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Chemistry > Optical Isomerism( Aldehydes And Ketones and Carbonyl Compounds ) > Flashcards

Optical Isomerism( Aldehydes And Ketones and Carbonyl Compounds ) Flashcards

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1
Q

Which compound forms optically active compounds on reduction?

A CH3CH2C(CH3)=CHCH3

B CH3CH2C(CH3)=CH2

C CH3COCH3

D CH3CH2COCH3

A
  • D ( When reduced, a chiral carbon is present )
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2
Q

Which one of the following can exhibit both geometrical and optical isomerism?

A ( CH3 )2C=CHCH( CH3 )CH2CH3

B CH3CH2CH=CHCH( CH3 )CH2CH3

C ( CH3 )2C=C( CH2CH3 )2

D CH3CH2CH( CH3 )CH( CH3 )C=CH2

A
  • B
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3
Q

Ibuprofen is a drug used as an alternative to aspirin for the relief of pain, fever and inflammation.
The structure of ibuprofen is shown below.

( Diagram shows, CH3CHCOOH )
( A Benzene ring is attached to the second carbon )
( At the end of the Benzene ring, there is a group, CH2CH( CH3 )2 )

Which one of the following statements is not correct?

A It has optical isomers.

B It liberates carbon dioxide with sodium carbonate solution.

C It undergoes esterification with ethanol.

D It undergoes oxidation with acidified potassium dichromate(VI).

A
  • D ( It’s a carboxylic acid, so cannot undergo oxidation with acidified potassium dichromate )
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4
Q

Which one of the following reaction mixtures would give a product capable of exhibiting optical isomerism?

A CH3CH=CH2 + HBr

B CH3CH2CH2Br + NaOH

C CH3CH2CH2OH + H2SO4

D CH3CH2CHO + HCN

A
  • D
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5
Q

The infra-red spectrum of compound A, C3H6O2, is shown below.

( Shows a spectrum with two peaks labelled x and y )

Identify the functional groups which cause the absorptions labelled X and Y.

Using this information draw the structures of the three possible structural isomers for A.

Label as A the structure which represents a pair of optical isomers.

A
  • x = O - H alcohols
  • y = C = O

Three possible structures:

  • CH2OHCOCH3
  • CH2OHCH2CHO
  • CH3CHOHCHO
  • ( All drawn in displayed formula )
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6
Q

Draw the structures of the three branched-chain alkenes with molecular formula C5H10.

Draw the structures of the three dibromoalkanes, C5H10Br2, formed when these three alkenes react with bromine.

One of these dibromoalkanes has only three peaks in its proton n.m.r. spectrum.
Deduce the integration ratio and the splitting patterns of these three peaks.

A

Three branches chains:

  • CH2 = C( CH3 )CH2CH3
  • CH3C( CH3 ) = CHCH3
  • CH3CH( CH3 )CH = CH2
  • ( All in displayed formula )

Three dibromoalkanes formed from these three alkenes which react with bromine:

  • CH2BrC( CH3 )BrCH2CH3
  • CH3C( CH3 )BrCHBr
  • CH3C( CH3 )CHBrCH2Br
  • ( All in displayed formula )

The dibromoalkane with 3 peaks is:

Integration ratio of these three peaks are:

  • 6:3:1
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7
Q

The structures of the amino acid alanine and glycine are shown below.

( Alanine = H2NCH( CH3 )COOH )
( Glycine = H2NCH2COOH )

Give the systematic name for alanine.

A
  • 2-amino-Propanoic acid
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8
Q

Alanine exists as a pair of stereoisomers.

( Alanine = CHNH2( CH3 )COOH )

Explain the meaning of the term stereoisomers.

A
  • Molecules with the same molecular structure

- but different atomic arrangement

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9
Q

State how you could distinguish between the stereoisomers.

A
  • Plane polarised light

- Rotates equally in opposite directions

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10
Q

Give the structural formula of the species formed by glycine at pH 14.

( Glycine = H2NCH2COOH )

A
  • H2NCH2COO^-

- ( In displayed formula )

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11
Q

When two amino acids react together, a dipeptide is formed.
Give the structural formulae of the two dipeptides which are formed when alanine and glycine react together.

( Alanine = H2NCH( CH3 )COOH )
( Glycine = H2NCH2COOH )

A

Dipeptide 1:

  • H2NCH( CH3 )CONHCH2COOH

Dipeptide 2:

  • H2NCH2CONHCH( CH3 )COOH
  • ( Both in displayed formula )
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12
Q

Give the structural formula of the organic compound formed when glycine reacts with methanol in the presence of a small amount of concentrated sulphuric acid.

( Glycine = H2NCH2COOH )

A
  • H2NCH2COOCH3
  • ( ( Amino )Carboxylic acid + Alcohol - > Product from Nucleophilic addition - elimination )
  • ( Mechanism = Nucleophilic addition - elimination )
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13
Q

Each of the parts ( a ) to ( e ) below concerns a different pair of isomers.
Draw one possible structure for each of the species A to J, using Table 2 on the Data Sheet where appropriate.

Compounds A and B have the molecular formula C5H10 A decolourises bromine water but B does not.

A
  • ( A must be an alkene )
  • ( B must be an ( no idea ) )

A:

  • CH3CH = CHCH2CH3

B:

  • ( Pentagon )
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14
Q

Compounds C and D have the molecular formula C2H4O2

Each has an absorption in its infra-red spectrum at about 1700 cm^-1 but only D has a broad absorption at 3350 cm^-1.

A
  • ( C has OH acids )
  • ( D has OH alcohols )
  • ( Both have C = O )

C:

  • CH3COOH

D:

  • CH2OHCHO
  • ( Both in displayed formula )
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15
Q

Compounds E and F are esters with the molecular formula C5H10O2
The proton n.m.r. spectrum of E consists of two singlets only whereas that of F consists of two quartets and two triplets.

A

E:

  • CHOOC( CH3 )2CH3 ( Ester )

F:

  • CH3CH2COOCH2CH3
  • ( Both in displayed formula )
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16
Q

Compounds G and H have the molecular formula C3H6Cl2 G shows optical activity but H does not.

A

G:

  • CH3CHClCH2Cl

H:

  • CH3CH2CHCl2
  • ( Both in displayed formula )
17
Q

Compounds I and J have the molecular formula C6H12
Each has an absorption in its infra-red spectrum at about 1650 cm^-1 and neither shows geometrical isomerism.
The proton n.m.r. spectrum of I consists of a singlet only whereas that of J consists of a singlet, a triplet and a quartet.

A

I:

  • C( CH3 )2 = C( CH3 )2

J:

  • CH2 = C( CH2CH3 )2
18
Q

Consider the reaction sequence shown below.

CH3CH2CHO - > CH3CH2CHOH( CN ) - > CH3CH2CHOHCOOH

( First arrow is labelled step 1 with HCN )
( Second arrow is labelled step 2 )
( First reactant is labelled propanal )
( Final product is labelled Q )

Name and outline a mechanism for the reaction in Step 1.

A

Name of mechanism:

  • Nucleophilic addition

Mechanism:

  • ( ^-CN, with lone pair of electrons, attacks the delta positive carbon attached to the O )
  • ( Arrows are drawn from the lone pair on CN to the carbon, then an arrow from the double bond in C = O to the oxygen )
  • ( The oxygen now has a lone pair of electrons and a negative charge )
  • ( The lone pair of electrons move to a H^+ which is freely in the “ Solution “ creating a O - H bond )
19
Q

Name compound Q formed in Step 2.

( CH3CH2CHO - > CH3CH2CHOH( CN ) - > CH3CH2CHOHCOOH

( First arrow is labelled step 1 with HCN )
( Second arrow is labelled step 2 )
( First reactant is labelled propanal )
( Final product is labelled Q ) )

A
  • 2-hydroxy-butanoic acid
20
Q

Two stereoisomers are formed by the dehydration of Q.
Give the structures of these two isomers and name the type of stereoisomerism shown.

( CH3CH2CHO - > CH3CH2CHOH( CN ) - > CH3CH2CHOHCOOH

( First arrow is labelled step 1 with HCN )
( Second arrow is labelled step 2 )
( First reactant is labelled propanal )
( Final product is labelled Q ) )

( Q = 2-hydroxy-butanoic acid )

A
  • ( Dehydration of Q results in alkenes forming )
  • ( Removal of OH alcohols )

Isomers:

  • CH( CH3 ) = CHCOOH ( Shows E - isomerism )
  • CH( CH3 ) = CHCOOH ( Shows Z - isomerism )

Type of isomerism:

  • Geometrical isomerism
21
Q

An isomer of Q which has the structure shown below is polymerised to form the biodegradeable polymerknown as PHB.

HOCH( CH3 )CH2COOH

( Isomers of Q = CH( CH3 ) = CHCOOH ( Shows E - isomerism ), and CH( CH3 ) = CHCOOH ( Shows Z - isomerism ) )

Draw the repeating unit of the polymer PHB.

A
    • OCH( CH3 )CH2CO - ( With trailing bonds )
22
Q

Suggest a reason why the polymer is biodegradeable.

Polymer = - OCH( CH3 )CH2CO - ( With trailing bonds )

A
  • Can be hydrolysed
23
Q

The amino acid R is shown below.

CH3CH2CH( NH2 )( COOH )

Draw the structure of the zwitterion formed by R.

A
  • CH3CH2CH( NH3^+ )( COO^- )
  • ( NH2 gains a H atom )
  • ( COOH loses a H atom )
24
Q

Draw the structure of the major organic product formed when an excess of R is reacted with bromomethane.

( R = CH3CH2CH( NH2 )( COOH ) )

A
  • CH3CH2CH( NHCH3 )( COOH )
25
Q

Name the mechanism of the reaction which results in the formation of the product given in part( ii ).

( CH3CH2CH( NH2 )( COOH ) + CH3Br - > CH3CH2CH( NHCH3 )( COOH ) )

A
  • Nucleophilic substitution
26
Q

P, Q and R have the molecular formula C6H12

All three are branched-chain molecules and none is cyclic.
P can represent a pair of optical isomers.
Q can represent a pair of geometrical isomers.
R can represent another pair of geometrical isomers different from Q.

Draw one possible structure for one of the isomers of each of P, Q and R.

A

Structure of P:

  • CH( CH3 )( CH2CH3 )( CH = CH2 )

Structure of Q and R:

  • CH( CH3 ) = C( CH3 )( CH2CH3 )
  • CH( CH3 ) = CH( CH( CH3 )2 )
27
Q

Butanone reacts with reagent S to form compound T which exists as a racemic mixture.
Dehydration of T forms U, C5H7N, which can represent a pair of geometrical isomers.

State the meaning of the term racemic mixture and suggest why such a mixture is formed in this reaction.

A

Meaning of racemic mixture:

  • Equal mixture of enantiomers

Why the mixture was formed:

  • Attack from either side of the C = O bond is equally likely
28
Q

Identify reagent S, and draw a structural formula for each of T and U.

( Butanone reacts with reagent S to form compound T which exists as a racemic mixture.
Dehydration of T forms U, C5H7N, which can represent a pair of geometrical isomers. )

A

Reagent S:

  • HCN

Structural formula for T:

  • CH3CH2COH( CN )CH3

Structral formula for U:

  • CH( CH3 )= C( CH3 )( CN )
29
Q

Hydrogen and carbon monoxide were mixed in a 2:1 mole ratio.
The mixture was allowed to reach equilibrium according to the following equation at a fixed temperature and a total pressure of 1.75 × 104 kPa.

2H2( g ) + CO( g ) < - > CH3OH( g )

The equilibrium mixture contained 0.430 mol of carbon monoxide and 0.0850 mol of methanol.

Calculate the number of moles of hydrogen present in the equilibrium mixture.

A
  • 0.430 x 2 = 0.86
30
Q

Hence calculate the mole fraction of hydrogen in the equilibrium mixture.

( 2H2( g ) + CO( g ) < - > CH3OH( g ) )

( Total Moles = 11 / 8 = 1.38 )
( Moles of H2 = 0.86 )

A

Mole Fraction:

  • H2 = 0.86 / 11 / 8 = 172 / 275 = 0.63
31
Q

Calculate the partial pressure of hydrogen in the equilibrium mixture.

( Moles fraction = 172 / 275 = 0.63 )
( Total pressure = 1.75 × 10^4 kPa )

A
  • pP = 172 / 275 x 1.75 x 10^4

- = 120400 / 11 = 10945.45 ( 1.09 x 10^4 )

32
Q

In a different mixture of the three gases at equilibrium, the partial pressure of carbon monoxide was 7550 kPa, the partial pressure of hydrogen was 12300 kPa and the partial pressure of methanol was 2710 kPa.

( 2H2( g ) + CO( g ) < - > CH3OH( g ) )

Write an expression for the equilibrium constant, Kp, for this reaction.

A
  • Kp = ( pPCH3OH ) / ( pPH2 )^2 ( pPCO )
33
Q

Calculate the value of the equilibrium constant, Kp, for the reaction under these conditions and state its units.

( Kp = ( pPCH3OH ) / ( pPH2 )^2 ( pPCO ) )
( In a different mixture of the three gases at equilibrium, the partial pressure of carbon monoxide was 7550 kPa, the partial pressure of hydrogen was 12300 kPa and the partial pressure of methanol was 2710 kPa. )

A
  • Kp = ( 2710 ) / ( 12300 )^2 ( 7550 )

- Kp = 2.37 x 10^-9 kPa^-2

34
Q

Two isomeric esters E and F formed from methanol have the molecular formula C6H12O2
Isomer E has only 2 singlet peaks in its proton n.m.r. spectrum.
Isomer F is optically active.
Draw the structures of these two isomers.

A

Structure of E:

  • C( CH3 )3COOCH3

Structure of F:

  • CH3CH2CH( CH3 )COOCH3

Chemistry (13 decks)

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