Q.1 Draw structures for all amines of molecular formula C 4 H 11 N. Classify them as primary, secondary or tertiary amines.

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1 AMIES Structure ontain the 2 group. lassification primary (1 ) amines secondary (2 ) amines tertiary (3 ) amines quarternary (4 ) ammonium salts + 1 2 3 4 Aliphatic Aromatic methylamine, ethylamine, dimethylamine 2 group is attached directly to the benzene ring (phenylamine) omenclature amed after the groups surrounding the nitrogen + amine Aliphatic amines Aromatic amines 2 5 2 ethylamine 6 5 2 phenylamine (aniline) ( 3 ) 2 dimethylamine ( 3 ) 3 trimethylamine Q.1 Draw structures for all amines of molecular formula 4 11. lassify them as primary, secondary or tertiary amines. Properties The presence of the lone pair in 1, 2 and 3 amines makes them... Lewis bases - they can be lone pair donors Brønsted-Lowry bases - can be proton acceptors ucleophiles - provide a lone pair to attack a positive (electron deficient) centre

2 F324 Amines Physical properties Boiling point Boiling points increase with molecular mass. Amines have higher boiling points than corresponding alkanes because of intermolecular hydrogen bonding. Quarternary ammonium salts are ionic - they exist as crystalline salts. intermolecular hydrogen bonding in amines Solubility Soluble in organic solvents. Lower mass compounds are soluble in water due to hydrogen bonding with the solvent. Solubility decreases as molecules get heavier. hydrogen bonding between amines and water Basic properties Bases The lone pair on nitrogen makes amines basic. 2 + + > 3 + Strength depends on the availability of the lone pair and thus its ability to pick up protons the greater the electron density on the, the better its ability to pick up protons this is affected by the groups attached to the nitrogen. electron withdrawing substituents (e.g. benzene rings) decrease basicity as the electron density on is lowered. 2 electron releasing substituents (e.g. 3 groups) increase basicity as the electron density is increased 3 2 draw arrows to show the electron density movement Measurement the strength of a weak base is depicted by its K b or pk b value

3 pk and pk b values the larger the K b value the stronger the base the smaller the pk b value the stronger the base. the pk a value can also be used - ( pk a + pk b = 14 ) the smaller the pk b, the larger the pk a. ompound Formula pk b omments ammonia 3 4.76 methylamine 3 2 3.36 methyl group is electron releasing phenylamine 6 5 2 9.38 electrons delocalised into the ring strongest base methylamine > ammonia > phenylamine weakest base eactions Amines which dissolve in water produce weak alkaline solutions 3 2 (g) + 2 (l) 3 3 + (aq) + (aq) Amines react with acids to produce salts. 6 5 2 (l) + l(aq) > 6 5 3 + l (aq) phenylammonium chloride This reaction allows one to dissolve an amine in water as its salt. Addition of aqueous sodium hydroxide liberates the free base from its salt 6 5 3 + l (aq) + a(aq) > 6 5 2 (l) + al(aq) + 2 (l) ucleophilic haracter Due to their lone pair, amines react as nucleophiles with haloalkanes forming substituted amines nucleophilic substitution acyl chlorides forming -substituted amides addition-elimination

4 F324 Amines aloalkanes Amines can be prepared from haloalkanes (see below and previous notes). eagent onditions Product Excess, alcoholic ammonia ucleophile Ammonia ( 3 ) eflux in excess, alcoholic solution under pressure Amine (or its salt due to a reaction with the acid produced) Equation 2 5 Br + 3 (aq / alc) > 2 5 2 + Br ( or 2 5 3 + Br ) Why excess ammonia? The amine produced is also a nucleophile and can attack another molecule of haloalkane to produce a secondary amine (see mecahnism below). This in turn can react further producing a tertiary amine and, eventually an ionic quarternary amine. By using excess ammonia, you ensure that all the haloalkane molecules react with ammonia rather than the newly produced haloalkane. Further reactions 2 5 2 + 2 5 Br > Br + ( 2 5 ) 2 diethylamine, a 2 amine 2 5 - Br + + Br + Br 2 5 2 5 ( 2 5 ) 2 + 2 5 Br > Br + ( 2 5 ) 3 triethylamine, a 3 amine ( 2 5 ) 3 + 2 5 Br > ( 2 5 ) 4 + Br tetraethylammonium bromide a quarternary (4 ) salt Uses Quarternary ammonium salts with long chain alkyl groups are used as cationic surfactants in fabric softening. eg [ 3 ( 2 ) 17 ] 2 + ( 3 ) 2 l Prepared from haloalkanes ucleophilic substitution using ammonia... see above nitriles eduction of nitriles using Li + Al 4 in dry ether eg 3 2 + 4[] > 3 2 2 2 aromatic nitro compounds eduction by refluxing with tin and conc. hydrochloric acid eg 6 5 2 + 6[] > 6 5 2 + 2 2

5 Structure α - AMI AIDS Amino acids contain 2 functional groups amine 2 carboxyl Amine arboxyl They all have a similar structure - the identity of 1 and 2 vary 1 2 3 2 2 2 general 2-aminopropanoic acid 2-aminoethanoic acid structure (Alanine) (Glycine) ptical Isomerism Amino acids can exist as optical isomers if they have different 1 and 2 groups optical isomers exist when a molecule contains an asymmetric carbon atom asymmetric carbon atoms have four different atoms or groups attached two isomers are formed one rotates plane polarised light to the left, one rotates it to the right no optical isomerism with glycine - two s are attached to the atom Zwitterions a zwitterion is a dipolar ion it has a plus and a minus charge in its structure a proton from the group moves to 2 amino acids exist as zwitterions at a certain p the p value is called the isoelectric point produces increased inter-molecular forces melting and boiling points are higher - 3 + 3 a zwitterion Acid/base properties amino acids possess acidic and basic properties due to their functional groups they will form salts when treated with acids or alkalis. Basic properties react with + 2 2 + + > 2 + 3 l 2 2 + l > 2 + 3 l Acidic properties react with 2 2 + > 2 2 + 2 a 2 2 + a > a + 2 2 + 2

6 F324 Amines Q.2 Describe the arrangement of bonds in the amino acid 2 2 around... the atom in the 2 the atom in the the atom in the 2 What change, if any, takes place to the arrangement around the if the amino acid is treated with dilute acid? Formation PEPTIDES α-amino acids can join up together to form peptides via an amide or peptide link the peptide link Structure Sequences of amino acids joined together by peptide links 2 amino acids joined 3 amino acids joined many amino acids joined dipeptide tripeptide polypeptide 2 3 a dipeptide ydrolysis Peptides can be broken down into their constituent amino acids by hydrolysis 2 3 + 2 > 2 + 3 2 attack takes place at the slightly positive of the = the - bond next to the = is broken hydrolysis with just water is not feasible hydrolysis in alkaline/acid conditions is quicker hydrolysis in acid/alkaline conditions (e.g. a) will produce salts with l 2 will become 3+ l + 2 will become + 3 a will become a + will become

7 Q.3 Draw structural isomers for the compounds produced when 2 2 ( 3 ) is hydrolysed by water 2 2 ( 3 ) 2 is hydrolysed in acidic solution 2 2 ( 3 ) is hydrolysed in alkaline solution Q.4 Write out possible sequences for the original peptide if the hydrolysis products are 1 mol of amino acid A, 1 mol of amino acid B and 1 mol of amino acid 1 mole of amino acid A, 2 mol of amino acid B and 1 mol of amino acid ow many possible sequences are there for the original peptide if hydrolysis yields 1 mol of amino acid A, 1 mol of B, 1 mol of, 1 mol of D and 1 mol of E? Proteins polypeptides with high molecular masses chains can be lined up with each other the = and - bonds are polar due to a difference in electronegativity hydrogen bonding exists between chains dotted lines ---------- represent hydrogen bonding