Chapter 1 alcohols
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Chapter 1 alcohols Chapter 1 alcohols Presentation Transcript

  • ORGANIC CHEMISTRY II CHM 301 CHAPTER 1 ALCOHOLS
  • ALCOHOLS  Alcohols: Organic compounds containing hydroxyl (-OH) functional groups. R  OH Phenols: Compounds with hydroxyl group bonded directly to an aromatic (benzene) ring. OH
  • NOMENCLATURE OF ALCOHOLS
  • IUPAC RULES 1. 2. 3. 4. Select the longest continuous chain of carbon atoms containing the hydroxyl group. Number the carbon atoms in this chain so that the one bonded to the –OH group has the lowest possible number. Form the parent alcohol name by replacing the final –e of the corresponding alkane name by –ol. When isomers are possible, locate the position of the –OH by placing the number (hyphenated) of the carbon atom to which the –OH is bonded immediately before the parent alcohol name. Name each alkyl branch chain (or other group) and designate its position by number.
  • This is the longest continuous chain that contains an hydroxyl group. Select this chain as the parent compound.
  • 4 3 2 1 This end of the chain is closest to the OH. Begin numbering here.
  • 4 3 2 1 IUPAC name: 3-methyl-2-butanol New IUPAC name: 3-methylbutan-2-ol
  • Example: This is the longest continuous chain that contains an hydroxyl group. Select this chain as the parent compound.
  • 5 4 3 2 1 This end of the chain is closest to the OH. Begin numbering here.
  • 5 4 3 2 1 IUPAC name: 3-methyl-2-pentanol New IUPAC name: 3-methylpentan-2-ol
  • NOMENCLATURE OF CYCLIC ALCOHOLS   Using the prefix cycloThe hydroxyl group is assumed to be on C1. H 5 4 6 3 1 2 Br IUPAC name: OH HO CH2CH3 1 H trans-2-bromocyclohexanol new IUPAC name: trans-2-bromocyclohexan-1-ol 3 2 1-ethylcyclopropanol 1-ethylcyclopropan-1-ol
  • NOMENCLATURE OF ALCOHOLS CONTAINING TWO DIFFERENT FUNCTIONAL GROUPS  Alcohol containing double and triple bonds: - use the –ol suffix after the alkene or alkyne name.  The alcohol functional group takes precedence over double and triple bonds, so the chain is numbered in order to give the lowest possible number to the carbon atom bonded to the hydroxyl group.  The position of the –OH group is given by putting its number before the –ol suffix.  Numbers for the multiple bonds were once given early in the name.
  • EXAMPLE 5 4 CH2 CH 3 CH2 2 CH 1 CH3 OH 1) Longest carbon chain that contains –OH group - 5 carbon 2) Position of –OH group - Carbon-2 3) Position of C=C - Carbon-4 COMPLETE NAME = 4-penten-2-ol
  • Some consideration:  - OH functional group is named as a hydroxy substituent when it appears on a structure with a higher priority functional group such as acids, esters, aldehydes and ketones. - Examples: OH 4 CH3 3 CH O 2 CH2 1 C OH 3-hydroxybutanoic acid 4 3 2 5 1 6 OH O 2-hydroxycyclohexanone
  • MAIN GROUPS decreasing priority Acids Esters Aldehydes Ketones Alcohols Amines Alkenes Alkynes Alkanes Ethers Halides
  • NOMENCLATURE OF DIOLS  Alcohols with two –OH groups  Naming of diols is like other alcohols except that the suffix diol is used and two numbers are needed to tell where the two hydroxyl groups are located. diols or lycols. 5 OH 3 CH3 IUPAC name 2 CH 1 CH2 propane-1,2-diol 1 4 OH OH 3 2 OH trans-cyclopentane-1,2-diol
  • NOMENCLATURE OF PHENOLS  The terms ortho (1,2-disubstituted), meta (1,3disubstituted) and para (1,4-disubstituted) are often used in the common names. OH O2N OH CH3CH2 Br IUPAC name: 2-bromophenol common name: ortho-bromophenol OH 3-nitrophenol meta-nitrophenol 4-ethylphenol para-ethylphenol
  •  Phenols may be monohydric, dihydric or trihydric - (number of hydroxyl groups) in the benzene ring. OH OH OH OH OH benzene-1,3-diol OH benzene-1,4-diol OH benzene-1,2,3-triol
  • COMMON NAMES   Derived from the common name of the alkyl group and the word alcohol. For examples: CH3 H3C C OH CH3 IUPAC name: 2-methyl-2-propanol Common name: tert-butyl alcohol CH2CHCH3 OH IUPAC name: 2-propanol Common name: isopropyl alcohol CH3CH2OH IUPAC name: ethanol Common name: ethyl alcohol CH3OH IUPAC name: methanol Common name:methyl alcohol
  • CLASSIFICATION OF ALCOHOLS
  • CLASSIFICATION  According to the type of carbinol carbon atom (C bonded to the – OH group). C  OH Classes: i) Primary alcohol - -OH group attached to a primary carbon atom - one alkyl group attached ii) Secondary alcohol - -OH group attached to a secondary carbon atom - two alkyl group attached iii) Tertiary alcohol - -OH group attached to a tertiary carbon atom - three alkyl group attached
  • TYPE i) ii) Primary (1 ) Secondary (2 ) STRUCTURE H R C H EXAMPLES CH3 OH R' R C OH H CH3CH2-OH CH3CHCH2 OH ethanol 2-methyl-1-propanol OH 2-butanol iii) Tertiary (3 ) R' R C OH R'' OH H3C CH CH2CH3 cyclohexanol CH3 H3C C OH CH3 2-methyl-2-propanol
  • Polyhydroxy Alcohols • Alcohols that contain more than one OH group polyhydroxy alcohols. • Monohydroxy: one OH group. • Dihydroxy: two OH groups. • Trihydroxy: three OH groups.
  • PHYSICAL PROPERTIES OF ALCOHOLS
  • PHYSICAL PROPERTIES  PHYSICAL STATES OF ALCOHOLS - aliphatic alcohols and lower aromatic alcohols liquids at room temperature. - highly branched alcohols and alcohols with twelve or more carbon atoms solids.
  •  BOILING POINTS i) Boiling points of alcohols are higher > alkanes and chloroalkanes of similar relative molecular mass. - For example: C2H5OH 46 78 C Relative molecular mass: Boiling point: CH3CH2CH3 44 -42 C CH3Cl 50.5 -24 C - Reason: * intermolecular hydrogen bonds R Ar δ+ O H δ- H O δ- hydrogen bonding H δ+ O R δ- H Ar Oδ- hydrogen bonding
  •  SOLUBILITY OF ALCOHOLS IN WATER i) Alcohols with short carbon chains (i.e. methanol, ethanol) dissolve in water. - dissolve in water (hydrogen bonds are formed). ii) Solubility decreases sharply with the increasing length of the carbon chain. iii) Higher alcohols are insoluble in water. - alcohol contains a polar end (-OH group) called ‘hydrophilic’ and a non-polar end (the alkyl group) called ‘hydrophobic’.
  • iii) Polyhydroxy alcohols are more soluble than monohydroxy form more hydrogen bonds with water molecule. iv) Branched hydrocarbon increases the solubility of alcohol in water branched hydrocarbon cause the hydrophobic region becomes compact.
  • ACIDITY OF ALCOHOLS AND PHENOLS   Alcohol weakly acidic. In aqueous solution, alcohol will donated its proton to water molecule to give an alkoxide ion (R-O-). R-O- + H3O+ R-OH + H2O Ka = ~ 10-16 to 10-18 alkoxide ion Example CH3CH2-OH + H2O  CH3CH2-O- + H3O+ The acid-dissociation constant, Ka, of an alcohol is defined by the equilibrium R-OH + H2O Ka Ka = [H3O+] [RO-] [ROH] R-O- + H3O+ pKa = - log (Ka) * More smaller the pKa value, the alcohol is more acidic
  • Acidity OF PHENOLS  Phenol is a stronger acid than alcohols and water. R-OH + H2O alcohol OH R-O- + H3O+ Ka = ~ 10-16 to 10-18 alkoxide ion H2O phenol H 2 O + H2 O O- H3O + Ka = 1.2 x 10-10 phenoxide ion HO- + H3O+ hydroxide ion Ka = 1.8 x 10-16
  •  Phenol is more acidic than alcohols by considering the resonance effect. i) The alkoxide ion (RO-) - the negative charge is confined to the oxygen and is not spread over the alkyl group. - this makes the RO- ion less stable and more susceptible to attack by positive ions such as H+ ions.
  • ii) The phenoxide ion - one of the lone pairs of electrons on the oxygen atom is delocalised into the benzene ring. - the phenoxide ion is more stable because the negative charge is not confined to the oxygen atom but delocalised into the benzene ring. - the phenoxide ion is resonance stabilised by the benzene ring and this decreases the tendency for the phenoxide ion to react with H3O+. O O O O
  • EFFECTS OF Acidity  The acidity decreases as the substitution on the alkyl group increase. - Ethyl group is an electron-donating group bond harder to release a proton. strengthens the –O-H - i.e: methanol is more acidic than t-butyl alcohol.  The present of electron-withdrawing atoms enhances the acidity of alcohols. - The electron withdrawing atom helps to stabilize the alkoxide ion. - i.e: 2-chloroethanol is more acidic than ethanol because the electron-withdrawing chlorine atom helps to stabilize the 2chloroethoxide ion. - Alcohol with more than one electron withdrawing atoms are more acidic. i.e. 2,2,-dichloroethanol is more acidic than 2-chloroethanol. - Example of electron-withdrawing atom/groups: Halogen atoms and NO2.
  • IMPORTANT OF ALCOHOL  Ethanol - solvent for varnishes, perfumes and flavorings, medium for chemical reactions and in recrystallization. Also is an important raw material for synthesis.  Medically, ethanol is classified as a hypnotic (sleep producer), it is less toxic than other alcohol.
  •  Ethanol is prepared both by hydration of ethylene and by fermentation of sugars. It is the alcohol of alcoholic beverages. acid CH2 =CH2 + H2O -----------> CH3CH2OH yeast C6H12O6 -----------> 2CH3CH2OH + 2CO2
  • PREPARATION OF ALCOHOLS    Grignard synthesis Hydrolysis of alkyl halides Industrial and laboratory preparations of ethanol
  • GRIGNARD SYNTHESIS  The grignard reagent (RMgX) is prepared by the reaction of metallic magnesium with the appropriate organic halide. This reaction is always carried out in an ether solvent, which is needed to solvate and stabilize the Grignard reagent as it forms. R-X + Mg (X = Cl, Br or I)   CH3CH2OCH2CH3 R-Mg-X organomagnesium halide (Grignard reagent) Grignard reagents may be made from primary, secondary, and tertiary alkyl halides, as well as from vinyl and aryl halides. Alkyl iodides are the most reactive halides, followed by bromides and chlorides. Alkyl fluorides generally do not react.
  • EXAMPLES ether Mg CH3I CH3CH2Br Br Mg Mg ether CH3MgI CH3CH2MgBr ether MgBr
  • Grignard reactions of carbonyl compounds  Formaldehyde (H2C=O) reacts with Grignard reagents giving primary alcohol. R-MgX + H H C O ether H R C O H MgX H3O+ R CH2 OH or R-MgX + H H i) ether C O ii) H3O+ R CH2 OH Example: CH3CH2CH2CH2-MgBr + H H butylmagnesium bromide i) ether C O + ii) H3O H CH3CH2CH2CH2-C OH H 1-pentanol (92%)
  •  Aldehydes reacts with Grignard reagents giving secondary alcohols. R-MgX R' + H C O ether R' R C O H MgX H3O+ or R-MgX + R' H i) ether C O ii) H3O+ R' R C OH H Example: CH3CH2-MgBr + H3C H i) ether C O ii) H3O+ CH3 CH3CH2-C OH H R' R C OH H
  •  Ketones reacts with Grignard reagents giving tertiary alcohols. R-MgX + R' R'' C O ether R' R C O R'' MgX H3O+ R' R C OH R'' or R-MgX + R' R'' i) ether C O ii) H3O+ R' R C OH R'' Example: O CH3CH2-MgBr + H3C C CH2CH2CH3 i) ether ii) H3O+ OH CH3CH2-C CH2CH2CH3 CH3
  • HYDROLYSIS OF ALKYL HALIDES  Hydrolysis of alkyl halides is severely limited as a method of synthesizing alcohol, since alcohol are usually more available than the corresponding halides;indeed, the best general preparation of halides is from alcohols.  For those halides that can undergo elimination, the formation of alkene must always be considered a possible side reaction.
  • Example: 1) Second-order substitution: primary (and some secondary halides) KOH (CH3)2CHCH2CH2-Br (CH3)2CHCH2CH2-OH  H2O 2) First-order substitution: tertiary (and some secondary) halides CH3 H3C C CH3 Cl acetone/water heat CH3 H3C C CH3 OH CH3 H2C C CH3
  • INDUSTRIAL AND LABORATORY PREPARATION OF ETHANOL  There are three principle ways to get the simple alcohols that are the backbone of aliphatic organic synthesis. These methods are: a) hydration of alkenes obtained from the cracking of petroleum b) the oxo process from alkenes, carbon monoxide and hydrogen c) fermentation of carbohydrate
  • REDUCTION OF ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Aldehydes and ketones can be reduced to alcohols using: a) lithium aluminium hydride (LiAlH4) b) sodium borohydride (NaBH4) c) catalytic hydrogenation  O- O OH + R C H LiAlH4 or NaBH4 or H2, Ni R aldehyde C H H R C H H H o 1 alcohol O R C R' O LiAlH4 or NaBH4 or H2, Ni ketone + H = diluted acid such as H2SO4 - OH + R C R' H R C R' H H o 2 alcohol
  • Examples: O- O OH + CH3 C H LiAlH4 CH3 ethanal C H H CH3 C H H H ethanol O- O OH + CH3 C CH3 propanone H2/Ni CH3 C CH3 H H CH3 C CH3 H 2-propanol
  • reduced  Carboxylic acids primary alcohols  Reducing agents: LiAlH4 in dry ether H O R C OH (1) LiAlH4 / ether (2) H2O R C OH H carboxylic acids primary alcohols examples: H O CH3 C OH ethanoic acid (1) LiAlH4 / ether (2) H2O CH3 C OH H ethanol - Benzoic acid can be reduced to phenylmethanol by using LiAlH4 in eth
  • CH3 C OH (1) LiAlH4 / ether (2) H2O CH3 C OH H  Benzoic acid can be reduced to phenylmethanol by using LiAlH4 ethanol in ether at low temperatures. to phenylmethanol by using LiAlH4 in ether at - Benzoic acid can be reduced  An alkoxide intermediate is formed first. low temperatures.  On adding water, hydrolysisformed first. - An alkoxide intermediate is of the intermediate yields the - On alcohols. primary adding water, hydrolysis of the intermediate yields the primary alcohols. ethanoic acid O C OH (1) LiAlH4 / ether (2) H2O H C OH H benzoic acid phenylmethanol  LiAlH4 has no effect on the benzene ring or the double bond.  -COOH is reduced to –CH2OH but the C=C bonds remains unchanged. CH3CH2CH=CHCOOH 1) LiAlH4 2) H2O CH3CH2CH=CHCH2OH
  • Question: i) Give the structural formulae of L, M and N OH A PBr3 M Mg ether i) L ii)H3O+ N ii) How to prepare alcohol A from the reduction process? OH
  • Answers i) L: M: Br N: MgBr O ii) O (1) LiAlH4 / ether (2) H2O OH
  • REACTIONS OF ALCOHOLS       Reaction with sodium Oxidation Esterification Halogenation and haloform reactions Dehydration Formation of ether (Williamson ether synthesis)
  • Reaction with sodium  Alcohols reacts with Na at room temperature to form salts (sodium alkoxides) and hydrogen. 2R-O-H + 2Na → 2R-O- Na+ + H2  For example: CH3CH2OH + Na → CH3CH2O-Na+ + 1/2H2 alcohol  sodium ethoxide Reactivity of alcohols towards the reactions with sodium: CH3 > 1 > 2 > 3
  • Oxidation H 1 alcohol R C OH Pyridinium chlorochromate (PCC) H CH2Cl2, 25oC R-C=O H 1o alcohol aldehyde H R C OH Cu or Cr3O/pyridine H R-C=O H o 1 alcohol Cr3O/pyridine = Collins reagent aldehyde H R C OH H o 1 alcohol KMnO4/H+ or K2Cr2O7/H+ or CrO3/H+ O R-C-OH carboxylic acid
  • Examples: 1 alcohol O CH3(CH2)4-CH2-OH PCC 1-hexanol CH3(CH2)4-CH2-OH 1-hexanol CH3(CH2)4-C-H hexanal KMnO4/H+ or K2Cr2O7/H+ or CrO3/H+ O CH3(CH2)4-C-OH hexanoic acid
  • H + 2 alcohol R + KMnO4/H or K2Cr2O7/H C OH O + or CrO3/H R' R-C-R' ketone 2o alcohol R" 3 alcohol + R + KMnO4/H or K2Cr2O7/H C OH no reaction + or CrO3/H R' 3o alcohol Example: OH CH3 CH CH2CH3 2-butanol + O + KMnO4/H or K2Cr2O7/H + or CrO3/H CH3 C CH2CH3 2-butanone
  • Esterification  Esterification: - the reaction between an alcohol and a carboxylic acid to form an ester and H2O. O R C O H H O carboxylic acid CH3CH2-O-H ethanol R' O CH3 C methanol O R' H2O O H+ O H ethanoic acid C OH benzoic acid C ester O CH3-O-H R alcohol EXAMPLES H+ = catalyst O H+ CH3 C OCH2CH3 ethyl ethanoate H+ O C OCH3 methyl benzoate H2O H2O
  • Esterification also occurs when alcohols react with derivatives of carboxylic acids such as acid chlorides O O CH3-O-H CH3 C Cl CH3 C OCH3 methanol ethanoyl chloride methyl ethanoate HCl
  • Halogenation and haloform reactions 1) Hydrogen halides (HBr or HCl or HI) R-OH + H-X → R-X + H2O Example: C2H5-OH + H-Br • • H+ C2H5-Br + H2O Reactivity of hydrogen halides decreases in order HI > HBr > HCl Reactivity of alcohols with hydrogen halides: 3 >2 >1
  • 2) Phosphorus trihalides, PX3 or phosphorus pentahalides, PX5 3R-OH + PX3 3R-X + H3PO3 (PX3 = PCl3 or PBr3 or PI3) Example: (CH3)2CHCH2-OH + PBr3 → (CH3)2CHCH2-Br isobutyl alcohol isobutyl bromide 3) Thionyl chloride (SOCl2) R-OH + SOCl2 → R-Cl + SO2 + HCl Example: CH3(CH2)5CH2-OH + SOCl2 → CH3(CH2)5CH2-Cl + SO2 + HCl 1-heptanol 1-chloroheptane
  • Dehydration  Dehydration of alcohols will formed alkenes and the products will followed Saytzeff rules. R-CH2-CH2-OH  conc. H2SO4 R-CH=CH2 + H2O Saytzeff rule: - A reaction that produces an alkene would favour the formation of an alkene that has the greatest number of substituents attached to the C=C group. + H CH3CH2-CH-CH3 OH 2-butanol CH3CH2-CH=CH2 + H2O + H 1-butene CH3CH=CH-CH3 + H2O 2-butene major product
  •   Reactivity of alcohols towards dehydration: 3 >2 >1 Reagents for dehydration: i) Concentrated H2SO4 CH3-CH2-OH conc. H2SO4 CH2=CH2 + H2O ii) With phosphoric (v) acid OH 85% H3PO4, 165-170oC iii) Vapour phase dehydration of alcohols CH3CH2OH Al2O3 heat CH2=CH2 + H2O H2O
  • Formation of ether (Williamson ether synthesis)   Involves the SN2 attack of an alkoxide ion on an unhindered primary alkyl halides. The alkoxide is made by adding Na, K or NaH to the alcohol. R-O- + R’-X → R-O-R’ + Xalkoxide (R’ must be primary)   The alkyl halides (or tosylate) must be primary, so that a back-side attack is not hindered. If the alkyl halides is not primary, elimination usually occurs to form alkenes.
  • EXAMPLES CH3CH2-OH + Na CH3CH2-O Na CH3CH2-O-CH3 CH3I NaI or CH3CH2-OH OH 1) Na 2) CH3I 1) Na CH3CH2-O-CH3 OCH2CH3 2) CH3CH2-OTs cyclohexanol ethoxycyclohexane NaI
  • REACTIONS OF PHENOLS     Reaction with sodium Esterification Halogenation of the ring Nitration of the ring
  • REACTION WITH SODIUM OH Na O- Na+ 1/2 H2(g) sodium phenoxide REACTION WITH AQUEOUS SODIUM HYDROXIDE OH NaOH O- Na+ sodium phenoxide ROH + NaOH no reaction H2O
  • ESTERIFICATION EXAMPLES O OH NaOH ONa H2O O CH3CCl NaCl OCCH3 NaOH sodium phenoxide O OH C OH O H+ OC phenyl benzoate H2O
  • HALOGENATION   More reactive towards electrophilic substitution than benzene. ortho-para director. OH OH X 3X2 X room temperature 3HX X EXAMPLES OH OH Br 3Br2 Br room temperature 3HBr Br 2,4,6-tribromophenol (white precipitate) OH OH Cl 3Cl2 Cl room temperature 3HCl Cl 2,4,6-trichlorophenol (white precipitate)
  • • Monobromophenols are obtained if the bromine is dissolved in a non-polar solvent such as CCl4. OH 2 OH OH Br 2Br2 (CCl4) 2HBr Br
  • NITRATION  Dilute nitric (v) acids reacts with phenol at room temperature to give a mixture of 2- and 4-nitrophenols. OH 2 OH OH 2HNO3 < 20oC NO2 2H2O NO2 2-nitrophenol 4-nitrophenol
  •  By using concentrated nitric (v) acid, the nitration of phenol yields 2,4,6-trinitrophenol (picric acid).  Picric acid is a bright yellow crystalline solid. It is used in the dyeing industry and in manufacture of explosives. OH OH 3HNO3 O2 N NO2 NO2 2,4,6-trinitrophenol (picric acid) 3H2O
  • Question: Alcohol W is a secondary alcohol with a molecular formula of C4H10O. Compound M Step 1 CrO3 / pyrridine C4H10O Alcohol W Step 2 H+ / heat Reagent A C4H10ONa a) Draw and give the IUPAC name for alcohol W. b) Draw the structural formula for the following compounds: i) Compound M ii)Compound N iii)Compound O Compound N (major) + Compound O (minor)
  • c) Give the correct name for the following: i) Step 1 ii) Step 2 iii)Reagent A
  • Answers a) Alcohol W OH name: butan-2-ol b) i) compound M ii) compound N O c) i) Step 1: Oxidation ii) Step 2: Dehydration (of alcohol) iii) Reagent A: Na Metal iii) Compound O
  • TESTS TO DISTINGUISH CLASSES OF ALCOHOLS 1) Lucas Test - The alcohol is shaken with Lucas reagent (a solution of ZnCl2 in concentrated HCl). - Tertiary alcohol - Immediate cloudiness (due to the formation of alkyl chloride). - Secondary alcohol - Solution turns cloudy within about 5 minutes. - Primary alcohol - No cloudiness at room temperature.
  • CH3 CH3 HCl/ZnCl2 CH3 C CH3 OH room temperature Cl o 3 alcohol CH3 CH CH2CH3 OH (cloudy solution almost immediately) HCl/ZnCl2 room temperature o (cloudy solution within 5 minutes) HCl/ZnCl2 room temperature o 1 alcohol CH3 CH CH2CH3 Cl 2 alcohol CH3CH2CH2CH2OH CH3 C CH3 no reaction
  • 2) Oxidation of alcohols - only primary and secondary alcohols are oxidised by hot acidified KMnO4 or hot acidified K2Cr2O7 solution. - the alcohol is heated with KMnO4 or K2Cr2O7 in the presence of dilute H2SO4. - 1o or 2o alcohol: → the purple colour of KMnO4 solution disappears. → the colour of the K2Cr2O7 solution changes from orange to green. - 3o alcohol do not react with KMnO4 or K2Cr2O7.
  • 3RCH2OH + Cr2O 1o alcohol + 7 + 8H 27 + 3RCOOH 3+ + 7H2O (green) R' 3 R CH OH + Cr2O2-7 + 8H+ o + 2Cr carboxylic acid R' 2 alcohol (green) aldehyde + 8H (orange) + 2Cr3+ + 7H2O 3RCHO (orange) 3RCHO + Cr2O aldehyde 2- (orange) 3R C ketone 3+ O + 2Cr + 7H2O (green)
  • HALOFORM TEST TO IDENTIFY METHYL ALCOHOL GROUP 1) Iodoform:  Ethanol and secondary alcohols containing the group methyl alcohol group which react with alkaline solutions of iodine to form triiodomethane (iodoform, CHI3).  Triiodomethane – a pale yellow solid with a characteristic smell. H CH3 C OH (methyl alcohol group)
  • H CH3 C R + 4I2 + 6NaOH OH CHI3 (s) + RCOONa + 5NaI + 5H2O triiodomethane (iodoform) yellow precipitate where R = hydrogen, alkyl or aryl group • The iodoform test can distinguish ethanol from methanol H CH3 C O H + 4I2 + 6OH - CHI3 (s) + 5I + 5H2O iodoform OH H C O methanoate ethanol positive iodoform test H H C H + 4I2 + 6OH OH methanol no reaction negative iodoform test
  • • The iodoform test can distinguish 2-propanol from 1-propanol CH3 CH3 C O H + 4I2 + 6OH 2-propanol CH3 C O ethanoate iodoform OH positive iodoform test H H H C C C H H CHI3 (s) + 5I- + 5H2O H OH H + 4I2 + 6OH no reaction negative iodoform test 1-propanol * TERTIARY ALCOHOLS DO NOT GIVE POSITIVE IODOFORM TEST
  • 2) BROMOFORM H CH3 C R + 4Br2 + 6NaOH CHBr3 (s) + RCOONa + 5NaBr + 5H2O bromoform OH where R = hydrogen, alkyl or aryl group reagent sample iodoform
  • Question: a) Classify each of the following alcohols as primary, secondary or tertiary. i) 2-Propanol ii) 4-methylpentanol iii)2,3-dimethylbutan-2-ol b) Name a simple test to distinguish 1°, 2°, 3° alcohol. State the reagents and conditions required for the test and write down the expected observations.
  • Answer: a) i) 2° ii) 1° iii) 3° b) Test: Lucas test Reagent and conditions : Lucas reagent / Mixture of HCl and ZnCl2 Observatios: - Clear homogenous solution change into 2 layers or cloudiness - Rate of reaction: 3° > 2° > 1° alcohol