Form 6 Chemistry – ORGANIC CHEMISTRY 1.2- AMINES

H/W

3.  Why is phenylamine more soluble in  than in H₂O?

4.      Define the term “base.” Write down structural formula of the following compounds and arrange them in order of increasing basic strength;

  Methylamine, ammonia, ethylamine, ethanamide, phenylamine. Give reasons.

5.    Give structural formula of the organic products that are obtained when;

(a)Ethylamine

(b)Phenylamine

(c)Phenylmethylamin

(d)N – methylphenylamine are treated with;

  (i)   Chloromethane

 (ii)   A mixture of NaNO₂ and HCl

(iii)   Benzoylchloride

ANSWERS

                            

      Phenylamine is more soluble in HCl since they will form a stable hydrated ion but it is less soluble in water due to the large hydrophobic part i.e. the benzene ring.

4.      A base is a substance which releases OH^– as the only negative ion.

 

QUESTIONS

1.An aromatic compound contains 80% carbon, 9.6% hydrogen and nitrogen. Its relative molecular mass is 135g/mol. It is slightly basic and form salts but given no other reaction nitric (III) acid. Identify the structure of this compound.

2.   An organic base A contains 61% carbon, 15.3% hydrogen, 23.7% nitrogen. When treated with nitric (III) acid A yield an alcohol B and nitrogen gas is evolved. The alcohol B contains 60% carbon, 13.33% hydrogen and on careful oxidation yields compound C, which has a vapour density of 29. Compound C forms an oxime with hydroxylamine (H₂ N – OH) but does not react with Fehling’s solution. Suggest structures for the compounds A, B and C and indicate the cause of the above reactions.

3.      How does the introduction of nitro group – 2 – and – 4 – positions of phenylamine affect its basicity? Give reasons

(a) Give the conditions under which the following reactions proceed

          

 (b) Write the structural formula for each of the organic compound formed when compound M reacts with each of the following;
        
          

ANSWERS:

1.       

∴ Structure of the compound is C₉ H₁₃ N

Molecular formula:

Probable structural formula:

2.

Empirical formula C₃ H₉ N

The organic base will be

3.      The nitro group attached to position – 4 – will make phenylamine to be more basic than the nitro group attached to position – 2

Reason:- This is because in – 2 – nitrophenylamine there might be hydrogen bonding between the hydrogen and oxygen hence basicity will decrease.

 

QUESTIONS

Give the equations for preparation of propanoic acid stating with;

(a)    Haloalkane

(b)   Ethene

Give equations (including reagents and conditions) for the following conversions;

                          

3.      Arrange the following compoounds in order of their increasing acidity. Explain the differences in their acidity.

        a)  ethanol

       b)   Ethanoic acid

       c)   Pentanoic acid

      d)   Ethanedioic acid

       e)   1, 2 – benzenedicarboxylic acid

        f)   Trichloroethanoic acid

       g)  Butenedioic acid

        4.      Convert the ethanoic acid into;

        a) Ethylethanoate

         b) Ethanoylchoride

       c) Ethanol

d)Trichloromethanoic acid

        e) Ethanamide

For each each conversion give a chemical equation

5.      A compound A of molecular formula C₄H₈O₂  gives propane when fused with alkali. When A reacts with Na₂CO_3(aq), CO₂  is evolved. When A reacts with molecular formula C₈H₁₄O₄   which on strong heating yield 2, 4 – dimethylpentanone. Establish the structural formula of A & B and give reason.

6.      (a) An unknown compound W undergoes ozonolysis and hydrolysis. This produces an acid and a compound of formula C₄H₈0  and Y which gives a positive Iodoform test. When X is reacted with LiAl₃ the product Z, is mixed with an equal amount of ethanol, a litter conc. H₂SO₄  is added and the mixture is heated to 140⁰C major products are formula of R.M.M 74, 88 and 107 called P, Q and R.

  (b) Why has an acid been produced on ozonolysis?

 7.    Descried, giving necessary experimental conditions, the reactions which occur between sulphuric (VI) acid and;

(a     a)Ethene

(b    b)Ethanol

(c    c)Benzene

(d      d)Methanoic acid

ANSWERS

             

    4.    (a) 

                     

      (b)

        

 

When you oxidize  2, 4 – dimethylpentanone you will get carboxylic acid A.

7.

                  (a)       

 

                   (b )               

               

Reason:- They are both acids.

EXERCISE

1.  Suggest the reagents that bring about the following conversions;-

i.    Pent – 2 – yne to pentan – 3 – one

ii.   Benzoylchloride to Benzaldehyde

iii  Ethanenitrile to ethanol

iv.1 – hexyne to hexan – 2 – one

v.   4 – florotoluene to 4 – florobenzaldehyde

2.How will you convert ethanal into the following compounds:-

3.  How will you bring about the following in not more than 2 steps:-

ANSWERS

1.

      

ii.    Benzoylchloride to Benzaldehyde

 

iii   Ethanenitrile to Ethanal

 

 

iv1 –  hexyne to hexan – 2 – one 

 

 4 – Florotoluene to 4 – florobenzaldehyde

   

1.    (i) By aldo – keto condensation

                              

           

 

              (iii) Ethanal to butan – 1 – ol

                      

                       

                 (iv) Ethanal to butanoic acid

 

 

                             

                   

                        (v) Ethanal to butan – 2 – ol

                                

                        (vi) Ethanal to but – 2- enoic acid

        

 

                  

                                                                              

        2.  (i) Propanone to propene

             

         (ii) Propanal to butanone

                    

               

             (iv) Ethanol to 3 – hydroxybutanal

Solution