base Acid Most structurally complex organic macromolecules. The building blocks of proteins are amino – acids.BaseAcid= side chain20 amino acids in all12 the body can synthesize termed non-essential amino acids.8 that we obtain from our diet termed essential amino acids.
19Tertiary structures - sheets - helix Has a 3D structure. Contains a combination of secondary structures& primary structures.Structure determined by side – chains.
20Cys • Lys • Tyr Quiz B- Proteins 1. Which of the amino-acids have a hydrophobic side-chain?2. Which of the amino-acids have an acidic side-chain?3. Which of the amino-acids have an basic side-chain?4. Which amino acids have a side chain which can become involved in proteinhydrogen bonding?5. Draw the zwitter – ion structure of the amino-acids glycine and serine.Draw the chemical structure of the tripeptide formed from the following sequenceof amino-acids.• Lys• TyrCysHighlight the acidic end of the molecule and the basic end of the molecule.
21Carbohydrates Monosaccharides Can be split into two groups : Aldoses & Ketoses.‘ oses ’ “CH2O”‘ Ald ’ Aldehyde‘ Ket ’ Ketone‘ C6H12O6’Aldoses e.g. GlucoseKetoses e.g. FructoseBoth Glucose & Fructose are Hex - ‘oses’.Sugars can be 3 to 7 carbons long.
22Carbohydrates In aqueous solution sugars form rings e.g. glucose. 1 2 6354415236
23Carbohydrates Sugars polymerize via glycosidic 1–4 linkages. C O 1 2 345COH12345COH12345HOGlycosidiclinkage2C6H12O6 C12H22O11 + H2ODehydration
24Carbohydrates glucose + glucose maltose glucose + fructose sucrosepolysaccharidemonosaccharidedisaccharidee.g. GlucoseMaltoseStarch (100s s of units)
25CarbohydratesThe resulting polysaccharide of 1–4 linakges of glucose is starch (storagepolysaccharide in plants).AmyloseUnbranched starch 1–4 linkagesStarchAmylopectinBranched Starch61–4 linkages &1–6 linkagesBranched 1–6 linkages occur every 30 linkages
26Carbohydrates Sugars polymerize via glycosidic 1–4 linkages as well. 2345COH12345COH12345HO1–4GlycosidiclinkageThe resulting polysaccharide is cellulose (structural polysaccharide in plants).
27Quiz C- Macromolecules The most abundant (by mass) class of organic macromolecules in the biologicalworld is :A CarbohydratesB ProteinsC Nucleic Acids (DNA & RNA)D LipidsD – ribose is a pentose. It can also be classified as an aldose. Write down a possiblestructure for D – ribose.The monosaccharide threose has 4 Os ! That is it has 4 oxygen atoms.What is the molar mass of threose?
28Quiz C- Macromolecules A comprehension exercise4. When one draws the structure of D – glucose there are 2 ways to do it :either a Fischer projection or a Haworth projection.123456The Fischer projection shows glucose as a linearmolecule and the numbering of the carbon atoms is shownon the structure. When it cyclizes it is the oxygen atomon carbon 5 which ends up in the ring.123456This ring structure is called the Haworth projection andat equilibrium between the two forms the linear form isonly 0.02% of the total glucose present.
29Quiz C- Macromolecules The Haworth projection has the atoms on the RHS of the Fischer projection ‘down’i.e. below the ring in the Haworth projection, and those on the LHS of the Fischerprojection are up (above the ring). The –OH group on carbon 1 (the so calledanomeric carbon) can either be up or down, (), or up, (). The form is abouttwice as common as the form.Draw the Haworth projection of the form of D – idose, given the Fischer projectionbelow.b) Draw the Fischer projection of D – galactose given the Haworth projection below.- D - galactose