Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester)
Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester)
6th Edition
ISBN: 9781260699098
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Chapter 18, Problem 37P

(a) Give the IUPAC name for A and B. (b) Draw the product formed when A or B is treated with each reagent: 1. NaBH 4 , CH 3 OH ; 2. CH 3 MgBr , then H 2 O ; 3. Ph 3 P=CHOCH 3 ;4. CH 3 CH 2 CH 2 CH 2 NH 2 , mild acid; 5. HOCH 2 CH 2 CH 2 OH , H + .

Chapter 18, Problem 37P, Problem 21.40 (a) Give the IUPAC name for A and B. (b) Draw the product formed when A or B is

Expert Solution
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Interpretation Introduction

(a)

Interpretation: The IUPAC name for A and B is to be determined.

Concept introduction: IUPAC nomenclature is a systematic way of naming the organic compounds. The basic principles of IUPAC naming for hydrocarbon are:

1. The hydrocarbon is named after the carbon chain containing higher number of carbon atoms.

2. For functional group such as aldehyde suffix ‘al’ and for ketone suffix ‘one’ is added to the name.

3. When the chain is substituted with different substituents, then the numbering is done according to priority.

Answer to Problem 37P

The IUPAC name for A and B is 3, 3-dimethylbutanal and 5-isopropyl-2-methylcyclohexanone respectively.

Explanation of Solution

The given compound is,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  1

Figure 1

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure is,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  2

Figure 2

The parent hydrocarbon is butane. The functional group present is aldehyde. Two methyl groups are present on third carbon atom. When same substituents are present, then prefix depends on the number of substituents. The IUPAC name of the compound is 3, 3-dimethylbutanal.

The given compound is,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  3

Figure 3

The red coloured balls have two bonds. So, these are the oxygen atoms. Black coloured atoms have four bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure is,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  4

Figure 4

The parent hydrocarbon is cyclohexane. The functional group present is ketone. Methyl group is present on fifth carbon atom and isopropyl group is present on third carbon atom. The IUPAC name of the compound is 5-isopropyl-2-methylcyclohexanone.

Conclusion

The IUPAC name for A and B is 3, 3-dimethylbutanal and 5-isopropyl-2-methylcyclohexanone respectively.

Expert Solution
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Interpretation Introduction

(b)

Interpretation: The products formed when B is treated with given reagents are to be drawn.

Concept introduction: The metal hydride reagents are good reducing agents such as NaBH4. They can be used to reduce carbonyl compounds, conjugated double or triple bonds and CX bond in alkyl halides and epoxides.

Grignard reagents are organometallic compounds which are prepared using alkyl halides in presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases.

Answer to Problem 37P

The products formed when B is treated with given reagents are,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  5

Explanation of Solution

1. In the presence of NaBH4,CH3OH, compound B is reduced to alcohol. The corresponding reaction is shown below.

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  6

Figure 5

The product formed is 5-isopropyl-2-methylcyclohexanol.

2. The Grignard reagent reacts with compound B to form secondary alcohol. The corresponding reaction is shown below.

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  7

Figure 6

The product formed is 5-isopropyl-1, 2-dimethylcyclohexanol.

3. When compound B is treated with Ph3P=CHOCH3, the new carbon-carbon double bond is formed between the negatively charged carbon atom of Wittig reagent and carbonyl carbon of aldehyde. The corresponding reaction is as follows,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  8

Figure 7

The product formed is (E)-4-isopropyl-2-(methoxymethylene)-1-methylcyclohexane.

4. The compound B reacts with CH3CH2CH2CH2NH2 in presence of mild acid to form imine. The corresponding reaction is as follows,

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  9

Figure 8

The product formed is (E)-N-(5-isopropyl-2-methylcyclohexylidene)propan-1-amine.

5. The product formed by the reaction of compound B with ethylene glycol in the presence of p-toluenesulfonicacid is shown below.

Package: Loose Leaf For Organic Chemistry With Connect Access Card (1 Semester), Chapter 18, Problem 37P , additional homework tip  10

Figure 9

Conclusion

The products formed when B is treated with given reagents are shown in Figure 5, Figure 6, Figure 7, Figure 8 and Figure 9.

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Chapter 18 Solutions

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