Chapter 22, Problem 22.38P

Interpretation Introduction

(a)

Interpretation: The acceptable names for the given compounds are to be predicted.

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

• The hydrocarbon is named after the longest carbon chain.

• The parent hydrocarbon containing ester group is named with suffix “ate”.

• The alkyl group bonded to oxygen through single bond forms the first part of the ester and alkyl group bonded to carbonyl group forms the second part.

Answer to Problem 22.38P

The acceptable name of the compound A is $2$-methylpropyl $2$, $2$-dimethylpropanoate and compound B is $2$-ethylhexanenitrile.

Explanation of Solution

The given compound A is,

Figure 1

The red coloured ball has two bonds. So, this is the oxygen atom. The black coloured atoms have four single bonds. So, these are the carbon atoms. The grey coloured balls have one bond. So, these are the hydrogen atoms. The molecular structure of the compound is,

Figure 2

This compound contains ester group. The alkyl group attached to carbonyl group has longest chain of three carbon atoms with two methyl groups substituted at the second carbon atom. The alkyl group bonded to oxygen through single bond has longest chain of three carbon atoms with one methyl group substituted at the second carbon atom. The acceptable name of the compound is $2$-methylpropyl $2$, $2$-dimethylpropanoate.

The given compound B is,

Figure 3

The blue coloured ball has three bonds. So, this is the nitrogen atom. The 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 of the compound is,

Figure 4

The given compound B contains longest chain of six carbon atoms with ethyl group substituted at the second carbon atom. The acceptable name of the compound is $2$-ethylhexanenitrile.

Conclusion

The acceptable name of the compound A is $2$-methylpropyl $2$, $2$-dimethylpropanoate and compound B is $2$-ethylhexanenitrile.

Interpretation Introduction

(b)

Interpretation: The products formed on treatment of A or B with given reagents: [1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$; [2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$; [3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess), then ${\text{H}}_{\text{2}}\text{O}$; [4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: The metal hydride reagents are good reducing agents such as ${\text{LiAlH}}_{\text{4}}$. They can be used to reduce carbonyl compounds, conjugated double bonds, amides or triple bonds and the $\text{C}-\text{X}$ bond in alkyl halides and epoxides.

When the esters and nitriles are hydrolyzed to carboxylic acids in acidic medium and in basic medium carboxylate ions are formed.

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

Answer to Problem 22.38P

The reaction of compound A with given reagents are shown below.

[1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$

Figure 5

 [2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$

Figure 6

 [3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess) then ${\text{H}}_{\text{2}}\text{O}$

Figure 7

 [4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$.

Figure 8

The reaction of compound B with given reagents are shown below.

[1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$

Figure 9

 [2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$

Figure 10

 [3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess) then ${\text{H}}_{\text{2}}\text{O}$

Figure 11

[4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$.

Figure 12

Explanation of Solution

The reaction of compound A with given reagents are shown below.

[1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$

The ester group undergoes acidic hydrolysis to form alcohol and carboxylic acid as shown below.

Figure 13

The products formed are $2$, $2$-dimethylpropanoic acid and $2$-methylpropanol.

[2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$

The ester group undergoes basic hydrolysis to form alcohol and carboxylic acid as shown below.

Figure 14

The products formed are $2$, $2$-dimethylpropanoate ion and $2$-methylpropanol.

[3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess) then ${\text{H}}_{\text{2}}\text{O}$.

The given ester compound reacts with excess Grignard reagent followed by hydrolysis to form $4$-tert-butylheptan-$4$-ol. The corresponding reaction is shown below.

Figure 15

The product formed is form $4$-tert-butylheptan-$4$-ol.

[4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$.

In presence of lithium aluminium hydride, the ester group is reduced to two alcohols. The corresponding reaction is shown below.

Figure 16

The products formed are $2$, $2$-dimethylpropanol and $2$-methylpropanol.

The reaction of compound B with given reagents are shown below.

[1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$

The cyano group undergoes acidic hydrolysis to form carboxylic acid as shown below.

Figure 17

The products formed is $2$-ethylhexanoic acid

[2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$

The cyano group undergoes basic hydrolysis to form carboxylate ion as shown below.

Figure 18

The product formed is $2$-ethylhexanoate ion.

[3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess) then ${\text{H}}_{\text{2}}\text{O}$

The nitrile compound reacts with Grignard reagent to form alcohol. The corresponding reaction is shown below.

Figure 19

[4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$.

In the presence of lithium aluminium hydride, the ester group is reduced to two alcohols. The corresponding reaction is shown below.

Figure 20

The product formed is $2$-ethylhexanamine.

Conclusion

The products formed on treatment of A or B with given reagents: [1] ${\text{H}}_{\text{3}}{\text{O}}^{+}$; [2] ${\text{OH}}^{-}$, ${\text{H}}_{\text{2}}\text{O}$; [3] ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{MgBr}$ (excess), then ${\text{H}}_{\text{2}}\text{O}$; [4] ${\text{LiAlH}}_4$, then ${\text{H}}_{\text{2}}\text{O}$ is shown in Figure 13-20.