Both aldehydes & ketones contain carbonyl group as their functional group.
Structure of carbonyl group
Both aldehydes & ketones have carbonyl group as the functional group. The carbonyl carbon is sp2 hybridised & it uses sp2 hybrid orbitals to form 3σ bonds, one with oxygen atom & remaining 2 with two other atoms or groups (R or H). All these 3σ bonds lie in same plane at the angle of 120°.
|The unhybridized p – orbital of carbonyl carbon form π – bond with oxygen atom by sidewise overlapping with half filled
p – orbital of oxygen atom.
Since carbon & oxygen have different values of electronegativity, the bond between carbon & oxygen is polar. Infact electron density around the oxygen atom is increased which causes the development of partial positive charge (δ+) on carbon & partial negative charge (δ−) on oxygen.
Thus the carbonyl carbon is an electrophilic & carbonyl oxygen is nucleophilic centre.
|Orbital picture of carbonyl group|
Illustration 1. Give the IUPAC name for
Solution: (i) Ethanal
(ii) 2, 2 – diphenyl ethanal
Illustration 2. Write structure of the following compounds:
(i) 3-phenyl 2-propenal
(ii) 1, 5 – pentanedial
Write the formula of the following compound:
Calculate the number of sigma bonds in diphenyl ketone.
GENERAL METHODS OF PREPARATION OF ALIPHATIC ALDEHYDES & KETONES
- From Alcohols
(a) By Direct oxidation: Aldehydes are prepared by oxidation of 1° alcohols. Refer to Alcohols
(b) By catalytic dehydrogenation
When vapours of 1° or 2° alcohols are passed over copper gauze, they get dehydrogenated to form aldehydes or ketones.
Dehydrogenation reaction is a better method of preparation because there is no risk of further oxidation of aldehyde.
(c) By using PCC
PCC stands for pyridinium chlorochromate. It is an equimolar mixture of CrO3, HCl and pyridine. It is used to oxidise 1° alcohol to aldehyde and 2° alcohol to ketones without affecting double or triple bond.
Illustration 3. Complete the following reaction by writing down the major product:
Why is it important to distil out the aldehyde as soon as it is formed by the oxidation of primary alcohol?
- From Acid chlorides
Aldehydes are prepared from acid chlorides by reaction with H2 in the presence of palladium catalyst supported on BaSO4.
Ketones are obtained by reacting acid chlorides with dialkyl cadmium.
Illustration 4. Prepare ethanal by reduction method.
Write the structures of compound A and B in the following reaction:
- From fatty acids
(a) By heating calcium salt of fatty acid
Aldehydes are obtained by heating calcium salt of fatty acids with calcium formate.
Ketones are formed by distilling calcium salt of fatty acids alone.
Similarly mixed ketones, can also be obtained by similar reactions:
(b) By passing vapours of fatty acids over manganese oxide
In this method, formic acid alone gives formaldehyde. Acetic acid gives acetone & the mixture of two acids gives acetaldehyde.
- From Alkynes: (Refer to hydrocarbons)
- By reductive ozonolysis of alkenes: (Refer to hydrocarbons)
Illustration 5. What happens when (give equation only)?
(i) Ethyne is treated with dilute H2SO4 in the presence of HgSO4.
(ii) propan-2-ol is treated with Cu at 573 K.
Identify A, B and C.
Convert 2 – chloro but-2-ene to 2-butanone.
The compound (C) gives yellow precipitate with I2 and NaOH. What is the structure of compound A?
(i) What is the hybrid state of carbon 1 & 2 in ethanal?
(ii) What product is obtained in the following reaction?
PREPARATION OF AROMATIC ALDEHYDES & KETONES
- By oxidation of alkyl benzene
Aromatic aldehydes are obtained by oxidation of side chain in the aromatic ring.
Illustration 7. How will you prepare benzaldehyde commercially?
Solution: It is prepared form toluene.
- By Friedel crafts reaction (Refer to hydrocarbons)
Illustration 8. In the following electrophilic substitution reaction.
(i) Identify the compound A.
(ii) Write the structure of the electrophite.
Illustration 9. Complete the following
- By Gattermann Koch Reaction
In this method aromatic aldehydes are prepared by formylation of aromatic ring with carbon monoxide.
- From Grignard’s reaction
Both aliphatic & aromatic aldehyde can be obtained by this method.
HCN on treatment with Grignard’s reagent & subsequently followed by hydrolysis yield an aldehyde.
PHYSICAL PROPERTIES OF ALDEHYDES & KETONES
(i) Physical state
Most of aldehydes (except formaldehyde which is a gas) are liquids at room temperature. The lower ketones are colourless liquids and have pleasant smell.
(ii) Boiling points
Aldehydes & ketones have relatively high boiling points as compared to hydrocarbons of comparable molecular masses due to polar carbonyl group, which bring stronger intermolecular dipole – dipole interactions between the opposite ends of C = O dipoles.
Ketones are relatively more polar than their corresponding isomeric aldehydes due to the presence of two electron repelling alkyl group around the carbonyl carbon.
The lower members of aldehydes & ketones (upto four carbon atoms) are soluble in water. It is due to their capability of forming hydrogen bonds with water molecules. The solubility of these compounds in water decreases with the increase in the size of alkyl group because of the increase in magnitude of non polar part in the molecule.
Aldehydes & ketones are highly reactive compounds, they undergo nucleophilic addition reactions. Their reactivity is due to presence of a polar carbonyl group. The positively charged carbon atom of carbonyl group is readily attacked by nucleophilic species for initiation of the reaction. This leads to formation of intermediate anion which further undergoes the attack of H+ ion or other positively charged species to form the final product. The reaction in general may be represented as:
Relative reactivity of aldehydes & ketones
In general ketones are less reactive than aldehydes on a account of following facts:
(i) Electron releasing effect of two alkyl groups, decreases the magnitude of positive charge on ketones.
(ii) Steric effect caused by two alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon.
Illustration 10. Draw (i) resonance structures and (ii) an atomic orbital representation of the
Illustration 11. Ethanal is more soluble in water than ethyl chloride. Explain.
Solution: This is due to the ability of ethanal to form hydrogen bonds with water.
Predict the value of the angle (a) and (b) in the following figure.
Explain the following fact:
Dipole moment of ethanal = 2.72 D
Dipole moment of diethyl ether = 1.18 D
Give the increasing order of boiling point among following compounds.
I : CH3CH2CH2OH
II : CH3CH2OCH3
III : CH3CH2CHO
IV : CH3CH2CH3
TYPE OF CHEMICAL REACTIONS IN CARBONYL COMPOUNDS
(i) Addition across C = O bond.
(ii) Replacement of carbonyl oxygen by other groups.
(v) Reaction with alkalies
(vi) Miscellaneous reactions
- Addition across C = O bond
|Sr. No.||Addition of||Substrate||Product|
|3.||Grignard reagent (RMgX) followed by hydrolysis||HCHO|
|Aldehydes (except formaldehyde)||2° alcohol|
|4.||Alcohols (R⎯OH)||Hemiacetal which finally converts to acetal|
Illustration 12. Write the structure of compound A and B
Give the product of the reaction of acetaldehyde with sodium hydrogen sulphite.
The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is
(A) Benzaldehyde (B) p – Nitrobenzeldehyde
(C) Phenyl acetaldehyde (D) p – Hydroxybenzeldehyde
- Replacement of carbonyl oxygen atom with other groups
(a) Reaction with ammonia derivatives
Aldehydes & ketones react with a number of NH3 derivatives such as hydroxyl amine, hydrazine, semicarbazide etc, in weak acidic medium. In general, if we represent these derivatives by NH2 ⎯G, then their reaction with aldehydes & ketones can be represented as follows:
Ammonia derivatives & their products with carbonyl compounds
|⎯G||Ammonia Derivative||Product obtained|
|2, 4 – dinitrophenyl hydrazine||2, 4 – dinitrophenyl hydrazone|
(b) Reaction with ammonia
Like ammonia derivatives, ammonia also reacts with aldehyde (except formaldehyde) & ketones to form the products, called imines.
However, formaldehyde reacts with NH3 to form hexamethylene tetramine, (CH2)6N4 also known as urotropine as shown below:
Acetone reacts with NH3 to form diacetonamine
(c) Reaction with primary amines
Aldehydes & ketones react with 10 amines to form Schiff;s bases. These compounds are also called imines.
(d) Reaction with PCl5 or SOCl2 (thionyl chloride)
Aldehydes or ketones with PCl5 or thionyl chloride to form geminal dihalides.
Illustration 13. Write the structural formula of the following ammonia derivatives:
(i) 2, 4 – dinitrophenyl hydrazine
(iii) Hydroxyl amine
Which NH2 group of semicarbazide undergo condensation with the carbonyl group?
Complete the reaction:
Aldehydes are easily oxidised to carboxylic acids containing the same number of carbon atoms, as in parent aldehyde.
The reason for this easy oxidation is the presence of a hydrogen atom on the carbonyl carbon, which can be converted into ⎯OH group without involving the cleavage of any other bond. Hence, aldehydes are oxidised not only by strong oxidizing agent but also by weak oxidizing agents. As a result, aldehydes act as strong reducing agents.
- Aldehydes reduce Tollen’s reagent to Ag & appear in the form of silver mirror. This test is called silver mirror test. It is given by all aldehydes & reducing sugars.
- Aldehydes (except benzaldehyde) reduce Fehling’s solution (Cu+2 reduced to Cu+) which is an alkaline solution of cupric (Cu2+) ion complexed with tartarate ion.
- Aldehydes also reduce Benedict’s solution (Cu2+ complexed with citrate ion) to Cu+.
- Aldehydes & ketones with a methyl or methylene group adjacent to the carbonyl group are oxidised by SeO2
- Ketones are also oxidised by caro’s acid (H2SO5) or peroxybenzoic acid (C6H5CO3H) to esters.
It is called Bayer villiger oxidation.
It is exactly oxygen insertion between carbonyl carbon & the larger of two groups attached to it.
Due to the formation of yellow ppt. of iodoform in this reaction, it is known as iodoform test & used in for characterizing compound containing CH3CO⎯ or a group like CH3CH2OH which can be easily oxidised to CH3CO⎯ group by halogens.
Illustration 14. Give a chemical test to distinguish between each of the following pair of organic compounds.
(i) propanal an propanol
(ii) propanone and propanal
Solution: (i) Propanal is an aldehyde and gives a silver mirror with Tollen’s reagent while propanol is an alcohol which do not respond silver mirror test positively.
(ii) (a) Propanone give yellow ppt. of iodoform on reaction with I2/NaOH while propanal does not react.
(b) Propanal gives silver mirror with Tollen’s reagent while propanone does not.
Illustration 15. What is Fehling’s solution?
Solution: Fehling solution is a mixture of alkaline copper sulphate (Fehling A) and sodium potassium tartrate (Fehling B).
How does the oxidation state of copper changes when Fehling solution is added to acetaldehyde?
Which of the following compound gives yellow precipitate with iodine and sodium hydroxide?
(i) 3-methyl-4-phenyl but–3–en–2–one
(ii) 1- phenyl ethanone
Carbonyl compounds can be reduced to 1° or 2° alcohol, by LiAlH4, NaBH4 or direct reduction with H2/Ni.
with LiAlH4 ⎯CHO group is reduced to ⎯CH2OH (1° alcohol) and C = C bond is also reduced when it is in conjugation with carbonyl groups.
|Group is reduced to||(2° alcohol)|
- LiAlH4 also reduces ester & acid chloride to alcohols.
(b) NaBH4 has similar function. But this reagent does not affect (C = C) double bond.
NaBH4 does not reduce ester & acid chloride
(c) Amalgamated zinc, Zn(Hg) & conc. HCl (Clemmensen reduction) & hydrazine (NH2 –NH2) followed by reaction with strong base like KOH in alkaline glycol (Wolf Kishner reduction) reduces carbonyl group to alkyl group.
(d) Reduction to pinacol
Illustration 16. Find A and B:
Illustration 17. Explain Clemmensen’s reduction.
Solution: In Clemmensen’s reduction, we reduce the carbonyl group to CH2 using zinc amalgam in concentrated HCl.
Illustration 18. Name the major product of the following reactions:
Convert propanone into propane.
Write the structures of A and B.
can be brought about by
(A) LiAlH4 (B) NaBH4
(C) Clemmensen’s reduction (D) Wolff−Kishner reduction
The appropriate reagent for the transformation:
(A) Zn – Hg / HCl (B) (i) NH2 – NH2 (ii)
(C) Both (A) & (B) (D) None of these
Under Wolff Kishner reduction conditions, the conversion which may be brought about is
- Reaction with Alkalies
(A) Aldol Condensation
Two molecules of an aldehydes or a ketone having atleast one α – hydrogen atom, condense in presence of a dilute alkali to give a β – hydroxyaldehyde or β – hydroxy ketone.
The products of aldol condensation when heated with dilute acids undergo dehydration to form α, β – unsaturated aldehydes or ketones.
In general all aldehydes & ketones which contain α – hydrogen can undergo this reaction. Those which do not contain α – hydrogen like HCHO, C6H5CHO etc, do not undergo this reaction.
Mechanism involves formation of carbanions (i) a nucleophile form first molecules which is condensed with second molecule.
Aldol product on dehydration give α, β – unsaturated ketones.
Illustration 19. Convert ethanal into 2 – butenal.
(B) Cannizzaro’s reaction
Aldehydes that have no α-hydrogen atom (or acidic hydrogen) undergo cannizzaro reaction (CR) in which disproportionation reaction takes place one being reduced to alcohol & other being oxidised to salt of the corresponding carboxylic acid. The reaction lakes place with 50% aqueous or ethanolic alkali solution.
When an aldehyde (showing CR) is treated with HCHO & 50% base, then HCHO undergo oxidation (rather than any other aldehyde). This reaction is called crossed CR.
CR involving different aldehydes or same aldehydes is proton (H+), hydride (H−) transfer reaction.
When the reaction is carried out in D2O instead of in H2O, it is found that there is no new C – D bond formation. This indicate that the hydrogen must come from aldehyde & not from the solvent.
Illustration 20. Identify aldehydes which can give cannizaro reaction (CR):
(a) CCl3CHO (b) (CH3)2CHCHO
(c) (CH3)3CCHCl2 (d) C6H5CHO
Solution: Aldehydes which do not have H at α – C give CR.
(i) a & d do not H at α – C hence give CR.
(ii) b has H at α – C but due to steric hindrance it gives CR.
(iii) c with OH− reactant is first converted to aldehyde which does not have H at
α – C hence give CR.
Identify A & B in the following reactions:
Write the structure of A and B in the following reaction.
Find product of the reaction.
where X and Y are:
In the Cannizzaro reaction given below:
(A) The attack of at the carbonyl group
(B) The transfer of hydride to the carbonyl group.
(C) The abstraction of proton from carboxylic acid
(D) The deprotonation of Ph – COOH.
(C) Perkin reaction
In this reaction aromatic aldehyde is heated with an acid anhydride & its corresponding sodium salt to form condensation products which on hydrolysis gives α, β – unsaturated acids. Acetic anhydride & sodium acetate are commonly used in this reaction.
- Miscellaneous reactions
(i) Formation of phorone
Three moles of acetone condense in the presence of dry HCl to form phorone.
(ii) Formation of mesitylene
Three moles of acetone on refluxing with conc. Sulphuric acid produces mesitylene as one of the products.
(iii) Reaction with alc. KCN
On heating with ethanolic solution of KCN, two molecules of aromatic aldehyde undergo condensation to form benzoin. It is called benzoin condensation.
(iv) Reaction with chloroform
Ketones condenses with chloroform in presence of alkali to form chloretone.
Illustration 21. Convert acetone into mesityl oxide.
What is trioxane? Give its structure.
ANSWER TO EXERCISES
The structure of diphenyl ketone is
Number of sigma bonds = 25
We remove aldehyde as soon as it is formed thus preventing its further oxidation to carboxylic acid.
A → HOCH2 ⎯ CH2 ⎯ CH2 ⎯CHO
The hydrocarbon (A) is 2, 3 – dimethylbut-2-ene.
|(i)||sp2 & sp3 respectively.|
The angle (a) is 120° while angle (b) is 116.5°, slightly less than 120°.
The large value of the dipole moment of ethanal is due to larger contribution of the dipolar ion structure (II).
The correct order of increasing boiling point is IV < II < III < I
The NH2 group labelled (b) undergo condensation with carbonyl group. The NH2 group (a) is involved in resonance with CO group and is deactivated.
The oxidation state of copper change from +2 to +1.
(i) 3-methyl-4-phenyl but – 3 – en – 2 – one
(ii) 1- phenyl ethanone
Cyclohexanone undergoes aldol condensation in presence of dil. NaOH.
Trioxane is a cyclic trimer of methanal. Its structure is
Exercise 1: Give the industrial preparation of ethanal.
Exercise 2: Write an equation for making aldehydes by the oxo process.
Exercise 3: What happens when ethanal reacts with excess of methanol in the presence of trace amount of dry HCl?
Exercise 4: Convert acetic acid into acetaoxime.
Exercise 5: Give some physical properties of acetone.
Exercise 6: Which aldehyde smells like bitter almonds? Give its one use.
Exercise 7: Give one difference between the paraldehyde and metaldehydes.
Exercise 8: What type of aldehyde undergo cannizzaro reaction?
Exercise 9: What happens when ethanal is distilled with dilute H2SO4 at 273 K?
Exercise 10: What is formalin? Give its one use.
ANSWER TO MISCELLANEOUS EXERCISES
Exercise 1: Wacker’s process
Exercise 2: Alkene on reaction with (CO + H2) in the presence of dicobalt octacarbonyl as a catalyst giving an aldehyde with one carbon more.
Exercise 5: (i) It is a colourless, pleasant-smelling liquid.
(ii) It is miscible in water.
Exercise 6: Benzaldehyde smells like bitter almonds. It is used in perfumes.
Exercise 7: Paraldehyde is a cyclic trimer of CH3CHO and is a pleasant smelling liquid.
Metaldehyde is a cyclic tetramer of CH3CHO and is a white solid.
Exercise 8: Aldehyde without α-hydrogen atom undergo cannizzaro reaction.
Exercise 9: Ethanal form metaldehyde (cyclic tetramer of ethanal).
Exercise 10: 40% aqueous solution of formaldehyde is called formalin. It is used as a disinfectant.
Board Type Questions
Prob 1. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reaction:
Benzaldehyde, p – tolualdehyde, p – nitrobenzaldehyde, acetophenone.
Sol. Acetophenone is ketone, all other are aldehydes, ∴it is less reactive. +M group increases electron density and so makes the compound unfit for nucleophilic addition & vice versa is true for ⎯M group.
Acetophenone < p – tolualdehyde < benzaldehyde < p – nitrobenzaldehyde
Prob 2. Identify A, B & C in the following
Prob 3. Complete the following:
Prob 4. What is the final product of?
Prob 5. Arrange the following in increasing extent of hydration:
Sol. The extent of hydration increases as groups tendency to form H – bonding increases. In general hydration of an aldehyde is greater than that of a ketone.
Prob 6. Identify A, B & C in the following
Sol. With Pt, only double bond is affected
With Raney Ni, groups as well as double bond is reduced.
NaBH4, CH3OH does not reduce double bond.
Prob 7. Compound A, having the empirical formula C7H8 is chlorinated in sunlight to give a product which is hydrolysed to produce B. B after oxidation reacts with acetic anhydride in the Perkin reaction to produce an acid C, which has an equivalent weight of 148. Give the name & structure of A, B & C.
Sol. Since B on oxidation undergoes Perkin reaction, ∴ the oxidised compound is benzaldehyde & A is toluene.
Prob 8. Which of the following ketone is more acidic? Give a reason.
This ketone is more acidic because the resulting enolate ion obeys Huckel’s rule & is thus more stable.
Prob 9. Explain why acetophenone but not benzophenone forms an adduct with NaHSO3.
Sol. Because of steric hinderance of two phenyl groups.
Prob 10. Which of the following compound gives positive iodoform test?
(a) 2 – butanol (b) 1 – pentanol
(c) Acetone (d) 3 – pentanone
Sol. Compounds containing & will give a positive iodoform test.
∴ 2 butanol & acetone will give positive iodoform test.
Prob 11. Suggest a chemical test that will distinguished the following pairs of compounds
(a) Acetic acid & propanal
(b) Acetaldehyde & acetone
Sol. (a) Propanal gives Tollens test.
(b) Acetaldehyde gives Tollen’s test.
Prob 12. Complete the following reaction by giving the principal organic product.
Prob 13. Starting with cyclohexanone suggest reasonable synthesis for the following:
Prob 14. With in each pair which compound will react faster in carbonyl addition reactions:
Sol. (a) Cyclopropanone
Prob 15. Complete the following reaction:
Prob 1. The correct order of reactivity towards nucleophilic addition reaction will be
Sol. Aldehydes are more reactive than ketones due to lesser number of +I group & again the reactivity of carbonyl group depends on the major nature of alkyl group attached to it.
So correct order is
Prob 2. Which of the following will not give aldol condensation?
(A) RCH2CHO (B) RCOCH2R
(C) C6H5CHO (D) C6H5CH2CHO
Sol. Aldol condensation is given by those carbonyl compounds, which possess at least one α – H.
(C) is not having α hydrogen hence it will not give aldol condensation
Prob 3. Aldehydes & ketones can be reduced to corresponding hydrocarbons by
(A) refluxing with strong acids
(B) passing the vapours under heated PbO2
(C) Refluxing with zinc amalgam
(D) Refluxing with strong alkali
Prob 4. Which of the following compound gives a positive iodoform test?
(A) pentanal (B) 1 – phenyl ethanol
(C) 2 – phenyl ethanol (D) 3 – pentanol
Sol. Haloform test is used to identify ⎯COCH3 group or CH3 ⎯CH(OH)⎯ group which can be oxidised to. ∴ 1 – phenyl ethanol will give this test.
Prob 5. Which of the following compound will give Cannizzaro’s reaction?
(A) CH3CH2CH2CHO (B)
(C) (CH3)2CHCH2CHO (D) (CH3)3C. CHO
Sol. The compounds which do not contain α – hydrogen atom undergo cannizzaro’s reaction.
Prob 6. Compound A(C6H10O) form a phenyl hydrazone and gives a negative Tollen’s reagent test & iodoform test. On reduction with Zn/Hg, HCl compound A gives
n – pentane. A is
(A) 1° alcohol (B) aldehyde
(C) 2° alcohol (D) ketone
Sol. As A forms phenyl hydrazone it must be a carbonyl compound again, it is giving negative test with Tollen’s reagent test, it is a ketone.
Prob 7. The end product of the reaction is
Prob 8. Arrange the following compounds in order of their decreasing activity towards Tollen’s reagent:
HCHO CH3CHO CH3COCH3 C6H5CHO
I II III IV
(A) I > II > IV > III (B) I > II > III > IV
(C) II > I > III > IV (D) none
Sol. Aldehydes are more reactive than ketones
Prob 9. The most reactive compound towards formation of cyanohydrin on treatment with KCN followed by acidification is
(A) benzaldehyde (B) p – nitrobenzaldehyde
(C) phenylacetaldehyde (D) p – hydroxybenzaldehyde
Sol. Presence of electron withdrawing group in para position increases nucleophilic addition.
Prob 10. Which of the following chemical system has most acidic hydrogen?
(A) 3 – hexanone (B) 2, 4 – hexanedione
(C) 2, 5 – hexanedione (D) 2, 3 – hexanedione
Sol. 2, 4 – hexanedione has ⎯CH2⎯ group surrounded on both sides of electronegative (>C=O) group, hence it is acidic in nature, the carbanions
, once formed stabilizes due to resonance.
Prob 11. Predict the product ‘B’ in the sequence of reaction:
Prob 12. Acetophenone can be prepared by
(I) Oxidation of 1 – phenylethanol
(II) Reduction of benzaldehyde with methyl magnesium bromide
(III) Friedel – crafts reaction of benzene with acetylchloride
(IV) Distillation of calcium benzoate
Which of above is correct?
(A) I & III (B) II & IV
(C) III & IV (D) I & II
Prob 13. An organic compound ‘A’ has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives ‘B’ of molecular formula C9H14O, ‘A’ & ‘B’ respectively are
(A) propanal & mesitylene
(B) propanone & mesityl oxide
(C) propanone & 2, 6 – dimethyl – 2, 5 – heptadien – 4 – one
(D) propane & meistylene oxide
Sol. Since the compound A has the molecular formula C3H6O & undergoes iodoform test, it should be CH3COCH3. Further reaction of B with HCl gives C9H14O, this should be trimer of acetone, i.e. It should be phorone.
Prob 14. What will be the product when 2-Butenal reacts with sodiumborohydride?
(A) Butanal (B) Butanol
(C) But-2-ene-1-ol (D) None
Prob 15. Which will give a yellow precipitate with iodine & alkali?
(A) 2 – hydroxyl propane (B) benzophenone
(C) o – methyl toluene (D) acetamide
Level – O
- Suggest a reason for large difference in the boiling points of butanol and butanal, although they have same solubility in water.
- Arrange the following in order of their increasing reactivity towards HCN?
- To what oxidation state does ethanal reduce Cu (II).
- Write the I.U.P.A.C. name of
- Identify A, B and C:
- Complete the following reaction:
- Find the product:
- Give the equation associated with Fehling’s Test given by CH3CHO.
- How will you convert acetone into 2 – methyl – 2 – propanol?
- Mention an industrial product manufactured from methanal.
- Account for the following:
(i) In the reactions of aldehyde/ketones with ammonia derivative, the pH has to be carefully controlled.
(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) Cyclohexanone forms cyanohydrin in good yiled but 2, 2, 6 – trimethyl cyclohexanone does not.
- Effect the following conversions:
(a) Acetone to propyne
(b) Acetaldehyde to acetaldehyde semicarbazone.
- Give one chemical test to distinguish following pair of compounds in solution:
Acetone and acetaldehydes.
- Give one chemical test to distinguish :
Acetaldehyde and benzaldehyde.
- Write the equations and conditions to show how the following conversions are carried out:
(a) Benzaldehyde to acetophenone
(b) Acetaldehyde to 2 – Butenal
Level – I
- Arrange the following in order of decreasing acidity:
- The following ketone is unusually basic. Explain.
- Name the alkyne which can give the following carbonyl compounds on acid catalysed hydration.
- Which alkene would give the following products on ozonolysis?
- Complete the following reaction
- Explain why only α-hydrogen atom in aldehydes & ketones is involved in the aldol condensation.
- Which of the following compound gives a positive iodoform test?
2 pentanone & 3 – hexanol
- Arrange the following compounds in order of increasing reactivity towards the addition of HCN.
Acetone, acetaldehyde, methyl t – butyl ketone & di – t – butyl ketone.
- Show the product of following reaction
- Identify products when PhCOCHCl2 undergoes Cannizzaro reaction:
- Give the aldol condensation product of:
(a) CH3CH = CHCHO
- What happens when cyclopentanone is treated with ethyl magnesium bromide & the product is hydrolysed?
- What starting compound would you use in an aldol cyclisation to prepare each of?
- What reagents would you use to carry out following reactions?
Ethyl bromide ⎯⎯→ 1-butyne ⎯⎯→ 2 – butanone
- Arrange the following in increasing extent of hydration:
Level – II
- Identify the product of
- Complete the following reaction:
- Write the final product of
- Complete the following reaction:
- Identify A, B & C in the following
- Complete the following reaction:
- Show the mechanism of following reaction:
- Supply the structure from A to E:
- Effect the following conversion:
- How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene
(ii) Propanal to butanone
(iii) Benzaldehyde to benzophenone
(iv) Ethanol to 3-hydroxybutanal
(v) Benzaldehyde to 3 – phenyl propan – 1- ol
(vi) Benzene to m-Nitroacetophenone
- Illustrate the mechanism of reaction of carbonyl group with an ammonia derivative, H2N⎯Z.
- Give simple chemical test to distinguish between the following pairs of compounds:
(i) Propanal and propanone (ii) Ethanal and propanal
(iii) Propanal and Diethyl ether (iv) Propanone and propanol
- Explain haloform reaction.
- An organic compound with the molecular formula C9H10O form 2, 4-DNP derivative, reduces Tollen’s reagent and undergo Cannizzaro reaction. On vigorous oxidation, it gives benzene-1, 2- dicarboxylic acid. Identify the compound and write the reactions involved.
- Compound A on treatment with gives compound B which on reduction with in presence of gives compound C. C gives Tollen’s test, Fehling’s test and iodoform test. When C is treated with dil. NaOH, D is obtained, which on heating gives crotonaldehyde. Identify A, B, C, D and complete sequence of reaction.
- Polarization of electrons in acrolein may be written as:
- Which reagent will perform the following reduction?
(A) LiAlH4 (B) NaBH4
(C) H2/Ni (D) Mg/Hg + H2O
- The ketone that does not form a cyanohydrin is
(A) C6H5COCH3 (B) CH3COCH3
(C) C6H5COC6H5 (D) CH3COC6H5
- The decreasing order of solubility of methanal (I), propanaldehye (II), benzaldehyde (III) & acetophenone (IV) is
(A) I, II, III, IV (B) IV, III, II, I
(C) IV, I, II, III (D) II, I, III, IV
- Which of the following compounds will undergo self aldol condensation in presence of dil. alkali?
(A) C6H5CHO (B) CH ≡ C ⎯ CHO
(C) CH3CH2CHO (D) CH2 = CHCHO
- Which of the following does not undergo benzoin condensation?
- In the reaction
What is X?
- Identify the final product Z in the following sequence of reaction:
(A) (CH3)2C(OH)COOH (B) CH2 = C(CH3)COOH
(C) HO ⎯ CH2CH(CH3)COOH (D) CH3CH = CHCOOH
- A dihalo alkane on hydrolysis produces a ketone with formula C3H6O. The dihalo alkane is
(A) 2, 2 – dichloropropane (B) 1, 1 – dichloropropane
(C) 1, 2 – dichloropropane (D) 1, 3 – dichloropropane
- Identify ‘X’ in the sequence
|(A)||CH3 ⎯ CH2 ⎯ CH2OH||(B)|
|(C)||CH3 ⎯ O ⎯ CH2CH3||(D)||CH3 ⎯ CH2CHO|
|(C)||Propanaldehyde dimethyl acetal||(D)||All the above|
- Aldol condensation will not be observed in
(A) Chloral (B) Phenyl acetaldehyde
(C) Hexanal (D) None of these
- Greatest amount of hydration is in
|(C)||(D)||All are equal|
- Identify Z in the series
(A) C2H5I (B) C2H5OH
(C) CHI3 (D) CH3CHO
- The product (s) obtained via Oxymercuration (HgSO4 +H2SO4) of butyne – 1 would be
- Which one of the following reactions can not be used for the reduction of
(A) Clemmensen reduction (B) Wolf – Kishner reduction
(C) Wurtz reaction (D) HI and red phosphorous at 2000C.
- An organic compound C3H6O does not give a precipitate with 2, 4 – dinitro phenyl hydrazine and does not react with metallic sodium. It could be
(A) CH3CH2CHO (B) CH3COCH3
(C) CH2 = CH – CH2OH (D) CH2 = CH – OCH3
- A ketone reacted with Grignard reagent followed by hydrolysis gave a product which on dehydration gave an alkene. The alkene on ozonolysis gave diethyl ketone and acetaldehyde. The ketone is
(A) Dimethyl ketone (B) Ethyl methyl ketone
(C) Diethyl ketone (D) Ethyl isopropyl ketone
- Iodoform test is applicable for
|(C)||(D)||All the above|
Level – II
- Acetaldehyde reacts with
(A) electrophiles only (B) nucleophiles only
(C) free radicals only (D) both electrophiles & nucleophiles
- Aldehydes & ketones give addition (not condensation) reactions with
(A) hydrazine (B) phenyl hydrazine
(C) semicarbazine (D) hydrogen cyanide
In the above sequential reactions A & B are
(A) (CH3)2C(OH)CN, (CH3)2C(OH)COOH (B) (CH3)2C(OH)CN, (CH3)2C(OH)2
(C) (CH3)2C(OH)CN, (CH3)2CHCOOH (D) (CH3)2C(OH)CN, (CH3)2C=O
- In which of the following reactions, aldehydes & ketones are distinguished
(A) reactions with phenyl hydrazine
(B) reactions with hydroxylamine
(C) reactions with semicarbazide
(D) reactions with silver nitrate mixed with ammonia
- The most probable structural formula for the compound whose empirical formula is C3H6O & which can react with Benedict reagent is
|(C)||CH3OCH = CH2||(D)||CH2 = CHCH2OH|
- Base catalysed aldol condensation occurs with:
(A) propionaldehyde (B) 2 – methyl propionaldehyde
(C) both (D) none
- Aromatic aldehydeso undergoes disproportionation in presence of sodium or potassium hydroxide to give corresponding alcohols & acid. The reaction is known as
(A) Wurtz reaction (B) cannizzaro reaction
(C) friedel crafts reactions (D) claisen reaction
- Compound ‘A’ (molecular formula C3H8O) is treated with acidified K2Cr2O7 to form a product (B) (molecular formula C3H6O). ‘B’ forms a shining silver mirror on warming with ammoniacal AgNO3. ‘B’ when treated with an aq. Solution to H2NCONHNH2, HCl & Sodium acetate gives a product ‘C’. Identify the product ‘C’.
|(A)||CH3CH2CH = NNHCONH2||(B)|
|(C)||(D)||CH3CH2CH = NCONHNH2|
- In a cannizzaro reaction, the intermediate that will be best hydride donar is
- Mixture of CH3CH2OH & CH3CHO can be separated by using
(A) NaHSO3 (B) NH2OH
(C) HCN (D) NH2 ⎯ NH2
|13.||X and Y are|
|(C)||(D)||None is correct|
- Following is hydrated maximum at the position:
(A) 1 (B) 2
(C) 3 (D) equal
|(C)||(D)||None is correct|
In above conversion E can be
(A) Propanol (B) Butanal
(C) Pentanal (D) Both (B) and (C)
- Identify the final product (Z) in the following sequence of reactions.
- The conversion; CH3–CH=CH–CHO ⎯⎯→ CH3–CH=CHCOOH can be effected by
(A) alk. KMnO4 (B) K2Cr2O7/H+
(C) ammoniacal AgNO3 (D) SeO2
- The abstraction of proton will be fastest, in which carbon in the following compound,
(A) x (B) y
(C) z (D) p
- Which of the following will be most acidic?
ANSWERS TO ASSIGNMENT PROBLEMS
Level – O
- Butanol undergoes hydrogen bonding but butanal does not.
- +1 oxidation state.
- 4 – Hydroxypentan – 2 – one
|11.||(i) The reaction
|At lower pH (solution is highly acidic), the protonation of N takes place thereby decreasing the concentration of free base available.|
The H3N+⎯ Z can not attack now at the carbonyl carbon.
At higher pH, (solution is basic), the base will remove H+ ion from OH in the intermediate formed thereby slowing the dehydration step.
(ii) In semicarbazide, the ⎯NH2 group closer to the group is resonance stabilized and is deactivated (lone pair of N is involved in resonance) compared to the other end NH2 group).
(iii) 2, 2, 6 – trimethyl cyclohexanone does not form cyanohydrin as the carbonyl carbon is highly crowded and the electron density is very high due to the presence of three e− donating methyl groups.
- Acetaldehydes on warming with Fehling’s solution gives red precipitate of copper oxide, where as acetone does not.
- These two compounds can be distinguished by Fehling’s test. Acetaldehyde gives red coloured ppt. with Fehling’s solution but benzaldehyde does not.
Level – I
- III > II > I
- The salt of ketone is resonance stabilized.
- Because an α-hydrogen atom is acidic & the resulting enolate anion is stabilized by resonance.
- 2 – pentanone
- di – t – butyl ketone < methyl – t – butyl ketone < acetone < acetaldehyde.
Level – II
- The reaction of carbonyl group with H2N⎯Z is an acid catalysed reaction.
|12. (i)||Test||Propanal (CH3CH2CHO)||CH3COCH3 (Propanaone)|
|(a) Tollen’s reagent||Silver mirror||−ve test|
|(b) Iodoform test (I2/NaOH)||−ve||Yellow ppt.|
|Iodoform test (I2/NaOH)||Yellow ppt.||−ve test|
|(a) Tollen’s test||Silver mirror||−ve test|
|(b) Fehling’s test||Reddish brown ppt.||−ve|
|Iodoform test||Yellow ppt.||−ve test|
- Methyl ketones on oxidation with X2/NaOH (X = Cl, Br, I) gives sodium salt of carboxylic acid with one carbon atom less. The CH3− group bonded to carbonyl group is converted to haloform, CHX3. With iodine in NaOH, a yellow precipitate of Iodoform CHI3, is obtained. (Iodoform test).
- The compound is a carbonyl compound (+ve, 2, 4 – DNP test)
The carbonyl group is an aldehyde (+) ve Tollen’s reagent.
The aldehyde group is directly bonded to benzene, no α – H (Cannizzaro reaction).
The other side chain is at the ortho position as it gives phthalic acid on oxidation. Hence the compound is
The reactions are
Level – I
- D 2. B 3. C
- A 5. C 6. D
- A 8. B 9. A
- C 11. B 12. D
- A 14. C 15. C
- A 17. C 18. D
- C 20. D
Level – II
- B 2. D 3. A
- D 5. B 6. C
- B 8. B 9. A
- D 11. A 12. B
- A 14. B 15. A
- D 17. B 18. C
- A 20. A