For each part, the compound with the higher boiling point is to be identified; and the reasons for the same are to be explained. Concept introduction: Unbranched alkanes havea higher boiling point than branched alkanes. As the branching increases in alkanes, the boiling point decreases. If the molecular weight is the same then aldehyde and ketone have higher boiling points than their corresponding hydrocarbons as result of polarity. An alcohol of the same molecular weight has a higher boiling point than its corresponding aldehyde and ketone due to hydrogen bonding in alcohols. Primary amines have higher boiling points than secondary amines and secondary amines have higher boiling points than tertiary amines.

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Answer 1

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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 2, Problem 25PP

Interpretation Introduction

Interpretation:

For each part, the compound with the higher boiling point is to be identified; and the reasons for the same are to be explained.

Concept introduction:

Unbranched alkanes havea higher boiling point than branched alkanes. As the branching increases in alkanes, the boiling point decreases.

If the molecular weight is the same then aldehyde and ketone have higher boiling points than their corresponding hydrocarbons as result of polarity.

An alcohol of the same molecular weight has a higher boiling point than its corresponding aldehyde and ketone due to hydrogen bonding in alcohols.

Primary amines have higher boiling points than secondary amines and secondary amines have higher boiling points than tertiary amines.

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1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…

Please draw, not just describe!

can you draw each step on a piece of a paper please this is very confusing to me

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 2, Problem 25PP

Interpretation Introduction

Interpretation:

For each part, the compound with the higher boiling point is to be identified; and the reasons for the same are to be explained.

Concept introduction:

Unbranched alkanes havea higher boiling point than branched alkanes. As the branching increases in alkanes, the boiling point decreases.

If the molecular weight is the same then aldehyde and ketone have higher boiling points than their corresponding hydrocarbons as result of polarity.

An alcohol of the same molecular weight has a higher boiling point than its corresponding aldehyde and ketone due to hydrogen bonding in alcohols.

Primary amines have higher boiling points than secondary amines and secondary amines have higher boiling points than tertiary amines.

Expert Solution & Answer

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Check out a sample textbook solution

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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 2, Problem 25PP

Interpretation Introduction

Interpretation:

For each part, the compound with the higher boiling point is to be identified; and the reasons for the same are to be explained.

Concept introduction:

Unbranched alkanes havea higher boiling point than branched alkanes. As the branching increases in alkanes, the boiling point decreases.

If the molecular weight is the same then aldehyde and ketone have higher boiling points than their corresponding hydrocarbons as result of polarity.

An alcohol of the same molecular weight has a higher boiling point than its corresponding aldehyde and ketone due to hydrogen bonding in alcohols.

Primary amines have higher boiling points than secondary amines and secondary amines have higher boiling points than tertiary amines.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

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 2, Problem 25PP

Interpretation Introduction