Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 165

Which of the following statements is correct? (a) Ethylene show conformational isomerism. (b) Hex-3-ene does not show peroxide effect on treatment with HBr in presence of peroxides. (c) When but-1-ene is treated with excess of bromine, the product obtained is 2 -bromobutane. (d) Reactivity order of the hydrocarbons is alkane < alkenes < alkynes.

Short Answer

Expert verified
a) Ethylene shows conformational isomerism b) Hex-3-ene reacts with HBr in the presence of peroxides to form only one product c) But-1-ene reacts with excess bromine to form 2-bromobutane d) The reactivity order towards electrophilic/nucleophilic substances is: alkanes < alkenes < alkynes Answer: The correct statement(s) is/are (b) and (d).

Step by step solution

01

statement (a) Analysis

Ethylene (C2H4) is an alkene with two carbons double-bonded to each other. Conformational isomers are different spatial arrangements of a molecule that results from the rotation around single bonds (such as C-C single bonds). Since ethylene has a double bond between carbons, it does not exhibit rotation around the C=C bond, and hence, it does not show conformational isomerism. Therefore, statement (a) is incorrect. #Step 2: Evaluate statement (b)#
02

Statement (b) Analysis

Hex-3-ene is an alkene-position isomer of hexene (C6H12). Peroxide effect (or free radical addition) occurs when an alkene reacts with HBr in the presence of peroxides to form an alkyl bromide. Usually, HBr would add to an alkene according to Markovnikov's rule, however, in the presence of peroxides, the addition of bromine occurs in an anti-Markovnikov fashion. But, in this specific instance, where hex-3-ene does not possess allylic hydrogens, peroxide effect doesn't actually matter, so the statement (b) is true. #Step 3: Evaluate statement (c)#
03

Statement (c) Analysis

When an alkene like but-1-ene (C4H8) reacts with bromine (Br2), the reaction is an electrophilic addition reaction, leading to the formation of a vicinal dibromide instead of a monobromide. So, when but-1-ene reacts with excess bromine, the product formed is 1,2-dibromobutane, not 2-bromobutane as stated in the question. Therefore, statement (c) is false. #Step 4: Evaluate statement (d)#
04

Statement (d) Analysis

To assess the reactivity order of hydrocarbons, we should compare alkanes, alkenes, and alkynes in terms of their relative stability, electron density, and electrophilic/nucleophilic characteristics. In general, the electron density is highest in alkynes, followed by alkenes and then alkanes. When comparing their reactivity towards electrophilic/nucleophilic substances, alkynes are more reactive than alkenes, which are, in turn, more reactive than alkanes. So, the correct reactivity order should be alkanes < alkenes < alkynes. Therefore, statement (d) is true. #Step 5: Select the correct statement#
05

Correct Statement

After analyzing and evaluating all the given statements, statement (b) and statement (d) are correct. Hence, the correct answer for this question is (b) and (d).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(A) and (B) are respectively (a) But-2-yne and but-2-ene (b) But-2-ene and but-2-yne (c) Butane and but-2-ene (d) But-2-ene and butane

An organic compound (A) \(\mathrm{C}_{5} \mathrm{H}_{8}\) does not react with Tollens reagent but reacts with \(\mathrm{H}_{2}\) in the presence of Lindlar catalyst to give (B). Aqueous chlorine water reacts with (B) to give (C) which when reacted with hot alcoholic KOH gives (D). The compound (D) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\) (c) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d)

A reagent that can form a hydrocarbon with a Grignard reagent is (a) \(\mathrm{HCOOC}_{2} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}\)

An oxidizing agent such as \(\mathrm{HNO}_{3}\) is also added in the iodination of \(\mathrm{n}\) -butane. This is to (a) make the reaction less violent (b) prevent the formation of poly iodinated product (c) prevent the reducing action of HI (d) increase the yield of the desired isomer

Complete the following equations showing the stereo chemistry of the product wherever applicable. (i) Cis-2-pentene \(\stackrel{\mathrm{H}_{2} / \mathrm{Pt}}{\longrightarrow}\) (ii) Cis-2-butene \(\stackrel{\mathrm{Br}_{2} / \mathrm{CCl}_{4}}{\longrightarrow}\) (iii) Cis-2-butene \(\stackrel{\mathrm{m}-\mathrm{CPBA}}{\longrightarrow}\) (iv)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free