Show how the following compounds might be prepared from the commercially available nitroalkanes obtained from the nitration of propane. (a) \(\mathrm{CH}_{2}=\mathrm{CHNO}_{2}\) (c) \(\mathrm{HOCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\) (b) \(\left(\mathrm{O}_{2} \mathrm{NOCH}_{2}\right)_{3} \mathrm{CNO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\)

Short Answer

Expert verified
To synthesize the following compounds from commercially available nitroalkanes, follow these steps: (a) \(CH_{2}=CHNO_{2}\) 1. Nitrate propane with \(HNO_3\). 2. Reduce nitro group with \(Zn/H^+\). 3. Form vinyl nitro compound with \(\mathrm{NOHSO_4}\). (c) \(HOCH_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\mathrm{NH}_{2}\) 1. Nitrate propane with \(HNO_3\). 2. Reduce nitro group with \(Sn/HCl\). 3. Add formaldehyde and hydrocyanic acid. (b) \(\left(\mathrm{O}_{2} \mathrm{NOCH}_{2}\right)_{3} \mathrm{CNO}_{2}\) 1. Form trinitromethane from nitromethane. 2. Add nitroform. (d) \(H_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\) 1. Nitrate propane with \(HNO_3\). 2. Reduce nitro group with \(Sn/HCl\). 3. Form amide with acetyl chloride. 4. Regain nitro group with \(HNO_3\). 5. Reduce nitro group with \(Zn/HCl\) to form the final product.

Step by step solution

01

Nitration of propane

Start with propane and add Nitric Acid (\(HNO_{3}\)) to give \(CH_{3}CH_{2}CH_{2}NO_{2}\). The reaction mechanism involves the electrophilic substitution of a hydrogen on the hydrocarbon chain by the nitro group \(NO_{2}\).
02

Reduction of nitro group to a carbonyl group

Add zinc in the presence of acid(\(Zn/H^{+}\)) to the \(CH_{3}CH_{2}CH_{2}NO_{2}\) molecule to reduce the nitro group to a carbonyl group, forming \(CH_{3}CH_{2}CHO\).
03

Formation of vinyl nitro compound

Add nitrosylsulfuric acid, \(\mathrm{NOHSO}_{4}\), to the aldehyde to form the vinyl nitro compound (the desired product), \(\mathrm{CH}_{2}=\mathrm{CHNO}_{2}\). (c)
04

Nitration of propane

Start with propane and add Nitric Acid (\(HNO_{3}\)) to give \(CH_{3}CH(NO_{2})CH_{3}\).
05

Reduction of the nitro group to the amino group

Reduce the nitro group \(NO_{2}\) to an amino group \(NH_{2}\) by adding a reducing agent like tin and hydrochloric acid (\(Sn/HCl\)). This forms \(CH_{3}CH(NH_{2})CH_{3}\).
06

Formation of the desired product

Add formaldehyde (\(HCHO\)) and hydrocyanic acid (\(HCN\)) to the amino compound obtained in the previous step; this leads to the formation of the desired product, \(\mathrm{HOCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\mathrm{NH}_{2}\). (b)
07

Formation of trinitromethane

Form trinitromethane (\(\left(\mathrm{O}_{2} \mathrm{NOCH}_{2}\right)_{3} \mathrm{CNO}_{2}\)) by the nitration of nitromethane (\(\mathrm{CH}_{3} NO_{2}\)).
08

Formation of the desired product

Add nitroform (\(CH(NO_{2})_{3}\)), which gives the desired product \(\left(\mathrm{O}_{2} \mathrm{NOCH}_{2}\right)_{3} \mathrm{CNO}_{2}\). (d)
09

Nitration of propane

Start with propane and add Nitric Acid (\(HNO_{3}\)) to give \(CH_{3}CH_{2}CH_{2}NO_{2}\).
10

Reduction of the nitro group to an amino group

Reduce the nitro group \(NO_{2}\) to an amino group \(NH_{2}\) by adding a reducing agent such as tin and hydrochloric acid (\(Sn/HCl\)), forming \(CH_{3}CH_{2}CH_{2}NH_{2}\).
11

Formation of an amide

Next, add acetyl chloride (\(CH_{3}COCl\)) to form an amide, \(\mathrm{CH}_{3}\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CONH}_{2}\).
12

Reformation of the nitro group

Treat the amide in step 3 with Nitric acid (\(HNO_{3}\)), which regains the nitro group at the alpha-position, giving \(\mathrm{CH}_{3}\mathrm{NHCONCH_{2} NO}_{2}\).
13

Formation of the desired product

Reduce the nitro compound from step 4 with a reducing agent such as \(Zn/HCl\) to form the final desired product, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\mathrm{NH}_{2}\).

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Most popular questions from this chapter

A colorless basic liquid \(\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{~N}(\mathrm{~A})\) reacts immediately with sodium nitrite in cold dilute sulfuric acid to give a neutral yellow oil (B) . Reaction of A with acetic anhydride in pyridine solution gives a neutral white solid (C), the nmr spectrum of which shows a singlet at \(\delta 1.9\), a singlet at \(\delta 3.5\), and a multiplet at \(\delta 7.3\) with intensities of \(1: 1: 1.67\), respectively. Assign structures to \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\).

An azo compound is cleaved at the azo linkage by stannous chloride, \(\mathrm{Sn} \mathrm{Cl}_{2}\), to form two amines, (a) What is the structure of the azo compound that is cleaved to 3-bromo-4-aminotoluene and 2-methy1-4-aminophenol? (b) Outline a synthesis of this azo compound, starting with toluene.

Give for each of the following pairs of compounds a chemical test, preferably a test-tube reaction which will distinguish +A between the two compounds. (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CNH}_{2}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NC}_{2} \mathrm{H}_{5}\) \(+A L\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}\) and \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) \(7 \mathrm{MO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\) and \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCl}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3} \mathrm{Cl}\) F ALh (e) \(\mathrm{CH}_{3} \mathrm{NHCOCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{NHCO}_{2} \mathrm{CH}_{3}\)

The \(\mathrm{pK}_{\mathrm{a}}\) of \(\mathrm{NH}_{4}^{+}\) is \(9.24\), that of \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\) is \(10.64\). Is methylamine a stronger or a weaker base than ammonia? How many times stronger or weaker? What do you think this difference is due to?

Propose a method for separating a mixture of cyclohexane- carboxylic acid, tri-t-butylamine, and decane.

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