ORGANIC CHEMISTRY LL PRINT UPGRADE
4th Edition
ISBN: 9781119810643
Author: Klein
Publisher: WILEY
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Question
Chapter 24.5, Problem 9PTS
(a)
Interpretation Introduction
Interpretation:
For each of the given bifunctional compounds, the cyclic hemiacetal formed when treated with aqueous acid should be drawn.
Concept introduction:
- Hemiacetal are formed for the compound having both hydroxyl group and
aldehyde group in acidic condition. - The oxygen atom in the –OH group of compound attacks the carbonyl group of the same compound, thereby a ring is formed by the carbon atoms and oxygen atom of –OH group with incorporating into the ring.
To draw: the hemiacetal formed for the given molecule when treated with aqueous acid.
(b)
Interpretation Introduction
Interpretation:
For each of the given bifunctional compounds, the cyclic hemiacetal formed when treated with aqueous acid should be drawn.
Concept introduction:
- Hemiacetal are formed for the compound having both hydroxyl group and aldehyde group in acidic condition.
- The oxygen atom in the –OH group of compound attacks the carbonyl group of the same compound, thereby a ring is formed by the carbon atoms and oxygen atom of –OH group with incorporating into the ring.
To draw: the hemiacetal formed for the given molecule when treated with aqueous acid.
(c)
Interpretation Introduction
Interpretation:
For each of the given bifunctional compounds, the cyclic hemiacetal formed when treated with aqueous acid should be drawn.
Concept introduction:
- Hemiacetal are formed for the compound having both hydroxyl group and aldehyde group in acidic condition.
- The oxygen atom in the –OH group of compound attacks the carbonyl group of the same compound, thereby a ring is formed by the carbon atoms and oxygen atom of –OH group with incorporating into the ring.
To draw: the hemiacetal formed for the given molecule when treated with aqueous acid.
(d)
Interpretation Introduction
Interpretation:
For each of the given bifunctional compounds, the cyclic hemiacetal formed when treated with aqueous acid should be drawn.
Concept introduction:
- Hemiacetal are formed for the compound having both hydroxyl group and aldehyde group in acidic condition.
- The oxygen atom in the –OH group of compound attacks the carbonyl group of the same compound, thereby a ring is formed by the carbon atoms and oxygen atom of –OH group with incorporating into the ring.
To draw: the hemiacetal formed for the given molecule when treated with aqueous acid.
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Chapter 24 Solutions
ORGANIC CHEMISTRY LL PRINT UPGRADE
Ch. 24.2 - Prob. 1CCCh. 24.2 - Prob. 2CCCh. 24.2 - Prob. 3CCCh. 24.2 - Prob. 4CCCh. 24.2 - Prob. 5CCCh. 24.2 - Prob. 6CCCh. 24.4 - Prob. 7CCCh. 24.4 - Prob. 8CCCh. 24.5 - Prob. 1LTSCh. 24.5 - Prob. 9PTS
Ch. 24.5 - Prob. 2LTSCh. 24.5 - Prob. 12PTSCh. 24.5 - Prob. 13PTSCh. 24.5 - Prob. 3LTSCh. 24.5 - Prob. 16PTSCh. 24.5 - Prob. 19CCCh. 24.5 - Prob. 20CCCh. 24.5 - Prob. 21CCCh. 24.5 - Prob. 22CCCh. 24.6 - Prob. 23CCCh. 24.6 - Prob. 24CCCh. 24.6 - Prob. 25CCCh. 24.6 - Prob. 26CCCh. 24.6 - Prob. 27CCCh. 24.6 - Prob. 28CCCh. 24.6 - Prob. 29CCCh. 24.6 - Prob. 30CCCh. 24.6 - Prob. 4LTSCh. 24.6 - Prob. 31PTSCh. 24.6 - Prob. 34CCCh. 24.6 - Prob. 35CCCh. 24.6 - Prob. 36CCCh. 24.6 - Prob. 37CCCh. 24.6 - Prob. 38CCCh. 24.7 - Prob. 5LTSCh. 24.7 - Prob. 39PTSCh. 24.7 - Prob. 41CCCh. 24 - Prob. 42PPCh. 24 - Prob. 43PPCh. 24 - Prob. 44PPCh. 24 - Prob. 45PPCh. 24 - Prob. 46PPCh. 24 - Prob. 47PPCh. 24 - Prob. 48PPCh. 24 - Prob. 49PPCh. 24 - Prob. 50PPCh. 24 - Prob. 51PPCh. 24 - Prob. 52PPCh. 24 - Prob. 53PPCh. 24 - Prob. 54PPCh. 24 - Prob. 55PPCh. 24 - Prob. 56PPCh. 24 - Prob. 57PPCh. 24 - Prob. 58PPCh. 24 - Prob. 59PPCh. 24 - Prob. 60PPCh. 24 - Prob. 61PPCh. 24 - Prob. 62PPCh. 24 - Prob. 63PPCh. 24 - Prob. 64PPCh. 24 - Prob. 65PPCh. 24 - Prob. 66PPCh. 24 - Prob. 67PPCh. 24 - Prob. 68PPCh. 24 - Prob. 69PPCh. 24 - Prob. 70PPCh. 24 - Prob. 71PPCh. 24 - Prob. 72PPCh. 24 - Prob. 73PPCh. 24 - Prob. 74PPCh. 24 - Prob. 75PPCh. 24 - Prob. 76PPCh. 24 - Prob. 77PPCh. 24 - Prob. 78PPCh. 24 - Prob. 79PPCh. 24 - Prob. 80PPCh. 24 - Prob. 86IPCh. 24 - Prob. 87IPCh. 24 - Prob. 88IPCh. 24 - Prob. 89IP
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