When propyne is treated with excess chlorine water, the final product obtained is (a) 1,1 -dichloro propanone (b) 2, 3 dichloro dihydroxy propane (c) 2,2 -dichloropropanal (d) 1 -hydroxy-3-chloro propanone

Short Answer

Expert verified
Answer: The final product obtained when propyne is treated with excess chlorine water is 2, 3 dichloro dihydroxy propane.

Step by step solution

01

Identify the reaction taking place

Propyne reacts with chlorine water to form an addition product, since alkynes undergo addition reactions. In this case, the reagent is excess chlorine water, which is a halogen addition reagent.
02

Predict the outcome of the reaction

When propyne reacts with excess chlorine water, the excess chlorine water leads to the complete addition of two chlorine atoms and two hydroxyl groups (OH) across the triple bond. The reaction can be represented as follows: CH3C≡CH + 2(Cl2 + H2O) -> CH3C(OH)Cl-CH(OH)Cl The final product is 2,3-dichloro-1,2-dihydroxypropane.
03

Match the product with the given options

Now, let's compare the obtained product with the given options: (a) 1,1 -dichloro propanone (b) 2, 3 dichloro dihydroxy propane (c) 2,2 -dichloropropanal (d) 1 -hydroxy-3-chloro propanone Based on our reaction, the correct option is: (b) 2, 3 dichloro dihydroxy propane So, the final product obtained when propyne is treated with excess chlorine water is 2, 3 dichloro dihydroxy propane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Propyne Chlorine Water Reaction
Understanding the interaction between propyne and chlorine water is essential when exploring alkyne addition reactions in organic chemistry. Propyne, a carbon-based molecule with the formula CH₃C≡CH, possesses a triple bond that is highly reactive. When treated with chlorine water, a chemical reaction occurs in which the triple bond opens to accommodate additional atoms.

In this reaction, propyne is exposed to excess chlorine water, implying that there's more than enough of the reagent to completely saturate the triple bond with chlorine and water. The addition follows a specific pattern where the chlorine (Cl) atoms and hydroxyl (OH) groups are added across the triple bond.

To visualize the reaction, consider the following steps:
  1. The triple bond in propyne reacts with chlorine (from Cl₂) and hydroxyl groups (from H₂O).
  2. Each component of the triple bond receives an atom or group: one carbon gets a Cl, and the other gets an OH.
  3. This process occurs twice due to the 'excess' condition, leading to two Cl and two OH groups adding on at the '2' and '3' positions of the resulting compound, respectively.
The resulting molecule is 2,3-dichloro-1,2-dihydroxypropane, which has a more complex structure due to the addition of these atoms/groups. Such reactions are foundational in synthetic organic chemistry where modifying compounds to achieve different chemical properties is common.
Halogen Addition Reagent
Halogen addition reagents are widely used in organic chemistry to add halogens, such as chlorine or bromine, across double or triple bonds in hydrocarbons. These reagents are particularly keen on reacting with alkenes and alkynes due to the high electron density of their multiple bonds.

When a halogen addition reagent like chlorine water is used, the reaction typically follows a mechanism that leads to the vicinal addition of halogen atoms and hydroxyl groups to the original molecule. Here are a few key points about these reagents:
  • Halogen addition is stereospecific, following anti-addition or syn-addition paths depending on the reagent and conditions.
  • The reagents can be used in different amounts, from limiting to excess, to control the degree of saturation.
  • They are used both in academic research and industrial applications to create a variety of halogenated organic compounds.
For students preparing for entrance exams like the IIT JEE, understanding the reactivity, selectivity, and reaction conditions for halogen addition is crucial to solve complex problems. Regular practice of such reactions helps students gain confidence in predicting product outcomes and understanding reaction mechanisms.
Organic Chemistry IIT JEE
Organic chemistry is a significant portion of the IIT JEE syllabus – a highly competitive entrance examination for engineering colleges in India. For students aiming to excel in this exam, mastering reactions like the propyne chlorine water reaction is critical since it encompasses concepts of reaction mechanisms and reactivity.

Questions on organic chemistry often test the student's ability to:
  • Understand and apply various reaction mechanisms.
  • Identify stereochemistry of the resulting compounds.
  • Relate reactivity of molecules to their structure.
  • Predict the products of chemical reactions.
Given the extensive syllabus, students are encouraged to develop a deep understanding of fundamental concepts such as alkyne reactions, halogen addition, and other functional group transformations. Solving exercises from textbooks, working through previous years' papers, and revisiting challenging topics are some of the effective strategies to prepare for the organic chemistry section of the IIT JEE.

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Most popular questions from this chapter

An alkene on reductive as well as oxidative ozonolysis gave only one product. The alkene is (a) propene (b) 2,3-dimethyl-2-butene (c) pent-2-ene (d) 2 -methyl propene

Which of the following reactions give an alkyne? (a) sodium fumarate \(\frac{\text { Kolbe's }}{\text { electrolysis }}\) (b) \(\mathrm{CH}_{3} \mathrm{CBr}_{2} \mathrm{CHBr}_{2} \longrightarrow \underset{\mathrm{Zn} \text { /alcohol } / \Delta}{\longrightarrow}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBr}_{2} \stackrel{\mathrm{NaNH}_{2} / \Delta}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{Br} \stackrel{\mathrm{NaNH}_{2}}{\longrightarrow}\)

Alkanes and alkenes can be prepared by (a) Wurtz reaction (b) Williamson synthesis (c) Dehydrohalogenation (d) Kolbe's electrolysis

A reaction in which an alkene is formed is (a) 1 -bromo-1-methylcyclohexane is treated with \(\mathrm{NaOH}\) in the presence of acetone. (b) 1 -bromo- 1 -methylcyclohexane is treated with \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N}\). (c) Chlorocyclohexane is heated with \(\left(\mathrm{CH}_{3}\right)_{3}\) CONa in t-butanol (d) 1 -bromo-1-phenylbutane is treated with aqueous \(\mathrm{NaOH}\)

Which of the following on hydration will give an aldehyde? (a) \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (c) \(\mathrm{HC} \equiv \mathrm{CH}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CH}\)

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