Aromatic Chemistry Flashcards

Question 1

Q

Ethanoyl chloride reacts with methylbenzene forming compound X according to the equation below.

( Diagram shows an equation )

( Methylbenzene + CH3COCl - > 4-methylphenylethanoate + HCl )

( 4-methylphenylethanoate is labelled X )

If the experimental yield is 40.0%, the mass in grams of X ( Mr = 134.0 ) formed from 18.4 g of methylbenzene ( Mr = 92.0 ) is

A 26.8

B 16.1

C 10.7

D 7.4

Answer

A

Moles methlybenzene = 18.4 / 92
= 0.2
1 : 1 ratio
Mass of 4-methlyphenylethanoate = 0.2 x 134
= 26.8
40 x 26.8 = 1072 / 100
= 10.72
= 10.7
C

Question 2

Q

In a reaction which gave a 27.0% yield, 5.00 g of methylbenzene were converted into the explosive 2,4,6-trinitromethylbenzene ( TNT ) ( Mr = 227.0 ).
The mass of TNT formed was

A 1.35 g

B 3.33 g

C 3.65 g

D 12.34 g

Answer

A

Mr of methylbenzene = 92
Moles methylbenzene = 5 / 92
= 0.05
Mass TNT = ( 5 / 92 ) x 227
= 1135 / 92
= 12.3
( 1135 / 92 ) x 27 x 10^-2
= 6129 / 1840
= 3.33
B

Question 3

Q

This question is about the following reaction scheme which shows the preparation of polymer P.

( There’s a diagram of a series of conversions )

( Benzene - > Propylbenzene - > Phenol - > Cyclohexanol - > Cyclohexanal - > 1, 6-Hexandioic acid - > Polymer P ( Using H2NCH2CH2NH2 on the arrow ) )

( Propylbenzene is labelled J )

( Phenol is labelled K )

( Cyclohexanol is labelled L )

( Cyclohexanal is labelled M )

( 1, 6-Hexandioic acid is labelled N )

If 1.0 kg of benzene gave 0.98 kg of J, the percentage yield of J was

A 64

B 66

C 68

D 70

Question 4

Q

In which one of the following reactions is the role of the reagent stated correctly?

Reaction:

A TiO2 + 2C + 2Cl2 → TiCl4 + 2CO

B HNO3 + H2SO4 → H2NO3^+ + HSO4^-

C CH3COCl + AlCl3 → CH3CO^+ + AlCl4^-

D 2CO + 2NO → 2CO2 + N2

Role of reagent:

A TiO2 is an oxidising agent

B HNO3 is a Brønsted-Lowry acid

C AlCl3 is a Lewis base

D CO is a reducing agent

Question 5

Q

The relative molecular mass ( Mr ) of benzene-1,4-dicarboxylic acid is

A 164

B 166

C 168

D 170

Question 6

Q

1,3-dinitrobenzene can be prepared by heating nitrobenzene with a mixture of fuming nitric acid and concentrated sulphuric acid.
The reaction can be represented by the following equation.

( Diagram shows Nitrobenzene + NO2^+ - > 1,3-dinitrobenzene + H^+ )

If the yield of the reaction is 55%, the mass of 1,3-dinitrobenzene produced from 12.30 g of nitrobenzene is

A 16.90 g

B 16.80 g

C 9.30 g

D 9.24 g

Question 7

Q

Which one of the following can react both by nucleophilic addition and by nucleophilic substitution?

A CH3COCH = CH2

B CH2ClCH2COH

C CH2ClCH = CH2

D CH3CObenzene

( All are in displayed formula )

Question 8

Q

Which one of the following does not contain any delocalised electrons?

A poly( propene )

B benzene

C graphite

D sodium

Question 9

Q

Equations for the hydrogenation of cyclohexene and of benzene, together with the enthalpies of hydrogenation, are shown.

( Diagram shows two equations )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 )

Use these data to show that benzene is 152 kJ mol−1 more stable than the hypothetical compound cyclohexa-1, 3, 5-triene.

Answer

A

3( -120 ) - ( -208 ) = - 152 KJ mol^-1

Question 10

Q

State, in terms of its bonding, why benzene is more stable than
cyclohexa-1, 3, 5-triene.

( Equations for the hydrogenation of cyclohexene and of benzene, together with the enthalpies of hydrogenation, are shown.

( Diagram shows two equations )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 ) )

Answer

A

Delocalised electrons

Question 11

Q

Three carbon-carbon bonds are labelled on the structures shown.
These bonds are of different lengths.

( Diagram shows two molecules )

( Cyclohexene and Benzene )

( The single bond between carbon-carbon is labelled w )

( The double bond between carbon-carbon is labelled x )

( One side of the Benzene is labelled y )

Write the letters w, x and y in order of increasing bond length.

Question 12

Q

The structures of two cyclic dienes are shown.

( Diagram shows two molecules )

( cyclohexa-1, 4-diene and cyclohexa-1 , 3-diene )

Use the enthalpy of hydrogenation data given opposite to calculate a value for the enthalpy of hydrogenation of cyclohexa-1, 4-diene.

( Diagram shows two equations )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 ) )

Answer

A

- 240 KJ mol^-1
( There are two double bonds compared to Cyclohexene having one double bond )
( So the first equation’s enthalpy change is doubled )

Question 13

Q

Predict a value for the enthalpy of hydrogenation of cyclohexa-1, 3-diene.

( Diagram shows two equations )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 ) )

( Enthalpy change of Cyclohexa-1, 4-diene = - 240 KJ mol^-1 )

Answer

A

Between - 239 and - 121 KJ mol^-1

- - 238 KJ mol^-1

Question 14

Q

Explain your answers to part ( i ) and part ( ii ) in terms of the bonding in these two dienes.

( Enthalpy change of Cyclohexa-1, 4-diene = - 240 KJ mol^-1 )

( Enthalpy change of Cyclohexa-1, 3-diene = - 238 KJ mol^-1 )

Answer

A

The double bonds between the C = C are closer together in Cyclohexa-1, 3-diene than in Cyclohexa-1, 4-diene
There’s some overlap of electrons in Cyclohexa-1, 3-diene
There’s more stability for the Cyclohexa-1, 3-diene isomer

Question 15

Q

Each of the following conversions involves reduction of the starting material.
Consider the following conversion.

( Diagram shows an equation )

( 1, 4-dinitrobenzene - > C6H8N2 )

Identify a reducing agent for this conversion.
Write a balanced equation for the reaction using molecular formulae for the nitrogen-containing compounds and [ H ] for the reducing agent.
Draw the repeating unit of the polymer formed by the product of this reaction with benzene-1, 4-dicarboxylic acid.

Answer

A

Reducing agent:

HCl

Balanced equation:

C6H4N2O4 + 12[ H ] - > C6H8N2 + 4H2O

Repeating unit of the polymer formed from the reaction with benzene-1, 4-dicarboxylic acid:

-CObezeneCONHbenzeneNH-
( Must contain trailing bonds )

Question 16

Q

Consider the following conversion.

( Diagram shows an equation )

( Benzene - > Cyclohexane )

Identify a reducing agent for this conversion.
State the empirical formula of the product.
State the bond angle between the carbon atoms in the starting material and the bond angle between the carbon atoms in the product.

Answer

A

Reducing agent:

H2

Empirical formula of the product:

CH2

Bond angle in Benzene ring:

120 degrees

Bond angle in Cyclohexane:

109 degrees

Question 17

Q

The reducing agent in the following conversion is NaBH4

( Diagram shows a formula )

( CH3COCH2CH3 - > CH3CHOHCH2CH3 )

( In displayed formula )

Name and outline a mechanism for the reaction.

Answer

A

Mechanism:

Nuclephilic addition

Mechanism:

( H^- ion ( with lone pair of electrons ) from NaBH4 attacks the delta positive carbon which is bonded to the oxygen atom )
( At the same time, the electrons from the double bond between C = O moves to the oxygen atom )
( The carbon and hydrogen ion have now bonded, CH bond )
( Oxygen atom now has a lone pair of electrons and a negative charge )
( The lone pair of electrons attack a H^+ ion )
( Making an OH molecule )

Question 18

Q

By considering the mechanism of this reaction, explain why the product formed is optically inactive.

( Diagram shows a formula )

( CH3COCH2CH3 - > CH3CHOHCH2CH3 )

( In displayed formula )

Answer

A

There’s a planar C = O bond
So the chance of attack is equal from both sides
Therefore a racemic mixture is formed

Question 19

Q

The hydrocarbons benzene and cyclohexene are both unsaturated compounds.
Benzene normally undergoes substitution reactions, but cyclohexene normally undergoes addition reactions.

The molecule cyclohexatriene does not exist and is described as hypothetical.
Use the following data to state and explain the stability of benzene compared with the hypothetical cyclohexatriene.

( Diagram shows two formulas )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 )

Answer

A

Benzene is more stable than cyclohexatriene
The expected enthalpy of hydrogenation of benzene is 3( - 120 )
= - 360 KJ mol^-1
Actual enthalpy of hydrogenation of benzene is - 152 KJ mol^-1, ( less exothermic )
Due to delocalisation

Question 20

Q

Benzene can be converted into amine U by the two-step synthesis shown below.

( Diagram shows a series of reactions )

( Benzene - > Nitrobezene - > Phenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( Phenylamine is labelled “ U “ )

The mechanism of Reaction 1 involves attack by an electrophile.
Give the reagents used to produce the electrophile needed in Reaction 1.
Write an equation showing the formation of this electrophile.
Outline a mechanism for the reaction of this electrophile with benzene.

Answer

A

Reagents used to produce the electrophile needed in Reaction 1:

Conc HNO3
Conc H2SO4

Equation showing the formation of the electrophile:

2H2SO4 + HNO3 - > 2HSO4^- + NO2^+ + H3O^+

Mechanism of the reaction of this electrophile and Benzene:

( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 21

Q

Cyclohexene can be converted into amine W by the two-step synthesis shown below.

( Diagram shows an equation )

( Cyclohexene - > Compound V - > Aminocyclohexane )

( Aminocyclohexane is labelled “ W “ )

( The first arrow is labelled “ Reaction 3 “ )

( The second arrow is labelled “ Reaction 4 “ )

Suggest an identity for compound V.
For Reaction 3, give the reagent used and name the mechanism.
For Reaction 4, give the reagent and condition used and name the mechanism.

Equations and mechanisms with curly arrows are not required.

Answer

A

Identity of V:

Bromocyclohexane / Chlorcyclohexane
( Can be drawn or named )

Reagent for reaction 3:

HBr

Mechanism for reaction 3:

Electrophilic addition

Reagent for reaction 4:

Ammonia

Condition used for reaction 4:

Excess ammonia

Mechanism for reaction 4:

Nucleophilic substitution

Question 22

Q

Explain why amine U is a weaker base than amine W.

( Amine U = Phenylamine )

( Amine W = Aminocyclohexane )

Answer

A

There’s a lone pair of electrons on the “ N “ atom ( in amine U )
They are delocalised into the ring in amine U
So they are less available to accept protons

Question 23

Q

Use the following data to show the stability of benzene relative to the hypothetical cyclohexa-1,3,5-triene.

( There are two diagrams )

( One showing cyclohexa-1,3,5-triene )

( The other shows two equations )

( Cyclohexene + H2 - > Cyclohexane, enthalpy change = - 120 KJ mol^-1 )

( Benzene + 3H2 - > Cyclohexane, enthalpy change = - 208 KJ mol^-1 )

Give a reason for this difference in stability.

Answer

A

Cyclohexane evolves - 120 KJ mol^-1
Expected cyclohexa-1,3,5-triene to evolve - 360 KJ mol^-1
- 360 - ( - 208 ) = - 152 KJ mol^-1
Benzene is more stable due to delocalisation

Question 24

Q

Consider the following reaction sequence which starts from phenylamine.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > ( 1, 4 )NHCOCH3bezeneNO2 - > ( 1, 4 )NH2benzeneNO2 )

( First arrow is labelled Step 1, with CH3COCl )

( Second arrow is labelled Step 2 )

( Third arrow is labelled Step 3 )

State and explain the difference in base strength between phenylamine and ammonia.

Answer

A

Phenylamine is weaker
The lone pair on the “ N “ atom is less avaliable
The lone pair is delocalised into the ring

Question 25

Q

Name and outline a mechanism for the reaction in Step 1 and name the organic product of Step 1.

( Consider the following reaction sequence which starts from phenylamine.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > ( 1, 4 )NHCOCH3bezeneNO2 - > ( 1, 4 )NH2benzeneNO2 )

( First arrow is labelled Step 1, with CH3COCl )

( Second arrow is labelled Step 2 )

( Third arrow is labelled Step 3 ) )

Answer

A

Nucleophilic addition-elimination

Mechanism:

( Phenylamine attacks the carbon within the C = O bond )
( At the same time a lone pair of electrons the the double bond ( C = O ) moves to the oxygen atom )
( The oxygen atom now gains a negative charge )
( The NH2 has a positive charge on the “ N “, causing a hydrogen atom to be lost )
( Shown by an arrow from the N - H bond to the N atom )
( At the same time a double bond is reformed between the C = O )
( And the Cl atom is lost, shown by an arrow from the C - Cl bond to the Cl atom )

Name:

N-phenyl ethanaminde

Question 26

Q

The mechanism of Step 2 involves attack by an electrophile. Give the reagents used in this step and write an equation showing the formation of the electrophile.
Outline a mechanism for the reaction of this electrophile with benzene.

( Consider the following reaction sequence which starts from phenylamine.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > ( 1, 4 )NHCOCH3bezeneNO2 - > ( 1, 4 )NH2benzeneNO2 )

( First arrow is labelled Step 1, with CH3COCl )

( Second arrow is labelled Step 2 )

( Third arrow is labelled Step 3 ) )

Answer

A

Reagents used in step 2:

Conc HNO3
Conc H2SO4

Equation showing the formation of the electrophile:

HNO3 + 2H2SO4 - > NO2^+ + 2HSO4^- +H3O^+

Mechanism between Benzene and the electrophile:

( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 27

Q

Name the type of linkage which is broken in Step 3 and suggest a suitable reagent for this reaction.

( Consider the following reaction sequence which starts from phenylamine.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > ( 1, 4 )NHCOCH3bezeneNO2 - > ( 1, 4 )NH2benzeneNO2 )

( First arrow is labelled Step 1, with CH3COCl )

( Second arrow is labelled Step 2 )

( Third arrow is labelled Step 3 ) )

Answer

A

Linkage that is broken in step 3:

Peptide

Reagent for the reaction:

NaOH

Question 28

Q

Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ )

Name the type of mechanism for reaction 1.

Answer

A

Elctrophilic substitution reaction

Question 29

Q

Compound J is formed by reduction in reaction 2.

Give a reducing agent for this reaction.

( Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ ) )

Question 30

Q

Write an equation for this reaction. Use [H] to represent the reducing agent.

( Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ ) )

( Reducing agent = HCl )

Answer

A

CH3C6H4NO2 + 6[ H ] - > CH3C6H4NH2 + 2H2O

Question 31

Q

Give a use for J.

( Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ ) )

Answer

A

Making dyes

Question 32

Q

Outline a mechanism for the reaction of bromomethane with an excess of compound J.
You should represent J as RNH2 in the mechanism.

( Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ ) )

Answer

A

( The lone pair on the “ N “ in the RNH2 molecule attacks the carbon in the CH3Br molecule )
( This causes the “ Br “ atom to be broken off, shown by an arrow from the C - Br bond to the Br atom )
( The “ N “ gains a positive charge when bonded to the carbon, therefore breaking off one of the “ H “ atoms )
( This is shown by an arrow from the bond between the N^+ - H to the “ N “ )

Question 33

Q

Compound K ( C6H5CH2NH2 ) is a structural isomer of J.
Explain why J is a weaker base than K.

( Consider the following reaction sequence starting from methylbenzene.

( Diagram shows a series of reactions )

( Methylbenzene - > 4-methylnitrobenzene - > 4-methylphenylamine )

( First arrow is labelled “ Reaction 1 “ )

( Second arrow is labelled “ Reaction 2 “ )

( 4-methylphenylamine is labelled “ J “ ) )

Answer

A

There’s a lone pair of electrons on the N
In J, the electrons are delocalised into the ring
So they are less avaliable to accept protons

Question 34

Q

1,4-diaminobenzene is an important intermediate in the production of polymers such as Kevlar and also of polyurethanes, used in making foam seating.
A possible synthesis of 1,4-diaminobenzene from phenylamine is shown in the following figure.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

A suitable reagent for step 1 is CH3COCl

Name and draw a mechanism for the reaction in step 1.

Answer

A

Mechanism name:

Nucleophilic addition-elimination

Mechanism:

( The “ N “ with a lone pair of electrons from the “ C6H5NH2 “ attacks the carbon within the C = O bond )
( At the same time a lone pair of electrons the the double bond ( C = O ) moves to the oxygen atom )
( The oxygen atom now gains a negative charge )
( The NH2 has a positive charge on the “ N “, causing a hydrogen atom to be lost )
( Shown by an arrow from the N - H bond to the N atom )
( At the same time a double bond is reformed between the C = O )
( And the Cl atom is lost, shown by an arrow from the C - Cl bond to the Cl atom )

Question 35

Q

The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.

A small sample was then used to find the melting point.
Give reasons for each of the following practical steps.

The minimum quantity of hot water was used.

Answer

A

To ensure the hot solution would be saturated

Question 36

Q

The flask was cooled to room temperature before the crystals were filtered off

( The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.

A small sample was then used to find the melting point.
Give reasons for each of the following practical steps. )

Answer

A

The yield of cyrstals formed would be lower if warm

Question 37

Q

The crystals were compressed in the funnel

( The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.

A small sample was then used to find the melting point.
Give reasons for each of the following practical steps. )

Answer

A

Compressed because air passes through the sample, not just around it

Question 38

Q

A little cold water was poured through the crystals

( The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.

A small sample was then used to find the melting point.
Give reasons for each of the following practical steps. )

Answer

A

To was away soluble impurities

Question 39

Q

The melting point of the sample in part ( b ) was found to be slightly lower than a data-book value.
Suggest the most likely impurity to have caused this low value and an improvement to the method so that a more accurate value for the melting point would be obtained.

( The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.

A small sample was then used to find the melting point.
Give reasons for each of the following practical steps. )

Answer

A

The most likely impurity to cause this low value:

Water

Improvement to method in order to make a more accurater value:

Press the sample of cyrstals between the filter paper

Question 40

Q

The figure above is repeated here to help you answer the following questions.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

In an experiment starting with 5.05 g of phenylamine, 4.82 g of purified product were obtained in step 1.
Calculate the percentage yield in this reaction.
Give your answer to the appropriate number of significant figures.

Answer

A

Mr of NHCOCH3benzene ( Product ):

135

Mr of phenylamine:

93
Moles phenylamine:
5.05 / 93
= 101 / 1860

Theoretical mass of product:

( 101 / 1860 ) x 135
= 909 / 124
= 7.33

Percentage yield of product:

4.82 / ( 909 / 124 ) x 100
= 59768 / 909
= 65.8%

Question 41

Q

A reagent for step 2 is a mixture of concentrated nitric acid and concentrated sulfuric acid, which react together to form a reactive intermediate.
Write an equation for the reaction of this intermediate in step 2.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

Answer

A

NHCOCH3benzene + NO2^+ - > 4-NHCOCH3nitrobenzene + H^+

- ( All in displayed formula )

Question 42

Q

Name a mechanism for the reaction in step 2.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

Answer

A

Electrophilic substitution

Question 43

Q

Suggest the type of reaction occurring in step 3.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

Answer

A

Hydrolysis

Question 44

Q

Identify the reagents used in step 4.

( Diagram shows a series of reactions )

( Phenylamine - > NHCOCH3benzene - > 4-NHCOCH3nitrobenzene - > 4-aminonitrobenzene - > 1, 4-diaminobenzene )

( The arrows are labelled Step 1, 2, 3 and 4 correspondingly )

Question 45

Q

Benzene reacts with ethanoyl chloride in a substitution reaction to form C6H5COCH3.
This reaction is catalysed by aluminium chloride.

Write equations to show the role of aluminium chloride as a catalyst in this reaction.
Outline a mechanism for the reaction of benzene.
Name the product, C6H5COCH3.

Answer

A

Equations to show the role of aluminum chloride:

CH3COCl + AlCl3 - > CH3CO^+ + AlCl4^-
AlCl4^- + H^+ - > AlCl3 + HCl

Mechanism of reaction with benzene:

( An arrow comes from the Benzene ring, to the COCH3 )
( The positive charge is on the “ C “ atom which is bonded to the “ O “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the COCH3 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - COCH3 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Name of product:

Phenylethanone

Question 46

Q

The product of the substitution reaction ( C6H5COCH3 ) was analysed by mass spectrometry.
The most abundant fragment ion gave a peak in the mass spectrum
with m / z = 105.
Draw the structure of this fragment ion.

Answer

A

Benzene with a CO function group
” C “ contains a positive charge
On the second carbon from the top

Question 47

Q

When methylbenzene reacts with ethanoyl chloride and aluminium chloride, a similar substitution reaction occurs but the reaction is faster than the reaction of benzene.
Suggest why the reaction of methylbenzene is faster.

Answer

A

The methyl group increases the electron density on the benzene ring
So the electrophile has a stronger attraction to the methylbenzene

Question 48

Q

Kevlar is a polymer used in protective clothing.
The repeating unit within the polymer chains of Kevlar is shown.

( Reapeting unit shows -CObenzeneCONHbenzeneNH- )

( With trailing bonds )

Name the strongest type of interaction between polymer chains of Kevlar.

Answer

A

Hydrogen bonding

Question 49

Q

One of the monomers used in the synthesis of Kevlar is

H2NbenzeneNH2

An industrial synthesis of this monomer uses the following two-stage process starting from compound X.

( Stage 1 shows an equation )

( ClbenzeneNO2 + 2NH3 - > H2NbenzeneNO2 + NH4Cl )

( ClbenzneNO2 is labelled “ X “ )

( Stage 2 shows an equation )

( H2NbenzeneNO2 - > H2NbenzeneNH2 )

Suggest why the reaction of ammonia with X in Stage 1 might be considered unexpected.

Answer

A

Ammonia is a nucleophile

- benzene repels nucleophiles

Question 50

Q

Suggest a combination of reagents for the reaction in Stage 2.

( One of the monomers used in the synthesis of Kevlar is

H2NbenzeneNH2

An industrial synthesis of this monomer uses the following two-stage process starting from compound X.

( Stage 1 shows an equation )

( ClbenzeneNO2 + 2NH3 - > H2NbenzeneNO2 + NH4Cl )

( ClbenzneNO2 is labelled “ X “ )

( Stage 2 shows an equation )

( H2NbenzeneNO2 - > H2NbenzeneNH2 ) )

Answer

A

HCl and H2

Question 51

Q

Compound X can be produced by nitration of chlorobenzene.
Give the combination of reagents for this nitration of chlorobenzene.
Write an equation or equations to show the formation of a reactive
intermediate from these reagents.

( One of the monomers used in the synthesis of Kevlar is

H2NbenzeneNH2

An industrial synthesis of this monomer uses the following two-stage process starting from compound X.

( Stage 1 shows an equation )

( ClbenzeneNO2 + 2NH3 - > H2NbenzeneNO2 + NH4Cl )

( ClbenzneNO2 is labelled “ X “ )

( Stage 2 shows an equation )

( H2NbenzeneNO2 - > H2NbenzeneNH2 ) )

Answer

A

Reagents:

Conc HNO3
Conc H2SO4

Equation:

HNO3 + 2H2SO4 - > NO2^+ + H3O^+ + HSO4^-

Question 52

Q

Name and outline a mechanism for the formation of X from chlorobenzene and the reactive intermediate in part ( iii ).

( One of the monomers used in the synthesis of Kevlar is

H2NbenzeneNH2

An industrial synthesis of this monomer uses the following two-stage process starting from compound X.

( Stage 1 shows an equation )

( ClbenzeneNO2 + 2NH3 - > H2NbenzeneNO2 + NH4Cl )

( ClbenzneNO2 is labelled “ X “ )

( Stage 2 shows an equation )

( H2NbenzeneNO2 - > H2NbenzeneNH2 ) )

Answer

A

Name of mechanism:

Electrophilic substitution

Mechanism:

( Our reactant is Chlorobenzene, NO2^+ attacks from the opposisite side of the chlorine )
( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 53

Q

This question is about acylium ions, [ RCO ]^+

The acylium ion H3C - C^+ = O is formed in a mass spectrometer by
fragmentation of the molecular ion of methyl ethanoate.
Write an equation for this fragmentation.

Include in your answer a displayed formula for the radical formed.

Answer

A

[ H3C - C = O - O - CH3 ]^+^. - > H3C - C^+ = O + O^. - CH3
( Dot represents the free radical )
( All in displayed formula )

Question 54

Q

The acylium ion H3C - C^+ = O can also be formed from ethanoyl chloride.
The ion reacts with benzene to form C6H5COCH3

Write an equation to show the formation of this acylium ion by the reaction of ethanoyl chloride with one other substance.

( This question is about acylium ions, [ RCO ]^+

The acylium ion H3C - C^+ = O is formed in a mass spectrometer by
fragmentation of the molecular ion of methyl ethanoate. )

Answer

A

AlCl3

- CH3COOl + AlCl3 - > CH3CO^+ + AlCl4^-

Question 55

Q

Name and outline a mechanism for the reaction of benzene with this acylium ion.

( This question is about acylium ions, [ RCO ]^+

The acylium ion H3C - C^+ = O is formed in a mass spectrometer by
fragmentation of the molecular ion of methyl ethanoate. )

Answer

A

Name of mechanism:

Electrophilic substitution

Mechanism:

( An arrow comes from the Benzene ring, to the COCH3 )
( The positive charge is on the “ C “ atom which is bonded to the “ O “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the COCH3 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - COCH3 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 56

Q

Ethanoic anhydride also reacts with benzene to form C6H5COCH3
Write an equation for this reaction.

Answer

A

( CH3CO )2O + C6H6 - > C6H5COCH3 + CH3COOH

Question 57

Q

Many aromatic nitro compounds are used as explosives.
One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared from methylbenzene by a sequence of nitration reactions.

( Diagram shows a molecule of TNT )

The mechanism of the nitration of methylbenzene is an electrophilic substitution.

Give the reagents used to produce the electrophile for this reaction.
Write an equation or equations to show the formation of this electrophile.

Answer

A

Reagents:

Conc HNO3
Conc H2SO4

Equation:

2H2SO4 + HNO3 - > 2HSO4^- + NO2^+ + H3O^+

Question 58

Q

Outline a mechanism for the reaction of this electrophile with methylbenzene to produce 4-methylnitrobenzene.

( Many aromatic nitro compounds are used as explosives.
One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared from methylbenzene by a sequence of nitration reactions.

( Diagram shows a molecule of TNT )

The mechanism of the nitration of methylbenzene is an electrophilic substitution. )

Answer

A

( The attack of NO2^+ occurs on the opposite side of the methyl group )
( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 59

Q

Deduce the number of peaks in the 13C n.m.r. spectrum of TNT.

( Many aromatic nitro compounds are used as explosives.
One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared from methylbenzene by a sequence of nitration reactions.

( Diagram shows a molecule of TNT )

The mechanism of the nitration of methylbenzene is an electrophilic substitution. )

Question 60

Q

Deduce the number of peaks in the 1H n.m.r. spectrum of TNT.

( Many aromatic nitro compounds are used as explosives.
One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared from methylbenzene by a sequence of nitration reactions.

( Diagram shows a molecule of TNT )

The mechanism of the nitration of methylbenzene is an electrophilic substitution. )

Question 61

Q

Using the molecular formula ( C7H5N3O6 ), write an equation for the decomposition reaction that occurs on the detonation of TNT.
In this reaction equal numbers of moles of carbon and carbon monoxide are formed together with water and nitrogen.

( Many aromatic nitro compounds are used as explosives.
One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, can be prepared from methylbenzene by a sequence of nitration reactions.

( Diagram shows a molecule of TNT )

The mechanism of the nitration of methylbenzene is an electrophilic substitution. )

Answer

A

2C7H5N3O6 - > 5H2O + 3N2 + 7C + 7CO

Question 62

Q

Consider compound P shown below that is formed by the reaction of benzene with an electrophile.

( Diagram shows a compound )

( benzeneCOCH2CH3 )

Give the two substances that react together to form the electrophile and write an equation to show the formation of this electrophile.

Answer

A

Substances that react together to form the electrophile:

CH3CH2COCl
AlCl3

Equation to show how the electrophile is formed:

CH3CH2COCl + AlCl3 - > CH3CH2CO^+ + ALCl4^-

Question 63

Q

Outline a mechanism for the reaction of this electrophile with benzene to form P.

( Consider compound P shown below that is formed by the reaction of benzene with an electrophile.

( Diagram shows a compound )

( benzeneCOCH2CH3 ) )

( Electrophile = CH3CH2CO^+ )

Answer

A

( An arrow comes from the Benzene ring, to the CH3CH2CO^+ )
( The positive charge is on the “ C “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the CH3CH2CO and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - CH3CH2CO bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Question 64

Q

Compound Q is an isomer of P that shows optical isomerism.
Q forms a silver mirror when added to a suitable reagent.
Identify this reagent and suggest a structure for Q.

( Consider compound P shown below that is formed by the reaction of benzene with an electrophile.

( Diagram shows a compound )

( benzeneCOCH2CH3 ) )

Answer

A

Reagent:

Tollens’

Structure of Q:

BenzeneCH( CHO )CH3

Answer 54

A

Reaction between benzene and CH3CH2COCl:

C6H6 + CH3CH2COCl → C6H5COCH2CH3 + HCl

Name of organic product:

Phenylpropanone

Catalyst required for this reaction:

AlCl3

Reaction to show how the catalyst is used to form a reactive intermediate:

CH3CH2COCl + AlCl3 → CH3CH2CO^+ + AlCl4^-

Reaction to show how the catlasyt is reformed:

AlCl4^- + H^+ → AlCl3 + HCl

Answer 55

A

Name of mechanism:

Electrophilic substitution

Mechanism:

( An arrow comes from the Benzene ring, to the CH3CH2CO^+ )
( The positive charge is on the “ C “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the CH3CH2CO and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - CH3CH2CO bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Answer 56

A

Equation to show the reaction of propanal with HCN:

CH3CH2CHO + HCN → CH3CH2CH( OH )CN

Name of product:

2-hydroxybutanenitrile

Answer 57

A

Name of mechanism:

Nucleophilic addition

Mechanism:

( CN^- ( with lone pair of electrons ) from HCN attacks the delta positive carbon which is bonded to the oxygen atom )
( At the same time, the electrons from the double bond between C = O moves to the oxygen atom )
( The carbon and CN^- ion have now bonded, CCN bond )
( Oxygen atom now has a lone pair of electrons and a negative charge )
( The lone pair of electrons attack a H^+ ion )
( Making an OH molecule )

Answer 58

A

Propanal has a faster rate of reaction

- As it is easier for a nucleopile to attack at the end of the chain

Answer 59

A

Mr of N-phenylethanamide = 135
Mr of C6H5NH2 = 93
Mr of CH3COCl = 78.5
135 / ( 2 x 93 ) + 78.5 x 100
= 51.0%

Answer 60

A

Moles of C6H5NH2:

10 000 / 93
= 10 000 / 93
= 107.5

Moles of C6H5NHCOCH3:

( 10 000 / 93 ) / 2
= 5000 / 93
= 53.8

Mass of C6H5NHCOCH3:

5000 / 93 x 135
= 225000 / 31
= 7258.06 g
= 7.26 kg

Percentage yield of C6H5NHCOCH3:

5.38 / 7.26 x 100
= 16678 / 225
= 74.1%

Answer 61

A

Although the yield of N-phenylethanamide appears satifactory ( 74% )
Atom economy is only 51.0%
So nearly half of the material produced is waste and must be disposed

Answer 62

A

Name of mechanism:

Nucleophilic addition-elimination

Mechanism:

( The “ N “ with a lone pair of electrons from the “ C6H5NH2 “ attacks the carbon within the C = O bond in the CH3COCl molecule )
( At the same time a lone pair of electrons the the double bond ( C = O ) moves to the oxygen atom )
( The oxygen atom now gains a negative charge )
( The NH2 has a positive charge on the “ N “, causing a hydrogen atom to be lost )
( Shown by an arrow from the N - H bond to the N atom )
( At the same time a double bond is reformed between the C = O )
( And the Cl atom is lost, shown by an arrow from the C - Cl bond to the Cl atom )

Answer 63

A

Equation to show the formation of the electrophile:

HNO3 + 2H2SO4 - > 2HSO4^- + H3O^+ + NO2^+

Mechanism:

( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Answer 64

A

Name of mechanism:

Nucleophilic additon-elimination

Mechanism:

( The “ N “ with a lone pair of electrons from the “ CH3NH2 “ attacks the carbon within the C = O bond in the CH3CH2COCl molecule )
( At the same time a lone pair of electrons the the double bond ( C = O ) moves to the oxygen atom )
( The oxygen atom now gains a negative charge )
( The NH2 has a positive charge on the “ N “, causing a hydrogen atom to be lost )
( Shown by an arrow from the N - H bond to the N atom )
( At the same time a double bond is reformed between the C = O )
( And the Cl atom is lost, shown by an arrow from the C - Cl bond to the Cl atom )
( Show the product )

Answer 65

A

Equation to show the role of aluminium chloride as a catalyst:

CH3CH2COCl + AlCl3 → [ CH3CH2CO ]^+ + AlCl4^-
ALCl4^- + H^+ - > AlCl3 + HCl

Mechanism ( Electrophilic substitution ):

( An arrow comes from the Benzene ring, to the CH3CH2CO^+ )
( The positive charge is on the “ C “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the CH3CH2CO and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - CH3CH2CO bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Answer 66

A

Equation of reaction of water with propanoyl chloride:

CH3CH2COCl + H2O → CH3CH2COOH + HCl

Mr of CH3CH2COCl:

= 92.5

Moles of CH3CH2COCl

1.48 / 92.5
= 2 / 125
= 0.016
( Due to a 2:1 ratio, moles of NaOH would be double of CH3CH2COCl )

Moles NaOH:

2 x 2 / 125
= 4 / 125
= 0.032
Volume = Moles / Concentration
Volume = ( 4 / 125 ) / 0.42
= 8 / 105
= 0.076 dm^3
= 76.2 cm^3

Answer 67

A

Equation to show the formation of the reactive species:

CH3COCl + AlCl3 → CH3CO^+ + AlCl4^-

Name of mechanism:

Electrophilic subsitution

Mechanism:

( An arrow comes from the Benzene ring, to the CH3CO^+ )
( The positive charge is on the “ C “ atom )
( The arrow points to the “ C “ atom )
( Then show the bonding of the CH3CO and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - CH3CO bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Answer 68

A

Name of mechanism:

Nucleophilic addition

Mechanism:

( The “ H “ with a lone pair of electrons from the attacks the carbon within the C = O bond in the Phenylethanone ( CH3CObenzene ) molecule )
( At the same time a lone pair of electrons the the double bond ( C = O ) moves to the oxygen atom )
( The oxygen atom now gains a negative charge )
( The lone pair of electrons on the “ O “ attacke a H^+ atom from the “ Solution “ )
( Creating a OH molecule )
( Show the product )

Name of product:

1-phenylethanol

Answer 69

A

Dehydration / elimination reaction

Answer 70

A

Reagents:

Conc H2SO4
Conc HNO3

Equation:

2H2SO4 + HNO3 - > H3O^+ + NO2^+ + 2HSO4^-

Answer 71

A

Name of mechanism:

Electrophilic substitution

Mechanism:

( An arrow comes from the Benzene ring, to the NO2^+ )
( The positive charge is on the “ N “ atom )
( The arrow points to the “ N “ atom )
( Then show the bonding of the NO2 and the H on the carbon atom )
( A horseshoe is formed, contains a positive charge, that doesn’t go beyond the reacting carbon )
( The C - H bond next to the C - NO2 bond is attracted to the positive charge withing the horseshoe )
( This is represented by an arrow coming from the C - H bond to within the horseshoe )

Answer 72

A

Reagent suitable for step 2:

Sn or Fe

Answer 73

A

Reagent:

NH3

Condition:

An excess of ammonia

Answer 74

A

Name of the mechanism:

Nucleophilic substitution

Mechanism:

( Ammonia contains a lone pair of electrons on the “ N “ atom )
( These lone pairs attack the “ C “ atom adjacent to the “ Cl “ atom on the C6H5CH2Cl molecule ( shown by an arrow ) )
( At the same time, the C - Cl bond is broken off, shown by an arrow from the bond to the Cl atom )
( Once the ammonia has bonded, it “ N “ atom will gain a positive charge )
( Due to a positive charge, the electrons on the hydrogen adjacent to the ammonia will be attracted to the ammonia )
( This is shown by an arrow from the N - H bond to the ammonia )
( Brekaing off the hydrogen )

Answer 75

A

Formula for the reactive intermediate:

[ CH3CO ]^+

Answer 76

A

( Nucleophilic substitution )
( Benzene attacks the positive “ C “ with the C = O bond )
( Then show the “ H “ and “ COCH3 “ on the benzene and the horseshoe with the positive charge )
( Due to the positive charge, the “ H “ is attracted to the benzene, shown by an arrow from the “ benzene - H “ bond to the positive charge within the benzene )

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