Carboxylic Acid MCQs With Solutions By Arvind Arora€¦ · Subscribe Madeejee Chemistry No. 1 NEET JEE CHANNEL ON YOUTUBE 2 Carboxylic Acid MCQs With Solutions By Arvind Arora

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1 Carboxylic Acid MCQs With Solutions By Arvind Arora

Q.1.

O

O ⎯⎯⎯⎯ →⎯

OHCH3 A A is :-

(A) CH2—CH2

OH COOCH3

(B) CH2—CH2

OCH3COOH

(C)

both are correct

(D)

None is correct

Q.2. End product of the following sequence of reaction is

O

C—CH3

O

++⎯⎯⎯⎯⎯⎯ →⎯

,H .2

,NaOHI .1 2

(A) Yellow ppt. of CHI3,

O

COOH

(B) Yellow ppt. of CHI3,

O

CHO

(C) Yellow ppt. of CHI3,

O

(D) Yellow ppt. of CHI3, COOHCOOH

Q.3. End product of this conversion HCOCHCHCHCCH 22223

||O

−−−−−+

⎯⎯⎯⎯ →⎯H,OH.2

NaBH.1

2

4 is

(A)

(a) =O

CH3

O

(B)



(b) =O

CH3

O (C)

(c) =OO

(D)

CH3—CH—CH2CH2CH2CO2H

OH

Q.4. When acetic acid reacts with ketene, product formed

(A) ethyl acetate

(B) aceto - acetic ester

(C) acetic anhydride

(D) no reaction

Q.5. R—CH2—CH2OH can be converted in R—CH2CH2COOH. The correct sequence of reagents is

(A) PBr3, KCN, H+

(B) PBr3, KCN, H2

(C) KCN, H+

(D) HCN, PBr3, H+

Q.6. The pKa of acetylsalicylic acid (aspirin) is 3.5 . The pH of gastric juice in human stomach is about 2-3 and

pH in the small intestine is about 8. Aspirin will be.

(A) Unionized in the small intestine and in the stomach

(B) Completely ionized in the stomach and almost unionized in the small intestine.

(C) Ionized in the stomach and almost unionized in the small intestine

(D) Ionised in the small intestine and almost unionised in the stomach

Q.7. On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which

acid is obtained?

(A) (CH3)2C=CH—CH2COOH

(B) (CH3)2CH—COOH

(C) (CH3)2C=CH—COOH

(D) (CH3)2C=CH—CO—COOH

Q.8. The ease of alkaline hydrolysis is more for

(A)

COOCH3

NO2

(B)

COOCH3

Cl

(C)

COOCH3



(D)

COOCH3

OCH3

Q.9. Which of the following does not undergo Hell - Volhard Zelinsky reaction?

(A) HCOOH

(B) CCl3COOH

(C) C6H5COOH

(D) All

Q.10. Consider the following sequence of reactions :

X

(C11H14O2)

OH–

Y + CH3CH2COO–

1. KMnO4/OH–

2. H3O+

COOH

COOH The compound ‘X’ is -

(A)

O

O

(B)

O

O

(C)

O

O

(D)

O

O

Q.11.

SO3H

OH

⎯⎯⎯⎯ →⎯COONaCH3 products, the

products are -

(A)

SO3H

ONa

+ CH3COOH



(B)

SO3Na

ONa

+ CH3COOH

(C)

SO3Na

OH

+ CH3COOH

(D) No reaction

Q.12. CH2 – COOK |

CH2 – COOK | CH2

Electrolysis P (Predominant)

P can be ?

(A) CH3 – CH2 – CH3

(B)

(C)

(D) none of these

Q.13. An ester (A) with mol. formula C9H10O2 was treated with excess of CH3–Mg Br and

the complex. So formed was treated with H2SO4 to give on olefin (B). Ozonolysis of (B) gave

a ketone with molecular formula C8H8O which shows the iodoform test the structure of (a) is

-

(A) C6H5 COOC2H5

(B) CH3COCH2COC6H5

(C) P CH3O–C6H4–COCH3

(D) C6H5COOC6H5

Q.14. In the following sequence of reaction, what is D ?

CH3

[O] A

SOCl2 B

NaN3 C

Heat D

(A) Primary amine

(B) An amide

(C) phenyl isocynate

(D) A chain lengthened hydrocarbon

Q.15. Arrange the following in order of increasing reactivity towards nucleophilic substitution

reaction -

CH3–C–OCH2CH3

O

(1)

CH3–C–Cl

O

(2)

CH3–C–NH–CH3

O

(3) (A) 1 < 2 < 3

(B) 3 < 1 < 2

(C) 1 < 3 < 2

(D) 2 < 1 < 3



Q.16.

O

OMe +

OH Con.HCl

O

O

+

MeOH

Above reaction is an example of -

(A) Esterification

(B) Saponification

(C) Hydrolysis

(D) Trans Esterification

Q.17. Acrylic acid react with HBr to give -

(A) Br–CH2–CH–COOH

Br (B) Br–CH2–CH2–COOH

(C) CH2=CH–COBr

(D) CH3–CH–COOH

Br Q.18. An optically active compound 'X' has molecular formula C4H8O3. It evolves CO2 with

NaHCO3. 'X' reacts with LiAlH4 to give an Achiral compound 'X' is -

(A) CH3–CH2–CH–COOH

OH (B) CH3–CH–COOH

Me (C) CH3–CH–COOH

CH2OH

(D) CH3–CH–CH2–COOH

OH Q.19. What is compound Z ?

CH3–CH2–CH2–Br NaCN

X H3O

Y

CH3–CH2–OH H

Z

(A)

CH3–CH=CHCOH

O

(B) CH3–CH2–CH2CH=NOCH2CH3

(C) CH3–CH2–CH2CH(OCH2CH3)2

(D) CH3–CH2–CH2–

C–OCH2CH3

O

Q.20. End product of the following sequence of reaction is -

CHCH

CH3–Mg–Br CHC–Mg–Br (i) CO2(ii) H3O

HgSO4–H2SO4

Ag2O

(A)

CH3–C–COOH

O



(B)

CH2

COOH

COOH

(C)

CH3–C–CHO

O

(D)

H–C–CH2–COOH

O

Q.21. On vigorous oxidation by permanganate solution (CH3)2C = CH–CH2–CH3 gives

(A)

CH3–C––CH–CH2–Cl3

OH

CH3

OH

(B)

CH–CO2H+CH3–CH2–COOH CH3

CH3

(C)

CHOH+CH3–CH2–CH2–OH CH3

CH3

(D)

C=O+CH3–CH2–COOH CH3

CH3 Q.22. Formic acid is obtained when -

(A) Calcium acetate is heated with conc. H2SO4

(B) Calcium formate is heated with calcium

acetate

(C) Glycerol is heated with oxalic acid at 373 K

(D) Acetaldehyde is oxidised with K2Cr2O7 and

H2SO4

Q.23. Which of the following carboxylic acid undergoes decarboxylation easily ?

(A) C6H5–CO–CH2–COOH

(B) C6H5–CO–COOH

(C) C6H5– CH

OH

–COOH

(D) C6H5– CH–COOH

NH2 Q.24. The strongest acid among the following is -

(A) CF3COOH

(B) CCl3COOH

(C) CBr3COOH

(D) CH3COOH

Q.25. Which of the following does not give benzoic acid salt on oxidation with hot alkaline

KMnO4.

(A) Ph–CH3

(B) Ph–CH=CH–CH3

(C) Ph–CC–CH3

(D) Ph–C(CH3)3

Q.26. The acid D obtained through the following sequence of reactions is

C2H5Br ⎯⎯⎯ →⎯ KOH.AlcA

4

2

CCI

Br⎯→⎯ B)excess(

KCN⎯⎯ →⎯ C ⎯⎯ →⎯+OH2 D



(A) Succinic acid

(B) Malonic acid

(C) Maleic acid

(D) Oxalic acid

Q.27. Formic acid can be distinguish from acetic acid because formic acid :

(A) Release H2 with sodium

(B) Gives ester with alcohol

(C) Reduces AgNO3

(D) Turns red litmus to blue

Q.28. Which of the following methods is not used for the conversion of carboxylic acids into

acid halides ?

(A) RCOOH + SOCl2 ⎯→⎯

(B) RCOOH + PCl5 ⎯→⎯

(C) RCOOH + Cl2 ⎯→⎯

(D) RCOOH + PCl3 ⎯→⎯

Q.29. Which of the following will not undergo Hell-Volhard Zelinsky (HVZ) reaction ?

(A) HCOOH

(B) CH3 COOH

(C) CH3 CH2 COOH

(D) CH3 CHBrCOOH.

Q.30. CH3–CH2–CH2–COOH ⎯⎯⎯ →⎯ + 2BrPdRe

Br

COOHHCCHCH|

23 −−− This reaction is

called

(A) Cannizzaro reaction

(B) Schmit reaction

(C) Hell Volhard Zelinsky reaction

(D) Reimer tiemann reaction

Q.31. The decreasing order of reactivity towards nucleophilic acyl substitution is

(A) CH3COCl (B) CH3COOC2H5,

(C) CH3CONH2 (D) (CH3CO)2O

(A) (i) > (iv) > (iii) > (ii)

(B) (i) > (iv) > (ii) > (iii)

(C) (iv) > (iii) > (i) > (ii)

(D) (iii) > (i) > (iv) > (ii)

Q.32. Acetic anhydride is prepared in the laboratory by heating sodium acetate with

(A) Ethyl chloride

(B) Acetyl chloride

(C) Conc. H2SO4

(D) Zinc dust

Q.33. Synthesis of an ester involves the reaction of alcohols with

(A) A ketone

(B) An amide

(C) CH3MgBr



(D) RCOCl

Q.34. Benzoyl chloride on treatment with ammonia gives

(A) Benzamide

(B) Acetamide

(C) Benzylamine

(D) Benzoic acid

Q.35. The reaction, RCOOR + ROH (excess) ⎯⎯⎯ →⎯−+ OHorH

RCOOR + R OH is called.

(A) Esterification

(B) Trans-esterification

(C) Saponification

(D) Hydrolysis

Q.36. Ethanol on heating with acetic acid in the presence of a few drops of sulphuric acid

gives the smell of

(A) Oil of wintergreen

(B) Oil of mustard

(C) Fruity smell

(D) Oil of bitter almonds

Q.37. ⎯⎯ →⎯ 3NaN

X−

⎯⎯ →⎯

OH)2(

)1(Product.

Product is/are

(A) only

(B) , HCOO–

(C)

(D)

Q.38. Consider the following statements for hydrolysis reaction :

(I) is more reactive than C6H5COOC2H5

(II) , is more reactive than

(III) is more reactive than

Of these the correct statements are

(A) I and II

(B) I, II and III

(C) II and III

O

Cl

CONH2

– C – OH + NH4

O+



(D) I and III

Q.39. An organic compound X (C4H8O2) gives positive test with NaOH and phenopthalein.

Structure of X will be :

(A)

O

OHCCHCHCH| |

223 −−−−

(B)

O

CHC

O

CCH 3| || |

3 −−−

(C)

O

HCOCCH 52| |

3 −−−

(D)

O

OCHCCH 3| |

3 −−

Q.40. Which one of the following esters is the most reactive for saponification ?

(A)

(B) 3

3

|

| |

3 CHH

CH

CO

O

CCH −−−−

(C) 3

3

3|

|

| |

3 CH

CH

CH

CO

O

CCH −−−−

(D)

Q.41. A compound with molecular formula C4H10O4 on acylation with acetic anhydride

gives a compound with molecular formula C12H18O8. How many hydroxyl groups are

present in the compound ?

(A) One

(B) Two

(C) Three

(D) Four

Q.42. CH3CH2CH2CONH2 is boiled with aqueous NaOH, then the reaction mixture is

acidified with HCl. The products obtained are

(A) CH3CH2CH2COO– + NH3

(B) CH3CH2CH2COONa + NH3

(C) CH3CH2CH2COOH + NH4Cl

(D) CH3CH2CH2COO– + NH4Cl.

Q.43. When excess of chlorine is passed through acetic acid in presence of red phosphorus, it

forms

(A) Acetic anhydride

(B) Chloral

(C) Trichloroacetic acid



(D) Methyl chloride.

Q.44. SAcetamide and ethyl acetate can be distinguished by reacting with

(A) Aqueous HCl and heat

(B) Aqueous NaOH and heat

(C) Acidified KMnO4

(D) Bromine water.

Q.45. + 3 NaOH ⎯→⎯ (Q) + salt of fatty acid.

Product (Q) of the reaction is

(A) Ethylene glycol

(B) Glycerol

(C) Glyceryl tri nitrate (explosive)

(D) Cumene hydrogen peroxides

Q.46. Starting from propanoic acid, the following reactions were carried out

Propanoic acid ⎯⎯ →⎯ 2SOCI

X ⎯⎯ →⎯ 3NH

Y ⎯⎯⎯ →⎯ +KOHBr2 Z

What is the compound Z

(A) CH3–CH2–Br

(B) CH3–CH2–NH2

(C)

(D) CH3–CH2–CH2–NH2

Q.47. Identify the final product in the following sequence of reaction.

⎯⎯ →⎯ 2SICI ⎯⎯ →⎯ 3NH⎯⎯ →⎯ 52OP

(A)

(B)

(C)

(D)

Q.48. Which of the following acids remains unaffected on heating

(A) Malonic acid

(B) Malic acid

(C) Fumaric acid

(D) Succinic acid



Q.49. Identify the final product in the following sequence of reaction.

⎯⎯ →⎯ hv/Br2A

⎯⎯⎯ →⎯OH)2(

KCN)1(

3

(A)

(B)

(C)

(D)

Q.50. Esterification of the acid (P) with the alcohol (Q) will gives

+

(R– configuration) (±)

(A) Only one enantiomer

(B) A mixture of diastereomer

(C) A mixture of enantiomer

(D) Only one fraction on fractional distillation

Answers with Solution for these mcqs

Ans. 1. [A]



O

OThis bond breaks hence intermediate is

O–

O

CH3OH+

O

OCH3

OH Ans. 2. [C]

Intermediate is

O

COOH

Which loses CO2 on heating (-keto acid)

Ans. 3. [A]

NaBH4 reduces reactant to

CH3—CH—CH2—CH2—CH2—CO2H which forms ester

OH Ans. 4. [C]

)Ketene(23 OCCHCOOHCH ==+ ⎯⎯→ acetc anhydride

Ans. 5. [A]

RCH2CH2OH ⎯⎯⎯ →⎯ 3PBr RCH2CH2Br ⎯⎯⎯ →⎯

KCN RCH2CH2CN ⎯⎯⎯⎯ →⎯

+H/OH2 RCH2CH2COOH

Ans. 6. [D]

More ionized in basic medium and less ionized in acidic medium, common ion effect

Ans. 7. [C]

CH3

C=CH—C—CH3 ⎯⎯⎯⎯ →⎯ 2I/NaOHCH3

O

C=CH—C—OH + CHI3CH3

CH3

O

mesityl oxide

Ans. 8. [A]

COOCH3

NO2

Then is more electron deficiency oncarbonyl carbon.

Ans. 9. [D]

None of these contain alpha H atom

Ans. 10. [B]

O

O

OH–

(C11H14O2)

CH3

OH

CH3

KMnO4/OH–



COOH

COOH

Ans. 11. [C]

SO3H is most acidic

Ans. 12. [C] CH2

CH2

CH2

Ans. 13. [A]

C6H5 COOC2H5 (i) 2CH3–Mg–Br

(ii) H/H2O

C6H5 C OH

CH3

CH3

–H2O H2SO4

A MF. C9H10

HCHO + C6H5COCH3 O3/Zn

C6H5–C=CH2

CH3

MFC8H8O Ans. 14. [C] CH3

[O] SOCl2

–SO2–HCl

COOH

(A)

COCl

(B)

NaN3 –NaCl

CO–N–NN

. .

. .

Benzoylazide (C)

CO–N

. .

. . N=C=O

(Benzoyl nitrene) (Phenyl isocynate)

(D) Ans. 15. [B]

Leaving group tendency

(–Cl > – OEt > – NH–CH3)

Ans. 16. [D]

Trans esterification – reaction in which ester convert into another ester.

Ans. 17. [B]

1, 4 Addition occurs to , unsaturated carbonyl compound.

CH = CH–C=O + HBr

OH

+ – 1,4–Addition

Br–CH2–CH = C–OH

OH Tautomerises

Br–CH2–CH2–COOH Ans. 18. [A]



CH3–CH–COOH

CH2–OH

LiAlH4 CH3–CH–CH2–OH

CH2–OH

(X)

NaHCO3

*

CH3–CH–COO

CH2–OH + CO2

*

Ans. 19. [D]

X = CH3–CH2–CH2–CN

Y = CH3–CH2–CH2–CO2H

Z = CH3–CH2–Cl2–

C–OEt

O

Y ⎯→ Z is esterificatin reaction

Ans. 20. [B] CHCH

CH3–Mg–Br CHC–Mg–Br

CO2

HCC–COOMgBr

H3O

CHC–COOH HgSO4+H2SO4

Addition of H2O H–C=CH–COOH

OH

Tauetomerise H–O–CH2–COOH

O Ag2O

HO–C–CH2–C–OH

O O

Ans. 21. [D]

C=CH–CH2–CH3

CH3

CH3 KMnO4/H

+

C=O+OCH–CH2–CH3

CH3

CH3 Ans. 22. [C]

(COOH)2 373K/Glycerol

H– COOH

Ans. 23. [A]

–keto Acid on heating easily undergo decarboxylation via a six-membered. Cyclic transition

state.

C

CH2

C

C6H5

O

H

O

O

–CO2

C6H5 C

OH

CH2

Tauto

merises

C6H5–C–CH3

O

Ans. 24. [A]

+ CF3 COOH –I +

CCl3 COOH –I +



CBr3 COOH – CH3 COOH +I

Ans. 25. [D]

Ph–CH3 ⎯⎯⎯⎯⎯⎯ →⎯ 4KMnOAlkalineHot

Ph–COO– Ph–CH=CH–CH3⎯⎯⎯⎯⎯⎯ →⎯ 4KMnOAlkalineHot

Ph–COO– Ph–

CC–CH3 ⎯⎯⎯⎯⎯⎯ →⎯ 4KMnOAlkalineHotPh–COO–

Ans. 26. [A]

C2H5Br ⎯⎯⎯ →⎯ KOH,AlcCH2=CH2

4

2

CCI

Br⎯→⎯

Br

Br

HCHC 2

|

2|

−)excess(

KCN⎯⎯ →⎯

CN

CN

HCHC 2

|

2|

− ⎯⎯ →⎯+OH3

COOH

COOHCHHC 22|

−−

Ans. 27. [C]

HCOOH ⎯⎯⎯⎯⎯ →⎯+ OHNHAgNO 43 CO2 + H2O + Ag

Ans. 28. [C]

RCOOH + SOCl2 ⎯→⎯ RCOCl + SO2+ HCl

RCOOH + PCl5⎯→⎯

RCOCl + POCl3

RCOOH + PCl3 ⎯→⎯ RCOCl + H3PO3

Ans. 29. [A]

-halogenation reaction [-H must present].

Ans. 30. [C]

It is Hell Volhard Zelinsky reaction.

Ans. 31. [B]

Reactivity Partial positive charge on C=O carbon.

Ans. 32. [B]

CH3COONa + CH3COCl ⎯→⎯ CH3–COO–COCH3

Ans. 33. [D]

R–CH2OH + RCOCl ⎯→⎯ R–CH2–O–CO–R

Ans. 34. [A]

Ph–COCl ⎯⎯ →⎯ 3NHPh–CONH2 + HCl

Ans. 35. [B]

When an ester converts into another ester, it is called trans esterification.

Ans. 36. [C]

CH3–COOH + C2H5OH ⎯→⎯H

CH3COOC2H5 (ester)

Ans. 37. [B]

⎯⎯ →⎯ 3NaN⎯⎯ →⎯ − 2N

Ph – N = C = O ⎯⎯ →⎯−OH

+

HCOO–

O

Cl



Ans. 38. [A]

Electron withdrawing groups increase the rate of SN2Th reaction.

Ans. 39. [C]

Ester gives positive test with NaOH and phenopthalein.

Ans. 40. [A]

Saponification is a base catalysed hydrolysis (SN2Th) reaction which is a sterically

controlled reaction too. The least crowded ester around acyl carbon will be most reactive

towards this reaction.

Ans. 41. [D]

C4H10O4 ⎯⎯ →⎯ OAc2 C12H18O8

Ans. 42. [C]

CH3CH2CH2CONH2⎯⎯ →⎯NaOH

CH3CH2CH2COO– ⎯⎯→⎯HCICH3CH2CH2COOH + NH4Cl

Ans. 43. [C]

HVZ reaction.

Ans. 44. [B]

CH3CONH2 ⎯⎯ →⎯NaOH

NH3 + CH3COO–

52

| |

3 HOC

O

CCH −−⎯⎯ →⎯NaOH

HO

O

CCH| |

3 −− + C2H5OH

Ans. 45. [B]

+ 3 NaOH ⎯→⎯ + RCOOH\

Ans. 46. [B]

CH3CH2COOH ⎯⎯ →⎯ 2SOCICH3CH2COCl ⎯⎯ →⎯ 3NH

CH3CH2CONH2 ⎯⎯⎯ →⎯ +KOHBr2

CH3CH2NH2

Ans. 47. [C]

⎯⎯ →⎯ 2SOCI⎯⎯ →⎯

••

3HN ⎯⎯ →⎯ 52OP

Ans. 48. [C]

In fumaric acid both COOH groups are present on opposite side so it is unaffected on heating.

Ans. 49. [D]



⎯⎯ →⎯ hv/Br2

⎯⎯⎯ →⎯OH)2(

KCN)1(

3 Ans. 50. [B]

+ ⎯→⎯ (Diastereomers)

+ ⎯→⎯

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Carboxylic Acid MCQs With Solutions By Arvind Arora€¦ · Subscribe Madeejee Chemistry No. 1 NEET JEE CHANNEL ON YOUTUBE 2 Carboxylic Acid MCQs With Solutions By Arvind Arora