Wurtz Reaction Questions

Alkanes are the principal components of the crude oil. However, these can also be obtained from natural gas or even prepared in the laboratory. One of the methods used in the laboratory is the Wurtz Reaction.

Definition: In a Wurtz Reaction, an alkyl halide reacts with Na metal in the presence of dry ether to form a symmetrical alkane which has twice the number of carbons that are present in the alkyl halide.

Wurtz Reaction Chemistry Questions with Solutions

Q1: Write the product of the reaction:

\(\begin{array}{l}CH_{3}Br + CH_{3}CH_{2}Br + 2Na \xrightarrow[ether]{Dry}\end{array} \)

Answer: This is a Wurtz reaction and a mixture of 3 alkanes is obtained namely ethane, propane and butane. This is because the two different alkyl halides not only react with each other but also react among themselves.

\(\begin{array}{l}CH_{3}Br + CH_{3}CH_{2}Br + 2Na \xrightarrow[ether]{Dry} CH_{3}CH_{3}+ CH_{3}CH_{2}CH_{3} + CH_{3}CH_{2}CH_{2}CH_{3}\end{array} \)

Q2. Why only alkyl bromide and alkyl iodide are used in the Wurtz Reaction?

Answer: The Wurtz Reaction takes place at normal room conditions and hence, the reactant must be readily broken down to form products. Only the iodide and bromide groups are easily separable from RX. Hence, only RI and RBr are used in this reaction.

Q3. What is the difference between Wurtz Reaction and Wurtz Fittig Reaction?

Answer: In Wurtz Reaction, two alkyl halides (preferably the same) react with the Na metal in the presence of dry ether to form a symmetrical alkane having even number of C-atoms. While Wurtz Fittig reactions involve an alkyl halide and an aryl halide that react with the Na-metal in the presence of dry ether to form substituted aromatic compounds.

Q4. Which mechanism takes place in the Wurtz reaction?

  1. Ion-exchange mechanism
  2. Free Radical
  3. Addition-elimination
  4. Concerted

Answer: (b.)

Explanation: Wurtz reaction proceeds via free-radical mechanism.

Q5. Why Wurtz Reaction only forms alkanes with even number of carbons

Answer: The alkane formed in the Wurtz reaction has double the number of C-atoms that are present in the alkyl halide. Even if the two alkyl halides containing the odd number of C-atoms are taken, a mixture of products of alkanes is obtained. And hence, this reaction is only useful to form alkanes with even numbers of C-atoms.

Q6. Which of the following solutions will decolourize the cold alkaline KMnO4 solution?

  1. CH3CH3
  2. CH3Cl
  3. (CH3)4C
  4. CH2=CHCH3

Answer: (d.)

Explanation: The KMnO4 solution gives the test for unsaturation of organic compounds. This test is also called Baeyer’s test. In this test, the deep violet coloured solution of KMnO4 turns colourless.

Q7. The Wurtz reaction results in the formation of an even number of C-atoms containing alkanes. Which other reaction also gives the alkanes with an even number of carbons?

Answer: Kolbe’s reaction also results in the formation of alkanes with even no. of carbons. Kolbe’s reaction involves the electrolysis of the Na- or K-salts of carboxylic acids which result in the formation of symmetric even numbered carbon alkanes.

Q8. The melting points of some alkanes is shown hereunder.

Alkane Melting point (K)
C3H8 85.9
C4H10 138
C5H12 143.3
C6H14 178.5
C7H16 182.5

Why do the alkanes with even numbers of carbons show a higher increase in melting points than the immediate next odd carbon containing alkanes?

Answer: The even numbered carbon alkanes show higher increase in the melting points than the odd numbered carbon alkanes. This is because the even numbered carbon alkanes have symmetrical structure which result in the close-packing in the crystal structure. Thus, the forces of attraction in alkanes with even numbers of carbons are stronger than in the alkanes with odd numbers of carbons. And hence, the melting point varies accordingly.

Q9. Hybridization in C3H4 (allene) molecule is

  1. 2 sp2 and 1 sp hybrid carbon
  2. 1 sp2 and 2 sp hybrid carbon
  3. 2 sp2 and 1 sp3 hybrid carbon
  4. None of the above

Answer: The structure of C3H4 (allene) molecule is CH2=C=CH2. The central carbon is bonded to two other carbon atoms by two double bonds. Hence, it has two pi and two sigma bonds. The pi-bonds are not involved in the hybridization. Thus, the hybridization of the central carbon is sp.

Both the terminal carbons are attached to two hydrogen atoms and 1 C each by 3 sigma and 1 pi bond. Thus, the hybridization of terminal carbons is sp2.

Q10. Write the order of halogenation of alkanes in the presence of heat or UV light..

Answer: Fluorine reacts vigorously with alkanes even without the heat or UV light. Chlorine and Bromine readily react with alkanes in UV light. Iodine reacts with alkanes upon heating. Thus the order of halogenation of alkanes is F2 > Cl2 > Br2 > I2.

Q11. Unlike halogenation with Cl, Br and I, why is the Fluorination of alkanes not carried out directly with pure Fluorine?

Answer: Fluorination of alkanes with pure Fluorine is a highly vigorous reaction. This reaction often involves the cleavage of C-C bonds and hence results in a number of products. Due to this reason, pure F2 is not reacted with alkanes. Instead, Fluorine diluted with an inert gas like nitrogen or inorganic fluoride such as AsF3 are used.

Q12. In which conformation does the ethane exist at absolute zero temperature?

Answer: Ethane exists in staggered conformation at absolute zero temperature.

Q13. Write the reagents used for the isomerization of alkanes.

Answer: N-alkanes upon reaction with AlCl3 (anhyd.) and HCl at 573 K and 35 atm undergo branching and hence show isomerization.

Q14. Can Br2-water test be used for differentiating between the ethene and ethyne solutions?

Answer: The Bromine water is a reddish orange coloured liquid. It turns colourless upon reaction with unsaturated organic compounds. This is because the Br2 forms a bond at the place of unsaturation of carbon. Hence, Br2 cannot differentiate between ethene and ethyne.

CH2=CH2 + Br2/H2O (orange) → CH2BrCH2Br (colourless)

C2H2 + Br2/H2O (orange) → CHBr2CHBr2 (colourless)

Q15. Can pure staggered ethane and pure eclipsed ethane be separated at room temperature?

Answer: The only energy difference between the staggered and eclipsed forms of ethane is 12.55 kJ/mol. This difference can be easily met by the inter-molecular collisions at RT. Hence, pure staggered and eclipsed ethane cannot be isolated at RT.

Practise Questions on Wurtz Reaction

Q1. What is the alternation effect?

Q2. Arrange the following in increasing order of boiling point.

  1. 2-methylpentane
  2. 2,3-dimethylbutane
  3. 2,2-dimethylbutane

Q3. During the cracking of alkanes, why do the C-C bonds break instead of the C-H bonds?

Q4. Compare the melting points of n-pentane, isopentane and neopentane.

Q5. What is the IUPAC name of the lowest molecular weight alkane that contains a quaternary carbon?

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