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Chapter 1: Problem 3

Total number of structures in dihalogen derivative of propane is (A) 3 (B) 2 (C) 4 (D) 5

Short Answer

Expert verified
The total number of dihalogen derivative structures of propane is 5, which includes: 1. Both halogens on the first carbon. 2. One halogen on the first carbon and one halogen on the second carbon. 3. Both halogens on the second carbon. 4. One halogen on the second carbon and one halogen on the third carbon. 5. Both halogens on the third carbon. Hence, the correct answer is (D) 5.

Step by step solution

01

Draw the structure of propane

First, draw the structure of propane, which has three carbon atoms in a row, with each carbon atom bonded to the appropriate number of hydrogen atoms to satisfy the octet rule: \(C_3H_8\) H H H | | | H--C--C--C--H | | | H H H
02

Add two halogens to propane

Now, we have to explore the possibilities of adding two halogens (X) to propane, in such a way that each dihalogen derivative is unique. The possible locations for halogen attachments are on the first carbon, second carbon, or third carbon. The possible dihalogen derivatives are: (1) Both halogens are on the first carbon. H H H | | | X--C--C--C--H X | | H H (2) One halogen on the first carbon and one halogen on the second carbon. H H H | | | X--C--C--C--H |X| H H (3) Both halogens on the second carbon. H H H | | | H--C--C--C--H |X| H X (4) One halogen on the second carbon and one halogen on the third carbon. H H H | | | H--C--C--C--H |X| H |X (5) Both halogens on the third carbon. H H H | | | H--C--C--C--H | | | H H X X
03

Count the unique dihalogen derivatives of propane

Now, count the unique dihalogen derivatives that we have found in Step 2: 1. Both halogens on the first carbon. 2. One halogen on the first carbon and one halogen on the second carbon. 3. Both halogens on the second carbon. 4. One halogen on the second carbon and one halogen on the third carbon. 5. Both halogens on the third carbon. Hence, the total number of dihalogen derivative structures of propane is 5. So, the correct answer is (D) 5.

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Key Concepts

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

Propane Structure
Propane is a simple hydrocarbon with the chemical formula \(C_3H_8\). It belongs to the alkane family and contains three carbon atoms connected in a straight chain. Each carbon atom forms single covalent bonds with hydrogen atoms to satisfy the octet rule. This structure can be visualized as a straight line of carbon atoms, each flanked by hydrogen atoms to complete its valency: \[\begin{align*}H &- C - C - C - H \ | & \; | \; | H & H \; H \\end{align*}\] The structure is symmetrical when viewed at the center carbon, making it a good starting point for exploring chemical derivatives like dihalogen compounds.
Organic Chemistry
Organic chemistry focuses on the study of carbon-containing compounds and their behavior. It covers structures, reactions, and properties of organic molecules, and is central to understanding dihalogen derivatives of propane. In organic chemistry, the arrangement of atoms within molecules plays a crucial role in determining their physical and chemical properties. Compounds with the same molecular formula but different structures are known as isomers. These isomers can have distinct characteristics, making the study of organic structures essential for predicting reactivity and stability.
Halogen Attachment
Halogen attachment refers to the substitution of hydrogen atoms in hydrocarbons with halogen atoms. These halogens, typically represented by the symbol \(X\), include fluorine, chlorine, bromine, and iodine. For propane, halogen atoms can attach to different carbon atoms, altering the molecule's properties. The positions for halogen attachment in propane include:
  • Both on the first carbon atom
  • One on the first and one on the second carbon atom
  • Both on the second carbon atom
  • One on the second and one on the third carbon atom
  • Both on the third carbon atom
These arrangements lead to the formation of dihalogen derivatives, significantly affecting the chemical behavior and nomenclature of the new compounds. By understanding where halogens can attach, students can predict and grasp the differences in isomer formations.
Unique Isomers
Isomers are compounds with the same molecular formula but different arrangements of atoms. Unique isomers arise when the attachment positions of atoms or groups differ. In the case of dihalogen derivatives of propane, they emerge from various placement combinations of two halogen atoms on the three carbons. Each arrangement provides a different structural isomer:
  • Both halogens on the same carbon atom
  • Halogens on two different carbon atoms
The exercise identified five unique isomers by placing two halogen atoms at different positions on the propane backbone. These unique structures exemplify how small changes in atomic placement can lead to distinct chemical entities with unique properties and reactivities.

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Most popular questions from this chapter

A compound with no tertiary hydrogen is (A) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}\left(\mathrm{CH}_{3}\right)_{2}\) (B) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}_{3}\) (C) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (D) \(\left(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}\right)_{4} \mathrm{C}\)

The heat of hydrogenation of 1-hexene is \(126 \mathrm{kJmol}\). When a second double bond is introduced in 1 -hexene, the heat of hydrogenation of the resulting compound is found to be \(230 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The resulting compound (diene) is (A) 1,5 -hexadiene (B) 1,4 -hexadiene (C) 1,3 -hexadiene (D) 2,4-hexadiene

The maximum angle strain is seen in (A) Cyclopentane (B) Cyclobutane (C) Cyclopropane (D) Cyclohexane

Nitration of resorcinol and catechol maximum lowers their boiling point if nitration is carried out at position respectively (A) \(\mathrm{C}-2, \mathrm{C}-2\) (B) C-2, C-3 (C) C-3, C-2 (D) C-4, C-3

The correct order of acidic strength is (A) \(\mathrm{HCOOH}>\mathrm{ClCH}_{2}-\mathrm{COOH}\) (B) \(\mathrm{PhCOOH}>\mathrm{HCOOH}\) (C) \(\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{PhCOOH}\) (D) None of these
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