CAIE A-Level Chemistry 34. Nitrogen compounds

[Maximum number: 10]

(a)

Phenylamine, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ , and propylamine, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ , can be produced by different reduction reactions.

[ 6 ]

(i)

Identify an organic compound that can be converted into $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ by a reduction reaction. State the reagents and conditions for this reaction.
organic compound
reagents
conditions

[ 2 ]

nitrobenzene / $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$

- Sn and HCl

- concentrated HCl and heat / boil / reflux

Two correct point [1] all three correct points [2]

(ii)

Identify an organic compound that can be converted into $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ by a reduction reaction. State the reagent for this reaction.
organic compound
reagent

[ 2 ]

M1: propylamide / propenamide / $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CONH}_{2} \mathrm{OR}$ propanenitrile / $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CN}$

M2: $\mathrm{LiAlH}_{4}$

(iii)

Identify a single test that will distinguish between $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ by producing a white precipitate with only one of these amines.

Draw the structure of the compound that is precipitated.
testing reagent
amine that gives a precipitate
structure of the compound that is precipitated:

[ 2 ]

Figure for Answer to Question (iii) — CAIE A-Level Chemistry A2

M1: bromine (aq) AND phenylamine / $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$

M2: correct structure

(b)

Describe the relative basicities of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ and $\mathrm{NH}_{3}$ .

Explain your answer. < <
least basic
most basic

[ 4 ]

M1: phenylamine ammonia propylamine

M2: a base has a lone pair that can accept a proton / $\mathrm{H}^{+}$

M3: the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / $\pi$ -system

M4: propylamine has an electron donating alkyl group

[Maximum number: 4]

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.

(a)

$\quad \mathrm{KI}$ is used as a source of $\mathrm{I}^{-}$ ions in organic synthesis.

One example of this is shown in the synthetic route in Fig. 1.1.

Fig. 1.1

[ 4 ]

(i)

Identify the reagents required for steps 1 and 2 .
step 1
step 2

[ 2 ]

reaction 1 concentrated $\mathrm{HNO}_{3}$ AND concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$ (concentrated seen once)

reaction 2 Sn AND (concentrated) HCl

(ii)

Step 3 occurs in two stages.
stage I $\mathrm{NaNO}_{2}$ and HCl undergo an acid-base reaction to produce $\mathrm{HNO}_{2}$ .
stage II $\mathrm{HNO}_{2}$ reacts with $\mathbf{C}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$ , to produce $\mathbf{D}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}{ }^{+}$ .
Complete the equations for stage I and for stage II.

$stage I $\mathrm{NaNO}_{2}+\mathrm{HCl} \rightarrow$$

stage II

[ 2 ]

step $1 \quad \mathrm{NaNO}_{2}+\mathrm{HCl} \rightarrow \mathrm{HNO}_{2}+\mathrm{NaCl}$

step $2 \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HNO}_{2}+\mathrm{H}^{+} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

[Maximum number: 10]

Amino acids are molecules that contain $-\mathrm{NH}_{2}$ and -COOH functional groups.
Glycine, $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH}$ , is the simplest stable amino acid.

(a)

The isoelectric point of glycine is 6.2.

[ 2 ]

(i)

Define isoelectric point.

[ 1 ]

(ii)

Draw the structure of glycine at pH 4 .

[ 1 ]

(b)

Fig. 6.1 shows two syntheses starting with glycine.

Fig. 6.1

[ 6 ]

(i)

State the essential conditions for reaction 1.

[ 1 ]

(ii)

Identify the reagent used in reaction 2 .

[ 1 ]

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl} /$ benzoyl chloride / benzoyl anhydride

(iii)

Draw the structure of the organic product U that forms when hippuric acid reacts with an excess of $\mathrm{LiAlH}_{4}$ in reaction 3 .

[ 2 ]

- carboxylic acid to primary alcohol COOH to $\mathrm{CH}_{2} \mathrm{OH}$

- amide to amine CONH to $\mathrm{CH}_{2} \mathrm{NH}$

- rest of the molecule is correct (carbons and benzene ring) mark as •

(iv)

A molecule of phenylalanine, R, can react with a molecule of glycine to form two dipeptides, S and T.

S and T are structural isomers.

Figure for Question (iv) — CAIE A-Level Chemistry A2

Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed.
S
T

[ 2 ]

(c)

A student proposes a synthesis of hippuric acid by the reaction of benzamide, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONH}_{2}$ , and chloroethanoic acid, $\mathrm{ClCH}_{2} \mathrm{COOH}$ .

The reaction does not work well because benzamide is a very weak base.

[ 2 ]

(i)

Explain why amides are weaker bases than amines.

[ 2 ]

M1 nitrogen lone pair in amides is delocalised with C=O

M2 lone pair less available for donation / to accept $\mathrm{H}^{+}$

OR less electron density on N / NH2 so less able to accept H ${ }^{+}$

[Maximum number: 8]

Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1.

Table 8.1

(a)

Define isoelectric point.

[ 1 ]

(b)

Draw the structures of asparagine and aspartic acid at pH 2 .

Table for Question (b) — CAIE A-Level Chemistry A2

[ 2 ]

Figure for Answer to Question (b) — CAIE A-Level Chemistry A2

$\begin{aligned} \text { M1: }\left[\mathrm{HOOCCH}\left(\mathrm{NH}_{3}\right) \mathrm{CH}_{2} \mathrm{CONH}_{2}\right]^{+} \text {M2: }\left[\mathrm{HOOCCH}\left(\mathrm{NH}_{3}\right) \mathrm{CH}_{2} \mathrm{COOH}^{+}\right. \end{aligned}$

(c)

Propanedioic acid, $\mathrm{HOOCCH}_{2} \mathrm{COOH}$ , is treated with an excess of thionyl chloride, $\mathrm{SOCl}_{2}$ . Propanedioyl chloride, $\mathrm{ClOCCH}_{2} \mathrm{COCl}$ , is formed.

[ 3 ]

(i)

Write an equation for this reaction.

[ 1 ]

$\mathrm{HOOCCH}_{2} \mathrm{COOH}+2 \mathrm{SOCl}_{2} \rightarrow \mathrm{ClOCCH}_{2} \mathrm{COCl}+2 \mathrm{SO}_{2}+2 \mathrm{HCl}$

(ii)

Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula $\mathrm{C}_{11} \mathrm{H}_{16} \mathrm{~N}_{4} \mathrm{O}_{8}$ .

Each molecule of compound G has four amide groups.
Draw the structure of compound G.

\text { Compound G, } \mathrm{C}_{11} \mathrm{H}_{16} \mathrm{~N}_{4} \mathrm{O}_{8}

[ 2 ]

Figure for Answer to Question (ii) — CAIE A-Level Chemistry A2

$\mathrm{HOOCCH}\left(\mathrm{CH}_{2} \mathrm{CONH}_{2}\right) \mathrm{NHCOCH}_{2} \mathrm{CONHCH}\left(\mathrm{CH}_{2} \mathrm{CONH}_{2}\right) \mathrm{COOH}$

M1: three molecules joined, with four correct amide groups only

M2: whole structure correct as above, displayed formula accepted, skeletal formula as below accepted

(d)

Asparagine is hydrolysed with an excess of hot NaOH(aq).

Draw the structure of the organic product of this reaction.

[ 2 ]

M1: hydrolysis of amide to carboxylate ion

M2: hydrolysis of -COOH to carboxylate ion and rest of molecule

$-\mathrm{OOCCH}\left(\mathrm{NH}_{2}\right) \mathrm{CH}_{2} \mathrm{COO}^{-}$

CAIE A-Level Chemistry 34. Nitrogen compounds Question Bank

Covered syllabus points

Covered syllabus points

34. Nitrogen compounds question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

34. Nitrogen compounds question 2

Question (a)

Question (i)

Question (ii)

34. Nitrogen compounds question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question (c)

Question (i)

34. Nitrogen compounds question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)