a) Interpretation: A structure for the molecule which meets the descriptions given is to be assigned. C 6 H 12 O; IR: 1715 cm -1 ; 13 CNMR: 8.0 δ (1 0 ), 18.5 δ (1 0 ), 33.5 δ (2 0 ), 40.6 δ ( 30 ), 214.0 δ (4 0 ) Concept introduction: Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm -1 . Aldehydes show two characteristic C-H absorptions between 2700-2760cm -1 and 2800-2860 cm -1 . Saturated aldehydes absorb near 1730 cm -1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm -1 . Saturated ketones and cyclohexanones absorb near 1715 cm -1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm -1 . Cyclopentanones absorb around 1750 cm -1 . In 13 CNMR, carbons in double bond which are sp 2 hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups. To assign: A structure for the molecule which meets the descriptions given. C 6 H 12 O; IR: 1715 cm -1 ; 13 CNMR: 8.0 δ (1 0 ), 18.5 δ (1 0 ), 33.5 δ (2 0 ), 40.6 δ ( 30 ), 214.0 δ (4 0 ) Answer:

Question

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Answer 1

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

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For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. ? NH2 MgBr Will the first product that forms in this reaction create a new CC bond? ○ Yes ○ No MgBr ? Will the first product that forms in this reaction create a new CC bond? O Yes O No Click and drag to start drawing a structure. :☐ G x c olo Ar HE

Predicting As the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C - C bond as its major product: H₂N O H 1. ? 2. H3O+ If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. 0 If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. فا Explanation Check Click and drag to start drawing a structure.

Highlight the chirality (or stereogenic) center(s) in the given compound. A compound may have one or more stereogenic centers. OH OH OH OH OH OH

Answer 2

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

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Answer 3

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

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Answer 4

Question

Definition Definition Organic compounds that have a carbonyl group, C=O, as their functional group. The carbonyl group in aldehydes is placed at the end of the molecular structure, which means the C=O is attached to one hydrogen atom and an alkyl group or a benzene ring. Just like all the other homologous series in organic chemistry, the naming of aldehydes uses the suffix “-al”. The general molecular formula is C n H 2n O.

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

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Answer 5

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

Answer 6

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Answer 7

Chapter 11.SE, Problem 69AP

Interpretation Introduction

a)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₁₂O; IR: 1715 cm^-1; ¹³CNMR: 8.0 δ (1⁰), 18.5 δ (1⁰), 33.5 δ (2⁰), 40.6 δ (³⁰), 214.0 δ (4⁰)

Answer:

Answer 8

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms (14-12) less than the parent and hence has one unsaturation unit perhaps a C=O.

The IR absorption at 1715 cm^-1 and ¹³CNMR peak at 214.0 δ indicate that the compound is a ketone. The ¹³CNMR further indicates that the carbonyl group is flanked by a secondary and tertiary carbon indicated by the two downfield signals at 33.5 δ and 40.6 δ. The remaining three carbons has to be primary, of which one is attached as methyl(8.0 δ) to the secondary carbon and two others are attached as two methyl groups (18.5δ) on the tertiary carbon. Hence the compound is 2-methyl-3-pentanone.

Answer 13

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 3

Answer 14

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Answer 15

Interpretation Introduction

b)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ. The signals will be observed at downfield if these alkyl groups are attached to electron withdrawing groups.

To assign:

A structure for the molecule which meets the descriptions given.

C₅H₁₀O; IR: 1730 cm^-1; ¹³CNMR: 22.6 δ (1⁰), 23.6 δ (3⁰), 52.8 δ (2⁰), 202.4 δ (3⁰).

Answer 16

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 5

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The molecular formula of the compound is C₆H₁₂O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has two hydrogen atoms less than the parent and hence has one unsaturation unit, a carbonyl group.

The IR absorption at 1730 cm^-1 and ¹³CNMR peak at 202.4 δ indicate that the compound is a saturated aldehyde. The ¹³CNMR further indicates that the carbonyl group is attached to a secondary carbon indicated by the downfield signal at 52.8 δ. Of the remaining three carbons one has to be tertiary(23.6 δ) which is attached to two methyl groups(22.6 δ). Hence the compound is 2-methylbutanal.

Answer 21

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 6

Answer 22

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.

Conclusion

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 8

Answer 23

Interpretation Introduction

c)

Interpretation:

A structure for the molecule which meets the descriptions given is to be assigned.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Concept introduction:

Aldehydes and ketones show a strong absorption band in IR from 1660-1770 cm^-1. Aldehydes show two characteristic C-H absorptions between 2700-2760cm^-1 and 2800-2860 cm^-1. Saturated aldehydes absorb near 1730 cm^-1 while aromatic aldehydes and α, β- unsaturated aldehydes absorb near 1705 cm^-1. Saturated ketones and cyclohexanones absorb near 1715 cm^-1 while aromatic ketones and α, β- unsaturated ketones absorb near 1685-1690 cm^-1. Cyclopentanones absorb around 1750 cm^-1.

In ¹³CNMR, carbons in double bond which are sp² hybridized absorb from 110 to 220 δ. Saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ While aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region. The primary alkyl carbon absorbs in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

A structure for the molecule which meets the descriptions given.

C₆H₈O; IR: 1680 cm^-1; ¹³CNMR: 22.9 δ (2⁰), 25.8 δ (2⁰), 38.2 δ (2⁰), 129.8 δ (3⁰) 150.6 δ (3⁰), 198.7 δ (4⁰).

Answer 24

Expert Solution

Answer to Problem 69AP

A structure for the molecule which meets the spectral descriptions given is

Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th, Chapter 11.SE, Problem 69AP , additional homework tip 7

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The molecular formula of the compound is C₆H₈O. Ignoring oxygen the molecular formula of the parent compound becomes C₆H₁₄. Thus the compound has six hydrogen atoms (14-8= 6) less than the parent and hence has three unsaturation units.

The IR absorption at 1680 cm^-1 and ¹³CNMR peak at 198.7 δ indicate that the compound is an α, β- unsaturatedketone. The compound has three unsaturated units. α, β- unsaturatedketone accounts for two unsaturation units. Hence the compound contains a six membered ring. The signals at 129.8 δ (30) 150.6 δ (30) can be attributed to the carbons in double bond. The other three signals at 22.9 δ (20), 25.8 δ (20), 38.2 δ (20) can be attributed to three methylene groups. The signal at 38.2 δ (20) can be assigned to methylene carbon attached to carbonyl carbon, that at 25.8 δ (20) to the methylene carbon carbon attached to C4 while that at 22.9 δ (20) is due to methylene carbon C5.