Day -1

DAY-1 Functional Groups and Their Properties

Carboxylic Acid O || —COOH or — C — OH or carboxylic acid Alkanoicacid

Some Key Points 1. Acid turns Blue litmus paper red 2. Acid gives CO 2 with bi-carbonate 1.20

NaHCO 3 + R COOH ¾ ¾® R COONa

3.

+ H 2 CO 3 ¯ H 2 O + CO 2(­)

Cyclic acid is – HO

O Dibasic Cyclic acid

HO 4.

O

Acid reacts with alcohol to give ester; In this case, in presence of mineral acid :—OH of —COOH group and H + of alcohol are removed. R CO OH + H OR'

H+

R COOR' + H2O H+ CH COOR' + H O 3 2

CH3CO OH + H O*R' H+

R OH + HNO 3 ¾ ¾ ¾® Ester ( Acid)

R — O H + HO — N

O O

R—ONO2 Ester

If any acid gives –OH and alcohol gives H + in the reaction then ester is formed. CH2O H

HO NO2

CH2O H

H ONO2

CH2O H

HO NO2

3HNO3

CH2ONO2

CH2ONO2 Trinitro glycerene CH2ONO2 1.21

5. 6. 7. 8.

Formic Acid in Vapour phase is known as Formaline. CH 3COOH on solidification is known as Glacial acid. In vinegar, CH 3COOH is 7% In case of carboxylic acid it forms dimer due to H-bonding. 2 CH 3COOH ¾ ¾®(CH 3COOH)2 Dimer

CH3—C

O

H—O C—CH3 O—H O

9. Peroxy acid is a weaker acid as compared to carboxylic acid. R COOOH O R—C

H

O—O Peroxy link

R COOH O R—C—O—H

Acidic Nature of peroxy acid is lesser than their carboxy acid

] Reasons

In case of peroxy-acid, intra-molecular H-bonding is there, so removal of H + is not easy. And, the 2nd reason is no resonance in Peroxy- anion. O O–––H R—C )120° R—C ) 120° O 109° O—H O O O R—C ) R—C—O– H + O—O– No resonance stability of O 1.22

10. Carboxylic acid reacts with carbene to give ester. R—COOH

CH2NH2

D :CH2 carbene

R—COO–

R COOCH3 + N2

R—COO–

H+

+

CH3

11. b- Keto acid or b- Unsaturated acid on heating gives CO 2 (de-carboxylation), O O || || D CH 3 — C — CH 2 — COOH ¾¾® CH 3 — C — CH 3 + CO 2 b

a

12. Soda lime performs the given reaction CaO R — COOH ¾NaOH ¾ ¾+¾¾ ®R —H CaO CH 3COOH ¾NaOH ¾ ¾+¾¾ ® CH 4 –CO 2

CaO CH 3COOD ¾NaOH ¾ ¾+¾¾ ® CH 4 –CO 2

NaOD + CaO

CH 3COOH ¾ ¾ ¾ ¾ ¾ ¾® CH 3 - D

In this reaction lime absorb the H2O 1.23

] Reason

NaOH*+CaO

CH3COOH

CH3—H*

NaOH*

CH3COONa + H2O CaO

NaOH*

(absorb water)

Na2CO3 + CH3—H* (main product) H2O

NaOH + H2CO3 CaO H COOH ¾NaOH ¾ ¾+¾¾ ®H2 Soda lim e

CaO H COOH ¾NaOD ¾ ¾+¾¾ ®H — D

Your Secrets Not to be Copied by Others

My Secrets

............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ...............................................................................................................................

1.24

Do Yourself -1.4 Q.1 Find out the products of the following reactions. O || (a) CH3 — C — CH

COOH

COOH — CH — COOH | CH2 — COOH

(b) O ||

COOH

(c)

A

NaOD

B

CaO

| COOH (d)

CH2 COOH 2 CH3OH | CH2 COOH

D

A

Ans:

(a)

O || CH3 — C — CH3 O ||

(c)

(A)

(b) O ||

(B) | COOH

(d) |

— CH2 | CH2COOH CH2 COOCH3 | CH2 COOCH3

D

1.25

Final Shoot - 3 Q.1. IUPAC Name of following compounds: (a) CH3COOH (b) CH3 — CH — COOH | CH3

(b) HCOOH CH2 — COOH (d) | CH2 — COOH

Q.2. Which will turn blue litmus paper red ?. (a) CH3COOH

(b) HCOOCH3

(c) HCl

(d) CH3CH2OH

Ans: 2. (a, c)

Sulphonic Acid SO 3H (Alkane Sulphonic Acid) CH 3 — SO 3H Methan Sulphonic Acid C H 3 C H 2 SO 3H Ethan sulphonic Acid 2

1

Some Key Points 1.

Acidic Nature of R — SO 3H >> R — COOH R — SO 3H ¾NaOH ¾¾ ¾® R — SO 3Na R SO 3Na ¾NaOH ¾¾ ¾® R — OH + Na2 SO 3 (always removed)

1.26

Na2 SO 3 is removed CH 3 — SO 3Na ¾NaCl ¾¾® CH 3Cl + Na2 SO 3 CH 3 — SO 3Na ¾NaX ¾¾® CH 3 X + Na2 SO 3 SO3Na

SO3H NaOH

OH NaOH

+ Na2SO3

SO3Na H2O/H+

3.

H 2O / H +

CH 3 — SO 3Na ¾ ¾ ¾ ¾® CH 3 — H

ESTER O || RCOOR ' ¾ ¾® Alkyl alkanoate R — C — OR ' Albanoate

Alkyl

C H 3 C OO CH 3 ¾ ¾® Methyl ethanote 2

1

C H 3 C H 2 C OO CH 3 ¾ ¾® Methyl propanoate 3

2

1

O 2 1 O 5 3 4

Cyclo pentanoate.

1.27

O || Example: CH 3 — O — C — O — CH 3

?

General / Inorganic name : Methyl carbonate IUPAC: Dimethyl methanoate (diester) O || ? Example CH 3 —O —C —OC 2H 5 ethyl methyl methanoate. 1

Some Key Points: 1. Ester is neutral in nature with fruity smell. 2. Esters are formed from acids and alcohols in which –OH group of acid and —H of alcohol are removed. H+

CH 3COOH + HOCH 3 ¾ ¾ ¾® CH 3COOCH 3 O C H+ COOH O* *OH –H O 2

.

Do Yourself -1.5 Q.1. Find out the products of the following reactions : SO3H | NaOH

(a)

excess

HO

1.28

A

| COOH COOH

(b)

H3O+

CH2 — OH

H+

B

O ||

H3O+

COOCH3

(c)

A+B+C

E

COOC2H5

Find out the sum of molecular weights of A and C. COOH | NaHCO3

(d)

gas

HO

A+B

NaHCO3 gas

C+D

| SO3H

Q.2 Which has Fruity smell ? (a) CH 3 COOCH 3 (b) CH 3 CH 2 COOH

SO3Na | Ans:

(a)

A– | COONa

HO

(d) 44 + 46 = 90 C

O

(b) CH2

| COOH O ||

* A – CH3OH B – C2H5OH C – A

O || C

OH | B–

NaO

(c)

(c) CH 3 CH 2 OH (d) CH 3 NH 2

2. (a)

O || COOH COOH *

E–

* + CO2 + CO2

Amines ¾® Alkane Amine —NH 2 ¾ 2

1

C H 3 — C H 2 — NH 2

Ethane amine

C H 3 — C H 2 — NH — Cl

N-chloro ethan amine

2

1

C H 3 —C H 2 —C H 2 —NH —CH 3 N-methyl propan amine 3

2

1

1.29

CH3—N

C2H5 CH3

—NH2 1 NH 5 2

3

4

N,N-dimethyl ethan amine

Cyclo propan amine

N-cyclo pentan amine (N is in cyclopentane)

NH2 2 3

1

5

Cyclo pentan amine

4

¾® 1° a min e (Primary) CH 3 — CH 2 — CH 2 — NH 2 ¾

? ?

CH 3 — CH 2 — NH — CH 3

¾ ¾® 2 ° a min e (Secondary)

CH 3 —N — CH 3 | CH 3

¾ ¾® 3° a min e (Tertiaray)

P,S and T amine are functional isomers shawing different properties. P,S and T- (alcohol and halides) are not functional isomers.

Some Key Points: 1. Primary amine give "Mustard oil test" 2. 1° -amine give Carbylamine test in which chloroform reacts with 1°–amine in presence of KOH to give isocyanide (bad smell) 1.30

CHCl +KOH

CH 3NH 2 ¾ ¾ ¾3 ¾ ¾ ¾® CH 3 — NC (Isocynide)

3.

HNO

2 ¾ ® R — OH R — NH 2 ¾ ¾ ¾

(NaNO 2 +HCl ) HNO

2 CH 3CONH 2 ¾ ¾ ¾ ¾ ® CH 3COOH

NH2

N2Cl NaNO2 + HCl

CH2 — NH2

CH2 — OH

Double bonded carbon with —NH 2 gives diazonium chloride and single bonded one gives alcohol. Imine: —CH = NH : 2

1

Alkanal imine

C H 3 — C H = NH :

ethanal imine

Your Secrets Not to be Copied by Others

My Secrets

............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... 1.31

Do Yourself -1.6 Q.1

Find out the Products of the reactions? O || HNO2 (a) NH 2 — CH 2 — C — NH 2 ¾ ¾ ¾¾ ®A NaNO +HCl

D

2 (b) NH 2 — CH 2 — CH — CH = CH — NH 2 ¾ ¾ ¾ ¾ ¾¾® A ¾¾® B | CONH 2

a

b

Ans: (a) HO — CH 2 — COOH (b) HO — CH 2 — CH — CH = CH— N 2 Cl | COOH (c) HO — CH 2 — CH 2 — CH = CH — N 2 Cl

1.32

Final Shoot - 4 Q.1. Which is gives mustard oil test (a) CH3 — CH2 — NH2

(b) CH3 — NH — CH3

(c) CH3 — CH2 — NH — CH3 )

(d) CH3 —N — CH3 | CH3

Q.2. Which shows formation of alcohol by HNO 2 . (a) CH3CH2COOH (b) CH3 NH2 (c) CH3CONH2 (d) CH3 — NH — CH3

Q.3. If R — CN on reduction gives CH3CH2 NH2 , then R — CN is ? (a) CH3CH2CN

Ans: 1. (a) 2. (b)

(b) CH3CN

(c) HCN

(d) None

3. (b)

Amide O || RCONH 2 ® R — C —NH 2

Alkan amide

O || C H 3 — C — NH 2

ethan amide

C H 3 C H 2 C H 2 C ONH 2

Butan amide

O || C H 3 — C — NH — CH 3

N-methyl ethan amide

O || NH 2 — C — NH 2 :

Methan diamide

2

4

2

1

3

2

1

1

(Urea)

1.33

Some Key Point Hoffmann Degradation: It is also known as Hoffmann bromide reaction, in this, in presence of KOH + Br 2 amide gives P-amine. KOH + Br 2 CH 3CONH 2 ¾ ¾ ¾ ¾¾ ® CH 3 — NH 2 1°a min e

P2 O 5

CH 3CONH 2 ¾ ¾ ¾ ¾ ¾® CH 3CN dehydration

Alcohol R—OH

:

Alkanol

CH 3CH 2 OH

:

Ethanol

CH 3 —CH — CH 3 | OH CH 2 — OH | CH 2 — OH

:

Propane-2-ol

:

Ethan-1, 2-diol

Some Key Points : 1.

CH2—OH CH2—OH

Glycol or ethan glycol or ethan-1, 2-diol

CH2—OH CH—OH

Propan glycol

CH3 CH2—OH

It is not

CH2

glycol

CH2—OH 1.34

In case of glycol two -OH groups are near each other. 2. Glycol have Oxidative cleavage with HIO 4 : (Periodic acid) CH2—OH

HIO4

CH2—OH CH2—OH CH—OH

HCHO HCHO

HIO4

CH3 For oxidative process.

HCHO CHO CH3

0] 0] 0] * Alcohol ¾[¾ ¾ ® Aldehyde ¾[¾ ¾ ® acid ¾[¾ ¾ ® CO 2 + H 2 O or Ketone

CH2—OH (1 cleavage) CH—OH (2 cleavage) CH2—OH (1 cleavage) CH2—OH

HIO4

HCHO HIO4

HCOOH HCHO HCHO

CHO

HCOOH

CHO

HCOOH

CH—OH (2)

HIO4

HCOOH

CH2OH

HCHO

CH2—OH

HCHO

C=O (2 cutting) CHO

HIO4

CO2 +H2O HCOOH 1.35

CH2

No cutting

CH—OH

In this 3HIO4 are used.

(1)

CH—OH (2)

3HIO4

CHOH

(3)

CHOH

No cutting

CH2

3. 100% ethyl alcohol ; is known as, absolute alcohol 4. 95% absolute alcohol + 5% water ; is known as, Sprit. 5. If methyl alcohol is added in spirit, then it is known as denatured sprit. 6. 80% petrol and 20% ethyl alcohol are known as power Alcohol Used as Fuels. 7. Ethyl alcohol works on nervous system and cause of unconsciousness but methyl alcohol is poisonous and it is the cause of blindness on drinking. 8. Alcohol with group– OH | CH 3 — CH — R

and

OH | CH 3 — CH — H

Give haloform test (Iodoform, chloroform ; Bromoform test). Alcohols give Lucas test T > S > P Reagent is HCl (conc.)/ZnCl 2 1.36

Example: Which will give haloform test? OH | (a) CH 3 —C — CH 3 | CH 3

OH | CH 3 —CH — CH 3

(b)

OH | (c) CH 3 —CH — CH 2 — CH 3

OH | (d) CH 3 — CH 2 — OH or CH 3 —CH — H

Ans: b, c, d

Do Yourself -1.7 Q.1. Which of the following is not on alcohol ? (a)

(b)

OH

CH2OH

Q.2. Product of the reactions are ?

OH

OH HIO4

(a)

OH

OH

OH

HIO4

OH

O OH HIO4

(c) OH

OH

OH (b)

OH

OH

(c) CH3 — CH2 — OH (d) All

CH2 — NH2 (d)

HNO2

CH2 — NH2

A

HIO4

B

O Ans: 1. (a) 2. (a)

CHO CHO

(b) s HCOOH (c) 4HCOOH + 2CO2 (d) A–

CH2 — OH

B– 2HCHO

CH2 — OH

1.37

Your Secrets My Secrets

Not to be Copied by Others

............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... ............................................................................................................................... .............................................................................................................................

Final Shoot - 5 Q.1. Which gives lucas test fast ? CH3 | (b) CH3 — CH — OH

(a) CH3CH2OH CH3 | (c) CH3 —C — OH | CH3

(d) CH3OH

Q.2. Which gives Idoform tast (a) 2-butanol

(b) 2-propanol

(c) ethanol

(d) all

Q.3. Power alcohol has -- % alcohol (a) 20%

(b) 80%

(c) 100%

(d) 10%

Ans. 1. (c)

1.38

2. (d)

3. (a)

Lecture-4

Ether R — O — R Alkoxy Alkane Alkoxy of smaller and alkane of bigger. CH 3 — CH 2 —O — CH 3 Methoxy ethane C 2 H 5 —O — C 3H 7 Ethoxy propane CH 3 —O — CH 3 Methoxy methane

Some Key Points 1. Ethers are explosive in nature in air, due to peroxy formation. 2. They are used as solvents. Thio Ether: R — S — R : Alkyl alkyl thio ether CH 3 — S — CH 2 CH 3

: Ethyl, methyl thioether

CH 3 —CH 2 —CH 2 — S—CH 2 —CH 3 : Ethyl propyl thioether

]

Cyclic Ether Epoxy alkane —C—C— O CH3—CH—CH2 : epoxy propane O 4

3

2

1

CH3—CH2—CH2—CH3 : 2, 3-epoxy butane 1 2 3 4 O 1.39

Aldehyde and Ketone R\ R\ C = O - Alkanal or R — CHO / C = O Alkanone H/ R Aldehyde Starts with 1 Carbon but ketone Starts with 3 carbons O || CH 3 — CHO : Ethanal CH 3 —C —CH 3 CHO | CHO

Pr opanone

ethandial

Some Key Points: 1. Q

Aldehydes without a-Hydrogen have Cannizaro reaction. It is a redox reaction in presence of a base. Which will give Cannizaro reaction? (a) HCHO ¾ ¾® Do not have a - Carbon and a-hydrogen. So, gives cannizaro reaction. (b) CH — CHO ¾ ¾® Have a -Hydrogen, performs no 3 Cannizaro reaction

1.40

(c)

—CHO

Do not have a-Hydrogen, Performs cannizaro reaction.

(d)

—CHO

Have a-Hydrogen, will not Perform cannizaro reaction

CH 3 | 2. (a) CH 3 —C —CHO Cannizaro reaction is given by this due | CH 3 to absence of a-hydrogen. (b) CH 3 —CH 2 —CHO Cannizaro reaction is not given by this.

]

Cannizaro Reaction: Oxidation 2HCHO

?

NaOH*

HCOONa + CH3OH* Reduction

Cannizaro Reaction is Performed by those which does not have 'a' hydrogen. But only aldehyde with ‘ a ’ Hydrogen

CH3 CH3

CH—CHO

Gives Cannizaro reaction even with having a-Hydrogen.

a-Carbon with a-Hydrogen

]

Aldol Condensation:

If aldehyde or ketone has a-hydrogen then it gives aldol condensation in presence of base or acid. O O O OH || || || | NaOH CH 3 —C — H + CH 3 — C — H ¾ ¾ ¾ ¾® CH 3 —C — CH 2 — C — H or H + | Aldol H 1.41

Aldehydes except Benzaldehyde give Fehling solution test and tollen's test (silver mirror test).

?

O OH || | a-Hydroxy ketone R — C — CH 2 also gives these tests.

?

Q . Which will undergo aldol condensation ? (a) CH 3CHO O

(b) CH 3 COCH 3

(c) CH3—C—

(d) HCHO

Ans: — a, b, c Aldehyde or ketone with following groups will give haloform test. O O || || CH 3 —C —H or CH 3 —C — R

?

R=—CH3 — C2H5,

etc.

Any alkyl group Q

Which will give haloform test : O O || || (a) CH 3 —C —CH 3 (b) CH 3 —C —NH 2 or CH 3CONH 2 O O || || (c) CH 3 —C —OH or CH 3COOH (d) CH 3 —C —CH 2 CH 3 Ans: a and d only because b and c are acid amides and acids.

? 1.42

Carbonyl compounds gives Crystal with NaHSO 3 (Sodium bisulphite)

Do Yourself -1.8 Q.1. Find out the products of the reactions ? NH2

HNO2

(a)

A

HIO4

B

NH2 (b)

(c)

NH2

NH2

CHO

NaoD

CHO (d)

NaoD

HCHO

HNO2

B

A A+B

OH (a) A-

Ans:

[O]

A

BOH

(b) A- HO

CHO CHO

OH B- O

O

(c) COONa Cannizaro reaction (d) HCOONa + CH3OD CH2OD

Cyanide / Isocyanide —CN

Alkane nitrile 2

1

C H3— CN ethan nitrile

3

2

—NC

Alkan isonitrile or Carbyla min e 1

C H 3 — NC

Methan isonitrile

1

C H 3 C H 2 C H 2NC : Propan isonitrile 1.43

Some Key Points: 1.

Isocyanids have bad ( irritating) smell.

2.

R—NC

R—CN

Reduction

Reduction

R—NH—CH3

R—CH2NH2

KCN

3. 4.

R—Cl

R—CN

AgCN

R—CN R—NC

R—NC

H2 O / H+ H2 O / H+

R—COOH R—NH2 + HCOOH

Do Yourself -1.9 KCN

HNO

Reduction

2 (a) CH 3 CH 2 Cl ¾ ¾ ¾ ® A ¾ ¾ ¾ ¾¾® B ¾ ¾ ¾¾ ®C

H O+

AgCN

(b) CH 3 Cl ¾ ¾ ¾¾® A ¾ ¾3 ¾ ¾® B PCl

H O+

KCN

5 (c) CH 3 CH 2 OH ¾ ¾¾ ® A ¾ ¾ ¾ ® B ¾ ¾3 ¾ ¾® C

(d) CH 2

Cl

KCN

H O+

D

¾ ¾ ¾ ® A ¾ ¾3 ¾ ¾® B ¾¾® C

Cl Ans: (a) A- CH 3 CH 2 CN

C- CH 3 CH 2 CH 2 OH

(b) A- CH 3 NC

B- CH 3 NH 2 + HCOOH C- CH 3 CH 2 NH 2

(c) A- CH 3 CH 2 Cl

B- CH 3 CH 2 CN

(d) A- CH 2

CN CN

1.44

B- CH 3 CH 2 CH 2 NH 2

B- CH 2

COOH COOH

C- CH 3 CH 2 COOH C- O=C=C=C=O

Final Shoot - 6 Q.1. Give the test to separate CH3CHO and HCHO (a) Tollen’s test

(b) Cannizaro reaction

(c) Idoform test

(d) Fehling solution test

O || Q.2. CH3 — C — CH3 and CH3 — CHO are separated by (a) Tollen’s test

(b) Fehling solution test

(c) Both

(d) None

Q.3. Which will give Tollen’s test ? O || (a) CH3 — C — H

O OH | || (b) CH3 — C — CH2

O || (c) CH3 — C — CH3

(d) a and b

AgCN reduction Q.4. CH3 — Cl ¾ ¾ ¾¾® A ¾ ¾ ¾ ¾¾® B B is

(a) CH3CH2 NH2

(b) CH3 — NH2

(c) CH3CH2CH2 NH2

(d) None

KCN reduction Q.5. X ¾ ¾ ¾ ¾® A ¾ ¾ ¾ ¾¾® B B is CH3CH2 NH2 then ‘X’ is

(a) CH3 — Cl

(b) CH3CH2 — Cl

(c) H — Cl

Ans. 1. (c)

(d) None 2. (c)

3. (d)

4. (d)

5. (a)

1.45

Degree of Unsaturation It is the calculation of unsaturation in the compound in the form of = or º bond or cyclic structure formation. If the compound is C n 1 H n 2 2n 1 — n 2 + 2 ö 1. C n 1 H n 2 æç D.U = ÷ 2 è ø If D.U = 0 ,

Then all are single bonds.

If D.U = 1 ,

One double bond or one cyclic

If D.U = 2 ,

Two double bonds

One º bond or Two cyclics One (=) + one cyclic Ex- (1) C 4H 10 ® D.U . =

8 - 10 + 2 =0 2

D.U.= 0 means all single bonds. H3 C—CH2 —CH2 —CH3

2 structures

H3 C—CH—CH3 | CH3

2.

6-6+2 =1 2 D.U.= 1 means one (=) bond or one cyclic.

C 3H 6 ® D.U =

H2 C=CH—CH3 or 3. 1.46

C 4H 8 ® D.U =

8-8+2 =1 2

D.U.=1 means one (=) bond or cyclic C C—C=C C—C=C—C C=C—C—C

C

D.U. with oxygen: (a) If D.U.=0 with oxygen it is alcohol or ether 4 –6 +2 =0 2 CH 3OCH 3 or CH 3CH 2 OH

Ex- C 2 H 6O D.U.=

(b) If D.U.= 1 with oxygen then it is C=C, C=0 or cyclic Ex- C 3H 6O D.U. =1

O || CH 3 — CH 2 — CHO , CH 3 — C — CH 3

CH 2 = CH — CH 2 OH , CH 2 — O | | CH 2 — CH 2 (c) If two oxygen then (i) Acid (ii) Ester (iii) Keto-alcohal (iv) Keto-ether (v) Aldehyde-ether or alcohal Ex- C 3H 6O 2 D.U. =

6 –6 +2 =1 2

O || It is CH 3CH 2 COOH or CH 3COOCH 3 or CH 3 —C — CH 2 OH CH 2 — CH 2 — CHO etc . | OH 1.47

Do Yourself -1.10 Q.1

Find out which will have aldehyde formation (a) C6 H10 O

(b) C4 H10 O

(d) C3 H 6 O

(e) C n H 2n O

(c) C5 H10 O 2

Q.2. Which formula may have iodoform formation (a) C3 H 6 O

(b) C4 H10 O

(c) C5 H12 O

(d) C2 H 6 O

Ans: 1. (a, c, d, e) 2. (a, b, c, d)

Final Shoot - 7 Q.1. Find out D.U. of Benzene (a) 1

(b) 2

(c) 3

(d) 4

Q.2. Find out Cyclic structures of C4 H10 (a) 1

(b) 2

(c) 3

(d) None

Q.3. C4 H 8 has total number of cyclic structures (a) 1

(b) 2

(c) 3

(d) 4

Q.4. C3 H 8O has total number of alcohols (a) 1

(b) 2

(c) 3

(d) 4

Ans. 1. (d)

1.48

2. (d)

3. (b)

4. (b)

This content is a part of the Book “Organic for Doctors” with Video lectures and solutions by Er. Dushyant Kumar (B.Tech, IIT-Roorkee)

1.21