R a n ke rs S tudy Ma terial-45-P I V-A& K-CH 3'l

SOLVED PROBLEMS

Subjective:

Problem 1.

Solution:

Problem 2.

(A)

OH,intEmolEulrraldol con&nsaton

An organic compound (A), C6H1yO, on reaction withtreatment gives compound (B). The campoundcompound (C) which in presence of a base givesWrite the structures ot (A), (B), (C) and (D).

The reactions suggest that (A) isoll

tt\.,-

CftMgBr foilowed by acid(B) on ozonolysis gives

1 -acetylcyclopentene (D).

A U''X*,

(-)*(-lCH, CH.I L)o

r\ &ono,Ets /' .rot ) --:--'L J\_.- \_.(B)

HOH

cocH3

.\\-/

(D)

ldentify A, B, C and D, in the following schernes and write their structures.

( ,ts\ etlccr" r14; Na^rH, ,F)--Is:9LtflLr(C)

rbH,

NH2NHCONH'

(c)

Solution:

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G^";:- Gr'''u*'"' h,* el"o'"n',CH./\/\ r-\.N

/HN\r=

H,N

"t

R a n ke rs Study Ma teria/-45- P I V-A & K-C Hs$\\\\\N\\\\\\\\\\\

Cyclobutylbromide on treatment magnesium in dry ether forms an organometallic

compound (A). The organometallic compound reacts with ethanol to give an

alcohol(B) after mild acidification. Prolonged treatment of alcohol (B) with an

equivalent amount of HBr gives 14romo-+nethylcyclopentane (C). Write the

sfrucfures of (A), (B) and explain how C is obtained by (B).

.,8, /MsBr HoXcH' (d1"''

ff u"** ' il "tB"' ( '' --"-fIj" --"u-

&",An alkene (A) (CrcH16) on ozonolysis gives only one product (B) CaHaO.

Compound (B) on reaction with NaOH/lz yields sodium benzoate. Compound (B)

reacts with KOH/NHz-NHz yielding a hydrocarbon (C) CaHro. Write the structure

of (B) and (C). Based on this information give two isomeric structures of (,A)

Problem 3.

Solution:

Problem 4.

Solution:

,-eL6n3 &,oc;.

oil

oznoFs,rO^"*.

T.H"

rcL rrs"-Wo I

(A) has two isomeric structuresH,C.o ,CoHu HrCo -cH,"'-L/"' '\-J"''

and

H.c cH. H.c boHu

/Problem 5/ An organic compound (A) CyH4O3 in dry benzene in the presence of anhydrous

\ / AtCft gives compound (B), the compound (B) on treatment with PCts foltowed byreaction with Hy'Pd (BaSOq) gives compound (C) which on reaction with

hdyrazine gives a cyclized compound (D) CqH$NI. ldentify (A), (B), (C) and (D).

Explain the formation of (D) from (C).

Solution:

H2Pd/BSO4 .

(B) oH

NH.-M"

(c)

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Problem 6.

Solution: i)

Prob'em 7.

Solution:

il

[iY^t'. NH2oHHcr .''"(.-./o'\) *,rF, _

anti

K-CH

compound (A) (caHso) on treatment with NH2)HHC]gives (B) and (c). (B) and(c) rearrange to give (D) and (E) respectivery, on treatment with acid. (B), (c),(D) and (E) are all isomers of molecular formula csHsNo. r,uhen (D) is boited withalcoholic KoH an oily liquid (F) (c6u?N) separfes out. F reacts rapidry withcHscocl to give back (D). on the other hand (E) on boiting with atkafi fottowedby acidification gives white sotid (G) C7H6O2. tdentify (A) to (G).

33N'N

HuC.

H.C

syn

N

OH

. (oxime)

ii) Oxime undergo Beckmann rearrangement as followso//-----'----'> Huc. {\NH/

H.c(E)

NH,

D *o* rclt - COOK +

NH^t'I

,r\ s-cm, r\t-*Vouov J".(F) (D)

E ^o'>CH3NH+cuHucooH(G)-white sotid

How would you bring about the foilowing conversion?i) Ethanal to 24ydroxy-34utenoic acidii) 2tnethyl propanalto

HuCo pH H"C^\ / -\FN and F*.H.c H.i bn

o//+ H^c-4,\

NH/HuCu

(D)

CH.

H.C

iii)

iv)24utanone from ethyt atcohol34exanon from n-propyt alcohol

ot-l

4c-cHo -#-- 4c< --A-+ r-Lc1 * )

CHO CHO

FLC=1to , \_oFl

/HOOC

cH./"H"C--<

cHo

FLC:r

r-.*lNC

ii) cH"

H-|]cHoI

cH3

dil. NaOH-_---)

CH.

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iii)

34N

H,C-

9H.i

-5o----) H.c-.-AoH --!l--

H.c*-,rAon socl' ) H.c--,2\cl

PzHa ,CrHuH,o* > H,c< Iol r H,Cr-(

o//

H.C cH.butan-2-one

iv) Mq ,c"Ho

-#- H.c\,,Augcl --qgg-- H,c5-(

fCiMs

Problem 8.

Solutian:

Problem 9.

H.C. pH

0 H",sO"

--1-=_>

oFr

-------+cooc2Hs

ii)

CH, i) NaOH,Iii) Fr

iii)

iv)

NaOCI

-ffi" >A+B

n.l lfc(c7H12o)

v) O-HOOC-CuHo -CH, -CuHu "o"L ,(Al=mal"**(B)-+;ls--+(C)

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H

socr' > H.c--\ -;ffi-* H.c-l cHscHo > n'"X..-^,

bt Mgcr H,J -Mgcl

Sy nthe si ze 2 -+nethyl cyclohexanone from cyclohexane

ctI-,^'-

"r,nu I I r)nrrox-----' \-.,

_iixof_,

?*' cH'

i)BFr"/rHF /-y'o' oo /tu ,)lHplo*+ \rf ' oo.,*,

ry'Predict products of the following reactions

D,PH,c{ +

cH.

oil

o

crf

Ra n kers Study Ma terial-45-P I V-Ae K-CH 35N

=#'H-@oProblem 10. Predict the products in the following reaction

6

//+ H3C--{ bc. }\

cHs

H'cy1,-cH' Koc, )

cHa oOH

I

,."-^\r-"" IAs(NHs)'lNo3

>

IIo

Solution: i)

Solution: a)

cooc2H5

o

'."/\H.C^

aTr)r-o

ii)

iii)

iv)

cH./H.c-d.\/

HOOCo.ll cn,o

aY( ar

(, '",, ffi

oll .+r./'o-l-"oo*n(,

1'..v

fl "r..Y;",

\-,/

Br

* Fe'.Br

--coz

oll cn"a\L;

I I (Easy decarboxylation)

v)

a)

b)

c)

lll * Base+ l-i-l - -^ /i\-/ ' t'ase +

\-,lJ . *'"-i\ -------)

i /"'.\_-/

HO- +

,-l()_to-c-cH3

uH3

jnoH

l-\v/I

H3C-C-CH3

OH

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,r,1

R a n ke rs Study Ma terial-45-P I V-A & K-CH 36N

b) Haloform reaction. So product is,."YYO'

c)

CH"

It is Tollen's Reagent. lt does not oxidise ketone.present, that will be oxidised to ketone.

But if 2' OH group is

Problem 11. ldentify A,

a)

e)

D

o

B, C, D ln the following

Raney Ni, H2 . trr_U

NaBH 4, CH 3Ol

b)

c)

d)

oil

CHs- C- CH2- CHe

oil

CHs- C- CHz- CHs

olt

CHs- C- CH2- CHso

NaBD4 . HZO .n

--U

NaBD, DcO

-...------------1-__---,1-

NaBH4 . DZO .E'-'------------- ---------------- t

s)

h)

AI I equivatont NaBt-t4

> GI I cH3oH

\cnoolt

I I exc6NaBH4 >HI I CH:OH

-\cro

fl

n -!Er--+r Naa& ,J Hao* >KI L JoIt, ,* cH3oH\crto

Solution: a) Carbonyl group as well as doubl+ bond is reduced

$, =

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Hc)

37N

b)-i-, U-

OHI

d)

0

OD

I

E= CHs-C-CH2-CH3

to

I

-cH'oH

D= CHs-9-Cnr-CHa

te) ?o

F= CHs-f-arr-CH,

H

s) OH

I

H =a)tt\cHroH

OHh) ,9 oHAA'=l [p(o\, = [- l"ro.

t;J ;?Prohlem 12.

Solution:

Problem 13.

Solution:

Which of the following ketones is more acidic. Give a reasono_ lI .lltJ--\[]-'

c+*This ketone is more acidic because the resulting enolate ion Obey's Huckel's ruleand is thus more stable.

Two moles of an ester A are condensed in the presence of sodium ethoxide togive a P-keto esfe4 B, and ethanol. on heating in an acidic solution, B givesethanol and a p-keto acid, c. on decarboxylation c gives 3-pentanone. ldentifyA, B and C with proper reasoning.

The reaction of 2 mol of an ester giving B-keto ester and alcohol in the presenceof sodium ethoxide is known as Claisen condensation.Let the given reactions may be depicted as shown in the following.RCHz-fr-OO,+RCHz-fi -eft,

-oczH. t

o(A)

o(A)

R- CHz - C - CH - C- OR, + R,OHlt tiloRop-keto ester ethanol

(B)

R - cHz- fi - ?r - fi - on -$, R - cHz-uc - F, -u" - oH + q,eg

OROOAOathanolp-keloacid

(c)

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ii.Irl:l

R-CHz-C-CH-C-OHllllloRo

From these reactions, it is obvious thatR' : - CH2CH3

R=-CHaHence, the compounds A, B and C are

R-CHz-C-CHzillOR

haat

------)

A: CHsCHzfi - OCHzCHg

o

B: cHscHzfi-?H-3-oczHsCHs

(B)

alcohol

ooC: CHgCHzC -lt

o/

Problem 14. /An organic compound A (C6H12O) forms an oxime but does not reduce Tollen's\ jl reagent. A on reduction with sodium-amatgam forms an alcohol B which ondehydration forms chiefly a single alkene C. The ozonolysis of C produces D andE. The compound D reduces Tollens reagent buf does not answer iodoform test.What are the structures of fhe above compounds? Explain the reactions.

Solution: The compound A must be a ketone as it forms oxime but does not reduceTollen's reagent.The compound D must be an aldehyde. lts structure does not include thefragment CH..Q as it does not answer iodoform test.

IU

The compound E must be a ketone containing cH. c _ fragment."8

Let the compounds D and E be RCH2CHO and R'COR", respectively, where R,R' and R" are all alkyl groups. From these, we get

cH - cooHI

CH:

7R, 7R,RCI-|zCHO*O=C( ( 03 RCHzCH=Cr.R,, \R,,

(D) (E), p, (c) ./R,

RCHzccH (*,, --'*'*"*,:x- "*(*,,o(A) (B)

/c{z rHr TcHzCHgCHzCCI-|\ --lll+ CHsCHzCHCHTll \cn. | \cn.

ooH

Since the molecular formula of A is CoHrzO, it follows that R = R' = R" = CHe.Hence, the structures of molecules (A) to (E) and the reactions are as follows.

(A)

Forms oxime but does not reduceTollens reagent

o. //cH3CHgCHzCHO + CHgCOCH3

-!-+ CHgCHzCH = C,.

(D) (E) (c) \ cHe

Reduce Tollens Does not reducereagent but no Tollens reagentiodoform test but gives iodoform

test

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R a n ke rs Study Ma terial-45- P I V-A& K-C H 39

Solution:

Problem 15. A compound A reduces Fehling's solution and gives positive silver mirror fesf. On

warming with dilute alkali, followed by dehydration it gives a product B which also

responds to both the above fesfs. /n addition, it decolourises the colour ofbromine water. On treatment with hydrogen in the presence of nickel underpressure the compound B is converted to C which does nof give any one of the

above fhree test Molar mass of the compound C is 74g mof1. Deduce the

structures of A and B giving the chemical equations involved.

The compound A must be an aldehyde as it reduces Fehling's solution and gives

positive silver mirror test. The compound A also contains ct-hydrogen atom as it

undergoes aldol condensation, the dehydration of which gives B. The latter

contains unsaturation as bromine water is decolourised. Let the structure of A be

RCH2CHO. The reactions involved are

RCHzcHo + HzccHo NaoH , RCHz cH - cH - cHotttOHR

I

| -H2CtRCHzCH=C-CHO(B)

I

R

The reduction of B with H2lNi under pressure would be

RCHzcH = ccHo-H2/Ni > RCHzcHzcHCHzoHttRR

(c)

Since the molar mass of C is 74 g mol-l, R in the compound C must be hydrogen

atom. Hence, the structures of A, B and C are.

A = CHgCHO

C = CHgCHzCH2CH2OH

B=CHgCH=CHCHO

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Ra n kers Study Material'45-Flv-Ae K-CHN.\i\xss:Hr$.r\r\$s\i.r:*N\\"X.!u

Objective:

Problem 1. tn the Cannizzaro reaction given below:

Ph*cHo oH- >Ph-cH, -oH+PhCOO-fhe s/ouresf sfeP is

(A) the attack of Ol{ at the carbonyl group.

(B) the transfer of hydride to the carbonyl group

(C) the abstraction of proton from ca;rboxylic acid

(D) the deprotonation of of Ph - CAOH

Solution: Transfer of hYdride ion todetermining steP.

is the slowest or the rate

4AN.N

Problem 2.

Solution:

Problem 3.

Solution:

G on-!"),p onV ..* ,pnl oH , l,-* l srowesr

\ *o \-,

o-l.onx*l

Hlo

- ,n-4 + Pr\-o- oH

Hence (B) is the correct answer.

A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH

solution gives

(A) Benzyl alcahol and sodium formate-

(B) Sodium benzoate and methylalcohol

(C) Sodium benzoate and sodium formate

(D) None

C6H'CHO + CHrO NaoH ,C6H.CH2OH + HCOONa t

ln cross Cannizzaro reaction of formaldehyde, formaldehyde is always oxidised'

Hence (A) is the correct answer'

Hence (A) is the correct answer'

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Ra n kers Study Ma teria!45-Plv-A& K-CH 41N

(B) H.c ,/cHo

CH2O @) H.CACOOH + HCOOH

Solution:

H.c--,r-\cH + Hro Hssoo+H,Soo ,

Hence (A) is the correct answer.

Problem 6. Which one of the following reaction cannot be used for the reductionR

-------> \/

R

Wurtz reaction is not applicable for above reduction.Hence (C) is the correct answer.

The most reactive compound towards formation of cyanohydric on treatment withKCN followed by acidification is(A) Benzaldehyde(C) Phenylacetaldehyde

Due to electron-withdrawing effect of No2 group, the partial +ve charge on catom of the c = o group increases hence it becomes more susceptible tonucleophilic attack by CN-.Hence (B) is the correct answer.

The compound that will not formiodine is(A) Acetone(C) Diethylketone

Problem 4. Aldolcondensation will not be obserued in(A) Chloral(C) Hexanal

(B) Phenylaceialdehyde(D) None of these

Solution: chlorral does not contain an o-hydrogen atom hence does not undergo aldolcondensation. lnstead it undergoes hydrolysis to give CHC|3.

Hence (A) is the conect answer.

Problem 5. The product (s) obtained via oxymercuration (HgSOa + HzSOq) of butyne wouldbe

(A) *'tal-"t'o

(c) n.cncHo +

o__1/

/\H.C CH.

butan-2-one

R

F"R

Solution:

Problem 7.

Solution:

Problem 8.

(A) Clemmensen reaction(C) Wurtz reaction

(B) Wolf-Kishner reaction(D) Hl and red phosphorus at 200"C

(B) p-nitrobenzaldehyde(D) p Aydroxybenzaldehyde

iodoform on treatment with alkali and

(B) Ethanol(D) lsopropyl alcohol

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Solutian: Diethyl ketone CHg - CHz - CO - CH2 - CH.. does not contain CH3CO grouplinked to carbon and hence does not give iodoform test.Hence (C) is the correct answer.

Problem 9. The appropriate reagent for the transformation isoti/1--{

\ I cH. ----------->

HO

iA) Zn iHs)/HCt(C) Both (A)and (B)

HO

Solution: Both Zn(Hg)/HCl and NHz - NHz / OH- can reduce COCH3 group toCHz - CHa but HCI will aiso bring about replacement of OH group by Cl.Therefore the most appropriate reagent is NHz - NH2/OH-.

Hence (C) is the correct answer.

Problem 10.

Solution:

Problem'11.

Solution:

Problem 12.

IJ

li

u.c1\r-cuH' ",- ,ilo

The above reaction is known as(A) Beckmann rearrangement(C) benzoin condensation

(A) CtuCOOH(c) crucoNH2

(B) Benzilic acid rearrangement(D) AIdol condensation

(B) CfuCH2NHAH(D) CruCHO

OHHrCu_-,/

,/ 'coo-HrCu

Benzilic acid rearrangement.

Hence (B) is the correct answer.

Oppenaur oxidatian is the reverse process ofiA) Wolt-Kishner reduction(B) Rosenmund reduction(C) Clemmensen reductian(D) MeerweinPondort-Verley reduction

Oppenauer oxidation is reverse of Meenrvein Pondorf Verley reduction.Hence (D) is the correct answer.

ldentlfy the product C in the serles

CfuCN Na/ceHsoH ,4 HNoz ,3 KMnoa, H+ ,g

1^\--\\/CH.r(E) NHz- NH2/ Of{(D) None of these

Solution: CHs - CN + 4H Na/czHsoH , CH3CH2NH2

{A)

CH3CHzNH2 + HO - N = O----+ CHu - CH2OH + Nz + HzO

CH3CH3CH2 ---lE+ CH3COO HKMno4 (c)

Hence (A) is the correct answer.

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42

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Rankers Study Material-45-PtV-A&K-CH 4A$\\r\\\\\\Ni$N\\rrd.\n$$\r\\\\r\rhsr\tN N\N\\\\N

Problem 13. ln the reaction

C,H, - CH = CH - CHo .j:*r-, CH, _ CH = CH _ CeoH, the oxidising' agent

agent can be(A) alkaline KMnOq(C) Benedict's solution

(B) acidified KzCrzOt(D) all of the above

Solution: CH3CH = CH - "rO

Benedict (Cu**) solution , ar.Benedict solution (solution of CuSO+, sodium carbonatespecific for oxidation of aldehydes.

- CH = CH - COOF.|

and sodium citrate) is

Problem 14. ln a cannizaro reaction, the intermediate that will be best hydride donor is

(B)

(D) Both (A) and (B)

NOzH

I

(D) @1--",Problem 15. Benaldehyde on reaction wrth CHz = CH - CHppfu forms

ln a cannizaro reaction, the intermediate that witt be best hydride donor isf-'--/CH = CH - CH = CHz

-/\_-/CH = CH - CHg(A)(' (A)l€J

. (C)

U"H = CH - CHg

Solution: Wittig reaction

(D) (y-cH = cH - cH -= cHz

PPh.I

CoHu-C- H CH -CH = CHz----+ C6Hs-CH = CH-CH = CH2+ pOpha

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