Question 1
What product(s) would you expect from the following reaction? Provide a mechanism showing the formation of the product(s) you selected.
Question 2
What products would you expect from the reaction of ethylmagnesium bromide (CH3CH2MgBr)(CH3CH2MgBr) with each of the following reagents?
Question 3
What products would you expect to be formed when each of the following amines reacts with aqueous sodium nitrite and hydrochloric acid?
(a) Propylamine
(b) Dipropylamine
(c) N-Propylaniline
(d) N,N-Dipropylaniline
(e) p-Propylaniline
Answer to question 1
The structure of the product formed in reaction (a) is shown below
The structure of the product formed in reaction (b) is shown below
Answer to question 2
The products are in the figure below.
In a and b, the major products are the ethyl from the Grignard reagent with either H or D added.
The products in c and d result from the ethyl from the Grignard reagent attacking the carbonyl carbon and the workup adds the hydrogen to give the neutral product.
In e, Grignard reaction occurs twice with methoxy as the leaving group in the first step generating yet another carbonyl group for the second Grignard reaction to take place.
In f, the reaction mechanism is similar to c and d.
In g, the ethyl is a Lewis base, grabbing the terminal hydrogen on the alkyne (Lewis acid). The alkyne anion generated is stabilized by the +MgBr+MgBr counter cation. The other product in the reaction is ethane.
Answer to question 3
The reactivity of amines with aqueous sodium nitrite and hydrochloric acid depends upon the type of amines. Different amines show a different types of reactions with this reagent. The mixing of sodium nitrite and hydrochloric acid results in the formation of nitrous acid. Nitrous acid is highly unstable. It is, therefore, produced during the reaction and is not stored. The reaction of various amines with nitrous acid is as follows:
(a) Propylamine
Propylamine is a primary amine. It reacts with sodium nitrite and hydrochloric acid to give diazonium salt. The diazonium salts are very reactive, and they readily change into the carbocation in the presence of any nucleophile in the solution. The primary amines can further change into a mixture of alcohols, alkyl halides, and alkenes. The reaction is shown as follows:
CH3CH2CH2NH2+NaNO2+HCl→CH3CH2CH2N+≡NCl−1CH3CH2CH2NH2+NaNO2+HCl→CH3CH2CH2N+≡NCl−1
Propylamine gives propyldiazonium chloride, which is highly unstable and can further change into the following products:
CH3CH2CH2N+≡NCl−1→[CH3CH2CH2]++N2CH3CH2CH2N+≡NCl−1→[CH3CH2CH2]++N2
[CH3CH2CH2]++H2O→CH3CH2CH2OH[CH3CH2CH2]++H2O→CH3CH2CH2OH
[CH3CH2CH2]++Cl−1→CH3CH2CH2Cl[CH3CH2CH2]++Cl−1→CH3CH2CH2Cl
[CH3CH2CH2]+→CH3CH2=CH2[CH3CH2CH2]+→CH3CH2=CH2
In the presence of water, the diazonium salts form alcohols. In the presence of halides, they produce haloalkanes, and they can also change into alkenes.
(b) Dipropylamine
Dipropylamine is a secondary amine, and it will react with sodium nitrite and hydrochloric acid to give n-nitrosodipropylamine, as shown in the following way:
CH3CH2CH2NHCH2CH2CH3+NaNO2+HCl→(CH3CH2CH2)2−N−N=OCH3CH2CH2NHCH2CH2CH3+NaNO2+HCl→(CH3CH2CH2)2−N−N=O
(c) N-propylaniline
N-propylamine has the following structure:
It is a secondary aromatic amine. The alkyl group on the amine will participate in the reaction. It will react with sodium nitrite and hydrochloric acid and will lead to the formation of N-nitroso-N-propylaniline, as shown in the following reaction:
C6H5NHCH2CH2CH3+NaNO2+HCl→C6H5(CH3CH2CH2)N−N=OC6H5NHCH2CH2CH3+NaNO2+HCl→C6H5(CH3CH2CH2)N−N=O
(d) N,N-dipropylaniline
N,N-dipropylaniline is a tertiary aromatic amine. It has the following structure:
It will react with sodium nitrite and hydrochloric acid to form N-nitrosoamonium compounds, as shown in the following reaction:
C6H5N(CH2CH2CH3)2+NaNO2+HCl→(CH3CH2CH2)2C6H5N+HCl−+(CH3CH2CH2)2C6H5N+=OCl−C6H5N(CH2CH2CH3)2+NaNO2+HCl→(CH3CH2CH2)2C6H5N+HCl−+(CH3CH2CH2)2C6H5N+=OCl−
(e) p-propylaniline
p-propylaniline has the following structure:
It will react with sodium nitrite and hydrochloric acid to form diazonium salt of benzene.
CH3CH2CH2(C6H5)NH2+NaNO2+HCl→CH3CH2CH2(C6H5)N2Cl−