A2 Chemistry F324 - Rings, Acids and Analysis

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A2 Chemistry F324 - Rings, Acids and Analysis Carbonyl Compounds Plymstock School P.J.McCormack

Learning Objectives. By the end of the lesson I will be able to.... 13 November, 2018 Learning Objectives. By the end of the lesson I will be able to.... All.. State the two types of carbonyl compounds Most.. Draw the displayed and molecular formula of carbonyl compounds Some.. State with an equation the oxidation reaction of carbonyl compounds along with the reagents and conditions required Low High Key Words: Aldehyde, Ketone, reduction, oxidations P.J.McCormack

CARBONYL COMPOUNDS - STRUCTURE Structure carbonyl groups consists of a carbon-oxygen double bond the bond is polar due to the difference in electronegativity Difference ALDEHYDES - at least one H attached to the carbonyl group KETONES - two carbons attached to the carbonyl group C = O H C = O H CH3 C = O CH3 C = O C2H5 CH3

CARBONYL COMPOUNDS - FORMULAE Molecular C3H6O Structural C2H5CHO CH3COCH3 Displayed Skeletal C = O H C2H5 C = O CH3 H C C C H H O H H H H C C C O H H H H H O O

CARBONYL COMPOUNDS - NOMENCLATURE Aldehydes C2H5CHO propanal Ketones CH3COCH3 propanone CH3CH2COCH3 butanone CH3COCH2CH2CH3 pentan-2-one CH3CH2COCH2CH3 pentan-3-one C6H5COCH3 phenylethanone

CARBONYL COMPOUNDS - FORMATION ALDEHYDES Oxidation of primary (1°) alcohols RCH2OH + [O] ——> RCHO + H2O beware of further oxidation RCHO + [O] ——> RCOOH Reduction of carboxylic acids RCOOH + [H] ——> RCHO + H2O

CARBONYL COMPOUNDS - FORMATION ALDEHYDES Oxidation of primary (1°) alcohols RCH2OH + [O] ——> RCHO + H2O beware of further oxidation RCHO + [O] ——> RCOOH Reduction of carboxylic acids RCOOH + [H] ——> RCHO + H2O KETONES secondary (2°) alcohols RCHOHR + [O] ——> RCOR + H2O

CARBONYL COMPOUNDS - IDENTIFICATION Method 1 strong peak around 1400-1600 cm-1 in the infra red spectrum Method 2 formation of an orange precipitate with 2,4-dinitrophenylhydrazine Although these methods identify a carbonyl group, they cannot tell the difference between an aldehyde or a ketone. To narrow it down you must do a second test.

Preparation of Carbonyl Compounds 13 November, 2018 Preparation of Carbonyl Compounds A primary alcohol reacts with acidified potassium dichromate(VI) to form an aldehyde: RCH2OH + [O]  RCHO + H2O K2Cr2O7 H+ , distil Also RCHO + [O]  RCOOH P.J.McCormack

Preparation of Carbonyl Compounds 13 November, 2018 Preparation of Carbonyl Compounds If a secondary alcohol is refluxed with acidified potassium dichromate(VI), a ketone is formed: R’ R’ CHOH + [O]  C=O + H2O R K2Cr2O7 R /H+, reflux P.J.McCormack

Aldehydes and Ketones The dichromate colour change is from orange to 13 November, 2018 Aldehydes and Ketones The dichromate colour change is from orange to green. P.J.McCormack

Reduction of the Carbonyl Compounds 13 November, 2018 Reduction of the Carbonyl Compounds Since aldehydes and ketones are made by the oxidation of primary and secondary alcohols, alcohols can be made by the reduction of aldehydes and ketones. The reduction of an aldehyde results in the formation of a primary alcohol: RCHO + 2[H]  RCH2OH NaBH4/water Room temp. P.J.McCormack

Reduction of the Carbonyl Compounds 13 November, 2018 Reduction of the Carbonyl Compounds The reduction of a ketone results in the formation of a secondary alcohol: R R C=O + 2[H]  CHOH R’ NaBH4/water R’ room temp The reducing agent used is sodium tetrahydridoborate(III), NaBH4 in aqueous solution. P.J.McCormack

Learning Objectives. By the end of the lesson I will be able to.... 13 November, 2018 Learning Objectives. Low By the end of the lesson I will be able to.... State two chemical tests used to distinguish an aldehyde from a ketone; State the chemical equation for the oxidation of aldehydes to carboxylic acids with Tollens reagent; Draw the reaction mechanism for reduction of alcohols; High Key Words: Aldehyde, Tollens, Benedict’s, reduction P.J.McCormack

The best silver mirror will win an super duper chemistry prize!! 13 November, 2018 News Flash The best silver mirror will win an super duper chemistry prize!! P.J.McCormack

13 November, 2018 Tollens’ Reagent. Tollens' reagent contains ammonical silver nitrate, [Ag(NH3)2]+. This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to re-dissolve the precipitate. Ketone No change (colourless solution) Aldehyde Grey/silver precipitate, or a silver mirror on the test tube. P.J.McCormack

13 November, 2018 Tollens’ Reagent. P.J.McCormack

Fehling’s Solution or Benedict’s Solution 13 November, 2018 Fehling’s Solution or Benedict’s Solution Fehling's solution contains copper(II) ions complexed with tartrate ions in sodium hydroxide solution. Benedict's solution contains copper(II) ions complexed with citrate ions in sodium carbonate solution. Both solutions are used in the same way. A few drops of the aldehyde or ketone are added to the reagent, and the mixture is warmed gently in a hot water bath for a few minutes. P.J.McCormack

Fehling’s Solution or Benedict’s Solution 13 November, 2018 Fehling’s Solution or Benedict’s Solution Both solutions are used in the same way. A few drops of the aldehyde or ketone are added to the reagent, and the mixture is warmed gently in a hot water bath for a few minutes. Results. Ketone No change (blue solution) Aldehyde Dark red precipitate of copper (I) oxide P.J.McCormack

Learning Objectives. By the end of the lesson I will be able to.... 13 November, 2018 Learning Objectives. By the end of the lesson I will be able to.... All.. State two test to distinguish an aldehyde from a ketone State the observations in the above reaction Most.. State the equation for the reaction of an aldehyde with Tollen’s reagent Some.. Explain with an equation what happens when an aldehyde and a ketone react with 2,4, DNP Low High Key Words: Aldehyde, Ketone, reduction, oxidations P.J.McCormack

13 November, 2018 2,4, DNP C2 C3 C4 C1 P.J.McCormack

13 November, 2018 2,4, DNP P.J.McCormack

2,4,-dinitrophenylhydrazone. 13 November, 2018 2,4, DNP A water molecule is lost and the oxygen from the carbonyl compounds joins to the nitrogen to form 2,4,-dinitrophenylhydrazone. P.J.McCormack

Nucleophilic Addition Mechanism CH3COCH3 + Nu- CH3C(OH)(Nu)CH3 + - H OH CH3 C O Nu CH3 C O Nu H CH3 C O Nu P.J.McCormack

Nucleophilic Addition addition of hydrogen cyanide to carbonyls to form hydroxynitriles RCOR + HCN RC(OH)(CN)R RCHO + HCN RCH(OH)CN Conditions / Reagents HCN (aq) and NaOH(aq) to form the CN- nucleophile HCN + OH- CN- + H2O Room temperature and pressure P.J.McCormack

Nucleophilic Addition Mechanism hydrogen cyanide with propanone CH3COCH3 + HCN CH3C(OH)(CN)CH3 HCN / NaOH (aq) is a source of cyanide ions C N CN + - H CN CH3 C O CN CH3 C O CN H CH3 C O CN 2-hydroxy-2-methylpropanenitrile P.J.McCormack

The electronegative oxygen polarises the double bond, leaving the carbon electron-deficient. It is therefore susceptible to attack by a nucleophile. The electron-rich nucleophile approaches the electron-deficient side of the carbonyl. 13 November, 2018 P.J.McCormack

The electronegative oxygen polarises the double bond, leaving the carbon electron-deficient. It is therefore susceptible to attack by a nucleophile. The electron-rich nucleophile approaches the electron-deficient side of the carbonyl 13 November, 2018 P.J.McCormack

The electronegative oxygen polarises the double bond, leaving the carbon electron-deficient. It is therefore susceptible to attack by a nucleophile. The electron-rich nucleophile approaches the electron-deficient side of the carbonyl 13 November, 2018 P.J.McCormack

The electronegative oxygen polarises the double bond, leaving the carbon electron-deficient. It is therefore susceptible to attack by a nucleophile. The carbonyls bonds change angle 13 November, 2018 P.J.McCormack

The electronegative oxygen polarises the double bond, leaving the carbon electron-deficient. It is therefore susceptible to attack by a nucleophile. The electron-rich nucleophile is losing electron density The electron-deficient carbon is gaining electron density. 13 November, 2018 P.J.McCormack

The electron-rich nucleophile is losing electron density The electron-deficient carbon is gaining electron density. 13 November, 2018 P.J.McCormack

The electron-rich nucleophile is losing electron density The electron-deficient carbon is gaining electron density. 13 November, 2018 P.J.McCormack

The electron-rich nucleophile is losing electron density The electron-deficient carbon is gaining electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon is gaining electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon has gained electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon has gained electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon has gained electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon has gained electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

The electron-rich nucleophile has lost electron density The electron-deficient carbon has gained electron density. The oxygen has pulled a the p bond back as a lone pair 13 November, 2018 P.J.McCormack

A proton is then taken up by the lone pair on the oxygen 13 November, 2018 P.J.McCormack

A proton is then taken up by the lone pair on the oxygen 13 November, 2018 P.J.McCormack

A proton is then taken up by the lone pair on the oxygen 13 November, 2018 P.J.McCormack

A proton is then taken up by the lone pair on the oxygen 13 November, 2018 P.J.McCormack

A proton is then taken up by the lone pair on the oxygen 13 November, 2018 P.J.McCormack

ALDEHYDES & KETONES CONTENTS Prior knowledge Bonding in carbonyl compounds Structural differences Nomenclature Preparation Identification Oxidation Nucleophilic addition Reduction 2,4-dinitrophenylhydrazine

Before you start it would be helpful to… ALDEHYDES & KETONES Before you start it would be helpful to… know the functional groups found in organic chemistry know the arrangement of bonds around carbon atoms recall and explain the polarity of covalent bonds

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar PLANAR WITH BOND ANGLES OF 120°

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar the unhybridised 2p orbital of carbon is at 90° to these P ORBITAL PLANAR WITH BOND ANGLES OF 120°

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar the unhybridised 2p orbital of carbon is at 90° to these it overlaps with a 2p orbital of oxygen to form a pi (p) bond P ORBITAL PLANAR WITH BOND ANGLES OF 120°

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar the unhybridised 2p orbital of carbon is at 90° to these it overlaps with a 2p orbital of oxygen to form a pi (p) bond P ORBITAL ORBITAL OVERLAP PLANAR WITH BOND ANGLES OF 120°

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar the unhybridised 2p orbital of carbon is at 90° to these it overlaps with a 2p orbital of oxygen to form a pi (p) bond P ORBITAL ORBITAL OVERLAP PLANAR WITH BOND ANGLES OF 120° NEW ORBITAL

CARBONYL COMPOUNDS - BONDING PLANAR WITH BOND ANGLES OF 120° Bonding the carbon is sp2 hybridised and three sigma (s) bonds are planar the unhybridised 2p orbital of carbon is at 90° to these it overlaps with a 2p orbital of oxygen to form a pi (p) bond as oxygen is more electronegative than carbon the bond is polar P ORBITAL ORBITAL OVERLAP PLANAR WITH BOND ANGLES OF 120° NEW ORBITAL

CARBONYL COMPOUNDS - STRUCTURE Structure carbonyl groups consists of a carbon-oxygen double bond the bond is polar due to the difference in electronegativity Difference ALDEHYDES - at least one H attached to the carbonyl group C = O H C = O H CH3

CARBONYL COMPOUNDS - IDENTIFICATION Differentiation to distinguish aldehydes from ketones, use a mild oxidising agent Tollen’s Reagent ammoniacal silver nitrate mild oxidising agent which will oxidise aldehydes but not ketones contains the diammine silver(I) ion - [Ag(NH3)2 ]+ the silver(I) ion is reduced to silver Ag+(aq) + e¯ ——> Ag(s) the test is known as THE SILVER MIRROR TEST

CARBONYL COMPOUNDS - IDENTIFICATION Differentiation to distinguish aldehydes from ketones, use a mild oxidising agent Tollen’s Reagent ammoniacal silver nitrate mild oxidising agent which will oxidise aldehydes but not ketones contains the diammine silver(I) ion - [Ag(NH3)2 ]+ the silver(I) ion is reduced to silver Ag+(aq) + e¯ ——> Ag(s) the test is known as THE SILVER MIRROR TEST Fehling’s Solution contains a copper(II) complex ion giving a blue solution on warming, it will oxidise aliphatic (but not aromatic) aldehydes the copper(II) is reduced to copper(I) a red precipitate of copper(I) oxide, Cu2O, is formed The silver mirror test is the better alternative as it works with all aldehydes Ketones do not react with Tollen’s Reagent or Fehling’s Solution

CARBONYL COMPOUNDS - CHEMICAL PROPERTIES OXIDATION • provides a way of differentiating between aldehydes and ketones • mild oxidising agents are best • aldehydes are easier to oxidise • powerful oxidising agents oxidise ketones to a mixture of carboxylic acids ALDEHYDES easily oxidised to acids RCHO(l) + [O] ——> RCOOH(l) CH3CHO(l) + [O] ——> CH3COOH(l)

CARBONYL COMPOUNDS - CHEMICAL PROPERTIES OXIDATION • provides a way of differentiating between aldehydes and ketones • mild oxidising agents are best • aldehydes are easier to oxidise • powerful oxidising agents oxidise ketones to a mixture of carboxylic acids ALDEHYDES easily oxidised to acids RCHO(l) + [O] ——> RCOOH(l) CH3CHO(l) + [O] ——> CH3COOH(l) KETONES oxidised under vigorous conditions to acids with fewer carbons C2H5COCH2CH3(l) + 3 [O] ——> C2H5COOH(l) + CH3COOH(l)

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Mechanism occurs with both aldehydes and ketones involves addition to the C=O double bond unlike alkenes, they are attacked by nucleophiles attack is at the positive carbon centre due to the difference in electronegativities alkenes are non-polar and are attacked by electrophiles undergoing electrophilic addition Group Bond Polarity Attacking species Result ALKENES C=C NON-POLAR ELECTROPHILES ADDITION CARBONYLS C=O POLAR NUCLEOPHILES ADDITION

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Reagent hydrogen cyanide - HCN (in the presence of KCN) Conditions reflux in alkaline solution Nucleophile cyanide ion CN¯ Product(s) hydroxynitrile (cyanohydrin) Equation CH3CHO + HCN ——> CH3CH(OH)CN 2-hydroxypropanenitrile Notes HCN is a weak acid and has difficulty dissociating into ions HCN H+ + CN¯ the reaction is catalysed by alkali which helps produce more of the nucleophilic CN¯

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Mechanism Nucleophilic addition Step 1 CN¯ acts as a nucleophile and attacks the slightly positive C One of the C=O bonds breaks; a pair of electrons goes onto the O STEP 1

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Mechanism Nucleophilic addition Step 1 CN¯ acts as a nucleophile and attacks the slightly positive C One of the C=O bonds breaks; a pair of electrons goes onto the O Step 2 A pair of electrons is used to form a bond with H+ Overall, there has been addition of HCN STEP 1 STEP 2

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Mechanism Nucleophilic addition Step 1 CN¯ acts as a nucleophile and attacks the slightly positive C One of the C=O bonds breaks; a pair of electrons goes onto the O Step 2 A pair of electrons is used to form a bond with H+ Overall, there has been addition of HCN STEP 1 STEP 2

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Mechanism Nucleophilic addition Step 1 CN¯ acts as a nucleophile and attacks the slightly positive C One of the C=O bonds breaks; a pair of electrons goes onto the O Step 2 A pair of electrons is used to form a bond with H+ Overall, there has been addition of HCN STEP 1 STEP 2

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION ANIMATED MECHANISM

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Watch out for the possibility of optical isomerism in hydroxynitriles CN¯ attacks from above CN¯ attacks from below

CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION Watch out for the possibility of optical isomerism in hydroxynitriles CN¯ attacks from above CN¯ attacks from below

ANIMATED MECHANISM TO SHOW HOW DIFFERENT ISOMERS ARE FORMED CARBONYL COMPOUNDS - NUCLEOPHILIC ADDITION ANIMATED MECHANISM TO SHOW HOW DIFFERENT ISOMERS ARE FORMED

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion)

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion) Product(s) Alcohols Aldehydes are REDUCED to primary (1°) alcohols. Ketones are REDUCED to secondary (2°) alcohols. Equation(s) CH3CHO + 2[H] ——> CH3CH2OH CH3COCH3 + 2[H] ——> CH3CHOHCH3 Notes The water provides a proton

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion) Product(s) Alcohols Aldehydes are REDUCED to primary (1°) alcohols. Ketones are REDUCED to secondary (2°) alcohols. Equation(s) CH3CHO + 2[H] ——> CH3CH2OH CH3COCH3 + 2[H] ——> CH3CHOHCH3 Notes The water provides a proton Question NaBH4 doesn’t reduce C=C bonds. WHY? CH2 = CHCHO + 2[H] ———> CH2 = CHCH2OH

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion)

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion) Water is added

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion) Aldehyde Primary alcohol Water is added

CARBONYL COMPOUNDS - REDUCTION WITH NaBH4 Reagent sodium tetrahydridoborate(III) (sodium borohydride), NaBH4 Conditions aqueous or alcoholic solution Mechanism Nucleophilic addition (also reduction as it is addition of H¯) Nucleophile H¯ (hydride ion) ANIMATED MECHANISM

THE TETRAHYDRIDOBORATE(III) ION BH4 each atom needs one electron to complete its outer shell H atom has three electrons to share boron shares all 3 electrons to form 3 single covalent bonds H H B B H H H H H The hydride ion forms a dative covalent bond H

CARBONYL COMPOUNDS - REDUCTION WITH HYDROGEN Reagent hydrogen Conditions catalyst - nickel or platinum Reaction type Hydrogenation, reduction Product(s) Alcohols Aldehydes are REDUCED to primary (1°) alcohols. Ketones are REDUCED to secondary (2°) alcohols. Equation(s) CH3CHO + H2 ——> CH3CH2OH CH3COCH3 + H2 ——> CH3CHOHCH3 Note Hydrogen also reduces C=C bonds CH2 = CHCHO + 2H2 ——> CH3CH2CH2OH

CARBONYL COMPOUNDS - REDUCTION Introduction Functional groups containing multiple bonds can be reduced C=C is reduced to CH-CH C=O is reduced to CH-OH CN is reduced to CH-NH2 Hydrogen H• H2 H+ (electrophile) H¯ (nucleophile)

CARBONYL COMPOUNDS - REDUCTION Introduction Functional groups containing multiple bonds can be reduced C=C is reduced to CH-CH C=O is reduced to CH-OH CN is reduced to CH-NH2 Hydrogen H• H2 H+ (electrophile) H¯ (nucleophile) Reactions Hydrogen reduces C=C and C=O bonds CH2 = CHCHO + 4[H] ——> CH3CH2CH2OH Hydride ion H¯ reduces C=O bonds CH2 = CHCHO + 2[H] ——> CH2=CHCH2OH Explanation C=O is polar so is attacked by the nucleophilic H¯ C=C is non-polar so is not attacked by the nucleophilic H¯

CARBONYL COMPOUNDS - REDUCTION Example What are the products when Compound X is reduced? COMPOUND X H2 NaBH4

CARBONYL COMPOUNDS - REDUCTION Example What are the products when Compound X is reduced? COMPOUND X H2 NaBH4 C=O is polar so is attacked by the nucleophilic H¯ C=C is non-polar so is not attacked by the nucleophilic H¯

2,4-DINITROPHENYLHYDRAZINE C6H3(NO2)2NHNH2 Structure Use reacts with carbonyl compounds (aldehydes and ketones) used as a simple test for aldehydes and ketones makes orange crystalline derivatives - 2,4-dinitrophenylhydrazones derivatives have sharp, well-defined melting points also used to characterise (identify) carbonyl compounds. Identification / characterisation A simple way of characterising a compound (finding out what it is) is to measure the melting point of a solid or the boiling point of a liquid. C6H3(NO2)2NHNH2

2,4-DINITROPHENYLHYDRAZINE C6H3(NO2)2NHNH2 The following structural isomers have similar boiling points because of similar van der Waals forces and dipole-dipole interactions. They would be impossible to identify with any precision using boiling point determination. CHO Cl

2,4-DINITROPHENYLHYDRAZINE C6H3(NO2)2NHNH2 The following structural isomers have similar boiling points because of similar van der Waals forces and dipole-dipole interactions. They would be impossible to identify with any precision using boiling point determination. Boiling point 213°C 214°C 214°C CHO Cl

2,4-DINITROPHENYLHYDRAZINE C6H3(NO2)2NHNH2 The following structural isomers have similar boiling points because of similar van der Waals forces and dipole-dipole interactions. They would be impossible to identify with any precision using boiling point determination. Boiling point 213°C 214°C 214°C Melting point of 2,4-dnph derivative 209°C 248°C 265°C By forming the 2,4-dinitrophenylhydrazone derivative and taking its melting point, it will be easier to identify the unknown original carbonyl compound. CHO Cl

2,4-DINITROPHENYLHYDRAZINE C6H3(NO2)2NHNH2 The following structural isomers have similar boiling points because of similar van der Waals forces and dipole-dipole interactions. They would be impossible to identify with any precision using boiling point determination. Boiling point 213°C 214°C 214°C Melting point of 2,4-dnph derivative 209°C 248°C 265°C By forming the 2,4-dinitrophenylhydrazone derivative and taking its melting point, it will be easier to identify the unknown original carbonyl compound. CHO Cl