Chemistry | 11.43

Solved Examples

JEE Main/Boards

Example 1: Complete the following reactions: NBS Alc.KOH (B) (C ) (a) Me (A)

Me Br Alc.KOH (b) (Y) + (Z) (X) (Major) (Minor) Me

H (c ) (B)

(A)

Cl 1. Li -B 2 (B) (C ) (D) (d) Me at 500oC 2. CuI (A)

(e) HO KHSO 4 (Y) OH or (X) Al23O or PO25

Allylic Br Br 3 1 3 NBS Alc. KOH Sol: (a) 4 4 Me Allylic Me 2 -Hbr 2 But-1-ene Substitution 3-Bromo-but-1-ene Buta-1,3-diene (A)

(b) (Z) is formed by Saytzeff’s elimination, but the product (Y) is formed in major amount because product (Y) is a more stable conjugate diene than (Z), isolated diene 6 Me 4 2 -HBr Br 1 Me6 H at C-3 5 3 4 2 removed Me (b) 5 3 1 5-methly hexa-1,3-diene Me H H Alk.KOH (Y) Major

4-Bromo-5-methylhex-1-ene Me6 (X) 4 2 -HBr 5 1 H at C-3 3 is removed Me Saytzeff 5-methly hexa-1,4-diene elimination (Z) Minor 11.44 | Alkenes and Alkynes

(c) 4 1 H H 3 2 6 5 H -H ( c ) H 2 6 3 5 1 4

Less stable More stable isolated diene conjugated diene (A) (B) Cyclohexa-1,4-diene Cyclohexa-1,3-diene

2 o 2 Cl at 500 2 4 2 3 1. Li (d) 1 1 1 Me Allylic 2. Cu 3 (d) Prop-1-ene substitution Cl Corey-House Allylic synthesis Diallyl R (A) 3-Chloro- lithium cuperate prop-1-ene (B)

Cl 4 4 6 1 3 5 R’

H (e) (e) HO 2 4 KHSO4 OH Or 1 3 H Al23O or PO25 (X) (Y) (-H2O) Butan-1,4-diol Butan-1,3-diene

Example 2: Give the major products (not stereoisomers) of the following: 5 5 (a 1 4 (a)(a) 1 4

22 33 Cyclopenta-1,3-diene

1 1mo mol l 1 mo1 mol l 1 mo1l mol 11 mol of BrBr22 HBrHBr + + HBHBr r HS2 +HS + peroxide 2 peroxide peroperoxide xide

(b) 55 (b) 1 (b) Me 1 4 Me 4

2 3 3 1-Ethylcyclopenta-1,3-diene2 1-Ethylcyclopenta-1,3-diene 1 mol 1 mol 1 mol 1 mol of Br2 1 mol of Br HBr1 mo + l HBr1 mol 1 mol 2 HS2 + peroHBrxide + HBr peroHSxide2 + peroxide peroxide Chemistry | 11.45

Sol: 1

(a) (a) 2 5

3 4 (A) All 1,4-addition products HBr or HS+ Br2 2 1,4-Add. HBr+peroxide Peroxide

Br H H 3 5

2 4 1 4 1 4

1 2 2 5 Br 3 Br 3 SH 3,5-Dibromo- 3-Bromo- Cyclopent-1- cyclopent-1-ene cyclopent-1-ene ene-3-thiol

(b) 1 (b) 4 Me

2 3 All 1,4-addition products HBr + peroxide HBr + peroxide HS + peroxide 1 mol of Br2 2 Anti-Markonvikov’s Anti-Markonvikov’s Anti-Markonvikov’s add. add. add. 4 4 4 4 1 5 Br 3 5 Br 1 5 H 5 3 SH Me Br Me H Me Br Me H

2 3 2 1 2 3 1 2

3,5-Dibromo- 5-Bromo-3-ethyl- 3-Bromo- 5-Ethyl cyclo- 3-ethyl-cyclopent-1-ene cyclopent-1-ene 3-ethyl-cyclopent-1-ene pent-1-ene-3-thiol

Example 3: Write all the possible structures and give the structure of the products that are thermodynamically favoured.2 4 6 1 mol of 2 4 Products 1 3 Br6 Sol: 5 2 1 mol of 1 Products Hexa-1,3-5-triene3 5 Br2

Hexa-1,3-5-triene 2 4 6 1 3 2 5 4 6 Br Br 1 2 3 5 Br Br Br2 Br 2 Br 2 6 6 4 4 Br 1 3 5 1 3 5 2 1,4-AdditionBr 2 6 Br 6 4 4 Br 4 5 Br 1 3 6 5 Br 1 4 3 2 Br 1,4-Addition6 1 3 5 2 4 1 3 Br5 Br 6 Br 1,2-Addition 4 2 4 Br 6 Br 2 2 4 6 1 3 6 5 Br 2 1 3 5 1 3 Br 5 1 3 1,2-Addition5 (III) 2 4 1,6-Addition Br 2 4 6 6 Br 1 3 5 Br 5 Br 2 4 Br 2 1 3 (III) 6 2 1,6-Addition6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 Br 2 4 double bond 2 6 Br2 6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 double bond 2 4 6 1 mol of 1 Products 3 5 Br2 Hexa-1,3-5-triene

2 4 6 1 3 5

Br Br2 Br Br 2 Br 2 6 6 4 4 1 3 5 1 3 5 1,4-Addition 4 Br Br 6 Br 4 2 Br 6 1 3 5 2 1 3 5 1,2-Addition 2 4 Br 2 4 6 6 Br 1 3 5 Br 11.46 | Alkenes and Alkynes 1 3 5 (III) 1,6-Addition

Br 2 4 2 6 Br2 6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 double bond

More stable Products (II and III, conjugate dienes) are thermodynamically2 favoured4 products than2 I and 4IV (isolated dienes) (a) and 1 3 5 1 3 5 (A) (B) Example 4: Distinguish between the given pairs. 3 2 4 2 4 5 4 3 2 1 2 C = C=C COOH and (a) and (b) 6 4 1 1 3 5 1 3 5 H H COOH (A) (B) (C ) (D) Sol: (a) Compounds (A) and (B) are distinguished3 chemically by quantitative catalytic hydrogenation. Since 5 4 3 2 1 2 compound (A) has two doubleand bond whereas compound (B) has only one double bond .Compound (A) will require 6 C = C=C COOH 4 (b)2 moles of H for 1 mole of (A) while (B) will1 requires 1 mole of H per mole of (B). H2 H COOH 2 ( ) (D) C COOH Me C =C = C COOH H (C )

(b) Allene (C) with different groups on each of the double bonds is optically active and can be resolved into its enantiomers. Compound (D) is conjugate diene and does not show optical isomerism.

Example 5: Give the major and minor products. CH CH2 H 2 1 CH2 3 2 6 HBr 2 4 o 1 mol HBr at 45 C 3 o 5 at 45 C Br 1 5 4 6 (B) (major) (A) 3-Bromo-1-methyl- 3-Methylenecyclohex-1-ene cyclohex-1-ene Sol: At high temperature (45ºC), 1, 4-addition products is more favourable.

CH2 Br CH CH2 H CH 2 1 2 1 2 6 2 6 HBr HBr o at 45 C 3 5 3 5 Br H 4 4 (A) (B) (major) (C ) (major) 3-Bromo-1-methyl- 1-Bromo-1-methyl- cyclohex-1-ene cyclohex-1-ene

Example 6: Dehydration of (A) with conc.CH2 H BrSO gives a compound that exists in two CH 2 4 Me H isomeric forms.2 Give the structures of both1 the isomers. 2 6 H OH

HBr (A) 3 5 H (A) 4 (C ) (major) 1-Bromo-1-methyl- cyclohex-1-ene Chemistry | 11.47

Sol: 5 6 Me H Conc. Me 1 H Me H 4 + H OH HS24O H H 3 (H) H H H (Z) or cis (E) or trans

4-Methylclohex-1-ene

Example 7: Me H+ Lindlar’s Catalyst 2 (B) HS24O

(A) Me Sol: Addition of H atom by Lindlar’s catalyst is a syn-addition. It reduces (C ≡ C) bond to (C = C) bond so the product (B) is H H

O/3 Reduction Example 8: CH10 16(A) CH58(B) Not-reslovable Not-resolvable aldehyde

O/3 Reduction CH58O2(C ) Not-resolvable acid 3 equivalent of H+2 Pd at 120oC

(D)

Write all the possible structures of (A), (B) and (C).

Sol: Since the reductive and oxidative ozonolysis products are different, the alkene of the type HH is possible RR If , the possible structures of (A) can be: R = ( — C H2 —)

, the possible structures of (A) can be : R = ( CH2 ( 5 6 H H H 3 mol 4 7 3 8 of H2 H 2 9 1 (D) 10 Both optically inactive Decane 4 5 O/3 Reduction O3/Oxidation 3 6 2 CH = O 2 COOH Me 2 7 Me (B) (B) 1 8 Optically inactive Optically inactive (D1) 2,7-Dimethyloctane 11.48 | Alkenes and Alkynes

2 1 (D) is obtained when (C1 – C3) bond breaks H ( H 3 ( 2 1 (D1) is obtained when (C2 – C3) bond H breaks during hydrogenation of (A). H ( 3 ( Example 9: Give the number of stereoisomers of (A) in the following reactions.

O3 /oxidation (a) (A) (C7 H 12 ) Acetone + Oxalic acid + Acetic acid (B) (C) (D)

O3 /oxidation (b) (A) (C8 H 14 ) Butane-2-one + Oxalic acid + Acetic acid (B) (C) (D) Sol: (a) Me O HOOC - COOH HOOC - Me Me (B) (C ) (D) Convert acid to aldehyde Me OO==HC - CH OO= HC - Me Me Remove and join the double bond to obtain the structure of (A) Me 1 23456 CH CH = CH Me Me (A) (C H) 712 6 H H Me Here Compound (A) shows G.I around C — C H 5 4 5 4 5 4 double bond (since the two groups around C-4 C C and C C 3 6 3 and C-5 are different) but do not show G.I around CH Me CH H Me1 2 Me1 2 C2 — C3 double bond because the two groups around C-2 are the same. Me (I) Me (II) Number of stereoisomers for (A) — 2 4Z-2-Methylhexa-2,4-diene 4E-2-Methylhexa-2,4-diene

(b) Proceed in the same manner as in (a) Me

O HOOC - COOH HOOC - Me

Me (B) (C ) Convert acid to aldehyde Me

OO=HC CH= OO=HC Me

Me

Remove O and join the double bond to obtain the structure of (A) Me 5 4 3 2 1 CH CH = CH Me 7 Me 6 (A) (C81H)8 (5-Methylhepta-2,4-diene) Me

O HOOC - COOH HOOC - Me

Me (B) (C ) Convert acid to aldehyde Me

OO=HC CH= OO=HC Me Chemistry | 11.49

Me

Remove O and join the double bond to obtain the structure of (A) Me 5 4 3 2 1 CH CH = CH Me 7 Me 6 (A) (C81H)8 (5-Methylhepta-2,4-diene)

(A) shows geometrical isomers around both (C2 — C3) and (C4 — C5) double bonds (two of the groups around these double bonds are different) Me H Me H Moreover, terminal groups (—CH ) and (—C H ), around C and around C , respectively, are different. 3 2 5 2 H 5 4 H Me 5 C H 5 C Number of G.I. when the terminal groups are different = 2n, where 4n is the number of double bonds (two3 2 double1 7 C C 7 C C Me 6 3 Me 6 bonds). 2 Me H 1 G.I. = 22 = 4 (I) (II) (2Z,4Z) (2E,4Z) Number of stereoisomers of (A) = 4

Me H Me H Me Me H H 4 H Me 5 4 H 4 H Me 5 C H 5 C C H 5 C 4 3 2 1 2 3 2 1 7 C C 7 C C 7 C C 7 C C Me 6 3 Me 6 Me 6 H 3 Me 6 2 Me H Me H (I) 1 (II) (III) 1 (IV) (2Z,4Z) (2E,4Z) (2Z,4E) (2E,4E)

ExampleMe 10: Explain the formationMe of theH products giving the structures of the intermediates. 5 4 H 4 H Me (a) C HCl H 5 C 2 + 3 2Cl +1 other products (D)+(E) 7 OHC C 7 Cl C C Me 6 H 3 Me 6 Me H (A)(III) 1 (B) (IV) (2Z,4E) (2E,4E) Sol:

(a) H -H2O Cl OH Cl OH2 (A) (B) (B) (I)

In this reaction presence of double bond influences the reaction pathway and unusual product may also be formed. Due to the presence of e–-rich double bond, the intermediate (I) carbocation may also involve intra-molecular rearrangement to form a three membered cyclic ring as shown below:

1 2 2 Cl Cl 3 3 4 4 ( I ) (C )

Other products can be obtained due to carbocation (I) rearrangement. 11.50 | Alkenes and Alkynes

Cl 1 1,2H shift H Cl Cl H Cl Me o Me o 2 allyl C (1 C ) o (E) (More stable) (1 allyl C )

Example 11: Complete the following reaction: 6 5 3 1 (i) BH/THF NaOH 46 (B) (C ) Cl (ii) HO,OH 4 2 22 (A) D Al23O or PO26 Sol: It is an example of hydroboration-oxidation reaction, it is used in the formation of alcohol from an alkene. It follows anti-Markovnikov’s rule, and the addition of H and OH is syn (cis)(addition of H and BH2 takes place from the same side) 5 3 BH 1 5 3 26 - 2 (A) 2 HO22/ OH THF Cl Cl 4 B 4 OH ( H 3( H (5-Chloropentane-1-ol) (B)

Cl OH O 5 1 NaOH -HCl 4 2 H 3 Dihydropyran (C )

JEE Advanced/Boards

Example 1: Complete the following reaction with appropriate reagents.

? ? ? (a) (B) (C )

(A) (D)

OH Me D D ? ? (B) D H (b) Me Me H D Me (A) Me Me Me

Sol: (a) Compound (D) Contains a (C = C) bond in the ring, Double bond can be obtained by the dehydration of alcohol.So, (C) is an alcohol.Alcohol (C) is obtained by the reduction of (>C = 0) using reducing agent like LAH. ∴ So (B) contains a keto group (>C = O) , (> C = O) group can be obtained from (A) by ozonolysis. The different steps are as follows:

O/Red [H] (Conc.H SO ) 3 O OH 24 LAH -H O OO 2 (A) (B) (C ) (D) H H/+ HO 2 C - C

H 1. Hg(OAC), HO 22 C - C 2. NaBH4

Chemistry | 11.51

(b) Proceeding reverse: (Retrosynthetic approach)

Compound (D) contains one Me and one –OH group. It can be obtained by adding (Cl2+ H2O) (HOCl) to (A). Thus compound (B) will have one OH and one Cl group; now one Me group can be added to (B) by Corey-House synthesis or Wurtz reaction of (B).

D D OH Cl 2 Cl + HO CH3Cl + Na + dry ether 1 22 D D Me Wurtz Reaction Me HO Cl H Me H Me Me Me HO Me (A) (B) Due to + I effect (CH32) LiCu D D Me Electromeric effect Corey-House H takes place from synthesis Me Me (C ) CC21®

Example 2: Complete the following reactions: Ph Alc. KOH (B) + (C) (a) Ph Ph and D Alc. KOH (B) + (C) (a) Ph Br and D (A) Br (A) Me

(b) + CHBr3 + MeMe O K (B)

(b) + CHBr3 + MeMe O K (B) (A) Me (A) O O Cl + KOH + MeOH (B) ( c ) Ph Cl + KOH + MeOH (B) ( c ) Ph

Sol: H Ph H H H Ph Alc. KOH (a) andD H Ph Ph H (-HBr) Ph Ph (B) (C ) Br E or trans Z or cis (More stable)

1,2-Diphenyl ethene or styrene

Me Br Me (b) Me O HC Br Me OH + CBr D CBr3+ Br Dibromo carbene Me Br Me

3 2 2 Br 4 1 Br -Br 3 + :CBr2 5 6 Br Ring expansion 4 6 (A) (A) 5 (Bond breaks) 11.52 | Alkenes and Alkynes

(c) It is an example of -elimination. O O b Cl H g Ph Ph or H OH a HH Cl

O H H atom are more Ph acidic due to the presence Cl of EWG ( C = O ( group O

Ph C

Example 3: Identify the products in the following reactions giving their stereoisomers (if any). 5Me 1 2 3 D+2 Poisoned Pd H2+Pt (a) (A)Me 5 (B) C (No. of isomers) 3 4 Me D+Poisoned Pd H +Pt (A)Me1 2 2 (B) 2 C (No. of isomers) (a) 4 Baeyer’s (b) (B)(from Problem A) (D) + (E) (Colour of precipitate) Baeyerreagent’s (b) (B)(from Problem A) (D) + (E) (Colour of precipitate) reagent

PhCOH+H2O ( c ) (B) 3 (D) PhCOH+H2O ( c ) (B) 3 (D) Na+liq. ND3+ EtOH Alk.KMnO (d) (F) 4 (D) (A) Na+liq. ND + EtOH 3 Alk.KMnOor 4 (d) (A) (F) (D) OSO4/NaHSOor 3 OSO /NaHSO PhCO4 32H/H O3

PhCO32H/H O

Sol: 1 1 (a) Me 4 Me 4 2 5 Me D+Poisoned 2 3 5 H +Pt 3 2 2 * * Me Me Me Syn-addition Pd H H (A) DD DD ( CC ) ( C = C ) (B) º Þ (C ) 2Z-2,3-Dideuteropent-2-ene 2,3-Dideuteropentane or or 2Z-2,3-Di-d-pent-2-ene 2,3-Di-d-pentane

Since Addition of D2 by (D2+Pd) is a syn addition Compound (B) is a cis compound and the addition of H2 + Pt is also syn. But (B) has two different (R) groups (Me and Et), so the product (C) will not be Meso but a racemic compound. Compound (C) contains two chiral carbon, thus the number of optical isomers will be 22 = 4 (two pairs of enantiomer). These are: 1 1 1 1 Me Me Me Me

HD2 * D 2 * H HD* D * H 3 3 H * D D * H D * H H * D 4,5 Et Et Et Et I II III II Racemate Racemate Chemistry | 11.53

Pairs of enantiomers: I and II; III and IV (I, III); (I, IV); (II, III), Pairs of diastereomers: (II, IV) 1 (b) Me Me 4 Baeyer’s reagent 5 2 * Me + MnO2 3 * (b) Me (Cold. alk. KMnO ) D D 4 D OH OH D (Brown-black ppt) (B) sys-hydroxylation) (E) (D) cis-Compound 2,3-Dideutentropent-2,3-diol with two different or R group 2,3-Di-d-pent-2,3-diol

[Two chiral centres; four stereoisomers as in (a)]

(B) PhCOH32 +H O ( c ) cis-Compound Anti-hydroxylation (D) with two different [As in (b)] R group

Me1 D 2 Me Na+liq.ND+3 EtOH 3 (d) Me (D) 5 (Anti-addition of two D) D 4 Me (A) (trans or E) (2E-2,3-Deuteropent-2-ene) or 2E-2,3-Di-d-pent-2-ene

Alk.KMnO4 or D [as a in (b)] F Trans compound OsO4S/NaH O3 (sys-hydroxylation)

PhCO32H/H O

Example 4: Identify (A), (B) and (C) in the following reaction.

H+2 Pt (A) (C71H)0 (B) (C71H)4

H+22 Ni B No reaction

[O] Acidic KMnO4

2 3 1 COOH (4-Oxohexan-1,6-dioic acid) 4 COOH or 4-Ketoadipic acid) O 6 5 (C )

Sol: First of all we have to calculate degree of unsaturation: (2n+− 2) n (7×+ 2 2) − 10 D.U. in (A) = CH= = 3º 22 11.54 | Alkenes and Alkynes

(7×+ 2 2) − 14 D.U. in (B) = = 1º 2

Since (A) does not react with P-2 or Brown catalyst (Ni2B) that reduces alkyne to alkene Therefore, (A) does not contain (C ≡ C) bond. As we have calculated, (B) has 1.D.U., it must be a ring (because after hydrogenation of (A) no double bond will be left) From the above data it appears that, (A) has one ring and two (C = C) bonds; hence, (A) is a diene.

Total number of C atoms in (A)=7  one C atoms is lost as CO2 Total number of C atoms in (C)=6

Possible structure of (A) are as follows: i. 2 3 1 COOH 7 CO + 2 4 COOH O 6 5 Proceeding reverse

Racemate

2 6 3 1 1 CH 5 CH 4 CH 4 2 2 6 HC 5 2 3 (I) (4-Methyl cyclohexene) ii. Other two possibilities of the combination are as follows:

1 4 7 6

(a) (COOH) Groups combines with ( C == H ) group and CO2 combination with (COOH).

O 1 HOOC 2 CH 2 7 6 O Þ O CO + HOOC 2 3 3 4 7 6 4 CH2 HC 5 5 Þ

1 2 2 3

2' 7 6 3 4 4 1 HC2 CH 3' 5 1' (II) (II) (Numbering in accordance (Numbering in accordance with naming) with problem) 1-(prop-2-enyl)cyclobut-1-ene or 1-Allyl cyclobutene Chemistry | 11.55

6 1

(b) (COOH) Groups combines with (C=O) group and CO2 is in combination with (COOH) 2 7 1 7 1 3 3 CO3 + HOOC HC2 CH 4 O 4 5 Þ 5 6 6 COOH (Numbering in accordance with problem)

4 2

3 1

(Numbering in accordance with naming) (4-(Cyclopropenyl)but-1-ene)

Structure III is not possible, because on hydrogenation (A) gives (B) (C7H14), i.e., it absorbs 2 mol of H2, whereas III will absorb 3 mol of H2. Because of highly strained ring Structure III is unlikely. Two possible structures are as :- 5

6 4

1 3 HC2 Me 2 (I) (B)

(C71H)4 (1-Methyl cyclohexane)

3 2 2 mol H2

4 1 Me (II) (B) 1-Propyl-cyclobutane

Example 5: Complete the following reactions: O Ph Ph 4 5 H 1 + O O D (Z) + CO 2­ Ph 3 2 H Disappearance Ph O of purple (X) Purple colour (Y) colour 2,3,4,5,-Tetraphenyl cyclopenta- Maleic 2,4-dien-1-one anhydride

Sol: It is an example of Diels Alder reaction. The mechanism is a concerted type (bond breaking and bond making takes place simultaneously) 11.56 | Alkenes and Alkynes

O O Ph Ph H 5 Ph 4 5 H 1' Ph 1 4 1 1' O O O O 2' 2' 3 2 H 3 Ph Ph 2 H Ph O Ph O Ph Ph H O O 5 Ph Ph 1 4 1' O O D O 2' -CO 3 Ph 2 Ph H Ph O Ph O (Z) (Colourless compound)

Example 6: Identify (A) to (H)

(A)

Reagent I II III IV V 100 atm, D H+Ni NH NH +O or 2 222 Sia BH+ H+2 Ni 2 H22+Pt+Al O3 CH3COOH 35 atm, D (B) (C ) (D) + (E) (gas) (F) + (G) (H)

Sol:

(A)

Reagent I II III IV V 100 atm, D H+Ni NH NH +O or 2 222 Sia BH+ H+2 Ni 2 H22+Pt+Al O3 CH3COOH 35 atm, D (B) (C ) (D) + (E) (gas) (F) + (G) (H)

(A)

Reagent IReagent II Reagent III Reagent IV Reagent V H 6 5 Me Me Me + Me + Me 4 2 3 (D) (F) (B) (C ) BSia2 N2(gas) H (E) H (G)

(H) Chemistry | 11.57

Example 7: Write the product with its mechanism.

Conc. HS24O B+B+H HO2 O Me 2 HO Me Me (A) Sol: 2 2 5 1 6 3 1,2 Me Shift 3 Me 4 4 5 Me Me Me 2Co Me Me

3Co (Numbering in according with problem for understanding)

7 5 2 1 8 6 4 3 -H 3 Saytzeff elimination 4 9 5 6 1 2 Me 10 Me Me Me Me Me

2,2,3-Trimethyl bicyclo [4.4.0] dec-1-ene

Example 8: Which of the following show G.I.? Me Me Me (a) (b) ( c ) N OH Me

ON Me H Me 2 (d) N.OH (e) N NH (f) 2 N NH NO2 Me H Me

(g) Ph N = N Ph (h)

Me Me

Sol: In order to show geometrical isomerism the alkene must contain different substituents i.e. the substituents on both the carbon bearing double bond must have different substituents (a) It shows G.I. H H H H

or or

H H H H

cis trans 11.58 | Alkenes and Alkynes

(i) Two atoms on the dotted line (...... ) means that the groups are below the plane of the ring. (ii) Two atoms on the bold lines (...... ) means that the groups are above the plane of the ring. (iii) One atom on the dotted line (...... ) means that one group is below and another is above the plane of the ring. (b) It shows G.I. CH CH3 CH3 CH3 3 CH CH CH3 CH3 3 3

or or

cis trans (c) It does not show G.I.

Me OH

N:

Me

Two same groups

Priority of C2H5 > CH3 and OH > LP of e s

Me

N Two higher priority group on the same side, on two different OH C atom of the double bond Me

anti or E

(e) It does not show G.I

H NH2 N

H

Two different groups Two same groups

(f) It does not show G.I

ON2

Me NH NO2 N

Me

Two different groups Two same groups Chemistry | 11.59

(g) It shows G.I.

Ph Ph Ph HC3 CH3 HC3 H H CH3 N = N N = N (a) **** Ph

Two different HHTwo different H CHanti3 or CHE 3 H groups groups (+) or D (-) or L (b) Cis has a plane of (h) It shows G.I.symmetry so O.I.A Trans is optical active (Meso from) ( c ) HC3 CH3 HC3 H H CH3 Me (d) ****Br Br Me Me HHH CH3 CH3 H (a) (b) ( c ) Me (+) or D (-) or L Cis hasMe a plane of Me HO Me symmetry so O.I.A Trans is optical active Me Me (Meso from) Me Me

Example 9: Give IUPAC name to each of the following using E or Z designations. Me

Br Me Br Me (A) (B) (C ) Me Me Me Me HO Me Me Me Me

Sol: (a) (b) 2' Me

Br 1 Me Br 1' Me 2 3 Me 2 3 HO Me 4 5 1 4 5 6 Me Two higher priority groups on opposite side,1E, 1-Bromo-2,3-dimethyl pent-2-ene Priority of (-Br) > (-CH23OH) and [(CH )23CH-]>(-C H)7 . Two higher priory groups on the same side. 2Z,2-Bromo-3-(1-methyl ethyl) hex-2-en-1-ol

(c) 1 2 5 6

3 4 2' Me Me 1 1 2 Me Me Me

Priority of HC º C- > (CH33) C- > CH23=CH-B> (CH )2CH- Two higher priority groups on the same side. 3Z,3-(1-methyl ethyl)-4-(1-1-Dimethyl ethyl)-hexa-1,3-dine-5-yne 11.60 | Alkenes and Alkynes ALKYNES

1. INTRODUCTION

1.1 Nomenclature

Alkynes are unsaturated hydrocarbons that contain (C≡C) bond. Their general formula is CnH2n–2.

Condensed Stick Bond line IUPAC name Common or trivial name

2 1 C2H2 H—C≡C—H Ethyne Acetylene

321 C3H4 H3C—C≡C—H Propyne Methylacetylene

C H CH —CH —C≡CH 4 But-1-yne Ethylacetylene 4 6 3 2 3 21 CH —C≡C—CH But-2-yne Dimethy lacetylene 3 3 4 32 1

C H CH —CH —CH —C≡C—H 4 3-Methyl-but-1-yne Propylacety lene 5 8 3 2 2 3 2 1 5

4 3-Methyl-but-1-yne Isopropyl-acetylene or 3 2 1 CH3

CH3—CH—C≡C—H

5 Pent-2-yne Ethyl methylacetylene or 4 3 2 1 CH3—CH2—C≡C—CH3

1.2 Structure

Each carbon atom of ethyne is sp-hybridized. One sp-hybridized orbital of each carbon undergoes head on overlap with sp-hybrisized orbital of another carbon to form a sp-sp, C-C, σ-bond. The second sp-hybridized orbital of each carbon overlaps along the internuclear axis with 1 s-orbital of each of the two hydrogen atoms forming two sp-s,

C—H, σ-bonds. Each carbon is now left with two unhybridized p-orbitals (2px and 2py) which are perpendicular to each other as well as to the plane of the C—C sigma bond.

The two 2px– orbitals, one on each carbon, are parallel to each other and hence overlap sideways to form a π-bond.

Similar overlap between 2py– orbitals, one on each carbon, results in the formation of a second π-bond as shown in figure.

Sp-s Sp-s C-H, s-bond C-H, s-bond 1s sp sp 1s sp sp (a) H C C H

sp-sp, C-C s-bond Chemistry | 11.61

In the plane Below the plane of the paper of the paper

2px (b) p 2p p x

C C p

p

Above the plane In the plane of of the paper the paper

120 pm 107.9 pm H C C H

180o

Carbon-carbon triple bond consists of one strong σ-bond and two weak π-bonds. The total strength of C ≡ C bond in ethyne is 823 kJ mol–1. It is stronger than the C = C bond of ethane (599 kJ mol–1) and C – H bond of ethane (348 kJ mole–1). Further, due to the smaller size of sp-orbitals (as compared to sp2 and sp3–) and sideways overlap of p-orbitals, the carbon-carbon bond length in ethyne is shorter (120 pm) than those of C = C (134 pm) and C–C (154 pm).

1.3 Isomerism in Alkynes

(i) Position isomerism. The first two members, i.e., ethyne and propyne exist in one form only. However, butyne and higher alkynes exhibit position isomerism due to the different position of the triple bond on the carbon chain. For example,

CH3—CH2—C≡CH CH3—C≡C—CH3 But-1-yne But-2-yne (ii) Chain isomerism. Alkynes having five or more carbon atoms show chain isomerism. CH3 4 3 | 2 1 CH3—CH2—CH2—C≡CH CH3—CH—C≡CH

(iii) Functional isomerism: Alkynes are functional isomers of dienes 4 3 31 1 2 34 CH — CH — C≡ CH CH= CH— CH = CH 32 22 But-1-yne Buta-1,3-diene (iv) Ring chain isomerism: Alkynes show ring chain isomerism with cycloalkenes. For example,

CH3—C≡CH and are ring chain isomers. Cyclopropene Propyne Cyclopropene 11.62 | Alkenes and Alkynes

2. PREPARATION OF ALKYNES

2.1 From Calcium Carbide

Alkynes are prepared by the following general methods. 1. By the action of water on calcium carbide: Ethyne (acetylene) is prepared in the laboratory as well as on a commercial scale by the action of water on calcium carbide.

CaC2 + 2H2O → HC≡CH + Ca(OH)2

Calcium carbide Ethyne (Acetylene) Calcium carbide needed for the purpose is manufactured by heating limestone (calcium carbonate) with coke in an electric furnace at 2275 K. 2275K CaCO3    → CaO + CO2 2275K CaO + 3C    → CaC2 + CO 2. Procedure: Lumps of calcium carbide are placed on a layer of sand in a conical flask fitted with a dropping funnel and a delivery tube. The air present in the flask is replaced by oil gas since acetylene forms an explosive mixture with air. Water is now dropped from the dropping funnel and the acetylene gas thus formed is collected over water. 3. Purification: Acetylene gas prepared by the above method contains impurities of hydrogen sulphide and phosphine due to the contamination of CaS and calcium phosphide in calcium carbide. Phosphine is removed by passing the gas through a suspension of bleaching powder. Pure acetylene is finally collected over water.

2.2 Double Dehydrohalogenation of Dihalides

Double dehydrohalogenation of a geminal dihalide H X | | R—C—C—R’ + 2NaNH2 R—C≡C—R’ + 2NH3 + 2NaX | | H X

Geminal dihalide Sodium amide Alkyne Ammonia Sodium halide

Double dehydrohalogenation of a vicinal dihalide H H | | R—C—C—R + 2NaNH2 R—C≡C—R’ + 2NH3 + 2NaX | | X X

Vicinal dihalide Sodium amide Alkyne Ammonia Sodium halide The most frequent application of these procedures are in the preparation of terminal alkynes. Since the terminal alkyne product is acidic enough to transfer a proton to the amide anion. One equivalent of base in addition to the two equivalents required for double dehydrohalogenation is needed. Adding water or acid after the reaction is complete converts the sodium salt to the corresponding alkyne. Chemistry | 11.63

3NaNH →2 HO2 1. (CH3)3CCH2CHCl2 (CH3)3CC≡CNa   → (CH3)3 CC≡CH NH3 1,1-Dichloro-3,3- Sodium salt of alkyne 3,3-Dimethyl- Dimethylbutane product (not isolated) 1-butyne (56-60%)

3NaNH2 HO 2. CH3(CH2)2CHCH2Br →CH 3(CH2)7C≡CNa  2 → CH3(CH2)7C≡CH NH | 3 Br

1,2-Dibromodecane Sodium salt of alkyne 1-Decyne (54%) Product (not isolated)

Double dehydrohalogenation to form terminal alkynes may also be carried out by heating geminal and vicinal dihalides with potassium ter-butoxide in dimethyl sulfoxide. By heating with alcoholic solutio`n of KOH

CH —Br 2 CH | +2 KOH (alc.)  →∆ ||| + 2KBr + 2H2O CH —Br 2 CH 1,2-Dibromodecane Acetylene (Ethylene dibromide)

PLANCESS CONCEPTS

Misconception: The reaction, in fact, occurs in two steps and each step involves the loss of a molecule of HBr as shown below: ∆ BrCH2—CH2Br + KOH (alc.)  → CH2=CHBr + KBr + H2O Ethylene dibromide Vinyl bromide ∆ CH2=CH—Br + KOH (alc.)  → CH≡CH+KBr+H2O Vinyl bromide Acetylene Aman Gour (JEE 2012, AIR 230)

Explanation: In ethylene dibromide, Br is present on a saturated carbon atom. Therefore. Like alkyl halides, it is a reactive molecule. Consequently, on heating with alcoholic KOH, it readily eliminates a molecule of HBr to form vinyl bromide in good yield. In contrast, due to the presence of Br on a doubly bonded carbon atom, vinyl bromide is a highly unreactive molecule and hence on heating with alcoholic KOH, it does not easily lose a molecule of HBr to form acetylene. Thus, with alcoholic KOH, the yield of acetylene is low. Therefore, to obtain acetylene in fairly good yield from vinyl bromide, a much stronger base than alcoholic KOH such as NaNH2 in liquid NH2 is usually used. Thus, dehydrohalogeneation of ethylene dibromide to acetylene is preferably carried out in the following two stages. HH H KOH(alc.), NaNH /liq.NH RHCC D HHCC 23HHC C -HBr -HBr º Br Br Br Acetylene Vicinal dihalide Vinyl bromide

Θ With a strong base (NH2 ), isomerisation also taken place to give terminal alkyne. 11.64 | Alkenes and Alkynes

:

(i)NaNH2, 400K Me C º CMe Me C ºC Na

HO2

Me CCº H Prop-1-yne

Dehalogenation of tri/tetra halides Br

Me1 Br 3 (i) 2(Zn+EtOH) 4 321 Me C C Me + 2ZnBr 2 or º 2 (But-2-yne) or (ii) 2(Mg+EtOH) Br Me 2MgBr or 2 or Br (iii) 4(NaI in acetone) 4NaBr +2I 2,2,3,3,-Tetra bromobutane 2

3. By dehalogenation of tetrahalides.: Tetrahaloalkanes when heated with zinc dust in methanol undergo dehalogenation to yield alkynes. For example Br Br

CH3OH HCCH+ 2 Zn HCº C H + 2 ZnBr2 D Acetylene Br Br 1,1,2,2-Tetrabromoethane

2.4 Dehalogenation of Haloforms

By dehalogenation of haloforms: Chloroform and iodoform on heating with silver powder undergo dehalogenation to form ethyne. ∆ CH Cl33 + 6 Ag + Cl CH HC CH + 6 AgCl Ethyne Chloroform Chloroform ∆ CH l33 + 6 Ag + l CH HC CH + 6 AgI Iodoform Iodoform Ethyne

2.5 Kolbe’s Electrolytic Reaction

Kolbe’s electrolytic reaction: Acetylene can be prepared by electrolysis of a concentrated solution of sodium or potassium salt of maleic acid or fumaric acid. Thus,

H COOK H COOK

C Electrolysis C + 2H O CHº or 2 + 2CO+22 H+ 2KOH C C CH Acetylene H COOK KOOC H Pot. maleate Pot. fumerate (cis isomer) (trans-isomer)

This reaction is called Kolbe’s electrolytic reaction and is believed to occur by the following steps:

CHCOOK CHCOO + ionaization ; - + + 2 K 2 HO2 2OH + 2 H CHCOOK CHCOO CH CHCOO CHCOO - 2e- + 2CO2 CH CHCOO CHCOO unstable CHCOOK CHCOO + ionaization ; - + Chemistry | 11.65 + 2 K 2 HO2 2OH + 2 H CHCOOK CHCOO CH At anode: CHCOO CHCOO - 2e- + 2CO2 CH CHCOO CHCOO unstable

+− At cathode: 2H+ 2e  → [2H]  → H2

2.6 From α-diketo

α- Diketone reacts with hydrazine (NH2 – NH2) to give bis-hydrazone which on oxidation with HgO gives unstable bis(diazo) compound which decomposes to give alkyne.

O N NH2 4 2 1 3 Me 2NH22NH Me Me Me

O N NH2 Butan-2,4-dione Bis-hydrazone of (I) (1) HgO

N

N

-N2 Me Me Me But-2-yne Me

N

N Bis-(diazo) of (I) 2.7 Industrial Method

1773 K 2CH42( g) → CH ≡ CH( g) + 3H( g)

The reaction is highly endothermic, yet the optimum time for this reaction is 0.01 second.

∆n = nP – nR = 4 – 2 = 2 mol, and this causes a significant increase in∆ S. At this high temperature, the T∆S term in the equation ∆G = ∆H – T∆S predominates, making ∆G = –ve , although ∆H = +ve.

° –1 But ∆Hf for C2H2 = +227 k mol shows that acetylene is thermodynamically unstable, explodes readily, and gets converted into its elements.

° –1 C2H2(g)  → 2C(s) + H2 (g) , ∆H = – 227 kJ mole

2.8 Berthelot Synthesis

Mearcellin Berthelot synthesized acetylene from its elements (carbon and hydrogen) by striking an electric arc between two electrodes in an atmosphere of H2 gas.

2C + H2 CH º CH(g) Acetylene

2.9 Higher Alkynes from Acetylene

Synthesis of higher alkynes from acetylene: Acetylene is first treated with sodium metal at 475K or with sodamide in liquid ammonia at 196 K to form sodium acetylide. This upon treatment with alkyl halides gives alkyne. 11.66 | Alkenes and Alkynes

For example:

Liq.NH3 ,196K −+ HC≡ CH + NaNH23 →HC ≡ C Na + NH Ethyne(Acetylene) Sodiumacetylide

- + HC º C Na +BCH3 r HC º C CH3 + NaBr Sod. acetylide Bromomethane Propyne

- + HC º C Na + CH3 CH2 I HC º C CH23CH + NaI Sod. acetylide Idoethane But-1-yne

- + HC C CH CH CH + NaBr HC º C Na + CH3 CH2CH2 Br º 223 Sod. acetylide 1-Bromopropane Pent-1-yne (n-propyl bromide) An analogous sequence usually terminal alkynes as starting material yields alkynes of the type RC = CR’.

N23 NH CH3 Br (CH32 ) CHCH 2 C≡ CH → (CH32 ) CHCH 2 C≡ CNa → (CH32 ) CHCH 2 C≡ CCH 3 2− Methy −− 1 Pentyne NH3 5− Methyl −− 2 Hexyne − 181%

Dialkylation of acetylene can be achieved by carrying out the sequence twice.

1.NaOH,NH331.NaOH,NH HC≡ CH →HC ≡ CCH23 CH →CH3 C ≡ CCH 23 CH Acety lene 2.CH32 CH Br 2.CH3 Br 2− Pentyne(81%)

3. PROPERTIES OF ALKYNES

(i) No cis-trans isomerism: The sp hybrid orbitals are linear, ruling out cis-trans stereoisomers in which substituents must be on different sides of the multiple bond. (ii) Bonds: Bond length decreases in the following order: Alkane > Alkene > Alkyne

Bond length order: −CH − >= CH − >≡ CH − sp32 sp sp

Bond Strength order: ≡−>=−>CH CH –CH − sp sp23 sp

As a general rule ‘more the s – character of hybrid orbitals used by an atom, closer are the bonding electrons to the atom and shorter and stronger is any of its σ-bonds’. (iii) Solubility and Dipole moment

Decreasing solubility in H2O: Alkynes > Alkenes > Alkanes

Alkynes are slightly more soluble in H2O because they are somewhat more polar. The dipole moment of alkyne is greater than those in alkene because (C — C) sp bond is more polarized than (C — C) sp2 bond: this is because C with more s character is more electronegative. Dipole moment of but-1-yne is 0.8 D and that of but-1-ene is 0.3 D. Due to a slightly high polarity and dipole moment of alkynes than those of alkenes (and of alkenes higher than alkane) boiling point of alkynes > alkenes > alkanes, with the same number of C atoms. They go from gases to liquids to solids with increasing molecular weights. There is a little difference in boiling points of these hydrocarbons with similar C skeletons. Chemistry | 11.67

(iv) Smell: All alkynes are odourless except acetylene which has a garlic smell due to the presence of impurity, phosphine. (v) Physical state: The first three members are gases, the next eight are liquids, and the rest are solids. (vi) Melting and Boiling points: The m.p. and b.p. of alkynes are slightly higher than those of the corresponding alkenes and alkanes. This is probably due to the presence of a triple bond, alkynes have linear structures and hence their molecules can be more closed packed in the crystal lattice as compared to those of corresponding alkenes and alkanes. (vii) Densities: Densities of alkynes like those of alkenes and alkanes increase as the molecular size increases. However, they are all lighter than water, since densities lie in the range 0.69 – 0.77 g/cm2. ° (viii) Stability: Alkynes are less stable than isomeric dienes as is evident from their heat of formation ( ∆Hf ) values. 2 4

1 5 (1,4-Pentadiene) 3 +106.0 kJ mol-1 Me 5

1 Me (Pent-2-yne) +129.0 kJ mol-1 4 3 2 4 2 1 5 H (Pent-2-yne) +144.5 kJ mol-1 Me 3

(ix) Reactivity: However, alkynes are less reactive than the corresponding alkenes towards electrophilic addition (EA) reaction (except catalytic hydrogenation), even though they contain two π-bonds. This is due to: (a) Greater EN of sp-hybridised C atom of alkynes than sp2 –hybridised C atom of alkenes which hold the π - electrons of alkynes more tightly. (b) Greater delocalisation of π -electrons in alkynes (because of cylindrical nature of their π -electrons cloud) than in alkenes. As a result, π -electrons of alkynes are less easily available for addition reaction than those of alkenes. So alkynes are less reactive than alkenes towards electrophilic addition reaction. Catalytic hydrogenation, however, is an exception: Alkenes are adsorbed on the surface of catalyst only when the plane of π-bond approaches perpendicularly. In alkynes because of the cylindrical nature of π-bonds, approach by hydrogen along the axis of the cylinder is more effective

and thus transition state in alkynes is less strained. So alkynes react faster than alkenes with H2. (x) Acidity of alkynes: Terminal alkynes are more acidic than alkenes because alkyne C-atoms are sp hybridized and electrons in an s-orbital are more tightly held than in a p-orbital also s-electrons are closer to the nucleus. Furthermore, since the sp orbital has more s character (50% s) than sp2 (33% s) or sp2 orbital (25% s), the electrons in sp orbital are more tightly held by the nucleus than electrons in sp2 or sp3 orbital. In another words, sp-hybridised C is more EN than sp2, or sp3-hybridised C atom. Due to this greater EN, the electrons of (C — H) bonds are displaced more towards the C atom than towards the H atom. Therefore, the H-atom is less tightly held by the C atom and hence can be removed as a proton (H+ ion) by a strong base and consequently terminal alkynes behave as acids.

R — C C — H + strong base R — C C — Base — H As a rule, as the s character decreases from sp to sp2 to sp3 hybridised C atom, the acidic character of hydrocarbons decreases in the following order. Me− HC = O ← → Me − C⊕Θ H − O 11.68 | Alkenes and Alkynes

4. REACTIONS OF ALKYNES

4.1 Acidic Character

(i) Formation of alkali metal acetylides: Ethyne and other terminal alkynes (Alkynes in which the triple bond is at the end of the carbon chain) or 1-alkynes react with strong bases such as sodium metal at 475 K or

sodamide in liquid ammonia at 196 K to form sodium acetylides with evolution of H2 gas.

475K −+ 2HC≡ CH + 2Na   → 2HC ≡ C Na + H2 Ethyne Monosodiumeethynide (Acetylene) (monosodiumacetylide)

Liq.NH3 −+ R− C ≡ CH + NaNH23 →R − C ≡ C Na + NH (A terminalalkyne) 196K (A sod.alkynide) During these reactions, the acetylenic hydrogen is removed as a proton to form stable carbanions (acetylide ions) Sodium acetylide is decomposed by water regenerating acetylene. This shows that water is a stronger acid than acetylene and thus displaces acetylene from sodium acetylide.

−+ HC≡ C Na+ H2 O  → HC ≡ CH + NaOH (Monosodiumacetylide) (Acetylene)

(ii) Formation of heavy metal acetylides: Acetylenic hydrogens of alkynes can also be replaced by heavy metal ions such as Ag+ and Cu+ ions. For example. When treated with ammoniacal silver nitrate solution (Tollen’s reagent), alkynes form a white precipitate of silver acetylides.

+− CH≡ CH + 2[Ag(NH32 ) ] OH → AgC = CAg + 2H2 O + 4NH 3 Ethyne Tollensreagent Disilver ethynide

+− R−≡ C CH + [Ag(NH32 ) ] OH→−≡− R C C Ag + H2 O + 2NH 3 (Ter minalkyne) Tollensreagent Silver alkynide (Whiteppt.)

Similarly, with ammoniacal cuprous chloride solution, terminal alkynes form red ppt. of copper acetylides.

+− HC≡ CH + 2[Cu(NH32 ) ] OH → CuC ≡ CCu + 2H2 O + 4NH 3 (Ethyne) Dicopper ethynide

+− R−≡ C CH + [Cu(NH32 ) ] OH→−≡− R C C Cu + H2 O + 2NH 3 (Terminalalkyne ) Tollensreagent Monocopper alkynide (Redppt.)

Unlike alkali metal acetylides are not decomposed by water. They can however, be decomposed with dilute mineral acids to regenerate the original alkynes.

AgC≡ CAg + 2HNO33  → HC ≡ CH + 2AgNO (Disilver acetylide) Acetylene

CuC≡ CCu + 2HCl  → HC ≡ CH + 2CuCl (Dicopper acetylide) Acetylene

(iii) Formation of alkynyl Grignard reagents: Acetylne and other terminal alkynes react with Grignard reagents to form the corresponding alkynyl Grignard reagents. For example.

Dry HC≡ CH + RMgX  → HC ≡ CMgX + RH (Acetylene) Grignardreagent Ether Acetylenic Alkane Grignard reagent

R'−≡ C CH + RMgX →−≡Dry R' C CMgX + RH Terminalalkyne Grignardreagent Ether AlkynylGrignard reagent Alkane Chemistry | 11.69

Alkynyl Grignard reagents like usual Grignard reagents can be used to prepare a variety of organic compounds. Importance: The formation of metal acetylides can be used: (a) For the separation and purification of terminal alkynes from non-terminal alkynes, alkanes and alkenes. (b) To distinguish terminal alkynes from non-terminal alkynes or alkenes.

4.2 Hydrogenation / Reduction

cat RC≡ CR' + 2H2   → RCH22 CH R' Catalyst = Pt, Pd, Ni or Rh

Alkene is an intermediate

H2 Pt CH3CH2CH2CH3

HC3 CH H/24Pd/BaSO 3 HC CC CH 33º Quinoline C=C (Lindlar’s catalyst) H H

HC3 H Na,NH3(liq.) C=C

H CH3 The general case H H H H 2 C = C 2 HC3 C º C CH3 CH CH CH CH PVC PVC 3 2 2 3 HC3 CH3 The reaction takes place on the surface of Pt or Pd. Since the addition is twice, it is hard to see the syn addition. But the reaction can be stopped at the alkene stage by using the following reagents.

(a) H2 + Pd + BasO4 + S or quinoline in boiling xylene (called Lindlar’s catalyst). In this case, reaction proceeds

via syn addition of H2.

(b) H2 + P-2 or Brown catalyst (H2 + Ni + B) or (H2 + Ni2B). It also proceeds via syn addition of H2.

(c) Alkali metals (Na, K, Cs) + liq. NH3 and C2H5 OH (It is called Birch reduction). It proceeds via anti-addition of H2.

(d) LiAlH4(LAH). It reduces (C≡ C) to (C = C) via anti addition of H2; does not reduce (C = C) to (C — C) bond; reduces (C=C) to (C —C) only when the phenyl group is attached to the -C atom of the double bond. βα

(E.g. Ph−=− CH CH CH3 to Ph −−− CH 223 CH CH ) Reduction of alkynes to (Z) – alkenes

Lindlar’s catalyst: Pd, quinoline, Pb and CaCO3 poisons the metal catalyst. So that the H2 adds only to the alkyne. Not reactive enough for alkenes. H H H 2 C = C CH3(CH22) C º (CH22) CH3 Pd/CaCO3/Pb CH (CH ) (CH ) CH Lindlar catalyst 3 22 22 3 96% cis H (CH22) CH3 Na/NH 3 C = C CH3CH2 C º C (CH23)CH3 H CH3CH2 98% trans H H H 2 C = C CH3(CH22) C º (CH22) CH3 Pd/CaCO3/Pb 11.70 | Alkenes and Alkynes CH (CH ) (CH ) CH Lindlar catalyst 3 22 22 3 96% cis H (CH22) CH3 Na/NH 3 C = C CH3CH2 C º C (CH23)CH3 H CH3CH2 98% trans

Reduction of alkynes to (E) – alkenes. Reaction via Na radical dissolved in liquid NH3. ● + – Na + NH3 → Na + e [NH3]n Solvated electron +−°33 C +− Mechanism of Na/NH3 reductions: Na   → Na + NH3 (e )

R R H R H R H Na H NH Na NH H R C C R C C 2 C C C C 2 C C -NaNH + 2 + Na R R R H R NaNH2

4.3 Addition of X2 Like alkenes, 1 mol of alkynes adds 2 moles of halogens and proceeds via anti-addition of halogens, and orange or brown colour of Br2 is discharged (test for unsaturation)

H Br HBr HB+ r- 2 1 (I) HC CH HC=CH HC CH Markovnikov’s 2 (EA) Br Br 1,1-Dibromo ethane or Ethylidene bromide

(II) Me4 Me4 Br Mechanism: 2 1 HBr 2 1 HBr 3 2 1 H: CH Me ⊕ Like alkenes it proceed3 via the formationMarkovnik ofov’ s intermediate3 cyclic 2 bromoniumMarkovnik ov’ ion.s First, electrophile (Br ) add But-1-yne Θ 4 and then the addition of nucleophile (EA)(Br ) from anti-positionBr take place.(EA) That is whyMe this reaction is called EA (electrophilic addition) reaction. The reaction is stereospecific2-Brom (attacko by electrophile in a specificBr manner). 2,2-Dibromo Anti-Markovnikov’s But-1-ene HBr+ROOR butane free radical H (Peroxide)H Br addition H Br Br Br HC CH C = C C = C R.D.S H Me Br Slow4 Me 4 1 Br Br Br 3 2 1 HBr+ROOR 3 2 Br Br H H Br H Br Br C — C H C — C H 1,1-Dibromobutane 1-Bromobut-1-ene Br Br Br (ButylideneH bromide) Br

Cl Cl

CCl4 Cl24/CCl - HCl CHºCH + Cl3 CH = CH H - C - C - H Cl2C = CHCl Ba(OH)2 Cl Cl Cl Cl or Trichloroethylene Westron or 1,1,2,2,-tetrachloro or westrosol BaCl2 ethane or acetylene tetrachloride Chemistry | 11.71

4.4 Addition of Water

Hydration of alkynes O || H2 O/H 24 SO HC3− C ≡ C − H →HC33 − C − CH HgSO4 H H

+ H HO2 H3 C– C ≡ C–H H3 C– C = C – H H3 C– C = C – H + –H+ s-viniylcation OH H an enal

H3 C– C = C – H H33 C– C = CH

O O H Ketone enol

(Oxymercuration-demercuriation of triple bond)

2+ Hg OH2 Me – C C – H Me — C = C — H 2+ Hg

Attack of H2 O on H2O more substituted C atom

HO H ⊕ HO C = C HO3 C = C H 2+ Me H –Hg Me Hg Tautomerise

O

Me – C – Me

4.5 Addition of HX

H Br Br Br Br H CH3 CCº H C C C C CH CH3 2 HC CH CH3 CH H 3 3 CH3 HC3 Br 3 H Br HBr HB+ r- 2 1 (I) HC CH HC=CH HC CH Markovnikov’s 2 (EA) Br Br 1,1-Dibromo ethane or Ethylidene bromide

(II) Me4 Me4 Br 2 1 HBr 2 1 HBr 3 2 1 H: CH Me 3 Markovnikov’s 3 2 Markovnikov’s But-1-yne 4 (EA) Br (EA) Me 2-Bromo Br 2,2-Dibromo Anti-Markovnikov’s But-1-ene HBr+ROOR butane free radical (Peroxide) addition Br

Me 4 Me 4 1 Br 3 2 1 HBr+ROOR 3 2 Br H H H 1,1-Dibromobutane 1-Bromobut-1-ene (Butylidene bromide) H Br HBr HB+ r- 2 1 (I) HC CH HC=CH HC CH Markovnikov’s 2 (EA) Br 11.72 | Alkenes and AlkynesBr 1,1-Dibromo ethane or Ethylidene bromide

(II) Me4 Me4 Br 2 1 HBr 2 1 HBr 3 2 1 H: CH Me 3 Markovnikov’s 3 2 Markovnikov’s But-1-yne 4 (EA) Br (EA) Me 2-Bromo Br 2,2-Dibromo Anti-Markovnikov’s But-1-ene HBr+ROOR butane free radical (Peroxide) addition Br

Me 4 Me 4 1 Br 3 2 1 HBr+ROOR 3 2 Br H H H 1,1-Dibromobutane 1-Bromobut-1-ene (Butylidene bromide)

PLANCESS CONCEPTS

The order of reactivity of HX is HI > HBr > HCl > HF Aishwarya Karnawat (JEE 2012 AIR 839)

4.6 Addition of Acids

Addition of carboxylic acids: When acetylene is passed into warm acetic acid in presence of mercury salts, first vinyl acetate and then ethylidene diacetate is formed.

2+ CH COOH HC≡ CH + CH COO − H →Hg H C = CH − OCOCH →3 CH − CH(OCOCH ) 3 2 3 + 3 32 Acetylene Aceticacid 353K vinyl acetate Hg2 Ethylidenediscetate

Vinyl acetate is used for the manufacture of vinyl resin. Ethylidene diacetate, when heated rapidly to 573-673K, gives acetic anhydride and acetaldehyde.

OCOCH CH CO 3 573-673K 3 O + CH3CHO CH3 CH OCOCH3 CH3 CO Acetaldehyde Ethylidene diacetate Acetic anhydride

4.7 Addition of HCN The reaction is nucleophilic.. addition (NA) initiated by CN– from NaCN. HCN HC CH + CN CH = CH CH2 = CH CN +H Vinyl cyanide CN or Acrylonitrile

4.8 Addition to Other Compounds

CH2 O CH3 HSO CH O S OH 24 CH(OSO H) HC CH + Conc. HS24O 2 32 O Vinyl hydrogen sulphate Ethylidene hydrogen sulphate Chemistry | 11.73

CHCl HOCl CHCl -H O CHCl HC CH + HOCl 2 2 2 Hypochlorous CHOH D CH(OH)2 CH=O acid 2-chloro Dichloro acetaldehyde eth-1-en-1-ol or 2,2-Dichloro ehtanal

CH3OH CH CH CH3OK 2 CH3OK 3 HC CH + CH3OH 430-470K CH(OCH ) CH OCH3 32 under pressure Methyl vinyl Ethylidene ether methoxide

Hg+ HC CH + Conc. HNO CH(NO ) 3 oxidative 33 Nitroform cleavage

4.9 Ethyne + Methanal

The addition of ethyne to unsaturated link like (>C = O) is called ethinylation. Acetylene or terminal alkyne containing ≡ CH (a methine H-atom) reacts with a carbonyl group (>C = O) in the presence of sodium or potassium alkoxide

(RONa or ROK) or NaNH2 to give alkyndiol with a small amount of alkyneol. .. CH3O H C CH CH3OH + C CH .. 3 2 1 CH3OH H C C CH OH HC C CH2 = O HC C CH2 O 2 Propargyl alcohol or Formaldehyde Prop-2-yn-1-ol

CH O .. 3

C C CH3OH

CH2 = O 1 234 CH3OH HOCH C C CH2OH + CH3O 2 O HC2 C C CH2OH But-2-yn-1,4-diol

4.10 Nucleophilic Addition

Because of the greater electronegativity of the sp-hybridized carbons as compared to sp2 hybridized carbons, alkynes are more susceptible to nucleophilic addition reactions than alkenes. For example, when acetylene is passed into methanol at 433-473 K in presence of a small amount (1-2%) of potassium methoxide under pressure, methyl vinyl ether is formed.

−+ CH3 O K HC≡ CH + CH3 O − H →CH23 = CH − OCH − Acety lene Methanol 433 473K Methyl vinyl ether

−+ CH3 O K HC≡ CH + CH3 O − H →CH23 = CH − OCH − Acety lene Methanol 433 473K Methyl vinyl ether

Methyl vinyl ether is used for making polyvinyl ether plastics. 11.74 | Alkenes and Alkynes

4.11 Propargylic Halogenation

Alkenes undergo allylic substitution with NBS, whereas alkynes undergo propargylic halogenation with NBS, Cl2SO2

(sulfuryl chloride). Me3COCl(t-butyl hypchlorite), and Cl2 at 500°C. O

4Me NBS+CCl 4Me 2 1 4 3 2 1 H H + N-H 3 +h

But-1-yne n Br 3-Bromo but-1-yne O

Propargylic O-Cl 4Me

H-atom 2 3 1 OH H + +CCl + h 4 n Cl 3-Chlorobut-1-yne

Reaction proceeds via free radical mechanism to give propargylic radical, which is stabilized by resonance via extended π-bonding.

4.12 Hydroxylation of Alkynes

Hydroxylation of alkyne with aqueous or neutral KMnO4 solution (Baeyer’s reagent) test for unsaturation)

Pink colour of KMnO4 is discharged and brown black precipitate of MnO2 is obtained. This reaction converts alkynes first to enediols and then further gives tetraols, which being unstable lose 2H O to give diketones.

OH OH 1 234 aq. KMnO 1 2 3 4 OH (But-2-en 2,3 diol) Me C C Me 4 Me C C Me OH OH OH

But-2-yne aq. KMnO4 1 2 3 4 Me Me Me C C Me -2H2O H O CCOH OO OH O H Butan 2,3 dione Unstable

Me CCMe Me C C Me OO (Break two bonds and add O O O atom on both sides)

OH 1 OH OH 3 OH aq. KMnO H H 4 1 2 3 2 OO 4 O O

But-1-en-3-yne 3,4-Dihydroxy-2-oxo butan-1-al Chemistry | 11.75

4.13 Oxidation Reactions

4.13.1 Oxidative Cleavage (a) Alkynes are oxidatively cleaved in alkaline or acidic conditions at higher temperature, as mentioned.

RC R COOH

HC [H COOH] CO2

1 Me 3 (i) 6 Me 3 2 3 4 7 [O] 1 1 4 C C COOH + HOOC 5 Me 2 2 Me Hept-3-yne Propanoic Butanoic acid acid (ii) Me Me 3 3 2 1 [O] 1 C C H 2 COOH+[HCOOH] But-1-yne Propanoic acid CO2 6 3 (iii) Me Me 543 21 [O] 2 1 C CC C Me COOH + [HOOC COOH] Oxalic acid Hexa-2,4-diyne Propanoic acid +HOOC-Me 2CO2 Ethanoic acid

CrO3 in CH3COOH oxidizes alkenes but does not affect alkynes

4.14 Ozonolysis of Alkynes

Alkyne form ozonides with O3 and are decomposed by H2O to give diketones, which are then oxidised by H2O2 or

KMnO4/NaIO4 or peracids; given acids, and on reduction with metal / acid, LAH, or NaBH4 give diols. O

(i) HC CH+ O3 HCCH

OO

HO2

HC2 CH2 Reduction H C C+H HO22 Zn+CH3COOH OH OH or OO or HO HO OH LAH or NaBH4 ( HO ( Glyoxal Glycol HO22 [O] 2CO2 2HCOOH Formic or methanoic acid 11.76 | Alkenes and Alkynes

Terminal alkynes give HCOOH as one of the products which is further oxidized to CO2 These oxidative cleavage reactions are rarely used for synthesis because alkynes are not readily available. They are used to locate the position of a triple bond in an alkyne.

4.15 Hydroboration of Alkynes •• Terminal alkynes of HBO reaction give aldehyde, whereas internal alkynes give ketone.

•• Alkynes react with BH3 or B2H6 complexed with THF (tetrahydrofuran O) to give trivinyl borane, which upon

subsequent treatment with alkaline H2O2 gives alcohols corresponding to anti-Markovnikov’s addition of H2O to alkynes, which on tautomerisation give corresponding aldehydes or Ketones.

•• If trivinyl borane is treated with acid(CH3COOH), it gives cis-alkene (proceeds via syn-addition of H2).

Me

3BH +THF 3Me CCº Me 3 Me C = C B But-2-yne H

Syn-addition HO22/OH COOH CH 3 4 1 Me Me Me 3 2 C = C Me CH=C H H Z-But-2-ene OH

Tautomerise

421 1 Me CH2 C Me

O Butan-2-one 4.16 Polymerisation

1. Linear Polymerisation 1

Cu22Cl 2 3 HC CH + HC CH 4 CH CH + 1 2 3 4 5 6 NH4Cl or CH2 = CH C C CH = CH2 Cu Cl + NH Cl 2+ 1234 22 4 1 or Cu +O CH = CH C CH 6 2 ( 2 ( 2 3 4 5 Vinyl acetylene H C C C C H But-1-en-3-yne But-1,3-dyne Divinyl acetylene or 1mol 1mol Hexa-1,5-dien-3-yne HCl HCl Cl Cl 4 3 2 1 Peroxide/h H2 + Ni2B n HC C C= CH CH = CH C = CH 2 2 2 Polymerisation CH CH = C CH ( 2 2 n( Cl Chloroprene or Neoprene or 2-Chloro-buta-1,3-diene Polycholoropene Chemistry | 11.77

2. Cyclic Polymerisation CH CH 1 2 CH 8 3 CH CH CH CH Red hot tube Ni(CN)4 or 3 (HC CH) or 4 (HC CH) 4 or THF 7 CH CH CH CH Hot copper 6 5 Benzene HC CH CH Cycloocta 1,3,5,7 tetra-ene Me 3. Propyne Me CH CH CH or 3 (Me C CH) Red hot tube Me CH C Me Me Me HC 1,3,5-Trimethyl benzene or Mesitylene

4.17 Isomerization

When alkynes are heated with NaNH2 in an inert solvent such as kerosene oil or paraffin oil, they undergo isomerization. i.e. 2-alkynes isomerize to 1-alkynes and vice versa.

Uses of alkynes (i) Acetylene and its derivatives are widely used in synthetic organic chemistry for synthesis of cis-and trans- alkenes, methyl ketones etc. (ii) Oxyacetylene flame is used for cutting and welding of metals. (iii) Acetylene is used as illuminant in hawker’s lamp and in light houses. (iv) Acetylene is used for the ripening of fruits and vegetables. (v) Acetylene is used for manufacture of ethyl alcohol, acetaldehyde, acetic acid, vinyl plastics, synthetic rubbers such as Buna N and synthetic fibers such as Orlon.

PLANCESS CONCEPTS

Difference between Alkenes and Alkynes

Test Alkene Alkyne 1. Flame observed on comnustion Luminous Smoky

2. Br2/CCl4 solution Orange colour is discharged Orange colour is discharged

3. Cold aq. or alkaline KMnO4 Pink colour is discharged with the Pink colour is discharged with the solution (Baeyer's reagent) formation of diol compound formation of dicarbonyl compound 4. Ammoniacal AgNO3 solution No action Whote ppt. of silver alkynide (given [Ag(NH3)2]+ (Tollens reagent) by only terminal alkynes) 5. Ammoniacal CuCl solution No action Red ppt. of copper acetylide (given by only terminal alkynes) 6. Reactivity towards electrophilic Alkenes are more reactive than alkynes

addition reaction (e.g. HX, X2) 7. Reactivity towards cataytic Alkynes are more reactive than alkenes 8. Reactivity towards nuclephilic Alkynes are more reactive than alkenes addition reaction (e.g. CN+) 11.78 | Alkenes and Alkynes

POINTS TO REMEMBER

H HX C - C X H HBr C - C (Peroxides) Halogenation Br Reaction H X 2 C - C X H X22/H O(”HOX”) C - C (NBS, H O/DMSO) 2 X

H OH Anti-Markovikov 1. BH , THF 3 regiochemistry + C - C 2. HO,22 HO Syn addition H OH Syn 1. OOs 4 C - C Syn addition addition 2. NaHSO22, HO

HO OH KMnO , HO+/H O 42 C - C Syn addition Hydroxylation

C = C H Markovnikov H/+ HO 2 C - C regiochemistry OH Carbocation Anti intermolecular addition H 1. Hg(OAC), HO Markovnikov 22 C - C 2. NaBH regiochemistry 4 OH

H H H,2 Pt Reduction C - C Syn addition

Cl Cl

Dichloro methylene insertion C

CHCl3, KOH CC Ring forming Reaction HH

Methylene insertion C CHI, ZN(Cu), ether 22 CC Chemistry | 11.79

X H 2 HX Markovnikov C - C H R regiochemistry Halogenation X H Reaction X H 2 HX2 X R C - C X X

H H Complete reduction R C - C H H,2 Pt H H Reduction H H H, Lindlar catalyst 2 C = C Syn addition R Partial H reduction H H Li, NH (liq.) 3 C = C Anti addition R H

RC CH

O H + Markovnikov H/HO 2 C - C H regiochemistry HgSO4 Carbonyl R H Enol intermediate compound forming reaction H O Anti-Markovnikov 1. BH3, THF regiochemistry + R C - C 2. HO,22 HO Enol intermediate H H

Chain NaNH lengthening 2 R - CC - CH - R Alkyne anion intermediate RCHBr º 2 reaction 2

[O] RCOOH + CO Alkaline or 2 Acidic condition Oxidation

1. O3 + R - CH - CH22+ CO 2. Zn, HO3 OH OH 11.80 | Alkenes and Alkynes

Solved Examples

JEE Main/Boards

Example 1: Identify A to C.

Na+NH +EtOH HC CH 2Na (A) 3 (B)

2CO/23HO H/Pd/CaCO 23(C )

Sol: It is an example of ethinylation (addition of ethyne to unsaturated link like group). C First step is the reaction with the sodium metal to form disodium acetylide with evolution of hydrogen gas. This reaction shows the acidic character of alkynes. During this step acetylenic hydrogens are removed as a proton to form stable carbanion. Now the carbanion reacts with two moles of carbon dioxide and forms an adduct which on treatment with acid yields acid.

Next step is reduction using Lindlar Catalyst (H2/Pd-CaCO3) the addition is syn.

On treatment with (NH3+EtOH) trans product is formed thus the addition is anti.

O O 2Na H C CH Na C C Na O C C C C O

O C O

O C O

HO3

OOH 1 C4 HOOC HOOC H 2 3 H/23Pd/CaCO 1 23 4 Na + NH + EtOH 23 HOOC COOH 3 Syn-additon Anti- addition HH (A) H COOH (C ) But-2-yn-1,4-dioic acid Maleic acid or Fumaric acid (B) or cis-But-2-en-1,4-dioic acid trans-But-2-en-1,4-dioic acid

Example 2: Identify B to G.

2 3 Br 1. NaNH2 1 2 ( ) Me (B) 2. HO C Propene 2 NBS + CCl+ h (A) 4 n o Cl2+ 500 C (D) 1. Li (D) (E) 2. Cul (F) (G)

Sol:

•• Addition of Br2 across double bond forms 1,2-dibromo propane. •• This on treatment with soda lime undergoes double dehalogenation to form an alkyne

•• Now alkyne on the treatment with a brominating agent (NBS- N-Bromosuccinimide) and CCl4 as a solvent in presence of light. ◦ •• On the other side, the first step is Allylic chlorination using Cl2 at 500 C. Chemistry | 11.81

•• The reaction proceeds through radical formation. •• Now Allyl chloride undergoes chain lengthening reaction to form Hex-1-en-5-yne. •• The reaction is known as Corey house synthesis

Br 2 3 Br 3 2 2 Me 1 1. NaNH 1 Me 2 Me C CH 2. HO (A) Br 2 (B) (C ) 1,2-Dibromopropane Prop-1-yne

o Propar- Cl2 at 500 C Allylic gylic NBS + CCl+4 hn radical radical chlorination brominating

3 2 1 2 3 Cl H 1 Br (E) (D) Allyl chloride 3-Bromoprop-1-yne or or 3-Chloro prop-1-ene propargylic ( ( ( bromide ( Coerey 1. Li -House 2. Cul R’

2 3 CuLi 3 2 1 1 H 2 2 4 5 6 ( ( Br 1 3 (F) (D) Diallyl lithium R R’ cuperate (G) Hex-1-en-5-yne

Example 3: Alkenes are more reactive than alkynes towards electrophilic addition reaction, yet vinyl acetylene reacts with 1 mole of HBr at triple bond. Explain why.

Sol: •• Alkynes are less stable than isomeric diene. •• Compound (A) is a conjugate diene whereas compound (B) is an alkyne. •• As heat of formation of diene is less than the heat of formation of alkyne, Vinyl acetate reacts with 1 mole of HBr to form a conjugate diene and formation of compound (B) is not favoured. Br 4 3 2 1 HBr CH3 CH C CH HC2 CH C CH HC2 CH C CH2 4 3 2 1 Br Vinyl acetylene 3-Bromobut-1-yne 2-Bromobut-1,3-diene (B) (A)

The product A (Conjugated diene) is more stable than B (an alkyne) 11.82 | Alkenes and Alkynes

1 23 45 Example 4: There are two path (a) and (b) For the preparation of a compound (a) = Me ( Me ( (2-methylpent-1-en-3-yne), which path is correct and why? Also name the paths (a) and (b). Path (a)

Me Sol: Path (b) is correct because in path (a) compound vinyl halide which is not reactive, so the reaction Br does not occur. ( ( Path (a) is the alkylation of alkyne, whereas path (b) is propinylation (type of ethinylation), i.e., addition of propyne to unsaturated link like (> C = O) group. Me

Br Path (a) H Me NaC Me Me Me Me Path (b) O+ HO2 (A) Me Me Conc. HSO + Me 24D Me -H2O OH

JEE Advanced/Boards

Example 1: Complete the following missing reagents

Br C CH ? ? ? (a) (b) ( c ) ( 2( CuLi (A) (B) (C ) (D) Br ? ? (e) (d) Br (F) (E)

Sol: (i) (a) HBr + Peroxide First step is addition of HBr, it takes place according to Markovnikov’s rule. (b) (1) Li (2) CuI (Corey – House): Second step is Corey House synthesis.

(c) Br Third step is the reaction with alkyl halide.

(d) Br2/CCl4

Bromination using Br2 and CCl4 as a solvent (e) KOH(s), at 473 K Last step is double dehalogenation using base like KOH at 473 K Chemistry | 11.83

(ii) Cis compound on addition with Br2 (Anti addition) forms a racemic mixture or enantiomer.

Trans compound on addition with Br2 (Anti addition) produces Meso compound (it contains plane of symmetry)

HH H and H

cis trans

Br2(Anti-addition) Br2(Anti-addition) racemic or (+) (E) meso and (E)

HBr HBr HBr + Enantiomer Br H

Due to the plane of ( ( ( symmetry (

Example 2: Convert the following

? ? ? (B) C Cl (a) (b) ( c ) (D)

? (G)

Cl (d) CuLi ( 2( H (E) (F) (H)

Sol:

Cl Allylic Cl chlorination (a) 2 alc. KOH + h Cl n (b) (SO22Cl +)D (a) (A) (B) or ( c ) (D) ( O Cl + D (

Cl 1 2 2 1. Li Me (D) 3 H 3 2. Cul CuLi Cl (d) 2 (b) ( ( (E) (F) Less stable More stable Divinyl Vinyl isolable diene conjugated diene lithium chloride (G) (H) cuperate 3-Ethenyl cyclo 3-Ethylidene cyclo pent-1-ene pent-1-ene 11.84 | Alkenes and Alkynes

Example 3: Convert the following. 2 Br ? ? ? ? 3 (B) (C ) (D) Hept-2-yne Me 1 (E) Write the structure of (E).

Sol: The structure of E is 7 Me 5 3 2 1 Me 6 4

First prepare 3C – atom alkyne (Me−≡− C C H) from (A) and then react 4C – atom RX with alkynide ion to get (E).

Reaction: Br 2 2 Br alc.KOH Br2 3 3 1 Me 1 (B) Me Propyne (C )Br (A) 1,2-Dibromopropane

2NaNH2 Double dehalogenation Me H Propene 2NaNH 2 3 1 4 Me 2 Br 7 1 2 3 5 Me Butyl bromide 4 Me CC Me C C 6 Propynide ion Hept-2-yne (D) (E)

Example 4: Complete the following:

1 mol CHBr NaNH CH I HC CH B 25 C 2 D 3 (E) NaNH2 Ethyne (A) Sol:

1 mol CH25Br 12 4 HC CH HC CNa HC C C CH25 NaNH2 (A) Sodium ethynide But-1-yne (B) (C )

NaNH2

CH3I CH3 C C CH25 NaC C CH25 (E) (D) Pent-2-yne Sodium butynide Chemistry | 11.85

JEE Main/Boards

Exercise 1 (B) Starting with cyclohexyl ethyne, prepare acetyl cyclohexane. Q.1 What are alkenes? Discuss briefly the various methods used for the preparation of alkenes. Describe Q.9 (A) Complete the following reactions: with a labelled diagram the laboratory preparation of →NaNH2 →CH 22 ethane from ethanol. CH33 COCH AB

+ ∆  H → C →H2 (1equiv) D →Al23 O , E Q.2 Write the major products of the hydrocarbon and Pd 400ºC oxidation of: (A) 1-Ethyl cyclopentene (B) Complete the following reactions: NaNH CH –C≡CH →2 A (B) Methylene cyclopentene 3 CH22 CH Br

H CH I Q.3 Give the missing compounds in the following: →2 BC →22 Lindlar's catalyst Zn(Cu)

BH3.THF CH3COOH (A(a)) BH3.THF.THF ? CH3COOH ? (a) 3 ? 3 ? Q.10 Predict the product of the following: BH .THF CH COOH 3 3 Conc. HSO/ (a) ? ? 24D A BH3.THF CH3COOH (b) BH3.THF.THF ? CH3COOH ? (A) ((b)B) 3 ? 3 ? HCl B BH .THF CH COOH OH (b) 3 ? 3 ? + BH3.THF CH3COOH H O/H CH 2 ( c ) BH3.THF.THF ? CH3COOH ? 3 A ((C c )) 3 ? 3 ? BH .THF CH COOH CH2 (I) BH26 ( c ) 3 ? 3 - B (i) BH3.THF ? (ii) HO22(OH ) (i) BH .THF (B) HC3 (d) (i) BH3.THF ? (d) (ii) CH COOH ? CH (i) Hg(OAc)2 (i) BH 3.THF ? 3 C (ii) CH33COOH (ii) NaBH (D(d)) 3 ? 4 (ii) CH3COOH OH H+ RCOH HO (C ) 3 2 J Q.4 Complete the following equation:

H2 (A) CH3C ≡ CCH3   → (A) Pt Q.11 (A) Give reasons for the following: When 1-penten-4-yne is treated with HBr in  H2 → CH CH CH CH Pt 3 2 2 3 equimolecular proportion, the addition takes place on double bond and not on triple bond yielding thereby

Na+ C25 H OH the product (B) CH3 –C ≡ CCH3 → CH CH(Br)CH C≡CH. (B) + NaOC2H5 3 2 (B) Provide a suitable mechanism for the following Q.5 Give an account of physical and chemical reactions reaction. of alkenes. OH CH3 CH3 Q.6 What are alkynes? How are they prepared in the laboratory? Q.12 Three isomeric alkenes A, B and C, (C H ) are Q.7 Give an account of physical properties and chemical 5 10 hydrogenated to yield 2-methylbutane A and B gave reactions of alkynes. the same 3º ROH on oxymercuration-demercuration B and C give different 1º ROH’s on hydroboration- Q.8 (A) Starting with any alkyne, prepare ethyl oxidation supply the structures of A, B and C. cyclohexyl ethyne. 11.86 | Alkenes and Alkynes

Q.13 3,3-dimethyl-1-butene and HI react to give two Q.6 End product of the following sequence is: products, C H I. On reaction with alc. KOH one isomer, HO 6 13 CaO + C  Heat → (A)  2 → (C) (1) gives back 3,3-dimethyl-1-butene the other (J) Hg 2+ gives an alkene that is reductively ozonized to give → (C) H24 SO Me2C=O. Give the structures of (I) and (J) and explain the formation of the latter. (A) Ethanol (B) Ethyl hydrogen sulphate (C) Ethanal (D) Ethylene glycol Q.14 Identify X, Y and Z in the following sequence of reaction giving stereo chemical structure wherever Q.7 The treatment of C H MgI with water produces possible. 2 5 (A) Methane (B) Ethane (i) Na (i) H (I) CH CO H CCH (X) 2 (Y) 33 (Z) º + Pd BaSO4 (C) Ethanal (D) Ethanol (ii) Br (ii) H/HO2

(i) Na (i) H2 (I) CH33CO H CCH (X) (Y) (Z) º + Q.8 When isobutene is brominated, the percentage of Pd BaSO4 (ii) H/HO (ii) Br 2 Br

HC would be 3 CH3

Exercise 2 CH3

Single Correct Choice Type (A) 0% (B) 83% (C) 10% (D) 100%

Q.1 On heating CH3COONa with soda lime the gas evolved will be Q.9 Propene can be converted into 1-propanol by oxidation. Which set of the reagents is used to effect (A) C2H2 (B) CH4 (C) C2H6 (D) C2H4 the conversion?

(A) OsO4 — CHCl3 (B) O3/Zn — H2O Q.2 The addition of Br2 to trans-2-butene produces (C) Alkaline and cold KMnO (D) B H and alk. H O (A) (+) 2, 3-dibromobutane 4 2 6 2 2 (B) (–) 2, 3-dibromobutane Q.10 In the following sequence of reactions, identify (C) rac -2, 3-dibromobutane the product (d) (D) meso -2,3-dibromobutane HBr HBr alk.KOH CH≡CH   → (A)   → (B) → (C)

NaNH Q.3 Isomers which can be interconverted through →2 (D) rotation around a single bond are (A) Ethanol (B) Ethyne (A) Conformers (B) Diastereomers (C) Ethanal (D) Ethene (C) Enantiomers (D) Positional isomers Q.11 Which one is highly unstable?

Q.4 The olefin, which on ozonolysis gives CH3CH2CHO (A) CH2=CH–OH (B) CH2=CH–Cl and CH3CHO (C) CH =CH–CH –Cl (D) CH –C≡CH (A) But-1-ene (B) But-2-ene 2 2 3 (C) Pent-1-ene (D) Pent-2-ene Previous Years’ Questions Q.5 Which of the C — C bond is strongest? Q.1 Marsh gas mainly contains (1980) (A) Formed by sp3–sp3 hybridised carbon atoms (as in

alkanes) (A) C2H2 (B) CH4 (C) H2S (D) CO (B) Formed by sp2–sp2 hybridised carbon (as in alkenes) (C) Formed by sp–sp hybridised carbon atoms (as in Q.2 Which of the following will decolourise alkaline alkynes) KMnO4 solution? (1980)

(D) None of these (A) C3H8 (B) CH4 (C) CCl4 (D) C2H4 Chemistry | 11.87

Q.3 The compound 1,2-butadiene has (1983) (A) Both are highly ionic (A) Only sp-hybridised carbon atoms (B) One is oxidising and the other is reducing (B) Only sp2-hybridised carbon atoms (C) One of the steps is endothermic in both the cases (C) Both sp and sp2-hybridised carbon atoms (D) All the steps are exothermic in both the cases (D) sp, sp2 and sp3-hybridised carbon atoms Q.10 Which of the following compound will exhibit geometrical isomerism? (2015) Q.4 When propyne in treated with aqueous H2SO4 in presence of HgSO4, the major product is (1983) (A) 1-Phenyl-2-butene (A) Propanal (B) Propyl hydrogen sulphate (B) 3-Phenyl-1-butene (C) Acetone (D) Propanol (C) 2-Phenyl-1-butene (D) 1,1-Diphenyl-1-propane Q.5 Bayer’s reagent is (1984)

(A) Alkaline permanganate solution Q.11 The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate : (2016) (B) Acidified permanganate solution (A) CH – CH+ – CH –Cl (B) CH - CH(OH) - CH + (C) Neutral permanganate solution 3 2 3 2 (C) CH – CHCl - CH (D) CH – CH+ – CH – OH (D) Aqueous bromine solution 3 2 3 2

Q.12 The major organic compound formed by the Q.6 Acidic hydrogen is present in (1985) reaction of 1, 1, 1-trichloroethane with silver powder is (A) Ethyne (B) Ethene (2014) (C) Benzene (D) Ethane (A) Acetylene (B) Ethene (C) 2-Butyne (D) 2-Butene Q.7 Hydrogenation of the adjoining compound in the presence of poisoned palladium catalyst gives (2001) Q.13 In the reaction, H LiAlH4 PCl5 Alk.KOH HC3 CH COOH→ A → B →C CH3 3 the product C is (2014) HC3 H H (A) Acetaldehyde (B) Acetylene (C) Ethylene (D) Acetyl chloride (A) An optically active compound

(B) An optically inactive compound Q.14 In the following sequence of reactions, the alkene (C) A racemic mixture affords the compound ‘B’ O HO CH CH= CHCH →3 A →2 B. (D) A diastereomeric mixture 33 Zn The compound B is (2008) Q.8 The reaction of propene with HOCl proceeds via (A) CH CH CHO (B) CH COCH the addition of (2001) 32 33 (C) CH CH COCH (D) CH CHO (A) H+ in the first step 32 3 3

+ (B) Cl in the first step Q.15 The hydrocarbon which can react with sodium in – (C) OH in the first step liquid ammonia is (2008) – + (D) Cl and OH single step (A) CH322 CH CH C≡ CCH 223 CH CH

(B) CH32 CH C≡ CH Q.9 In the presence of peroxide, hydrogen chloride (C) CH CH≡ CHCH and hydrogen iodide do not give anti Markownikoff’s 33 addition to alkenes because (2001) (D) CH32 CH C≡ CCH 23 CH 11.88 | Alkenes and Alkynes

(B) Statement-I is true; statement-II is true; statement-II Q.16 The treatment of CH3 MgX with CH3 C≡− C H produces (2008) is not the correct explanation of statement-I.

(A) CH32−≡ CH CH (B) CH33 C≡− C CH (C) Statement-I is true; statement-II is false. HH (D) Statement-I is false; statement-II is true. || (C) CH−=− C C CH (D) CH 3 4 Q.19 Statement-I: Addition of Br2 to 1-butene gives 3 two optical isomers.

Q.17 The alkene that exhibits geometrical isomerism is: Statement-II: The product contains one asymmetric (2009) carbon. (1998) (A) Propene (B) 2-methyl propene Q.20 Statement-I: 1-butene on reaction with HBr in (C) 2-butene (D) 2- methyl -2- butane the presence of a peroxide produces 1 bromobutane.

Q.18 One mole of a symmetrical alkene on ozonolysis Statement-II: It involves the formation of a primary gives two moles of an aldehyde having a molecular radical. (2000) mass of 44 u. The alkene is (2010) Q.21 Statement-I: Dimethyl sulphide is commonly (A) Propene (B) 1–butene used for the reduction of an ozonide of an alkene to (C) 2–butene (D) Ethene get the carbonyl compound. Statement-II: It reduces the ozonide giving water Assertion and Reason Type soluble dimethyl sulphoxide and excess of it evaporates. (2000) Read the following questions and answer as per the direction given below: Q.22 Statement-I: Addition of bromine to trans-2- (A) Statement-I is true; statement-II is true; butene yields meso-2, 3-dibromo butane. statement-II is a correct explanation of statement-I. Statement-II: Bromine addition to an alkene is an electrophilic addition.

JEE Advanced/Boards

Exercise 1 (B) on treatment with alcoholic KOH followed by NaNH2 resulting in the formation of compound (C)

C6H10. Compound (C) on treatment with H2 (gas) over Q.1 Compound (X) (C5H8O) does not appreciably react with Lucas reagent at room temperature but gives a platinum forms 2-methyl pentane. Compound (C) gives ammoniacal AgNO test. Ozonolysis of (A) gives two precipitate with silver nitrate (ammoniacal). With Me 3 Mg Br, 0.42gms of (X) gives 22400 mL of CH at STP. aldehydes (D) and (E) where E is acetaldehyde. Identify 4 A, B, C, D and E. Treatment of H2 in the presence of Pt catalyst followed by boiling with excess HI, gives n-pentane. Suggest the structure for (X) and write equations involved. Q.4 A hydrocarbon (A) is treated with excess of HCl. A dihalogen derivative (B) is formed. Compound (B) on treatment with aqueous KOH gives C. Compound (C) Q.2 Compound (A) C5H10 decolourizes Br2. When (A) can be obtained by heating calcium salt of 2-methyl is treated with hot KMnO4, two acids (B) and (C) are formed. When (B) is treated with soda lime, methane propanoic acid with calcium acetate what are A, B and C? is formed. When (C) is heated with soda lime ethane is AlBr3 formed. What are the structure of (A), (B) and (C). Q.5 CF3CH = CH2 + HBr   → X

Q.3 An organic compound (A) C6H12 is treated with Cl2 Q.6 An organic compound (A) C9H12 gave (B) C8H6O4 on in the presence of CCl4 gives (B) C6H12Cl2. Compound oxidation by alkaline KMnO4. (B) on heating does not Chemistry | 11.89

form anhydride. Also (B) reacts with Br2 in the presence Q.15 A hydrocarbon (P) of the formula C8H10, on of iron to give only one monobromo derivative (C). ozonolysis, gives compound Q (C4H6O2) only. The

C8H5BrO4. What are A, B and C. compound (Q) can also be obtained from an alkyl

bromide, R (C3H5Br) upon treatment with Mg in dry

Q.7 A hydrocarbon (A) C9H10 adds Cl2 to give (B) ether, followed by CO2 and acidification. Identify (P), (Q)

C9H10Cl2. Hydrolysis of (B) gives (C) C9H12O2. Oxidation and (R) and give equations for the reactions. of (A) gave two acids identified as benzoic acid and acetic acids. What are A, B and C? Q.16 Two isomeric alkyl bromides (A) and (B) have

molecular formula C5H11Br. (A) on treatment with Q.8 An organic compound (A) C9H10Br2 forms C9H11OBr alcoholic KOH gives two isomers (C) and (D) of formula with caustic soda solution the later is resistant to further C5H10. (C) on ozonolysis gives formaldehyde and action of alkali. (B) on treating with H2SO4 converts 2-methylpropanal. (B) with alcoholic KOH gives (D) and into (C) with on ozonolysis gives ethanal and another (E). Catalytic hydrogenation of (C), (D) and (E) gives (F) compound (D). Identify (A) to (D) and give reactions. C5H12. Identify (A) to (F).

Q.9 An organic compound (A) C6H10, on reduction first Q.17 How would you bring about the following gives (B) C6H12 and finally (C) C6H14. (A) on ozonolysis conversion. followed by hydrolysis gives two aldehydes (D) C2H4O and (E), C H O . Oxidation of (B) with KMnO gives the (i) 1, 3-Butadiene to racemic - 1, 4-dibromo -2, 2 2 2 4 3-dideutrobutene acid (F) C4C8O2. Determine the structures of (A) to (F) with proper reasoning. (ii)  → Q.10 An organic compound (A) of molecular formula C H when treated with Na in liquid ammonia followed 5 8 Me H by reaction with n-propyl iodide yields (B) C H . (A) (iii) NBS A →(i) Li B 8 14 C C (ii) Cu P gives a ketone C5H10O when treated with dil. H2SO4 and H Me (iii) A HgSO . (B) on oxidation with alkaline. KMnO gives two 4 4 (iv) Z-2, 3-Dideutero-2-butene to racemic-erythro-2- isomeric acids (D) and (E) C H O . Give structures of 4 8 2 Bromo-2,3-Didenterobutane compounds (A) to (D) with proper reasoning. (v) 1-Methyl cyclopentene to trans-2-Methyl cyclopentane Q.11 An organic compound (A) C8H6, on treatment with dil. H2SO4 containing mercuric sulphate gives (vi) H a compound (B), which can also be obtained from a O reaction of benzene with acid chloride in the presence C of anhydrous AlCl3. The compound (B) when treated C Me with iodine in aq. KOH, yields C and a yellow compound H (D). Identify A, B, C and D with justification. Me to (S, S) – Glycol + R, S Glycol Q.12 An organic compound A (C5H8) on hydrogenation gives compound B(C5H12). Compound A on ozonolysis gives formaldehyde and 2-ketopropanal. Deduce Q.18 A compound (A) has C = 85.7%, H = 14.3%. Its molecular weight is 70. It does not react with Cl in dark structure of compound A. 2 but gave a substitution product C5H9Cl in presence of Q.13 What are the possible products from 1,4-addition light. What is (A)? CH of HBr on 2 Draw all possible carbocations and

Q.19 An organic compound (A) C6H10 does not react with ammoniacal AgNO but on reduction first gives (B) state which carbocation is more stable 3 C6H12 and then (C) C6H14. (A) on ozonolysis gives two

aldehydes C2H4O (D) and C2H2O2 (E). Oxidation of (B)

with alkaline KMnO4 gives acid (F) C3H6O2. What are (A) Q.14 ++CH CH2=CH–CH=CH= CH-CH=CH2 22 of (F)? H H Q.20 Two isomeric organic compounds (A) and (B) on A Predict the structure of A and explain. reduction with Zn-Cu couple give n-butane. On treating 11.90 | Alkenes and Alkynes

(A) and (B) with Na and ether separately, n-octane Q.6 Addition of halogen acid (HX) is least in (C) and 3, 4-diemthylhexane (D) are formed but if a (A) CH2=CHCl (B) CH2=CH2 mixture of (A) and (B) reacts with Na and ether, one (C) CH –CH=CH (D) (CH ) C=CH more product (E) in addition to (C) and (D) is 3 2 3 2 2

Q.7 Which of the following alkene will react fastest with Exercise 2 H2 under catalytic hydrogenation condition? R R R H Single Correct Choice Type (A) (B) Q.1 A sample of 1.79 mg of a compound of molar mass H H H R 90g mol–1 when treated with CH MgI releases 1.34 mL 3 R R of a gas at STP. The number of active hydrogen in the R R molecule is (C) (D) (A) 1 (B) 2 (C) 3 (D) 4 R H R R CH CH33 3 Q.2 Reaction of Br on ethylene in presence of NaCl + 2 + DO2 /D gives Q.8 Products of the reaction.DO2 /D HC3 CH2 HCHC33 CH2 CH (A) BrCH2-CH2Br (B) ClCH2-CH2Br 3 CH33 CH3 CH3 CH3 (C) Both of these (D) None of these CH3 CH3 CH3

(A) HC3 CH2D (B) HC3 CH2D (A)(A)(A) HC33 CH2D (B)(B) HC3 CH22D Q.3 The reaction conditions leading to the best yield of CH3 OD OD

C2H5Cl are CH3 OD OD CH CH3 H 3 UV (A) C H (excess) + Cl   → CH H CH3 2 6 2 light 3 (C ) HC3 CH3 (D) HC3 Ch2 ( ) HC CH (D) HC Ch dark (C)(C C ) HC 33 CH3 (D)(D) HC3 OD D Ch22 CH D OD (B) C H + Cl → 3 2 6 2 room temp. OD D D OD D CH3 D OD (C) C H + Cl (excess)  UV → 2 6 2 light Q.9 The C–H bond length is longest in (D) C H + Cl  UV → (A) C H (B) C H (C) C H (D) C H Br 2 6 2 light 2 2 2 4 2 6 2 2 2

Q.10 Point out (A) in the given reaction sequence: Q.4 The treatment of CH3C(CH3)=CHCH3 with NaIO4 or boiling KMnO4, produces O /H O (A) →32 (B)  →∆ 2CH COOH+CO (A) CH3COCH3 + CH2O (B) CH3CH2CHO + CH3CHO 3 2 (C) CH COCH + CO (D) CH COCH + HCOOH 3 3 2 3 3 Q.11 Hydrogenation of the compound HC 3 Me Q.5 What is the chief product obtained when n-butane Me is treated with bromine in the presence of light at 130ºC? H CH3 H H (A) H3C Br (B) H3C Br In the presence of poisoned palladium catalyst gives: CH3 (A) An optically active compound H C 3 (C) Br (D) H3C C CH2Br (B) An optically inactive compound CH3 CH3 (C) A racemic mixture (D) A diastereomeric mixture Chemistry | 11.91

Q.12 HC3 Q.17 Which one of the following heptanols can be CH2 + HBr A dehydrated to hept-3-ene only? CH 3 (A) Heptan-3-ol (B) Heptan-4-ol Br (Predominant). A is Br (C) Heptan-2-ol (D) Heptan-1-ol HC HC (A) 3 Br CH (B) 3 3 Br HC3 HC CH HC3 (A) 3 3 CH (B) Q.18 Aqueous solution of potassium propanoate is 3 HC 3 electrolysed. Possible organic products are: CH3

Br (A) n-Butane (B) C2H5COOC2H5

Br (C) CH3–CH3 (D) CH2=CH2 (C ) HC3 C CH2CH3 (D) Non of these (C ) HC CHC CH CH (D) Non of these 3 3 2 3 - OH CH Q.19 + 3 N: Q.13 Identify (B) in the following sequence of reactions

Cu22Cl +NH4Cl HCl CH CH (A) (B) The above compound undergoes elimination on heating to yield which of the following products?

(A) CH32CH Cl (B) CH22=CH-CH CH2Cl OH: Cl + (A) CH (B) ( ) (D) C N 2 N CH2 C CH CH3 HC2 CH2 Me Me Me Me Cl

Cl –H O Q.14 CH –CH=CH →22 Z . Z may be 3 2 (C) Me (D) N N Me Cl Me Me Me Me CH CH CH (A) 3 2 (B) CH3 CH CH2 Cl Cl Cl Multiple Correct Choice Type

(C) CH3 CH CH2 (D) CH3 CH CH2 NBS Q.20 CH2=CHCH2CH=CH2   → A, A can be Cl OH OH Cl (A) CH2 CHCHCH CH2 Br HCOOHHCOOH Q.15 (Peroxyformic(Peroxyformic acid ) acid ) Z. Z mayZ. Z be may be (B) CH2=CHCH=CH–CH2Br

(C) CH2=CHCH2CH=CHBr OH OHOH OH H H H (D) CH2 CHCH2C CH2 (A) (A) (B) (B) Br HHHH OH OHOH OH Q.21 Which of the following elimination reactions will H H H occur to give but-1-ene as the major product?  EtOH → (C ) (C ) (D) Non(D) ofNon these of these (A) CH3.CHCl.CH2.CH3 +KOH EtOH (B) CH3.CH.CH2.CH3 + NaOEt   → O O ∆

HIO NMe3  →4 + Q.16 CH3 CH C CH3 +[O] Z.Z may be (C) CH .CH .CHCl.CH +Me CO–K+  →∆ OH O 3 2 3 3

(D) CH3.CH2.CHOH.CH3+conc.H2SO4

(A) CH3–CHO (B) CH3–COOH

(C) CH3 C C CH3 (D) None of these O O 11.92 | Alkenes and Alkynes

Q.22 Select true statement(s): BH /THF HgSO Q.28 C →3 A →4 B – H SO H22 O /OH 24 (A) I2 does not react with ethane at room temperature even though I is more easily cleaved homolytically 2 B and C are identical when A is – than the other halogens. (A) H H (B) H (B) Regiochemical outcome of a free radical addition and an electrophilic addition reaction on propene is (C) (D) H identical. (C) The rate of bromination of methane is decreased if Assertion Reasoning Type HBr is added to the reaction mixture. Each of the questions given below consists of two (D) Allylic chloride adds halogens faster than the statements, an assertion (A) and reason (R). Select the corresponding vinylic chloride. number corresponding to the appropriate alternative as follows: Q.23 An alkene on ozonolysis yields only ethanal. There (A) If both Assertion and Reason are true and Reason is is an isomer of this which on ozonolysis yields: the correct explanation of Assertion . (A) Propanone (B) Ethanal (B) If both Assertion and Reason are true but Reason is (C) Methanal (D) Only propanal not the correct explanation of Assertion . (C) If Assertion is true but Reason is false. Q.24 CH –CH=CH–CH + CH N  → A 3 3 2 2 ∆ (D) If Assertion is false but Reason is true. A can be (A) (B) Q.29 Assertion: In α , β -unsaturated compounds with N C=C and C=O conjugated, attack of nucleophile takes N place on C=C.

(C) (D) Reason: The C=O bond is stronger than C=C. N N Q.30 Assertion: Alkyl iodides are more reactive than Q.25 Which of the following will give same product alkyl chlorides for elimination reactions. with HBr in presence or absence of peroxide. Reason: I– is a better leaving group than Cl– (A) Cyclohexene Q.31 Assertion: CH CH– is a stronger base than F–. (B) 1-methylcyclohexene 32 Reason: The negative charge density on carbon is (C) 1, 2-dimethylcyclohexene greater than the negative charge density of F–. (D) 1-butene

Q.32 Assertion: CH3–C≡C–CH3 is more reactive than CH≡CH towards HCl. Q.26 The ionic addition of HCl to which of the following compounds will produces a compound having Cl on Reason: The carbocation formed is more stable in the carbon next to terminal. case of CH3–C≡C–CH3 than CH≡CH. (A) CH .(CH2) .CH=CH (B) CH .CH=CH 3 3 2 3 2 Q.33 Assertion: CH≡CH reacts with HCl in the presence (C) CF .CH=CH (D) CF .CH=CH 3 2 3 2 of HgCl2 while CH2=CH2 does not. Reason: There is more unsaturation in CH≡CH than in Q.27 Which is/are true statements/reactions? CH2=CH2.  → (A) Al4C3 + H2O CH4  → Comprehension Type (34-36) (B) CaC2 + H2O C2H2  → (C) Mg2C3 + H2O CH3C≡CH n-Butane is produced by mono bromination of ethane followed by Wurtz reaction, as follows (D) Me3C–H + KMnO4  → Me3C–OH (i) CH CH  Br2 → CH CH –Br 3 3 hν 3 2 Chemistry | 11.93

Na (ii) CH CH Br   → CH CH CH CH + CH =CH + 14 Br /low conc. 3 2 3 2 2 3 2 2 Q.38 CH –CH=CH →2 3 2 high temperature CH –CH + NaBr 3 3 Product of the above reaction is

+ - 14 14 Mechanism: Na → Na + e ; (A) Br– CH2–CH =CH2 (B) CH2=CH–CH2–Br

CH3CH2Na+CH3CH2Br → (C) Both (a) and (b) (D) None of these

CH3CH2CH2CH3+CH2=CH2+CH3–CH3 • • Q.39 Trans-2-butene when reacts with Br2, the CH3CH2Br → CH32 CH +Br • product formed is CH CH + e– → CH CH ᛫ 32 3 2 (A) Racemic mixture (B) Meso form (C) Optically pure (D) No specific stereochemistry Q.34 The amount of ethyl bromide required to prepare 55g of butane would be Match the Columns (A) 106.72 (B) 206.72 (C) 20.67 (D) 2.067 Q.40 Match the entries in column I with entries in column II Q.35 If the yield of n-butane was 85%, then the actual amount of ethyl required to produce 55g of n-butane Column I Column II would be (A) CH3–C≡C–CH3cis-2-butene (p) Na/NH3(l) (A) 2.432g (B) 24.32g (C) 243.2g (D) 0.2432g

 → (q) H2/Pd/BaSO4 (B) CH3–C≡C–CH3 trans-2-butene Q.36 The other products which could be formed (r) Alc. KOH, ∆ (C) CH3C≡C–CH3  → 1-Butyne (A) CH2=CH2 (B) CH3CH2CH3 (D) CH –CH –C≡CH  → 2-Butyne (s) NaNH , ∆ (C) CH4 (D) CH3CH2CH2–CH2–CH3 3 3 2

Comprehension Type (36-38) Q.41 Match List-I with List-II and select the correct answer The functional group in alkenes is C–C double bond which is made of one 휎-bond and one 휋-bond. The List I (Reaction) List II (Reagents) weaker 휋-bond would tend to attract the electrophile and (A) CH –CH=CH CH –CHBr–CH (p) HBr convert itself to a stable carbonium ion which reacts with 3 2 → 3 3 nucleophile remainder to form the addition product. The (q) Br2 (B) CH3–CH=CH2 → CH3–CH2–CH2Br addition reaction of alkenes is stereo specific as well as (r) HBr/Peroxide stereoselective. i.e., the isomeric alkenes react differently (C) CH3–CH=CH2 → BrCH2–CH=CH2 with same addendum to give stereochemically different (D) CH –CH=CH → CH –CHBr– (s) NBS products. Where addition of HBr proceeds through 3 2 3 CH Br carbocation formation and hence racemisation is 2 expected. With low concentration of Br2 in vapour phase substitution predominates over addition. Q.42 Match the column Column I Column II Q.37 Reaction of 1,2-dimethylcyclohexene with (p) Corey-House bromine water gives the product →Electrolysis (A) RCOONa R–R reaction

(A) (B) Soda Lime (q) Kolbe electrolysis (A) Br CH (B) Br Br (B) R–CH –COOH → 3 2 ∆ Br CH3 Br Br R–CH3

CH3 Br CH3 CH3 CH Br CH CH (i) AgNO (r) Oakwood 3 3 3 (C) RCOOH →3 R–Cl (ii) Cl2 /∆ degradation (C ) (D) (C ) Br OH (D) Br CH3 Br OH Br CH3

CH3 CH3 CH3 OH CH3 CH3 CH3 OH 11.94 | Alkenes and Alkynes

Previous Years’ Questions through the following sequence of reactions, in which Q is an intermediate organic compound.

Q.1 (CH3)3CMgCl on reaction with D2O produces (1997) Q.7 The structure of compound P is (A) (CH3)3CD (B) (CH3)3OD (A) CH CH CH CH —C≡C—H (C) (CD3)3CD (D) (CD3)3OD 3 2 2 2

(B) H3CH2C—C≡C—CH2CH3 Q.2 In the compound, H the C2–C3 bond is of the type (1999) H 3C (C) H C C C CH (A) sp–sp2 (B) sp3–sp3 (C) sp–sp3 (D) sp2–sp3 3 H3C

Q.3 The product(s) obtained via oxymercuration (HgSO4 H 3C (D) + H2SO4) of 1-butyne would be H3C C C C H

O H3C

(A) CH3–CH2–C–CH3 (B) CH3 – CH2 – CH2 – CHO Q.8 The structure of the compound Q is (C) CH3 – CH2 – CHO + HCHO (i) dil. H24 SO (catalytic P amount) (–H O) (i) dil. H24 SO /HgSO 4 Q2→2 Q.4 Identify a reagent from the following list which → (ii)O /ethanol (C H ) (ii) NaBH4 /ethanol 3 can easily distinguish between 1-butyne and 2-butyne. 6 10 (iii)dil. acid (iii) Zn/H2 O (2002) O

(A) Bromine, CCl4 (B) H2 , Lindlar catalyst C (2011) (C) Dilute H2SO4 , HgSO4 (D) Ammoniacal CuCl2 solution H3C CH3

OH Q.5 Consider the following reaction HC3 • (A)(A) H C C CH2CH3 H 3C–CH–CH–CH3 + Br  → ’X’ + HBr HC3 H D CH3 OH HC Identify the structure of the major product X (2002) 3 (B)(B) HC3 C C CH3 (A) H 3C–CH–CH–CH2 (B) H 3C–CH–C–CH3 HC 3 H D CH3 D CH3 OH HC3 (C)(C ) (C) H 3C–C–CH–CH3 (D) H 3C–CH–CH–CH3 H C CH2 CHCH3

HC3 D CH3 CH3

(D) CH3CH2CH2CH(OH)CH2CH3 Hg 2+ Q.6 Ph–C≡C–CH    → A ; A is (2003) 3 H + Q.9 Give reasons for the following in one or two O sentences: (1983) (A) Ph (B) Ph

O (i) Methane does not react with chlorine in dark.

H3C H3C (ii) Propene reacts with HBr to give isopropyl bromide

OH but does not give n-propyl bromide. (C) Ph (D) Ph

OH Q.10 Following statements are true, only under some specific conditions. Write the condition for each sub Comprehension Based Questions (Q.7 & 8) question in not more than two sentences:

An acyclic hydrocarbon P, having molecular formula (i) 2-methyl propene can be converted into isobutyl bromide by hydrogen bromide C6H10, gave acetone as the only organic product Chemistry | 11.95

(ii) Ethyne and its derivatives will give white precipitate Q.11 The product X is (2014) with ammoniacal silver nitrate solution. (1984) (A) HC3 O (A) HC3 O (B) (B) H H Paragraph for Questions 11 and 12 Schemes 1 and 2 describe sequential transformation H H of alkynes M and N. Consider only the major products HHHH HC3 O HC3 O CH CH O CH CH O H H formed in each step for both schemes. (C ) 32(C ) 32 (D) (D)

1. NaNH2(excess) 1. NaNH (excess) H H 2. CH32CH2(1 equivalent) H 2. CH CH (1 equivalent) X Scheme-1 3. CH 32l(1 equivalent) HHHH H 3 X Scheme-1 CH32CH O CH32CH O 3. CH l(1 equivalent) HO M 4. H,2 Lindlar3 ’s catalyst HO M 4. H,2 Lindlar’s catalyst

Q.12 The correct statement with respect to product Y is 1. NaNH2(2 equivalent) 2.1. NaNH (2OH equivalent) H Br 2 (2014) 2. OH H Br Y Scheme-2 N + (A) It gives a positive Tollens test and is a functional 3. HO3 (mild) Y Scheme-2 N + 3. HO(mild) isomer of X. 4. H,23 Pd/C

5.4. CrOH,2 Pd/C 3 (B) It gives a positive Tollens test and is a geometrical 5. CrO3 isomer of X. (C) It gives a positive iodoform test and is a functional isomer of X. (D) It gives a positive iodoform test and is a geometrical isomer of X.

PlancEssential Questions

JEE Main/Boards JEE Advanced/Boards Exercise 1 Exercise 1

Q.3 Q.8 Q.13 Q.15 (b) Q.1 Q.8 Q.13 Q.18 Q.21 Q.20 (a) Exercise 2

Exercise 2 Q.5 Q.9 Q.14 Q.22 Q.26 Q.1 Q.7 Q.9 Q.14 Q.28 Previous Years’ Questions

Q.6 Q.8 Previous Years’ Questions

Q.10 Q.13 11.96 | Alkenes and Alkynes

Answer Key

JEE Main/Boards Exercise 2

Single Correct Choice Type Q.1 B Q.2 C Q.3 A Q.4 D Q.5 C Q. 6 C Q.7 B Q.8 D Q. 9 D Q.10 B Q.11 A

Previous Years’ Questions

Q.1 B Q.2.D Q.3 D Q.4 C Q.5 A Q.6 A Q 7 B Q.8 B Q.9 C Q.10 A Q.11 A Q.12 D Q.13 C Q.14 D Q.15 B Q.16 D Q.17 C Q.18 C Q.19 A Q. 20 B Q.21 A Q.22 B

JEE Advanced/Boards Exercise 2

Single Correct Choice Type

Q.1 C Q.2 C Q.3 A Q.4 B Q.5 B,C Q.6 B Q.7 D Q.8 B Q.9 C Q.10 C Q.11 B Q.12 C Q.13 D Q.14 D Q.15 C Q.16 A Q.17 B Q.18 A Q.19 B

Multiple Correct Choice Type

Q.20 A,B Q.21 B,C Q.22 A,C,D Q.23 A,C Q.24 A,B,C,D Q.25 A,C Q.26 A,B,D Q.27 A,B,C,D Q.28 A,C

Assertion Reasoning Type

Q.29 B Q.30 A Q.31 A Q.32 A Q.33 B

Comprehension Type

Q.34 A Q.35 A Q.36 A Q.37 D Q.38 C Q.39 B

Match the Columns Q.40 A → q; B → p; C → s; D → r Q.41 A → p; B → r; C → s; D → q Q.42 A → q; B → r; C → s; D → p

Previous Years Questions

Q.1 A Q.2 D Q.3 A Q.4 D Q.5 B Q.6 A Q.7 D Q.8 B Q.11 A Q.12 C BH3.THF (a) BH3.THF (a) 3 B 3 B

CH3.COOH Chemistry | 11.97 CH3.COOH

B H H H BH .THF B CH .COOH Solutions(b) 3 3 H BH .THF CH .COOH (b) 3 3 3 3

JEE Main/Boards BH3.THF CH3.COOH (C)( c ) BH .THF CH .COOH 3 3 H H ( c ) 3 Exercise 1 3 H H

Sol 1: Alkenes are substances which contain less (D) Same as (a) hydrogen than the maximum quantity of hydrogen (d) HH Same as (a) which a carbon atom can have (d) HH Structure:- Sol 4 H H H H H H2 H2 (A) H C C H C C (A) Pt Pt H H H H (B) (B) Na Preparation of Alkenes:- + NaOC25H SH OH (1) Dehydrohalogenation of Alkyl halide 5 H (2) Dehyration of Alcohols Sol 5: Physical properties: Refer theory pg.26 (3) Dehalogenation of vicinal dihalides Chemical properties: Refer theory pg.29 (4) Dehalogenation of geminal dihalide

Sol 6: Alkynes:- Sol 2: (A) In the family of hydrocarbon alkynes are the most unsaturated one with C≡C bond and are highly reactive. 1. BH ,THF 3 + mirrorimage Preparation:- 1. BH3,THF– + mirrorimage 2. H2O2,OH– 2. H2O2,OH H H (1) Dehydrohalogenation OH OH (2) From tetradehalogenation transadditon transadditon (3) From haloform Recemic mixture \ Recemic mixture \ (4) Kolbe’s electrolytic decarboxylation (B) OH (5) Berthlot’s reaction OHH H Sol 7: Physical properties: Refer theory pg.42 Antimarkov'saddition Antimarkov'saddition Chemical properties: Refer theory pg.44

BH3.THF BH3.THF(a) Sol 8: (a) 3 Sol 3: (A) 3 B (A) B (a)

NaNH BH3.THF CH .COOH 2 3 CH3 CCº H (a) CH .COOH 3 3 B B H Br H H B CH3 CC+ NH3 BH .THFH CH3.COOH º (B) 3 CH .COOH (b) 3 BH .THF CH3.COOH (b) 3 3 CH3 CCº 3 H B O H BH .THF CH3.COOH ( c ) 3 BH3.THF CH3.COOH (b) BH .THF CH3.COOH H (b) 3 3 H ( c ) 3 3 H H Hg2+ OH

HO2 BH3.THF CH3.C(d)OOH Same as (a) ( c ) HH (d) 3 Same as H(a) H HH

(d) Same as (a) HH (a)

NaNH2 CH3 CCº H

Br

CH3 CCº + NH3

11.98 | AlkenesCH3 CC ºand Alkynes

O (B) OH (b) H O/H (B) 2 (b) BH Hg2+ OH 22 HOOH- 22 OH HO 2 Anti-maekonikov’s addition Sol 9: Hg(OAc)2 (A) + NaBH4 - + O-Na OH O O Na - + - + NH2 O Na Markonikov’s (a) O O Na NaNH2 NH2H (A) (a) addition NaNH2 H (A) CH (C) OH OH 22 ( c ) + CH H RCOH3 22 + O-Na CHOH OH - + H+ O Na OH + (B) H -H O (B) 2

H (1 equivalent) 2 Br H2(1 equivalent) HBr Al23O Sol 11: (a)(A) Al O OH 23D OH D Doubled bonds are more electron donating than triple bond because of difference in electronegativity (B) (i) NaNIH H2 (b) 2 and because of which first addition will takes place on H H2 (i) NaNIH2 Cindlars (b) H (ii) Prv double bond rather than triple bond. catalystCindlars (ii) Prv catalyst CH22I (B) OH CH22I

(b) HS24O D

H+

Sol 10: (A) OH2

(a) Con. HS24O/D OH

Sol 12: A A hydrogenation B hydrogenation B C C OH CH 2 CH510 510 HBO A o B A 3Ro OH HBO different 3ROH B different Cl HBO o HCl B OMDM CHBO o 1ROH B OMDM C 1ROH OH Sol 13:

-H2O (B)(B) (A)(A) (C )(C )

o 3Ro3ROHOH Chemistry | 11.99

I Sol 5: (C) C ≡ C is strongest because it is shortest and + HI (2) more orbitals are overlapping. I (I) (J) Sol 6: (C)

D HO2 CaO + C CaO H-C C-H 2+ - - 2 º Alc. KOH Ca ( CCº ) 2+ HS24O Hg O

Ozonized CH3-C-H ethanal 2 O HO2 Sol 7: (B) C2H3MgI   → C2H6 Sol 14 Abstracts acidic hydrogen. (i) Na CCº H CCº (ii) Br Br (X) Sol 8: (D) HBr only product Na CCº Na H Br

H 2 (Y) (X) H Pd-BaSO4 H H OH O Sol 9: (D) Anti Markonikov’s addition (1) CH COOH 3 Anti-additon Anti Markonikov’s addition (2) H/+ HO 2 HBr Sol 10: (B) HC º CH HC2 º CH Br H OH Alc. KOH HO H HC º CH

Exercise 2 Sol 11: (A) Enol form is not net stable as keto is much more stable. Single Correct Choice Type

Sol 1: (B) Dicarboxylation of acid ion Previous Years’ Questions

– NaOH Sol 1: (B) Methane is produced due to the decay of CH3 C O Na CH3 H + CO2↑ vegetables or animal organisms present in swamps and O marsh, by the action of bacteria. Due to this method of CH 3 formation, methane is also known as marsh gas. Br Sol 2: (C) 2 H Br Br H + Mirror image CH Sol 2: (D) Unsaturated compounds which contain 3 C=C or C≡C, decolourises the purple colour of alkaline Recemic mixture KMnO4 solution.

OH Sol 3: (A) Definition of conformers. CH ­ HO- 2 CH2=CH2 + KMnO + MnO O 4 2 O H purple coloured CH OH Sol 4: (D) Olefin 3 2 Zn+H O + H 2 O Sol 3: (D) Structural formula of 1, 2-butadiene is:

pent-2-ene 11.100 | Alkenes and Alkynes

Sol 4: (C) Alkynes undergo Markownikoff’s addition of •  • water in presence of H2SO4/HgSO4: CH3− CH = CH 2 + Br → CH 32 − CH − CH Br ∆< H 0   OH  • • HgSO CH− CH − CH Br + HBr → CH − CH − CH Br + Br ∆< H 0 CH CC H + HSO 4  3 2 322 3 º 24 CH3 C=CH2 In case of addition of HCl and HI, one of the propagation unstable enol step is endothermic, reaction fail to occur. O C CH CH3 3 Sol 10: (A) acetone

C6H5CH2=CHCH3 has two geometrical isomers

Sol 5: (A) Bayer’s reagent is cold, dilute, alkaline CH65CH2 H CH65CH2 CH3 perman-ganate solution, used to detect presence of C=C and C=C olefinic bonds. H CH3 H H

Trans Cis Sol 6: (A) Terminal alkynes are slightly acidic, forms salt So it is 1-Phenyl-2-butene with very strong base like Na, NaNH2 etc. 1 H− C ≡ C −+ H Na  →� H−≡ C C−+ Na +H ↑ Sol.11: (D) CH – CH+ – CH –Cl 2 2 3 2 ethyne Sol 12: (A) Sol 7: (B) Hydrogenation with poisoned palladium LiAlH CH COOH 4 CH CH OH ‘A’ brings about cis hydrogenation of alkyne and does not 3 32 affect double bonds: PCl3

CH32CH Cl ‘B’ Sol 8: (B) + Alc. KOH HOCl HO- + Cl + Cl CH22=CH ‘C’ + Cl CH3 CH CH2+ Cl CH3 CH CH2 Ag 2Cl C CH3 CH33CCº CH +6AgCl Cl Cl HO- CH3 CH CH2 1, 1, 1- Trichloroethane

CH3 OH Sol 13: (C) CH3 H/2 Pd Sol 14: (D) HC3 HH CH 3 H O HC3 CH 3 O3 CH33CH=CH - CH CH3 - CH CH - CH3 O - O H H meso-compound HO2 (optically inactive) 2CH3CHO Zn + i.e. Reaction is initiated by Cl ( Chloronium ion Sol 15: (B) electrophile). Θ Na/Liq.NH CH CH CH≡ CH →3 CH CH C ≡ CNa⊕ 322 ∆ 32 Sol 9: (C) In addition of HBr to an alkene, in presence of It is a terminal alkyne, having acidic hydrogen. peroxide, both the propagation steps are exothermic: •• HBr+→+ HO H2 O Br Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with Na in liq. NH Propagation 3 Chemistry | 11.101

CH− MgX + CH −≡−→ C C H CH Sol 16: (D) 33 4 JEE Advanced/Boards

Sol 17: (C) H H H CH3 Exercise 1 C=C C=C CH CH HC H 3 3 3 Sol 1: (X) Lucas reagent

Cis Trans C5H8O AgNO 3 Ag¯ Sol 18: (C) 2–butene is symmetrical alkene H2 Pt (X5) O3 HI CH33− CH = CH − CH →2.CH 3 CHO Zn/H2 O Degreeof unsaturation : 2 Molar mass of CH₃CHO is 44 u.

no lucas test Sol 19: (A) • • OH Ag CH3− CH 2 − CH = CH 2 +→ B r CH 32 − CH − CH − CH 2 Br ¯ a secondary radical OH OH Therefore, Statement I is correct but Statement II is H2 incorrect. Pt X5 Sol 20: (B) Both Statement I and Statement II are correct HI and Statement II is correct explanation of Statement I. Sol 2: (A) Br2 Br dicolourizes (A) 2 dicolourizes C5H10 implies unsaturation Sol 21: (A) C H O 5 10 implies unsaturation hotKMnO4 C = C + O CH3 S CH3 3 C C hotKMnO4 (B)+(C) OO (B)+(C) ozonide Acids C = O + O = C + (CH )SO Acids 32 CaO CaCaO O CaO Both Statement I and Statement II are correct and Statement II is correct explanation of Statement I. CH4 C2H6 CH4 C2H6 O O Sol 22: (B) Statement I is correct. Statement II is also (B) ® CH3 C OH correct. Meso form of the product is due to anti addition (B) ® CH3 C OH - of Br on cyclic bromonium ion intermediate, hence O O Statement II is not correct explanation of Statement II. (C) C2 H5 C OH (C) ® C2 H5 ®C OH + (A) (A) ® Br ® HC3 H C CH C= C + Br2 3 Sol 3: CH C H 3 H Cl Alc.KOHNaNH Ammo H (A) 2 (B) 2 (C) CCl AgNO CH C H 4 3 3 6 12 C6H12Cl2 C H Br 6 10

CH3 Ozonolysis H2,Pt C - CH3 Br (D) +(E) H C H O Br CH3 C H Cl (B) Cl

Alc. Cl KOH

(C)terminalalkyne (D) +(E) O H + CH3 C H O Cl Alc.KOHNaNH Ammo (A) 2 (B) 2 (C) CCl4 AgNO3 C6H12 C6H12Cl2 C6H10

Ozonolysis H2,Pt

(D) +(E) O 11.102 | Alkenes and Alkynes

CH3 C H Cl Sol 7: (A) Cl2 H2O (B) Cl (B) C9H12O2 C9H10 C9H10Cl2 Alc. Cl KOH oxidation

COOH (C)terminalalkyne O (D) +(E) +CH3 C OH O H Cl Cl + CH3 C H (A) (B) O C C CH3 CH CH CH3 Sol 4: H H H X5 Aq. (A) (B) (C) H HCl KOH OH dihalogen O (C) OH D CH CH CH3 O– Cl Cl Ca2+ 2 NaOH NaOH O Sol 8: (A) (B) X 2+ – Ca + C O C9H10Br2 C9H10OBr CH3 2

H2SO4 AlBr Sol 5: CF CH CH + HBr 3 3 2 (C)

+ – Ozonolysis H [AlBr4 ] O

CH3 CH2 C H+ (D) CF3 CH2 CH2 Since only one –Br is hydrolysed ∴ One is attached to alkyl group and other one to aryl CF3 CH2 CH2 Br group. alk. Sol 6: (A) (B) D does not Br OH KMnO4 form anhydride (A) C9H12 C8H6O4 NaOH Fe Br2

onlyone monobromo Br Br derivative

H2SO4 Degree of unsaturation of (A) = 4 So it should contain benzene ring CH CHO O3 In the compound (B) both –COOH should be at para CH3 2 Zn/H2 position to give only one bromination product + O H Br (A) (B) COOH (C) COOH Br

Br COOH COOH Chemistry | 11.103

Reduction Sol 11: O Sol 9: (A) (B) (C) Dil. H2SO4 R C Cl C6H10 C6H12 C6H14 (A) (B) Hg2+ C8H6 Ozonolysis KMnO4 Friedel craft's I2+KOH alkylation (F) acid Iodoform Hydrolysis reaction C4H8O2 (C)+ (D) (D)+ (E)Aldehydes yellow

2CH2OC2H2O2 For reaction (B) → (C) it is iodoform so (D) → CHI3 O O O H yellow precipitate and (B) should have acetyl ketone. O O– O C So (A) is (B) (C) (B)is 1-4 addition

oxidation (C) KMnO4 (A) (F) OH OMDMfor alkyne 2 O give ketone

Na Sol 10: (A) (B) Hydrogenation liq.NH3 I Sol 12: (A) (B) C5H8 C8H14 C5H8 C5H12 Alk. Hg2+ H2SO4 KMnO4 dil. O3+Zn/H2O

H2SO4

(D)+ (E) C4H8 O O (A) C5H10O (C) isomeric acids O Ketone C H + (B) Degree of unsaturation of (A) = 2 ∴ Alkyne. Terminal alkyne because formation of (B) and Sol 13: (A) (B) branched because formation of ketone (C) H Br HBr + 1-4additon (A) 1. NaNH2 Br H 2. +

H

()C O

(D),(E) O COOH OH , H Compound (A) carbocation is more stable resonance is in both the compounds but (A) have 3 α – H (B) have 0 α – H 11.104 | Alkenes and Alkynes

Sol 14: Sol 16: (A), (B) C5H11Br alc.   → + isomeric alkyl bromides (A) KOH (C) + (D) isomers

C5H10

Sol 17: D Br 1-4 addtion (i) Br HH D

O d O Br –Br (C) 3 H C H + O d Zn+H2O d d H Hydrogenation H2/Ni Br Br \ (C) (B) Alc. (D)+(E) KOH

(C), (D), (E) (ii)

Catalytic hydrogenation alkyl carbon NBS gets brominated C5H12 \ (D) is isomer of (C) H Br Alc. KOH (D) not becabeuscaeuse \\ (D) not willwillnonotgtget byet byisomis eric omeric (iii) alkyalkylbrolbmirodemideof (Bof)(B) NBS (i)Li (A) (D) (E) (B)(B) (D) (E) (ii) CuP Br

Hydrogenation Hydrogenation Li

H H LiBr

H H (iv) D H Sol 15: Br D 1. Mg/Et O3 2 MBr D H (P) (Q) C3H5Br H D + Zn/H2O 2. CO2,H C8H10 C4H6O2 only Anti-addition

symmetric compound (v) + Q is an acid because after reaction with CO2, H of RMgBr we get a carboxylic acid. If O Br C OH (R) (Q) O H OH H (vi) HH + OH MgBr O=C=O H3O HO H H OH H+ CH3 CH3

\ (P) Hydrolysis H to satisfy the degree of unsaturation of (P) alkyl group OH –H+ H2O OH2 must be cyclic. Chemistry | 11.105

Cl2 Sol 18: (A) →C H Cl Exercise 2 hν 59 Single Correct Choice Type C = 85.7%; h = 14.3% total molecular weight = 70 Sol 1: (C) Releases 1.34 mL of a gas at STP C H 5 10 ∴ 3 moles of CH4 ∴ 3 active α - H hydrogen.

Sol 19: Amm. Amm. Br (A) (A) (X)Not(Xa)N terotminala terminal CH 2 Br AgNO Sol 2: (C) 2 CH2 Br CH2 CH2 AgNO3 3 NaCl C H Alkyne Alkyne C6H10 6 10 Cl Br CH2 CH2 Reduction Reduction CH CH 2 2 Br Alk. Alk. (B) C6H12 (F) C3H6O2 Br Cl (B) C6H12 KMn(F)OC4 3H6O2 KMnO4

(C) C6H14 hν (C) C6H14 CH → Sol 3: (A) 26 + Cl2 C2H5Cl (X5) (A) O3 O3 (D)+(E) O (A) Zn+H(D2O)+(E) Zn+H2O C2H4O C2H2O2 boiling C2H4O C2H2O2 O + CH C OH O Sol 4: (B) KMnO 3 O O O 4 O O H Oxidative cleavage H H H H H (A) (A) Br (B) 1-4 addition Sol 5: (B,C) 2 (B) 1-4 addition h n Br (C) (C) O + (F) O more stable (F) OH OH ⊕

Sol 20: (A) + (B) CH3 − CH2 Sol 6: (B) Carbon is least stable in ↑ isomers CH= CH 22 Zn-Cu Zn-Cu ∴ Addition will be least CH2=CH2

Sol 7: (D) Stability of complex that forms during (A) Na/Ether (C) reaction with H2 R R (B) Na/Ether (D)

But mixture gives (E) R R H Na/Ether is Wurtz reaction and dimerise the alkyl H bromide so they are symmetric. Sol 8: (B) D2O D D (A) Br OH– (E) OH (B) Br D

Sol 9: (C) C-C-H is sp3 hybridised, hence ‘s’ character is less, So bond length in C-H is longest 11.106 | Alkenes and Alkynes

Sol 10: (C) Sol 17: (B) O O O Dehydration CH322 CH CH CH(OH)CH 22 CH CH 3→ O −HO2 3 D C C OH (Heptan−− 4 ol) C CH2 C H2O CH CH CH= CHCH CH CH + CO 32 22 3 OH OH 2 (Hept−− 3 ene)

– + Sol 11: (B) Hydrogenate product Sol 18: (A) O K Me O Sol 19: (B)

H H OH same group + N N \ opticallyinactivecompound.

Br Sol 12: (C) HBr Multiple Correct Choice Type

Br Sol 20: (A,B) NBS

Br Br

Sol 13: (D) CH CH CH Cl 2 2 H H CH2 C CC Sol 21: (B,C) C NH4Cl CH H KOH HCl (A) Cl Cl

CH2 CH C CH2 NMe3 H (B) EtOH HC º CH Cl2 D Sol 14: (D) Cl H2O Cl OK (C ) OH D Cl Cl less hindered –Hwillget abstracted

Sol 15: (C) conc.H2SO4 O (D) O D H–C–O–O–H OH + HC– OH O Sol 22: (A,C,D) but the reaction will be reversible since O C–I is very week. O H C (B) × (C)  (D) more stable system H O Sol 23: (A,C) Sol 16: (A) O O ozonolysis Alkane CH3-CHO HIO4 + + [O] CH3 CH CH3 C OH HO Alkene : O Isomers : O

+ HCHO HCHO + O Chemistry | 11.107

Sol 24: (A,B,C,D) Sol 27: (A,B,C,D) D (A) Al C  HO2 → Al(OH) + 3CH 4Al3+ + 3C4– + CH22N 4 3 3 4 (B) Ca2+ –C≡C– + H O  → Ca(OH) + HC ≡ CH -N2 2 2 + N 2 2+ 4– (C) 2Mg + C3 + H2O  → Mg(OH)2 + H3C–C≡CH CH2-N=N N=N (D) Me3C–H+KMnO4  → Me3C–OH Same for trans form. Sol 28: (A,C)

Sol 25: (A,C) H2SO4 (A) H C C H CH4 CHO Br / F BH3 TH (A) – H2O2/OH CH CHO Br 3 Br (B) O Hg2+ CH3 C C H CH C CH absence presence (B) H+ 3 3 BH3/THF Br H2O2 (C ) CH3 CH2 CHO O ( ) CH C C CH (D) C 3 3 Br + CH3 C CH2 CH3 presence absence Hg2+ O Sol 26: (A,B,D) (D) H+ CH CH CH (A)(a) CF3 CH2 CH CH 2 CH CH2 2 O (A) CF3 CH2 CH2 2 2 BH3.THF

HCHCl l H2O2

CFCF33 (C(CHH22)3)3 CHCH2 2 CHCH3 3

Assertion Reasoning Type CH CH CFCF33 (C(CHH22)3)3 CH2 2 CH3 3 ClCl Sol 29: (B) When the C=C bond is conjugated with a Cl C=O group, the electron withdrawing nature of carbonyl Cl group polarizes the C=C bond so the nucleophilic (b(B)) HCHCll attack can take place at the C=C carbon.

Sol 30: (A) Alkyl iodides are more reactive than alkyl chlorides for elimination reactions as I– is a better HClHCl CH CH Cl leaving group than Cl– (c)(C CF) CF3 3 CHCH CHCH2 2 CFCF33 CH22 CH2 Cl

Sol 31: (A) is a stronger base than F–because the CF3 CH2 CH2 negative charge density on carbon is greater than the CF3 CH2 CH2 negative charge density of F–. (D) CH (d) 3 CH2 CH CH CH2 CH3 CH2 CH CH CH2 Sol 32: (A) CH3–C≡C–CH3 is more reactive than

HCl CH≡CH towards HCl because the carbocation HCl formed is more stable in thecase of CH3C≡C–CH3 than CH≡CH. Cl Cl 11.108 | Alkenes and Alkynes

Sol 33: (B) Alkynes are electron-rich nucleophiles with Match the Columns a cylindrical electron cloud formed of two bonds around a carbon-carbon bond. An electrophilic Sol 40: reagent can therefore easily react with the relatively (A) weak alkyne bond. cis addition H2/Pd/BaSO4 (B) Comprehension Type Birch reduction (C) Sol 34: (A) CH CH Br NaNH2, D 3 2 Isomerism 55 gm (D) Alc. KOH, D m 55 = Isomerism 208 58 A → q; B → p; C → s; D → r

Sol 35: (A) 2.432g Sol 41: (A) → (p) Markonikov (B) → (r) Anti-markonikov Sol 36: (A) Alkenes generally forms by elimination. (C) → (s) Alkylic carbon get brominated

(D) → (q) Br2 addition Sol 37: (D) Br2 Br CH3 H O 2 Sol 42: (A) → (q) Free radical mechanism

CH3 OH (B) → (r) Decarboxylation Anti addition (C) → (s) Free radical mechanism (D) → (p) Same as grignard H2O Br

Br OH Previous Years’ Questions

CH3 Br 14 H (CH ) – Sol 1: (A) 33CMgCl + DO2 CH3 C D + Mg (OD)Cl CH CH CH Br Br Sol 38: (C) 3 2 CH3 CH CH3 CH3 Br + Sol 2: (D) According to the IUPAC conventions, Br 14 compound can be numbered as: Br 12 3 4 56 CH3 CH CH2 CH2 CH CH3 14 H2 C=−−−≡− CH CH 22 CH C C H Br Br Here, C-2 is sp2 and C-3 is sp3 hybridised. Sol 39: (B) CH3 Sol 3: (A) Oxymercuration - demercuration brings about Markownikoff’s addition of water as Br2 H Br (B) Anti Addition H Br HgSO4 CH3 CH3 CH2 C C H +H24SO

OH O

CH3 CH2 C CH2 CH3 CH2 C CH3 butanone

unstable enol Chemistry | 11.109

Sol 4: (B) Ammonical CuCl2 forms red precipitate with terminal alkynes, can be used to distinguish terminal alkynes from internal alkynes: NH3 (aq) CH32− CH − C ≡ C − H + CuCl2 →

−+ CH32− CH −≡ C C Cu ↓ red− ppt

Sol 5: (B) Bromination is highly selective, occur at the carbon where the most stable free radical is formed:

Sol 6: (A) Reaction proceeds through carbocation intermediate:

++ Ph− C ≡ C − CH33 + H  → Ph − C = CH − CH OH O | || HO2 →−=Ph C CH − CH3 ←→−− Ph C CH23 − CH

Sol 7, 8: (D, B) The final ozonolysis product indicates that the alkene before ozonolysis is O HC3 CH3 O3 CC 2 C Zn-H2O

HC3 CH3 HC3 CH3

Also P(C6 H10) has two degree of unsaturation and oxymercuration demercuration hydration indicates that it is an alkyne. As alkyne, on hydration, gives a carbonyl compound which on reduction with NaBH4 gives a 2° alcohol. O OH

(i) NaBH4 C C +H2O C CH2 + C CH2 (ii) H

o H 2 alcohol

The secondary alcohol that can give above shown alkene on acid catalyzed dehydration is CH CH CH3 OH CH3 3 3 + + + + -H H Me-shift CH C C CH CH3 C CH CH3 CH3 C CH CH3 CH3 C CH CH3 3 3 -H2O CH CH CH3 CH3 3 3 2o carbocation

CH3 CH3 O HgSO (i) NaBH 4 CH C C CH 4 Q CH3 CC C H 3 3 + HS2 O4 (ii) H

CH3 CH3

Sol 9: (i) Free radical chorination of alkane require energy which is supplied either in the form of heat or radiation. + - CH CH + H+ CH Br CH3 2 ® CH3 CH3 CH3 CH CH3 2o Carbocation Br Isopropyl bromide 11.110 | Alkenes and Alkynes

(ii) Addition of HBr proceeds through carbocation intermediates.

Sol 10: (i) Peroxide CH3 C = CH2+HBr CH3 CH CH2Br

CH3 CH3 2-methyl propene Isopropyl bromide In the absence of peroxide, HBr would be added giving tertiary butyl bromide. (ii) Tertiary alkynes are slightly acidic, forms silver salt with ammoniacal solution of silver nitrate:

NH3 (aq) R− C ≡ C − H + AgNO3 →R − C ≡ C Ag ↓ white ppt.

Solution for the Q. No. 11 and 12: (A, C)

C CH CH l NaNH2 32 H O HO

HC3 O

H,2 Lindlar’s catalyst CH l 3 O H H X OH Br NaNH C H 2 1 equivalent NaNH2 O

OH O + H,Pd/C HO2 2 CrO3 mild OH Y