Quiz A- Basic Organic Chem. Principles 1.How many bonds do the following atoms form in a covalent compound? a) O b) C c) S d) N e) H f) P 2. Write down the formula of the functional group in each of the following compounds. a) amine b) amide c) alcohol d) carboxylic acid e) ketone f) aldehyde g) ester 3.Give the name for each of the following molecules. a) CHCH b) CH 3 CH 2 OH c) C 2 H 6 d) e) f) g)
Quiz A- Basic Organic Chem. Principles 4.Which of the following molecules might be unsaturated and which are definitely unsaturated? a) C 3 H 8 b) C 4 H 8 c) C 3 H 4 d) CH 3 CHCHCH 3 5 a) Name the following molecules. i) CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH ii) CH 3 CH 2 CH 2 CH 2 CH 2 OH iii) CH 3 CH 2 OH b) 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 point temperature? Explain your answer. temperature? Explain your answer.
Quiz A- Basic Organic Chem. Principles 6. Which type of reactions do you expect the following molecules to undergo? 7. 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 following formulae. formulae. a) C 3 H 8 O a) C 4 H 8 a) CH 2 CH 2 b) CHCH c) CH 3 CH 2 CH – CH 3 Cl
Quiz A- Basic Organic Chem. Principles 9. Identify the functional groups present in the following molecule.
Proteins The building blocks of proteins are amino – acids. Most structurally complex organic macromolecules. = side chain 20 amino acids in all 12 the body can synthesize termed non-essential amino acids. 8 that we obtain from our diet termed essential amino acids.
Primary/Tertiary structure of Proteins L–Alanine L–Cysteine
Secondary structures - -helix Clockwise 3.6 Amino acids per turn of the helix
Secondary Structures - Zwitter ion formation Some amino acids have extra reactive groups e.g. Carboxylic acid (COOH) – glutamic acid – aspartic acid – aspartic acid Amine (NH 2 ) – lysine H+ - Ionic bond
Secondary structures - -sheets Top view Side view Hydrogen bonding
Tertiary structures - sheets - helix Contains a combination of secondary structures & primary structures. Has a 3D structure. Structure determined by side – chains.
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 protein hydrogen bonding? 5.Draw the zwitter – ion structure of the amino - acids glycine and serine. 6.Draw the chemical structure of the tripeptide formed from the following sequence of amino-acids. Highlight the acidic end of the molecule and the basic end of the molecule.
Carbohydrates Monosaccharides ‘ oses ’ “CH 2 O” Can be split into two groups : Aldoses & Ketoses. ‘ Ald ’ Aldehyde ‘ Ket ’ Ketone Aldoses e.g. Glucose Ketoses e.g. Fructose ‘ C 6 H 12 O 6 ’ Both Glucose & Fructose are Hex - ‘ oses’. Sugars can be 3 to 7 carbons long.
Carbohydrates In aqueous solution sugars form rings e.g. glucose. 1 2 3 4 5 6 1 2 3 4 5 6
1 23 4 5 Carbohydrates Sugars polymerize via glycosidic 1–4 linkages. 1 23 4 5 Glycosidic linkage linkage Dehydration 2C 6 H 12 O 6 C 12 H 22 O 11 + H 2 O
Carbohydrates The resulting polysaccharide of 1–4 linakges of glucose is starch (storage polysaccharide in plants). Starch Amylose Unbranched starch Amylopectin Branched Starch 1–4 linkages 1–4 linkages & 1–6 linkages Branched 1–6 linkages occur every 30 linkages 6
1 23 4 5 1 23 4 5 Carbohydrates Sugars polymerize via glycosidic 1–4 linkages as well. 1–4 Glycosidiclinkage The resulting polysaccharide is cellulose (structural polysaccharide in plants).
Quiz C- Macromolecules 1.The most abundant (by mass) class of organic macromolecules in the biological world is : A Carbohydrates B Proteins C Nucleic Acids (DNA & RNA) D Lipids 2.D – ribose is a pentose. It can also be classified as an aldose. Write down a possible structure for D – ribose. structure for D – ribose. 3.The monosaccharide threose has 4 Os ! That is it has 4 oxygen atoms. What is the molar mass of threose ?
Quiz C- Macromolecules 4. When one draws the structure of D – glucose there are 2 ways to do it : A comprehension exercise either a Fischer projection or a Haworth projection. The Fischer projection shows glucose as a linear molecule and the numbering of the carbon atoms is shown on the structure. When it cyclizes it is the oxygen atom on carbon 5 which ends up in the ring. 12 3 4 5 6 1 2 3 4 56 This ring structure is called the Haworth projection and at equilibrium between the two forms the linear form is only 0.02% of the total glucose present.
Quiz 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 Fischer projection are up (above the ring). The –OH group on carbon 1 (the so called anomeric carbon) can either be up or down, ( ), or up, ( ). The form is about twice as common as the form. a)Draw the Haworth projection of the form of D – idose, given the Fischer projection below. below. b) Draw the Fischer projection of D – galactose given the Haworth projection below. - D - galactose