What is a Lucas test?

What is an iodoform test?

Answer:
Lucas test in alcohols is a test to differentiate between primary, secondary and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides.

The iodoform test or iodoform reaction is a qualitative chemical test for the detection of ketones carrying an alpha methyl group. The reagents are iodine and sodium hydroxide.

Only methyl ketones, or alcohols with the feature:

CH3CH(OH)-R
may undergo this reaction.


For the details pls refer to this web page:
http://en.wikipedia.org/wiki/lucas_test...
http://en.wikipedia.org/wiki/iodoform_te...
Lucas' reagent is a solution of zinc chloride in concentrated hydrochloric acid, used to classify alcohols of low molecular weight. The reaction is a substitution in which the chlorine replaces the hydroxy group. Even though this reaction is normally very unfavorable, the zinc ion complexes with the hydroxy group(by accepting a lone electron pair from O of -OH) , making it a better leaving group. The remaining carbonium ion then combined with chloride ion to from alkyl halide.
Lucas reagent is a mixture of concentrated Hcl and anhydrous Zinc Chloride.It is a test to distinguish between Primary,Secondary and Teritiary alcohols.If Lucas reagent reacts with primary alcohols then a cloudy appearance develops only on heating.Secondary alcohols give a cloudy appearance after 5 min on reacting with the Lucas reagent.Teritiary alcohols give a cloudy appearance immediatly on reacting with Lucas reagent.
Lancenigo di Villorba (TV), Italy

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LUCAS TEST)
This test interests some chemical compound of Organic Chemistry. These organic compounds binding the oxydril group to the aliphatic chain of the molecule are known as ALCOHOLS. I remember you organic compound binding oxydril groups to aromatic rings are NOT alcohols, they are Phenols. There are many alcohols reacting with Lucas test, some others do not.
Lucas test is a strong acidic aqueous solution of hydrochloric acid saturated by adding of zinc chloride salt. At room temperature, several drops of pure alcohol is added to the reactant put in a glass-tube. Positive remarks happens when you may see a cloudy aspect of the mixture.
Generally, the Molecular Structure of Alcohols determine their Reactivities versus Lucas test. "Tertiary alcohols" show HIGH REACTIVITIES similar to someothers alcohols (e.g. allylic or benzylic ones), instead "primary alcohols" show VERY LOW reactivities. Hence, allylic, benzylic and tertiary alcohols fastly react, secondary ones may do it, primary ones do not.
HOW DOES IT HAPPENS? In strong acidic solutions, alcohols act as a base and they become in "protonated forms", as follows

R-OH + H+ <---> R-OH2+

Some types of alcohols may undergo the succeding steps

R-OH2+ + Cl- <---> R-Cl + H2O

where chloride ions play an important role as in Nucleophilic Substitution Reactions. Alkyl chloride makes cloudy aspect.

IODOFORM TEST)
This test interests some chemical compound of Organic Chemistry. These organic compounds arounding their carbonyl group by means of two aliphatic chain of the molecule are known as KETONES. There are many ketones reacting with Iodoform test, some others do not. I remember Acetaldehyde is the lonely aldehyde reacting in Iodoform way, similarly acts Ethanol among the alcohols.
Iodoform test is an alkaline aqueous solution of iodine. Several drops of pure stuff is added to the reactant put in a glass-tube, hence you must warm the tube. Positive remarks happens when you may see a yellowish coloration by suspension of a yellowish bodies.
Generally, the Molecular Structure of Ketones determine their Reactivities versus Iodoform test. "Methyl-ketones" may react, instead other ketones do not.
HOW DOES IT HAPPENS? In alkaline solutions, iodine may form hypoiodite

I2 + 2 OH- <---> I- + IO- + H2O

which may reacts with methyl-ketones by means of Electrophilic Substitution Reactions

R-CO-CH3 + IO- ---> R-CO-CH2I + OH-
R-CO-CH2I + IO- ---> R-CO-CHI2 + OH-
R-CO-CHI2 + IO- ---> R-CO-CI3 + OH-
R-CO-CI3 + OH- <---> R-COO- + CHI3

It begins succeeding steps of substituting hydrogen atoms by means of iodine ones in methyl unit. The TRI-IODO-METHYLKETONE undergoes hydrolysis in alkaline media.
Thus, it forms IODOFORM as Tri-iodo-methane which is unsoluble and yellowish body.

I hope this helps you.

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