EBK ORGANIC CHEMISTRY-PRINT COMPANION (
EBK ORGANIC CHEMISTRY-PRINT COMPANION (
4th Edition
ISBN: 9781119776741
Author: Klein
Publisher: WILEY CONS
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Chapter 7, Problem 47PP

(a)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  1

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(a)

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Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  2

The systematic name of the given compound must be 2-chloropropane and the common name is isopropyl chloride.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  3

In the given organic compound, the parent longest chain contains 3 carbon atoms so the root word must be propane with one substituent; chloro at C2 position. Thus, the systematic name of the given compound must be 2-chloropropane and the common name is isopropyl chloride.

(b)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  4

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(b)

Expert Solution
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Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  5

The systematic name of the given compound must be 2-bromo-2-methylpropane and the common name is tert-butyl bromide.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  6

In the given organic compound, the parent longest chain contains 3 carbon atoms so the root word must be propane with two substituents: chloro and methyl at 2 positions. Thus, the systematic name of the given compound must be 2-bromo-2-methylpropane and the common name is tert-butyl bromide.

(c)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  7

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(c)

Expert Solution
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Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  8

The systematic name of the given compound must be 1-iodopropane and the common name is propyl iodide.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  9

In the given organic compound the parent longest chain contains 3 carbon atoms so the root word must be propane with one substituent; iodo at 1 position. Thus, the systematic name of the given compound must be 1-iodopropane and the common name is propyl iodide.

(d)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  10

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(d)

Expert Solution
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Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  11

The systematic name of the given compound must be (R)-2-bromobutane and the common name is (R)- sec-butyl bromide.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  12

In the given organic compound the parent longest chain contains 4 carbon atoms so the root word must be butane with one substituent; bromo at C2 position. Thus, the systematic name of the given compound must be (R)-2-bromobutane and the common name is (R)- sec-butyl bromide. The prefix (R) indicates the clockwise rotation of the groups bonded at the chiral C atom when arranged from 1 to 4 in the increasing order of their atomic masses.

(e)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  13

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(e)

Expert Solution
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Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  14

The systematic name of the given compound must be 1-chloro-2,2- dimethylpropane and common name is neopentyl chloride.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  15

In the given organic compound, the parent longest chain contains 3 carbon atoms so the root word must be propane with three substituents as methyl and chloro at C1 and C2 position. Thus, the systematic name of the given compound must be 1-chloro-2,2- dimethylpropane and the common name is neopentyl chloride.

(f)

Interpretation Introduction

Interpretation: The systematic name and common name of the given organic compound is to be interpreted.

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  16

Concept Introduction: Organic compounds are chemical compounds that are mainly composed of carbon and hydrogen atoms. The naming of organic compounds is followed the IUPAC rules. The longest chain of carbon atoms is considered the parent chain. The substituents must be given with the least locant number. The locant number must be written as the prefix in the IUPAC name.

(f)

Expert Solution
Check Mark

Answer to Problem 47PP

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  17

The systematic name of the given compound must be 4-bromo-3,3-dimethyl-4-propylhaptane.

Explanation of Solution

  EBK ORGANIC CHEMISTRY-PRINT COMPANION (, Chapter 7, Problem 47PP , additional homework tip  18

In the given organic compound the parent longest chain contains 7 carbon atoms so the root word must be haptane with four substituents as propyl, methyl and bromo. Thus, the systematic name of the given compound must be 4-bromo-3,3-dimethyl-4-propylhaptane.

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Chapter 7 Solutions

EBK ORGANIC CHEMISTRY-PRINT COMPANION (

Ch. 7.5 - Prob. 9CCCh. 7.6 - Prob. 10CCCh. 7.6 - Prob. 11CCCh. 7.7 - Prob. 12PTSCh. 7.7 - Prob. 13PTSCh. 7.7 - Prob. 14ATSCh. 7.7 - Prob. 4LTSCh. 7.7 - Prob. 16ATSCh. 7.7 - Prob. 17CCCh. 7.7 - Prob. 18CCCh. 7.7 - Prob. 5LTSCh. 7.7 - Prob. 19PTSCh. 7.7 - Prob. 20ATSCh. 7.8 - Prob. 21PTSCh. 7.8 - Prob. 22ATSCh. 7.8 - Prob. 23CCCh. 7.8 - Prob. 24CCCh. 7.8 - Prob. 25CCCh. 7.8 - Prob. 26CCCh. 7.8 - Prob. 27CCCh. 7.9 - Prob. 7LTSCh. 7.9 - Prob. 29ATSCh. 7.9 - Prob. 30ATSCh. 7.9 - Prob. 31ATSCh. 7.10 - Prob. 32CCCh. 7.10 - Prob. 33CCCh. 7.10 - Prob. 34CCCh. 7.11 - Prob. 8LTSCh. 7.11 - Prob. 35PTSCh. 7.11 - Prob. 36PTSCh. 7.11 - Prob. 37ATSCh. 7.11 - Prob. 9LTSCh. 7.11 - Prob. 40PTSCh. 7.11 - Prob. 41ATSCh. 7.12 - Prob. 42CCCh. 7.12 - Prob. 43CCCh. 7.12 - Prob. 44CCCh. 7.12 - Prob. 45CCCh. 7.12 - Prob. 46CCCh. 7 - Prob. 47PPCh. 7 - Prob. 48PPCh. 7 - Prob. 49PPCh. 7 - Prob. 50PPCh. 7 - Prob. 51PPCh. 7 - Prob. 52PPCh. 7 - Prob. 53PPCh. 7 - Prob. 54PPCh. 7 - Prob. 55PPCh. 7 - Prob. 56PPCh. 7 - Prob. 57PPCh. 7 - Prob. 58PPCh. 7 - Prob. 59PPCh. 7 - Prob. 60PPCh. 7 - Prob. 61PPCh. 7 - Prob. 64PPCh. 7 - Indicate whether you would use NaOEt or tBuOK to...Ch. 7 - Prob. 68PPCh. 7 - Draw a plausible mechanism for each of the...Ch. 7 - Prob. 70PPCh. 7 - Prob. 71PPCh. 7 - Prob. 72PPCh. 7 - Prob. 73PPCh. 7 - Prob. 74PPCh. 7 - Prob. 77PPCh. 7 - Prob. 78PPCh. 7 - Prob. 81ASPCh. 7 - Prob. 87ASPCh. 7 - Prob. 90ASPCh. 7 - Prob. 91IPCh. 7 - Prob. 92IPCh. 7 - Prob. 93IPCh. 7 - Prob. 94IPCh. 7 - Prob. 95IPCh. 7 - Prob. 96IPCh. 7 - Prob. 97IPCh. 7 - Prob. 98IPCh. 7 - Prob. 99IPCh. 7 - Prob. 100IPCh. 7 - Prob. 101IPCh. 7 - Prob. 102IPCh. 7 - Prob. 103IPCh. 7 - Prob. 105IPCh. 7 - Prob. 106IPCh. 7 - Prob. 107IPCh. 7 - Prob. 109IPCh. 7 - Prob. 110CPCh. 7 - Prob. 112CPCh. 7 - Prob. 114CP
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