EBK GET READY FOR ORGANIC CHEMISTRY
EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 9780321830555
Author: KARTY
Publisher: VST
bartleby

Concept explainers

Question
Book Icon
Chapter F, Problem F.21P
Interpretation Introduction

(a)

Interpretation:

The structure of the molecule hexanedinitrile is to be drawn.

Concept introduction:

The structure of a nitrile can be drawn on the basis of its IUPAC name as follows. The name consists of three parts, a prefix, a root, and a suffix. The suffix is ‘nitrile’. The root, which is the middle part of the name, shows the number of carbons in the longest continuous carbon chain that also contains the nitrile group. The nitrile carbon is a part of this chain. If the nitrile is attached to a ring, the nitrile carbon is not considered a part of the root. Any lower priority functional groups attached to this chain or ring are named in the prefix with their locants. The numbering of the carbons in the root starts at the nitrile carbon if the root is an open chain. If the root is a ring, the carbon with the nitrile group is numbered 1, and the numbering continues in the direction that will give the lowest possible numbers to any substituent groups.

Interpretation Introduction

(b)

Interpretation:

The structure of the molecule, (S)-4-nitroheptanenitrile is to be drawn.

Concept introduction:

The structure of a nitrile can be drawn on the basis of its IUPAC name as follows. The name consists of three parts, a prefix, a root, and a suffix. The suffix is ‘nitrile’. The root, the middle part of the name shows the number of carbons in the longest continuous carbon chain that also contains the nitrile group. The nitrile carbon is a part of this chain. If the nitrile is attached to a ring, the nitrile carbon is not considered a part of the root. Any lower priority functional groups attached to this chain or ring are named in the prefix, with their locants. The numbering of the carbons in the root starts at the nitrile carbon if the root is an open chain. If the root is a ring, the carbon with the nitrile group is numbered 1, and the numbering continues in the direction that will give the lowest possible numbers to any substituent groups.

If the molecule contains a chiral carbon, the absolute configuration is specified at the start of the name. The configuration is shown using a dash/wedge representation so that the priority groups 1 to 3 are arranged in a clockwise direction with the lowest priority group pointing away from the observer for an R configuration. The three groups are arranged in counterclockwise direction with the lowest priority group pointing away from the observer for an S configuration. The respective directions are reversed if the lowest priority group is pointing toward the observer.

Interpretation Introduction

(c)

Interpretation:

The structure of 4, 4-diethylcyclohexanecarbonitrile is to be drawn.

Concept introduction:

The structure of a nitrile can be drawn on the basis of its IUPAC name as follows. The name consists of three parts, a prefix, a root, and a suffix. The suffix is ‘nitrile’. The root, the middle part of the name shows the number of carbons in the longest continuous carbon chain that also contains the nitrile group. The nitrile carbon is a part of this chain. If the nitrile is attached to a ring, the nitrile carbon is not considered a part of the root. Any lower priority functional groups attached to this chain or ring are named in the prefix, with their locants. The numbering of the carbons in the root starts at the nitrile carbon if the root is an open chain. If the root is a ring, the carbon with the nitrile group is numbered 1, and the numbering continues in the direction that will give the lowest possible numbers to any substituents.

Blurred answer
Students have asked these similar questions
Don't used hand raiting
CS2(g) →CS(g) + S(g) The rate law is Rate = k[CS2] where k = 1.6 × 10−6 s−¹. S What is the concentration of CS2 after 5 hours if the initial concentration is 0.25 M?
CS2(g) → CS(g) + S(g) The rate law is Rate = k [CS2] where k = 1.6 × 10-6 s−1. S Calculate the half-life.
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning