Chemistry 3.5 - PowerPoint PPT Presentation

chemistry 3 5 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chemistry 3.5 PowerPoint Presentation
Download Presentation
Chemistry 3.5

play fullscreen
1 / 118
Chemistry 3.5
423 Views
Download Presentation
meghana
Download Presentation

Chemistry 3.5

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups

  2. You will notice the first thing the 3.5 standard says is : You will be expected to know the principles of Organic Chemistry you covered in AS 2.5 How well do you know your stuff from last year? The standard then goes on to say….

  3. Organic compounds described are limited to those containing no more than 8 carbon atoms Don’t think you get off that easy It then goes on to say ……… Larger organic molecules may be used in questions involving the application of organic principles (e.g. the identification of functional groups)

  4. So lets start with Alkanes 1. What’s the general formula for an alkane? CnH2n+2

  5. Can you draw the structural formula and molecular formula for the following? Methane Molecular Formula - CH4 Structural Formula

  6. Drawing a 3 dimensional Structure How about drawing the 3 dimensional structure for methane? Use the molymods to make a 3 D model Any ideas on how to draw it? These two lines represent bonds on the same plain as the paper These lines represent the bond going behind the paper This wedge represents the bond coming out from the paper

  7. What is the bond angle between each of the atoms? 109 o 109 o Describe the shape of this molecule Tetrahedral Is CH4 polar or non polar? Non polar Can you give a reason why?

  8. Alkane Nomenclature - Give the names, molecular and condensed structural formula for the first ten alkanes C2H6 (CH3)2 ethane C3H8 propane CH3CH2CH3 CH3(CH2)2CH3 C4H10 butane pentane C5H12 CH3(CH2)3CH3 hexane C6H14 CH3(CH2)4CH3 heptane C7H16 CH3(CH2)5CH3 octane C8H18 CH3(CH2)6CH3 nonane C9H20 CH3(CH2)7CH3 C10H22 decane CH3(CH2)8CH3

  9. Can you remember the alkane order? A simple mnemonic is- Many Elderly People Buy Pent Houses High Over North Dakota Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

  10. Cyclic Alkanes These are Alkanes with at least 1 ring of carbons eg cyclohexane Draw the structure of cyclohexane in your book? Cyclohexane Draw cyclopropane The general formula for an alkane with one ring is: CnH2n

  11. Do you remember how to name a branched alkane? Steps 1. Find the longest chain of continuous carbons (called the parent chain) and name it: ie 5 carbons – name pentane 2. Identify any side branches or functional groups ie methyl if there are more than one of the same type use prefixes di (2) , tri (3) etc. 2 methyls = dimethyl. So the name so far is dimethylpentane 3. Number the parent chain from the end that gives the side branch groups the lowest number methyl Pentane (parent chain) methyl

  12. Steps continued 3. Number the parent chain from the end that gives the side branch groups the lowest number ie from the left methyl 1 2 3 4 5 methyl 4. Separate numbers by comma’s and separate numbers from words with a dash – now you can name it 2,3-dimethylpentane

  13. Exercise – Draw the structures for the following • 3 – ethylheptane • 2,2,4-trimethylpentane

  14. Lets look at making some structure • Turn to Expt 1 in your booklet • In pairs make and draw the structures for Q 1 to Q 3 • Answer the questions

  15. CH3CH2C CCH3 CH3CH2CHCH2CH2CH2CH2CH3 OH CH3CHCH2CH3 Br CH3CH2CH2CH2CHCH2CH3 CH3 Structure Classification Name alkyne pent-2-yne alcohol (secondary) octan-3-ol O carboxylic acid 2-methylbutanoic acid CH3CH2CH C CH3 OH H H C C cis-pent-2-ene alkene H3C CH2CH3 haloalkane 2-bromobutane alkane 3-methylheptane

  16. Structure Classification Name CH3CH2 CH3 C C alkene 2,3-dimethylpent-2-ene H3C CH3 CH3 alcohol (tertiary) CH3CH2CH2C OH 2-methylpentan-2-ol CH3 O H C ester butyl methanoate O CH2CH2CH2CH3 CH3 CH3CH2CHCH2CHCH3 alkane 2,4-dimethylhexane CH3

  17. Structure Classification Name CHC CH CH3 3-methylbut-1-yne alkyne CH3 CH3CHCH2CHCH3 haloalkane 2-chloro-4-methylpentane CH3 Cl carboxylic acid CH3CH2CH2CH2CH2CH2COOH heptanoic acid O CH3CH2 C ester propyl propanoate O CH2CH2CH3

  18. Cholesterol is a major component of gallstones. From the following structure of the compound predict its reaction with .. C8H17 CH3 CH3 HO • Br2 • H 2 with a Pt catalyst • CH3COOH cholesterol

  19. The reaction with Br2results in addition of bromine to the double bonded carbons forming a single carbon bond. The solution would change colour from orange to colourless. C8H17 CH3 CH3 HO Br Br

  20. With H2 and a Pt catalyst a hydrogenation reaction would occur and H atoms would be added across the double bond forming a single C – C bond. C8C17 CH3 CH3 HO H H

  21. Ethanoic acid reacts with the hydoxy group to form an ester and water C8H17 CH3 CH3 O H3C C O + H2O

  22. We have some new functional groups to learn this year Title your page Functional groups Use the photocopied sheet and copy the complete the table neatly into your exercise book Yes the whole table! We must learn these!

  23. Complete the task on the handout , glue into your lab book Use your chart to help you classify and name the listed compounds (complete in pencil)

  24. Answers to the left hand column on handout CH3CH2CH2CH2CHCH2CH3 Cl ClasshaloalkaneName3-chloroheptane O CH3CH2CH2 C Cl Classacyl chlorideName butanoyl chloride CH3 HO C CH2CHCH3 O Classcarboxylic acidName3-methylbutanoic acid

  25. Answers to the left hand column continued CH3CH2CH2CH2NH2 ClassamineName 1- aminobutane O CH3CH2CH2CH2CH2C NH2 ClassamideName hexanamide CH3CH2CH2CH2 O C CH2CH2CH2CH3 O Class ester Namebutyl pentanoate

  26. Answers to the Right hand column CH3CH2CH2COCH2CH3 Class ketone Name hexan-3-one CH3CH2CH2CH2CH2OH Class alcohol Name pentan-1-ol CH3CH2CH2CH2CH2C O H Class aldehyde Name hexanal

  27. Answers to the right hand column continued CH3CH2CH2CH2CH2COCl Classacyl chlorideName hexanoyl chloride CH3CH2CHCH3 NH2 ClassamineName 2-aminobutane CH3CH2CH2CH2CH2CH2CH2CONH2 Class amide Nameoctanamide

  28. Structural Isomers (are also called constitutional isomers) These are molecules with the same molecular formula but different structural formula. The isomers of a particular molecule will have different physical properties e.g. melting and boiling points. They may also have different chemical properties. Draw and name the structural isomers of C4H10 Name: butane Name: 2-methylpropane Boiling point 36.1o C Boiling point -11.7o C

  29. Task – in pairs use the models to make hexane Draw the structural formula for hexane Then make as many structural isomers of hexane as you can Name and draw each one There should be 5?

  30. Structural Isomers of Hexane Hexane C6H14 H H H H H H C C C H H C C C H H H H H H H H H CH3 H 2-methylpentane C C H C C C H H H H H H

  31. Structural Isomers of Hexane H H H CH3 2,2-dimethylbutane C C H C C H H H H CH3 CH3 H H H CH2 3-methylpentane C C H C C H H H H H

  32. Structural Isomers of Hexane H H H CH3 2,3-dimethylbutane C C H C C H H CH3 H H

  33. Geometric (cis and trans) Isomers Geometric Isomers will only occur if…. The compound has a double or triple bond where there can be no rotation around the C C bond Remember alkanes don’t exhibit geometric isomerism because there is rotation around the single C C bond

  34. To exist as geometrical isomers the C atoms at both ends of the double bond must each have two different groups (or atoms attached). e.g. The Geometric Isomers of but-2-ene cis-but-2-ene trans-but-2-ene Bpt 3.7o C Bpt 0.88 o C * Geometric isomers have similar chemical properties but different physical properties

  35. Geometric isomerisim does not occur if one of the carbon atoms in the fixed (ie the double bond) has two identical atoms or groups of atoms attached But-1-ene does not have geometric isomers, because it has two groups, H atoms, attached to the C atoms on either side of the double bond flip it over and it’s the same as Therefore but-1- ene does not have geometric isomers

  36. Geometric isomers are a form of stereoisomerisim Stereoisomerism – are where the atoms are bonded in the same sequence but are arranged differently in space in a molecule. e.g. but-2-ene Same sequence of atoms Two different geometric isomers exist where atoms are arranged differently in space cis-but-2-ene trans-but-2-ene Bpt 3.7o C Bpt 0.88 o C

  37. Exercise • Draw the structures of the following alkenes and decide which of them can exist as cis-trans isomers • 2-methylbut-2-ene • b) 3 – methylpent-2-ene forms no cis/trans isomers Occurs as cis trans isomers cis-methylpent-2-ene trans-methylpent-2-ene

  38. Identify whether cis trans isomers occur with in the following molecules H3C CH3 H3C Cl HO Cl C C C C C C Cl CH3 H CH3 H CH3 No Yes Yes

  39. H3C H H3C CH3 C C C C H CH3 H H Geometric isomers have the same chemical properties, but different physical properties. Why? Because cis isomers have bulky side groups and cannot be packed closely together, this causes weaker intermolecular forces between molecules and therefore lower melting points than the trans version

  40. Cl H Cl Cl C C C C Cl H H H However cis forms are sometimes polar (as above) and therefore have stronger intermolecular forces between molecules causing higher melting points.

  41. Testing for Cis - Trans Isomers Weigh out 2 grams of maleic acid into a 50ml beaker Add 4mls of water and warm slightly to dissolve the acid Pour this into a pear shaped flask Carefully add 5mls of concentrated HCl Place a condenser on top of the flask and secure it in a retort stand with a water bath. Then warm the solution until a solid forms. Cool the solution to room temperature by placing the flask in a cold water bath ie a 250ml beaker of cold water Pour the solution into an evaporating dish, pouring off the excess liquid then carefully rinse with water – Then complete task B on page 159

  42. COOH COOH C C H H Maleic acid (cis isomer) H O O H O C C O C C H H

  43. Give the names and structural formula for the following substances from their condensed structural formulae. (c) CH3CH2CHCHCH2CH2CH3 H H H H H H H C C H C C C C H hept-3-ene C H H H H H (d) (CH3)3COH CH3 C OH CH3 2-methylpropan-2-ol CH3

  44. Give the names and structural formula for the following substances from their condensed structural formulae. (a) CH3CH2CHClCH2CH3 Cl H H H H 3-chloropentane C C H C C C H H H H H H (b) (CH3)2CHCH2CH2CHCH2 H H H CH3 H 5 -methylhex-1-ene C C C C C CH3 H H H H

  45. Another form of Stereoisomerism is Optical isomerism

  46. E,Z Nomenclature of bond geometry In cis and trans – nomenclature the ‘like” groups are identified and used to specify the type of isomer. With E,Z rules the pair of substituent's at each end of the double bond are given a priority. Highest priority = atom of highest atomic no attached directly to the double bond Eg 2-chloro-3-methylpent-2-ene This isomer has the highest priority groups on opposite sides of double bond therefore is the E isomer The Z isomer has the highest priority groups on the same side of the double bond CH3 CH2 -CH3 C C Cl CH3 1st pair Cl highest priority 2nd pair CH2CH3 highest priority

  47. Optical Isomerism Optical isomers involve an asymmetric carbon – a carbon bonded to four different atoms or groups of atoms such as: H, OH, CH3, C2H5, C3H7 etc A molecule with an asymmetric carbon is known as a chiral molecule. The two forms of the chiral molecule are known as enantiomers or optical isomers

  48. 3-D diagram of butan-2-ol Structural formula of butan-2-ol C2H5 C2H5 * CH3 * C OH C H3C OH H H C* = asymmetric carbon

  49. 3-D optical isomers of butan-2-ol C2H5 C2H5 * * C C H3C OH HO CH3 H H Mirror line Optical isomers are: mirror images of each other Are not superimposable on each other

  50. Optical activity Glucose is an optically active compound. On the straight-chain form of glucose shown here, all four of the carbons in the middle of the chain are chiral centres – they each have four different groups attached to them.