How haloalkanes are prepared from alcohols?
This is the common method for preparing haloalkanes in laboratory. Alcohols can be converted to haloalkanes by substitution of - OH group with a halogen atom. Different reagents can be used to get haloalkanes from alcohols as described below: Reaction by Halogen Acids: alcohols can be converted into haloalkanes by action with halogen acids. The reaction, in general, can be represented as Chlorolakanes are obtained by treating alcohols with HCl in the presence of anhydrous zinc chloride. The anhydrous ZnCl2 acts as dehydrating agent and thus favours the forward reaction (Le Chateliar principle). The mixture of HCl and anhydrous ZnCl2 is known as Lucas reagent. Bromoalkanes can also be obtained in an identical manner by heating alcohols with KBr or NaBr and conc. H2SO4.HBr is generated in situ (during the reaction) by reaction of KBr or NaBr with conc. H2SO4. KBr + H2SO4 KHSO4 + HBrC2H5OH + HBr C2H5Br + H2OThe reaction can also be carried out with conc. Solution (48%) of hydrobromic acid. Iodoalkanes are obtained by heating alcohols withKI and 95% H3PO4 (phosphoric acid). C2H5OH + HI C2H5I + H2OThe reaction can also be carried out with conc. solution (57%) of hydroiodic acid.Reactivity of halogen acids towards this reaction is: HI > HBr > HClIt is because of the fact that the bond dissociation energy of HI is less than that of HBr which in turn is less than that of HCl.Reactivity of alcohols towards this particular reaction is: tertiary > secondary > primaryIt is because of the fact that greater the number electron releasing groups on -carbon atom of alcohol more is the polarity of C-OH bond. Consequently greater is the ease with which it cleaves. Reaction by phosphorus halides: Chlorolakanes are obtained by reaction of alcohols with PCl3 or PCl5. Bromo and iodo alkanes are obtained by reaction of alcohols with a mixture of red phosphorus and Br2 or I2. The function of red phosphorus and Br2 or I2 is to get PBr3 and PI3 during the course of the reaction. PBr3 and PI3 being relatively unstable have to be prepared either in situ or just before use. They cannot be accumulated for long periods. For example, P4 + 6Br2 4PBr3 P4 + 6I2 4PI3 The alkyl halides are obtained in excellent yield (80%) in the reaction. The phosphorus halide method is better than halogen acid method because Certain alcohols such as, secondary and tertiary alcohols tend to dehydrate in the presence of halogen acids. Intermediate carbocation formed can undergo rearrangement in the presence of acids. Reaction by Thionyl Chloride: in this method the alcohol is refluxed with thionyl chloride in the presence of a small amount of pyridine. Refluxing is the process of heating a volatile liquid in a flask fitted with condenser so that the vapours are condensed back into the flask.The reaction of straight chain primary alcohols in the presence or absence of pyridine is called Darzen's procedure. Pyridine is used in the reaction to neutralize hydrogen chloride in the reaction.The above methods cannot be applied for conversion of phenols to aryl halides because carbon-oxygen bond in phenols has partial double bond character due to resonance and hence is stronger and difficult to break.
This is the common method for preparing haloalkanes in laboratory. Alcohols can be converted to haloalkanes by substitution of - OH group with a halogen atom.
Different reagents can be used to get haloalkanes from alcohols as described below: Reaction by Halogen Acids: alcohols can be converted into haloalkanes by action with halogen acids. The reaction, in general, can be represented as Chlorolakanes are obtained by treating alcohols with HCl in the presence of anhydrous zinc chloride. The anhydrous ZnCl2 acts as dehydrating agent and thus favours the forward reaction (Le Chateliar principle). The mixture of HCl and anhydrous ZnCl2 is known as Lucas reagent. Bromoalkanes can also be obtained in an identical manner by heating alcohols with KBr or NaBr and conc. H2SO4.HBr is generated in situ (during the reaction) by reaction of KBr or NaBr with conc. H2SO4. KBr + H2SO4 KHSO4 + HBrC2H5OH + HBr C2H5Br + H2OThe reaction can also be carried out with conc. Solution (48%) of hydrobromic acid. Iodoalkanes are obtained by heating alcohols withKI and 95% H3PO4 (phosphoric acid). C2H5OH + HI C2H5I + H2OThe reaction can also be carried out with conc. solution (57%) of hydroiodic acid.Reactivity of halogen acids towards this reaction is: HI > HBr > HClIt is because of the fact that the bond dissociation energy of HI is less than that of HBr which in turn is less than that of HCl.Reactivity of alcohols towards this particular reaction is: tertiary > secondary > primaryIt is because of the fact that greater the number electron releasing groups on -carbon atom of alcohol more is the polarity of C-OH bond. Consequently greater is the ease with which it cleaves. Reaction by phosphorus halides: Chlorolakanes are obtained by reaction of alcohols with PCl3 or PCl5. Bromo and iodo alkanes are obtained by reaction of alcohols with a mixture of red phosphorus and Br2 or I2. The function of red phosphorus and Br2 or I2 is to get PBr3 and PI3 during the course of the reaction. PBr3 and PI3 being relatively unstable have to be prepared either in situ or just before use. They cannot be accumulated for long periods. For example, P4 + 6Br2 4PBr3 P4 + 6I2 4PI3 The alkyl halides are obtained in excellent yield (80%) in the reaction. The phosphorus halide method is better than halogen acid method because Certain alcohols such as, secondary and tertiary alcohols tend to dehydrate in the presence of halogen acids. Intermediate carbocation formed can undergo rearrangement in the presence of acids. Reaction by Thionyl Chloride: in this method the alcohol is refluxed with thionyl chloride in the presence of a small amount of pyridine. Refluxing is the process of heating a volatile liquid in a flask fitted with condenser so that the vapours are condensed back into the flask.The reaction of straight chain primary alcohols in the presence or absence of pyridine is called Darzen's procedure. Pyridine is used in the reaction to neutralize hydrogen chloride in the reaction.The above methods cannot be applied for conversion of phenols to aryl halides because carbon-oxygen bond in phenols has partial double bond character due to resonance and hence is stronger and difficult to break.
Each symmetry operation can be represented by a transformation matrix.You have seen what happens when a molecule is subjected to the symmetry operation that corresponds to any of the symmetry elements of the point group to which the molecule belongs. The m
Select the right answer of the question. A solution of CaCl2 is 0.5 mol litre , then the moles of chloride ion in 500ml will be : (a) 0.25 (b) 0.50 (c) 0.75 (d)1.00
Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.
The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at
For motion in one dimension, the distribution of the molecules over quantum states, speeds, and energies can be deduced.Here we show that the energy of a macroscopic gas sample can be described on the basis of our knowledge of the quantum states allowed to
The infrared spectrum of gas samples shows the effect of rotational-energy changes along with the vibrational energy change.As we know from the interpretations given to thermodynamic properties of gases, gas molecules are simultaneously rotating and vibrating. It follows that an absor
Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent
The second law states that dS ≥ (dQ/T), where dS = dQ/T for a reversible process and dS > dQ/T for an irreversible process. a. Show that since dW12 = -dW21 (dWreverse = -dWforward) for a r
Help me to solve this Question. The number of moles of SO2Cl2 in 13.5 gm is in is : (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4
Provide solution of this question. Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution is: (a) 3.05 M (b) 1.35 M (c) 2.50 M (d) 4.50 M
18,76,764
1957314 Asked
3,689
Active Tutors
1417047
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!