Note on Analysis of Haloalkanes and Haloarenes

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Analysis of Haloalkanes and Haloarenes

1. Silver nitrate test.

In this test, the halo compound under test is warmed with aqueous or alcoholic KOH solution. The solution is then acidified with dil. HNO3 followed by adding of AgNO3 solution.

a. If the precipitate is formed, it indicates the alkyl, benzyl or allyl halides. The precipitate formed should be insoluble in HNO3.

b. If the precipitate does not form, it indicates vinyl or aryl halides.

2. Lassaigne's type.

This test is performed to identify the halogen present in the compound. In this test, Lassaigne's solution is prepared by heating the compound with sodium metal. The lasagne's solution is acidified with dilHNO3 and then treated with silver nitrate solution. Formation of precipitate shows the presence of halogen.

  1. Formation of white precipitate soluble in ammonium hydroxide shows the presence of chlorine element in the compound.
  2. Formation of pale yellow precipitate partially soluble in ammonium hydroxide shows the presence of bromine element in the compound.
  3. Formation of yellow precipitate insoluble in ammonium hydroxide indicates the presence of iodine.

Some important problem with the solution.

Q. Give the IUPAC name of X and Y.

$$CH_3CH_2Br\xrightarrow{AgCN\,,Δ}X\xrightarrow{LiAlH_4\,,Δ}Y$$

The reaction proceed as

$$CH_3CH_2I\xrightarrow{AgCN\,,alc}\underbrace{CH_3-CH_2-N≡C}_{X}\xrightarrow{LiAlH_4}\underbrace{CH_3-CH_2-NH-CH_3}_{Y}$$

Hence, the compound X and Y are.

$$X=\underbrace{CH_3-CH_2-N≡C}_{Ethanisonitrile}$$

$$Y=\underbrace{CH_3-CH_2-NH-CH_3}_{N-methylethanemine}$$

The only difference is that Br is presence in place of I in this molecule althrough it shows the same chemical behaviour.

Starting from iodomethane ,how ould you prepare ethane and ethene.

1, Ethane can be prepared from iodomethane by reacting iodomethane with Na -metal in presence of dry ether.

$$\underbrace{2CH_3-I}_{Iodomethane}+2Na\xrightarrow{dry\,ether}\underbrace{CH_3-CH_3}_{Ethane}+2NaI$$

2. Ethene can be prepared from iodomethane as follows.

$$\underbrace{CH_3-I}_{Iodomethane}\xrightarrow{Na\,,dry\,ether}\underbrace{CH_3-CH_3l}_{ethane}\xrightarrow{Cl_2\,hv}\underbrace{CH_3-CH_2Cl}_{Chloroethane}\xrightarrow{alc\,.KOH}\underbrace{CH_2=CH_2}_{Ethene}+$$

Identify the major products A and B in the following reaction.

$$C_2H_5Br\xrightarrow{Na\,,dry\,ether}A\xrightarrow{H_2O}B$$

The reaction proceeds as follows.

$$C_2H_5Br\xrightarrow{Mg\,,dry\,ether}C_2H_5MgBr\xrightarrow{H_2O}C_2H_6+(Mg(OH)Br$$

So, the product are.

$$A=\underbrace{C_2H_5MgBr}_{Ethyl\,Magnesium\,bromide}$$

$$B=\underbrace{C_2H_6}_{Ethane}$$

Q. Identify A abd B in the following reaction.

$$CH_3-CH_2-Br\xrightarrow{Alc.KOH}A\xrightarrow{(1.)\,O_3\,(2.)\,Zn/H_2O}B$$

The reaction proceeds as follows.

$$CH_3-CH_2-Br\xrightarrow{Alc.KOH}CH_2=CH_2\xrightarrow{(1.)\,O_3\,(2.)\,Zn/H_2O}HCOH$$

$$A=\underbrace{CH_2=CH_2}_{Ethene}$$

$$B=\underbrace{HCOH}_{Formaldehyde\,Methanol}$$

A primery haloalkane (X), if allowed to react with KCN yields a compound (Y), which on hydrolysis give propanoic acid. Identify the reaction X and Y.

The reaction proceeds as follows.

$$X\xrightarrow{KCN}Y\xrightarrow{H_2O\,/H^+}CH_3CH_2COOH$$

The compound Y contains a C-N group and on hydrolysis, it yields propionic acid so, it should be propane nitrile (CH3CH2CN). Again, the compound Y is obtained by reaction X with KCN .So X should be alkyl halide containing one carbon atom less than Yi,e CH2CH2X. Now the complete reaction is.

$$CH_3-CH_2-X\xrightarrow{KCN}CH_3-CH_2-CN\xrightarrow{H_2O\,/H^+}CH_3CH_2COOH$$

$$X=\underbrace{CH_3-CH_2-X}_{haloethane}$$

$$Y=\underbrace{CH_3-CH_2-CN}_{Propane\,nitrile}$$

Long question important question answer.

Starting from CH3MgI, how would you prepare ethanol. Convert ethanol into propanol.

2. ethanal into propanone.

Answer.

CH3MgI is reacted with HCHO to get an addition product which is hydrolysed to get ethanol.

$$CH_3MgI+HCHO→CH_3-CH_2-O^-Mg^+I\xrightarrow{H_2O\,H^+}\underbrace{CH_3-CH_2-OH}_{Ethanol}+Mg(OH)I$$

Conversion of Ethanol into propanol.

Ethanol can be converted into propanol as shown in the reaction sequence given below.

Etanol can be converted into propanol as shopwn in the reaction sequence foven below.

$$\underbrace{CH_3-CH_2-OH}_{Ethanol}\xrightarrow{PCl_5}\underbrace{CH_3-CH_2-Cl}_{Chlorothane}\xrightarrow{KCN}\underbrace{CH_3-CH_2-CN}_{Propanenitrile}\xrightarrow{Sn\,,HCl}\underbrace{CH_3-CH_2-CH_2-NH_2}_{Propan-1-amine}\xrightarrow{HNO_2}\underbrace{CH_3-CH_2-CH_2-OH}_{Propan-1-ol}$$

Reference.

Bahl, B S, Bahl, and Arun. Advanced Organic chemistry. S. Chand and company Ltd., n.d.

Sthapit, M K, R R Pradhananga, and K B Bajracharya. Foundations of chemistry. Taleju Prakashan, n.d.

Tewari, K S, S N Mehrotra, and N K Vishnoi. A textbook of organic chemistry. Vikash publishing House Pvt. ltd., n.d.

Verma, N K and S K Khanna. Compressive chemistry. 8th edition. Laxmi publications P. Ltd., 1999.

  1. If the precipitate is formed, it indicates the alkyl, benzyl or allyl halides. The precipitate formed should be insoluble in HNO3.
  2. Formation of pale yellow precipitate partially soluble in ammonium hydroxide shows the presence of bromine element in the compound.
  3. $$\underbrace{CH_3-I}_{Iodomethane}\xrightarrow{Na\,,dry\,ether}\underbrace{CH_3-CH_3l}_{ethane}\xrightarrow{Cl_2\,hv}\underbrace{CH_3-CH_2Cl}_{Chloroethane}\xrightarrow{alc\,.KOH}\underbrace{CH_2=CH_2}_{Ethene}+$$

     

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