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—COOH or —||C— O

OH or carboxylic acid Alkanoicacid

Some Key Points

1. Acid turns Blue litmus paper red 2. Acid gives CO₂ with bi-carbonate

C

arboxylic Acid

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NaHCO R COOH R COONa H CO H O CO 3 2 3 2 2 + ¾ ®¾ + + ­ ¯ ( ) 3. Cyclic acid is –

4. 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 OH HNO Ester

Acid

H

+ ₃ ¾¾¾+®

( )

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

Trinitro glycerene R CO OH + H OR' H+ R COOR' + H O₂ CH CO OH + H O*R'₃ H+ CH COOR' + H O_{3 2} R — O H + HO — N O O R—ONO₂ Ester CH O H₂ HO NO₂ CH O H₂ H ONO₂ CH O H₂ HO NO₂ 3HNO₃ CH ONO2 2 CH ONO_{2 2} CH ONO_{2 2} HO HO O O

5. Formic Acid in Vapour phase is known as Formaline. 6. CH COOH₃ on solidification is known as Glacial acid. 7. In vinegar, CH COOH₃ is 7%

8. In case of carboxylic acid it forms dimer due to H-bonding.

2CH COOH₃ CH COOH_{3 2}

Dimer

¾ ®¾ ( )

9. Peroxy acid is a weaker acid as compared to carboxylic acid.

Acidic Nature of peroxy acid is lesser than their carboxy acid

]

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.

No resonance stability of O CH —C₃ O O—HH—O_O C—CH3 R—C O—O O H Peroxy link R—C—O—H O R COOOH R COOH R—C )120° O–––H O 109° O R—C ) O—H O 120° R—C ) – O—O O – R—C—O O + H

10. Carboxylic acid reacts with carbene to give ester.

11. b- Keto acid or b- Unsaturated acid on heating gives CO₂ (de-carboxylation), CH C O CH COOH CH C O CH CO 3— 2 3 3 2 || — — —||— b _a D ¾ ®¾ +

12. Soda lime performs the given reaction R —COOH ¾NaOH CaO¾ ¾ ¾¾+ ®R —H

CH COOH NaOH CaO CH

CO 3 4 2 + ¾¾ ¾ ¾¾® –

CH COOD NaOH CaO CH

CO 3 4 2 + ¾¾ ¾ ¾¾® –

CH COOH₃ ¾NaOD CaO¾ ¾ ¾ ¾+ ¾®CH₃ -D

R—COOH – R—COO CH NH_{2 2} R COOCH₃ – R—COO + N₂ CH₃ + H+ CH₂ carbene

:

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H COOH NaOH CaO H Soda e

+

¾¾ ¾ ¾¾®

lim 2

H COOH ¾NaOD CaO¾ ¾ ¾¾+ ®H —D

... ... ... ... ... CH COOH₃ NaOH*+CaO CH —H*₃ NaOH* CH COONa + H O_{3 2} NaOH* CaO (absorb water) Na CO + _{2 3} CH —H*₃ H O₂ NaOH + H CO_{2 3} (main product)

Q.1 Find out the products of the following reactions. Ans: (a) CH — C — CH₃ (b) (c) (d) — CH — COOH COOH | CH — COOH₂ O || COOH COOH O || | COOH A NaOD D CaO B A | CH COOH₂ CH COOH₂ 2 CH OH₃

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Q.1. IUPAC Name of following compounds:

(a) CH COOH₃ (b) HCOOH

(b) CH — CH — CH COOH 3 | 3 (d) | CH — COOH CH — COOH 2 2

Q.2. Which will turn blue litmus paper red ?.

(a) CH COOH₃ (b) HCOOCH₃ (c) HCl (d) CH CH OH_{3 2}

Ans: 2. (a, c)

S

ulphonic Acid

SO H₃ (Alkane Sulphonic Acid)

CH₃—SO H₃ Methan Sulphonic Acid C H C H SO H

2 3 1 2 3 Ethan sulphonic Acid

Some Key Points

1. Acidic Nature of R — SO H₃ >> R —COOH R —SO H₃ ¾NaOH¾ ¾¾®R —SO Na₃

R SO Na₃ ¾NaOH¾ ¾¾®R —OH Na SO+ _{2 3} (always removed)

Na SO_{2 3} is removed CH₃—SO Na₃ ¾NaCl¾¾®CH Cl Na SO₃ + _{2 3} CH₃—SO Na₃ ¾NaX¾¾®CH X Na SO₃ + _{2 3} 3. CH_{3 —}SO Na_{3 ¾}H O H_{¾ ¾ ¾}2 / +_®CH_3—H

E

STER

C H C OOCH Methyl ethanote

2 3 1 3 ¾ ®¾

C H C H C OOCH Methyl propanoate

3 3 2 2 1 3 ¾ ®¾ + H O/H₂ SO Na₃ SO H₃ NaOH SO Na₃ NaOH OH + Na SO_{2 3} 2 1 3 4 5 O O Cyclo pentanoate.

?

General / Inorganic name : Methyl carbonate IUPAC: Dimethyl methanoate (diester)

?

OC H 3

1 2 5

— —||— ethyl methyl methanoate.

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.

CH COOH HOCH₃ + ₃ ¾H¾¾®CH COOCH_{3 3} +

.

Q.1. Find out the products of the following reactions :

COOH *OH + H –H O₂ C O O*

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Find out the sum of molecular weights of A and C.

Q.2 Which has Fruity smell ?

(a) CH COOCH_{3 3} (b) CH CH COOH_{3 2} (c) CH CH OH_{3 2} (d) CH NH_{3 2}

(d) 44 46 90 A + C = 2. (a)

A

mines

2 3— 1 2— — N-chloro ethan amine

C H C H C H NH CH

3 3—2 2—1 2— — 3 N-methyl propan amine

SO Na₃ | OH | O || C | COONa | COOH NaO HO (a) (c) (b) CH₂ O A – A – CH OH B – C H OH₃* _{2 5} C – E – B – O || O || COOH COOH * + CO + CO_{2 2}* Ans: + H O₃ (c) COOCH3 COOC H_{2 5} O || A + B + C E COOH | | SO H₃ HO NaHCO₃ gas A+B C + D (d) NaHCO3 gas

CH₃ —CH₂ —CH₂—NH₂ ¾ ®¾ 1 a° min (Primary)e CH₃—CH₂—NH —CH₃ ¾ ®¾ 2 a° min (Secondary)e CH N CH CH 3 3 3 — — | ¾ ®¾ 3 a° min (Tertiaray)e

?

different properties.

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Some Key Points:

1. Primary amine give "Mustard oil test"

2. 1° -amine give Car byl amine test in which chlo ro form re acts with 1°–amine in pres ence of KOH to give isocyanide (bad smell)

CH —N₃ C H2 5

CH₃ N,N-dimethyl ethan amine

—NH₂ Cyclo propan amine

NH

1 2

3 4

5 _{N-cyclo pentan amine}

(N is in cyclopentane) NH₂ 1 2 3 4 5

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CH CONH_{3 2 ¾}HNO_{¾ ¾¾}2_®CH COOH₃

Double bonded carbon with —NH₂ gives diazonium chloride and single bonded one gives alcohol.

Imine: —CH NH= : Alkanal imine

C H2 ₃—C H NH1 = : ethanal imine ... ... ... ... ... ... ... NH₂ N Cl₂ CH — NH_{2 2} CH — OH₂ NaNO + HCl₂

Q.1 Find out the Products of the reactions? (a) NH — CH — C || O — NH A 2 2 2 HNO₂ ¾¾ ¾¾® (b) NH — CH — CH — | CONH CH = CH — NH 2 2 2 2 NaNO +HCl₂ A B ¾¾ ¾ ¾ ¾¾® ¾¾D®

Ans: (a) HO — CH — COOH₂ (b) HO — CH — CH — | COOH CH = CH— N Cl 2 2 a b (c) HO — CH — CH — CH = CH — N Cl_{2 2 2}

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Q.1. Which is gives mustard oil test (a) CH — CH — NH_{3 2 2} (b) CH — NH — CH_{3 3} (c) CH — CH — NH — CH_{3 2 3}) (d) CH —N — CH CH 3 3 3 |

Q.2. Which shows formation of alcohol by HNO₂.

(a) CH CH COOH_{3 2} (b) CH NH_{3 2} (c) CH CONH_{3 2} (d) CH — NH — CH_{3 3}

Q.3. If R — CN on reduction gives CH CH NH_{3 2 2}, then R — CN is ?

(a) CH CH CN_{3 2} (b) CH CN₃ (c) HCN (d) None Ans: 1. (a) 2. (b) 3. (b)

A

mide

— — N-methyl ethan amide

NH C O NH Urea 2— 2 || — ( ) : Methan diamide

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Some Key Point

Hoffmann Degradation: It is also known as Hoffmann bromide reaction, in this, in presence of KOH + Br₂ amide gives P-amine. CH CONH KOH Br CH NH a e 3 2 3 2 1 2 + ° ¾¾ ¾ ¾¾® — min CH CONH_{3 2} ¾_dehydration¾ ¾ ¾ ¾P O2 5 ®CH CN₃

A

lcohol

Some Key Points :

2

Glycol or ethan glycol or ethan-1, 2-diol CH —OH₂ CH—OH CH₃ Propan glycol CH —OH₂ CH₂ glycol It is not

In case of glycol two -OH groups are near each other. 2. Glycol have Oxidative cleavage with HIO₄: (Periodic acid)

For oxidative process.

*Alcohol [ ] Aldehyde [ ] [ ] or Ketone acid C 0 0 0 ¾¾¾® ¾¾¾® ¾¾¾® O₂ +H O₂ CH —OH₂ CH —OH₂ HIO₄ HCHO HCHO CH —OH₂

CH—OH HIO4 HCHO

CH₃ CHO CH₃ CH —OH₂ CH—OH HIO4 HCHO HCOOH CH —OH₂ (1 cleavage) (1 cleavage) (2 cleavage) HCHO CH —OH₂ CHO HIO₄ HCHO HCOOH CHO CH—OH (2) HIO4 HCOOH HCOOH CH OH₂ HCHO CH —OH₂

C=O (2 cutting) HIO4

HCHO CO +H O_{2 2}

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–

CH CH OH R 3— | — and CH CH OH H 3— | —

Give haloform test (Iodoform, chloroform ; Bromoform test). Alcohols give Lucas test

T > S > P Reagent is HCl (conc.)/ZnCl₂ CH₂ CH—OH CH—OH CHOH CHOH CHNo cutting₂ No cutting (1) (2) (3)

In this 3HIO are used.₄ 3HIO₄

Ex am ple:

Ans:

b, c, d

Q.1. Which of the following is not on alcohol ?

Q.2. Product of the reactions are ?

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HIO4 HNO2 HIO4

OH O OH OH OH OH OH O CH — NH_{2 2} CH — NH_{2 2} OH CH OH₂ (a) (b) (c) CH — CH — OH_{3 2} (d) All

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

CHO (b) s HCOOH (c) 4HCOOH + 2CO2 (d) A– B– 2HCHO CH — OH₂

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Q.1. Which gives lucas test fast ?

(a) CH CH OH_{3 2} (b) CH — CH — CH OH 3 | 3 (c) CH —C — | CH | CH OH 3 3 3 (d) CH OH₃

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) 2. (d) 3. (a)

Lecture-4

E

ther

R — — Alkoxy AlkaneO R

Alkoxy of smaller and alkane of bigger. CH₃—CH₂— —O CH₃ Methoxy ethane C H_{2 5} — —O C H_{3 7} Ethoxy propane

CH₃— —O CH₃ 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 — — : Alkyl alkyl thio etherS R

CH₃— —S CH CH_{2 3} : Ethyl, methyl thioether

CH₃—CH₂—CH₂— —S CH₂—CH₃ : Ethyl propyl thioether

]

A

ldehyde and Ketone

H R

C O Alkanal

/\ = - or R —CHO R_R/\ = AlkanoneC O

Aldehyde Starts with 1 Carbon but ketone Starts with 3 carbons CH₃—CHO : Ethanal CH C O CH opanone 3— — 3 || Pr | CHO CHO ethandial

Some Key Points:

1. Aldehydes without a-Hydrogen have Cannizaro reaction. It is a redox reaction in presence of a base.

Q Which will give Cannizaro reaction?

_{(a) HCHO ¾ ®}¾ Do not have a - Carbon and a-hydrogen.

So, gives cannizaro reaction.

_{(b) CH3—CHO ¾ ®}¾ Have a -Hydrogen, performs no

Cannizaro reaction

—CHO Do not have a-Hydrogen,

Performs cannizaro reaction. (c)

—CHO Have a-Hydrogen, will not

Perform cannizaro reaction (d)

2. (a) CH C CH CH CHO 3 3 3 — |— |

Cannizaro reaction is given by this due to absence of a-hydrogen.

(b) CH₃—CH₂—CHO Cannizaro reaction is not given by this.

]

?

have 'a' hydrogen. But only aldehyde with ‘ ’a Hydrogen

]

If aldehyde or ketone has a-hydrogen then it gives aldol condensation in presence of base or acid.

CH C O H CH C O H C OH H CH C NaOH or H 3— 3 3 2 || — —||— —| |— — + ¾¾ ¾¾₊ ®CH ||— O H Aldol CH—CHO CH₃ CH₃

a-Carbon with a-Hydrogen

Gives Cannizaro reaction even with having

a-Hydrogen.

2HCHO NaOH* HCOONa + CH OH*₃ Reduction

?

?

O

CH OH

—||— | ₂ also gives these tests.

Q . Which will undergo aldol condensation ? (a) CH CHO₃ (b) CH COCH_{3 3}

(c) (d) HCHO

Ans: — a, b, c

?

CH C O H 3— — || or CH C O R 3— — ||

Q Which will give haloform test : (a) CH C O CH 3— — 3 || (b) CH C O NH 3— — 2 || or CH CONH_{3 2} (c) CH C O OH 3— — || or CH COOH₃ (d) CH C O CH CH 3— — 2 3 ||

Ans: a and d only because b and c are acid amides and acids.

?

Any alkyl group

CH —C—₃ O

Q.1. Find out the products of the reactions ?

C

yanide / Isocyanide

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(c) COONa Cannizaro reaction (d) HCOONa + CH OD₃

OH A- HO OH B- O CH OD₂ A- B-CHO (a) O

Some Key Points:

1. Isocyanids have bad ( irritating) smell.

(a) CH CH Cl_{3 2 ¾}KCN_{¾ ¾®}A _¾Reduction_{¾ ¾ ¾¾®}B_¾HNO_{¾ ¾¾}2 _®C (b) CH Cl_{3 ¾}AgCN_{¾ ¾¾®}A _¾H O_{¾ ¾}3 _¾+_®B (c) CH CH OH_{3 2 ¾}PCl_¾¾5 _®A _¾KCN_{¾ ¾®}B_¾H O_{¾ ¾}3 _¾+_®C (d) CH Cl Cl A B C 2 KCN H O3 + ¾¾ ¾® ¾¾ ¾¾® ¾¾D® Ans: (a) A- CH CH CN_{3 2} B- CH CH CH NH_{3 2 2 2} C- CH CH CH OH_{3 2 2} (b) A- CH NC₃ B- CH NH + HCOOH_{3 2} C- CH CH NH_{3 2 2} (c) A- CH CH Cl_{3 2} B- CH CH CN_{3 2} C- CH CH COOH_{3 2} (d) A- CH₂ CN B- CH₂ COOH C- O=C=C=C=O

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Q.1. Give the test to separate CH CHO₃ and HCHO

(a) Tollen’s test (b) Cannizaro reaction (c) Idoform test (d) Fehling solution test

Q.2. CH — C || O

— CH

3 3 and CH — CHO3 are separated by

(a) Tollen’s test (b) Fehling solution test

(c) Both (d) None

Q.3. Which will give Tollen’s test ?

(a) CH — C || O — H 3 (b) CH — C || O — CH OH 3 2 | (c) CH — C || O — CH 3 3 (d) a and b Q.4. CH — Cl₃ ¾¾ ¾¾AgCN ®A ¾¾ ¾ ¾¾reduction®B B is (a) CH CH NH_{3 2 2} (b) CH — NH_{3 2} (c) CH CH CH NH_{3 2 2 2} (d) None Q.5. X¾¾ ¾KCN¾® A¾¾ ¾ ¾¾reduction®B B is CH CH NH_{3 2 2} then ‘X’ is (a) CH — Cl₃ (b) CH CH — Cl_{3 2} (c) H — Cl (d) None Ans. 1. (c) 2. (c) 3. (d) 4. (d) 5. (a)

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D

egree 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 H_{n1 n2} 1. C H_{n1 n2} æD U. = n —n + è ç ö ø ÷ 2 2 2 1 2

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 H_{4 10} ® D U. . = 8 10 2- + =

2 0

D.U.= 0 means all single bonds.

2. C H_{3 6} ® D U. = 6 6 2- + =

2 1

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

3. C H_{4 8} ® D U. = 8 8 2- + = 2 1 or H C=CH—CH_{2 3} H C—CH —CH —CH_{3 2 2 3} 2 structures H C—CH—CH_{3 3} | CH₃

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

D.U. with oxygen: (a) If D.U.=0 with oxygen it is alcohol or ether

Ex- C H O_{2 6} D.U.= 4 6 2

2 0

– + ₌ CH OCH_{3 3} or CH CH OH_{3 2}

(b) If D.U.= 1 with oxygen then it is C=C, C=0 or cyclic

Ex- C H O_{3 6} D.U. =1 CH CH CHO CH C O CH 3 — 2 — , 3 — 3 || — CH₂ = CH —CH OH₂ , CH O CH CH 2 2 2 — | | —

(c) If two oxygen then

(i) Acid (ii) Ester (iii) Keto-alcohal

(iv) Keto-ether (v) Aldehyde-ether or alcohal

Ex- C H O_{3 6 2} D.U. =6 6 2+ = 2 1 – It is CH CH COOH_{3 2} or CH COOCH_{3 3} or CH C O CH OH 3 — 2 || — CH OH CH CHO etc 2 2 | — — . C—C=C C—C=C—C C C=C—C—C C

Q.1 Find out which will have aldehyde formation (a) C H_{6 10}O (b) C H_{4 10}O (c) C H_{5 10}O₂ (d) C H O_{3 6} (e) C H_{n 2n}O

Q.2. Which formula may have iodoform formation

(a) C H O_{3 6} (b) C H_{4 10}O (c) C H_{5 12}O (d) C H O_{2 6}

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

Q.1. Find out D.U. of Benzene

(a) 1 (b) 2

(c) 3 (d) 4

Q.2. Find out Cyclic structures of C H_{4 10}

(a) 1 (b) 2

(c) 3 (d) None

Q.3. C H_{4 8} has total number of cyclic structures

(a) 1 (b) 2

(c) 3 (d) 4

Q.4. C H O_{3 8} has total number of alcohols

(a) 1 (b) 2

(c) 3 (d) 4

Ans. 1. (d) 2. (d) 3. (b) 4. (b)

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