Organic Chemistry: Principles And Mechanisms
Organic Chemistry: Principles And Mechanisms
2nd Edition
ISBN: 9780393663549
Author: KARTY, Joel
Publisher: W. W. Norton and Company
Question
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Chapter B, Problem B.22P
Interpretation Introduction

(a)

Interpretation:

The name for the given structure is to be written.

Concept introduction:

In case of molecules containing a C=C  or CC bond, the longest continuous carbon chain or largest ring that contains the double or triple bond is considered as the root. It is possible that the longest carbon chain has more carbons than specified by the root, but the double or triple bond must be a part of the root. The root is named from the name of the analogous alkane or cycloalkane by replacing the suffix ane by ene if a double bond is present. The suffix ane is replaced by yne if a triple bond is present.

If the root is a chain, numbering begins from that end of the chain which encounters the C=C  or CC bond at the earliest. If there is a tie, the chain is numbered such that the substituents get the lowest locator numbers.

If the root is a ring, the double or triple bonded carbon atoms are designated as C1 and C2. This must be done such that the locator numbers for the substituents are minimized. The lower of the two locator numbers for the C=C  or CC is written before the ene or yne suffix. Prefixes and locator numbers are used for the substituents attached.

Interpretation Introduction

(b)

Interpretation:

The name for the given structure is to be written.

Concept introduction:

In case of molecules containing a C=C  or CC bond, the longest continuous carbon chain or largest ring that contains the double or triple bond is considered as the root. It is possible that the longest carbon chain has more carbons than specified by the root, but the double or triple bond must be a part of the root. The root is named from the name of the analogous alkane or cycloalkane by replacing the suffix ane by ene, if a double bond is present. The suffix ane is replaced by yne if a triple bond is present.

If the root is a chain, numbering begins from that end of the chain which encounters the C=C  or CC bond at the earliest. If there is a tie, the chain is numbered such that the substituents get the lowest locator numbers.

If the root is a ring, the double or triple bonded carbon atoms are designated as C1 and C2. This must be done such that the locator numbers for the substituents are minimized. The lower of the two locator numbers for the C=C  or CC bond is written before the ene or yne suffix. Prefixes and locator numbers are used for the substituents attached.

Interpretation Introduction

(c)

Interpretation:

The name for the given structure is to be written.

Concept introduction:

In case of molecules containing a C=C  or CC bond, the longest continuous carbon chain or largest ring that contains the double or triple bond is considered as the root. It is possible that the longest carbon chain has more carbons than specified by the root, but the double or triple bond must be a part of the root. The root is named from the name of the analogous alkane or cycloalkane by replacing the suffix ane by ene, if a double bond is present. The suffix ane is replaced by yne if a triple bond is present.

If the root is a chain, numbering begins from that end of the chain which encounters the C=C  or CC bond at the earliest. If there is a tie, the chain is numbered such that the substituents get the lowest locator numbers.

If the root is a ring, the double or triple bonded carbon atoms are designated as C1 and C2. This must be done such that the locator numbers for the substituents are minimized. The lower of the two locator numbers for the C=C  or CC bond is written before the ene or yne suffix. In case of cycloalkene or cycloalkyne, the locator number is not included explicitly. Prefixes and locator numbers are used for the substituents attached.

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8 00 6 = 10 10 Decide whether each of the molecules in the table below is stable, in the exact form in which it is drawn, at pH = 11. If you decide at least one molecule is not stable, then redraw one of the unstable molecules in its stable form below the table. (If more than unstable, you can pick any of them to redraw.) Check OH stable HO stable Ounstable unstable O OH stable unstable OH 80 F6 F5 stable Ounstable X Save For Later Sub 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy C ཀྭ་ A F7 매 F8 F9 4 F10
Just try completing it and it should be straightforward according to the professor and TAs.
The grading is not on correctness, so if you can just get to the correct answers without perfectionism that would be great. They care about the steps and reasoning and that you did something. I asked for an extension, but was denied the extension.
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