Download [1] Ans1.Dows-proc - Sacred Heart School Moga,Best ICSE School

SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Max. Marks: 30
Max. Time: 40min
Q1.write reaction to prepare phenol from chlorobenzene?
Ans1.Dows-process
C6H5Cl+NaOH
320atm,3600C

C6H5OH(phenol) +NaCl
[1]
Q2.C6H5CH2Cl and C6H5CHClC6H5 which is more easily hydrolysed aq.KOH?Why?[1]
Ans2. C6H5CH2Cl
[alkyl halides are more easily hydrolysed being more reactive].
Q3.Distinguish between benzylchloride and chlorobenzene?
C6H5CH2Cl + KOH (aq) boil  C6H5CH2OH + KCl
Benzyl chloride
[1]
Benzyl alcohol
KCl
+ AgNO3  KNO3 + AgCl  white ppt
C6H5Cl (chlorobenzene) + KOH (aq) Δ  no reaction
(Ar-X less reactive than R-X)
Q4. How is chloro benzene prepared from benzene and benzene diazonium
chloride?
Ans4. Raschig process or chlorination (Electrophylic substitution reactions)
C6H6 + Cl2
FeCl3  C6H5Cl
+
HCl
[2]
Q5.Which is more reactive alkyl halide or aryl halide? Explain all the factors with the
resonating structures?
[3]
Ans5. Alkyl halide (RX) are less reactive than aryl halide because of the following
reasons:
(i) Resonance Effect:
Aryl halide are resonance stabilised than alkyl halides. Lone pair of electrons are
delocalised over the benzene ring. As a result Carbon and halogen bond (C-X) in
haloarenes acquires double bond character C=Cl while the other hand C-X bond is
single in nature in alkyl halide (RX). Thus C-X bond in aryl halides are stronger than
alkyl halides, hence haloalkanes (RX) are less reactive. Resonance structures can
be written as:
(ii) Hybridisation: In RX alkyl halides C attached to halogen is sp3 hybridised
while in Ar-X C is sp2 hybridised. Therefore bond length in Ar-X is smaller than R-X.
Therefore C-Cl bond is stronger in Ar-X than R-X.
(iii) Dipole moment: C-X bond in aryl halide is less polar than alkyl halide (R-X)
e.g. Chlorobenzene dipole moment is 1.69D while of CH3Cl is 1.86D.
Due to above reasons halogen atom in aryl halide cannot be easily replaced by
nucleophilic.
Q6. Convert Benzene to diphenyl?
C6H6+Cl2
FeCl3 
C6H5Cl
+HCl
C6H5Cl
+ C6H5Cl
2Na Dry ether (fittig) C6H5-C6H5
[1]
Diphenyl
Q7. Convert chlorobenzene to phenol?
[1]
C6H5Cl
+NaOH
dil HCl 320 atm, 3600C Dow’s processC6H5OH + NaCl
Q8. Benzene to chlorobenzene?
C6H6+Cl2
anhyd FeCl3 
C6H5Cl
[1]
+HCl
Q9. Benzene to toluene?
C6H6+ CH3Cl
anhyd FeCl3 Friedel Craft reaction  C6H5CH3 + HCl
[1]
Q10. Give mechanism of reaction when benzene is chlorinated.
Electrophyllic substitution reaction
[2]
1
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Q11.convert benzene to toluene
C6H6+ CH3Cl
anhyd FeCl3 
C6H5CH3
+HCl or
C6H6+Cl2
anhyd FeCl3 
C6H5Cl
+HCl
C6H5Cl
+ CH3Cl
2Na Dry ether wurtz fittig C6H5CH3
[1]
+2NaCl
Q12. benzene to acetophenone
C6H6+ CH3COCl anhyd. FeCl3
+HCl
[1]
friedel craft acylation C6H5COCH3
Q13. benzene to bromo benzene
C6H6+Br2
anhyd. FeBr3 bromination C6H5Br
[1]
+HBr
Q14.benzene to nitrobenzene
C6H6+HNO3 H2SO4 nitration  C6H5NO2+H2O
[1]
Q15.benzene to benzene sulphonic acid
C6H6+ H2SO4 C6H5SO3H +H2O
[1]
Q16.convert aniline to phenyl isocyanide
C6H5-NH2 + CHCl3 + 3KOH (alc) carbylamine  C6H5NC
+ 3KCI +
[1]
3H2O
Q17.convert benzene to BHC or lindane or 666
C6H6+Cl2
hv,
C6H6Cl6
[1].
Q18.convert chlorobenzene to diphenyl
C6H6+Cl2
Anhyd. FeCl3 
C6H5Cl
+HCl
C6H5Cl
+ C6H5Cl
2Na dry ether  C6H5-C6H5
[1]
+2NaCl
Q19.convert aniline to benzene diazonium chloride.
[1]
C6H5NH2 + HONO (NaNO2+ HCl) diazotisation 273-278K  C6H5N2Cl + 2H2O
Q20.convert benzenediazonium chloride to bromo benzene
Gattermann reaction:
[1]
Q21.Convert chlorobenzene to DDT?
[1]
Q22.Convert chloroform to tear gas?
Cl3CH
+ OHNO2
 Cl3CNO2 + H2O
[1]
Q23.Convert toluene to benzyl chloride (C6H5CH2Cl)?
C6H5CH3
+ Cl2 hv
C6H5CH2Cl
+HCl
Q24.convert chlorobenzene to anisole
[2]
0
C6H5Cl
+ NaOH
Dow’s 320atm, 360 C  C6H5ONa
C6H5ONa
+ CH3Cl
Williamson synthesis C6H5OCH3
[1]
Chloroform
nitric acid
Q25. Ullmann reaction
C6H5I +Cu +C6H5I
--xxx—
chloro picrin
+
+HCl
NaCl
[1]

C6H5-C6H5
+CuI2
2
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Q1.Discuss briefly the following reaction
(a) Balz-Shiemann reaction?
C6H5N2Cl
[1]
+ HONO, 273-278K diazotisation 
C6H5NH2
C6H5N2Cl
Benzenediazonium chloride

+ HBF4 heat
C6H5F
Benzenediazonium chloride
+
N2 + BF3
Fluoro benzene
(b) Wurtz reaction.Why it is not suitable for preparation of odd number alkanes?
C2H5-Br + 2Na + Br-C2H5
dry ether wurtz  C2H5-C2H5 + 2NaBr
[2]
Ethyl bromide
Butane
Because mixture of products are formed. This is not a suitable reaction to prepare
higher unsymmetrical alkanes, as by-products are also formed during the reaction
and thus their separation becomes difficult.
(c) Friedel Craft acylation reaction for chlorobenzene?
C6H5Cl + C6H5COCl Anhyd. AlCl3
[1]
C6H4Cl COC6H5 o-p cholorobenzophenone
(d) How does Iodobenzene reacts with copper powder in a sealed tube? What is the name of
reaction?
[1]
C6H5I + Cu + IC6H5 Δ  C6H5-C6H5
Iodo benzene
Biphenyl
+ CuI2
Ullmann
(e)Chlorobenzene to phenol?
C6H5Cl + NaOH (3600C,320 atm P) Dow’s process
(f) Finkelstein reaction?
C2H5Cl
+ NaI
acetone

C6H5-OH+ NaCl
[1]
[1]
C2H5-I + NaCl
Ethyl Iodide
Ethyl Chloride
(g) Give 'Williamson Synthesis' reaction for sodium phenoxide with methylbromide. Write
the names of this compound formed.
[2]
C6H5-ONa
+
Sodium phenoxide
BrCH3

methyl bromide
C6H5-O-CH3
Anisole
+ NaBr
Q2. Give the name of the reaction, in which only haloarenes are treated with sodium,
forming diaryls?
[1]
Fittig reaction
Q3. Bromoalkanes can be easily prepared by refluxing the silver salts of a fatty acid. Name
and explain the reaction.
[1]
CH3COOAg + Br2 350K, CCl4 REFLUX Borodine HDCH3Br + CO2 ↑ + AgBr 
Silver salt of acetic acid
Bromo methane
Silver bromide
Borodine Hunsdiecker reaction
Q4. How haloarenes are prepared from diazonium salts. Explain, by giving names of the
reactions?
[1]
Ans Sandmeyer and Gattermann reaction
C6H5N2Cl
+
CuBr / HBr sandmeyer
Benzene diazonium chloride
C6H5N2Cl
 C6H5Br
+
Bromobenzene
+ Cu / HBr
Gattermann 
Benzenediazonium chloride
N2
C6H5Br
Bromobenzene
+ Cl-
+ N2
Q5. Convert benzene to chlorobenzene to phenol and chlorobenzene to phenol and phenol
to benzene(4step)?
[2]
C6H6 + Cl2 FeCl3 chlorination,Electrophyllic Substitution Reaction  C6H5Cl + HCl
C6H5Cl + NaOH (3600C,320 atm Pressure)Dow’ s process C6H5-OH+ NaCl
C6H5-OH+ Zn dust Dephenolation  C6H6
+ ZnO
Q6.Convert butyl bromide to butyl-iodide?
CH3CH2CH2CH2-Br + NaI acetone (Finkelstien)  NaBr + CH3CH2CH2CH2-I
Q7.What is Haloform reaction?
[1]
Q8.Give reactions of nitration of bromobenzene and bromination of nitrobenzene?
[1]
Q9. What is Hoffmann ammonolysis reaction?
[1]
CH3COCH3 + 3NaOI (I2+ NaOH) CI3COCH3 + 3NaOH
Acetone
Tri-Iodoacetone
NaOH + CI3COCH3  CHI3  (yellow) Iodoform + CH3COONa
[1]
Nitration C6H5Br
+ HNO3/H2SO4  o-p nitrobromobenzne +
H2O
Bromination : C6H5NO2 + Br2 /FeBr3  m- bromonitrobenzene + HBr
CH3CH2Br + H-NH2
C2H5OH  CH3CH2NH2 +HBr
CH3CH2Br + CH3CH2NH2 C2H5OH  (CH3CH2)2NH +HBr
CH3CH2Br + (CH3CH2)2NH  (CH3CH2)3N +HBr
3
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Q10. Write Fermentation reactions from sucrose?
[1]
C12H22O11 sucrose
Invertase enzyme + H2O  C6H12O6 (Glucose) + C6H12O6 (Fructose)
C6H12O6 (Glucose ) Zymase 2C2H5OH + 2CO2
Q11. Convert aniline to benzene diazonium chlroide?
[1]
(i) C6H5NH2+HONO(NaNO2+ HCl) diazotisation 273-278K C6H5N2Cl + 2H2O
Q12. Give one chemical test to distinguish between carbon tetrachloride and chloroform?[1]
Carbylamines test.
CHCl3
+ C2H5NH2 + 3KOH (alc)  C2H5NC + 3KCl + 3H2O
Chloroform
10 amine
Ethyl Isocyanide (bad smell)
Q13. Convert benzene to toluene?
C6H5-Br + 2Na + Br–CH3 wurtz fittig
Bromo benzene
(Dry ether)
Methyl bromide
 C6H5-CH3 + 2NaBr
[1]
(toluene)
Q14. Convert benzene to chlorobenzene by Raschig process and then convert to diphenyl?
[2]
2 C6H6+2HCl + O2(g) Raschig CuCl2 , 550K, heat  2C6H5Cl + 2H2O
C6H5Cl + 2Na + ClC6H5 dry ether Fittig 
C6H5-C6H5 + 2NaCl
Bromobenzene
Biphenyl
Q15. Write the reaction when 2-bromobutane is treated with alcoholic KOH?
Ans15. Saytzeff rule
[1]
CH3-CH2-CHBr-CH3 + KOH (alc) Δ CH3-CH=CH-CH3 + CH3-CH2-CH=CH2 + KBr + H2O
2-Butene (80%)
1-Butene (20%)
***
Max. Marks: 35
Q1.Which isomer of C4H9Cl will have the lowest boiling point?
Ans1. 30 alkylhalide i.e. (CH3)3Cl because it is having lesser surface area.
Reason: VanderWaal forces is directly proportional to surface area.
Max. Time: 40min
Q2. Write the reaction and formula of the main-product formed?
(CH3)2CHCl
Na,
Dry ether
?
+
?
Ans2. (CH3)2CHCl
Na, Dry ether ( Wurtz reaction)  (CH3)2CH-CH(CH3)2 +
[1]
[1]
2NaCl
Q3. How will you convert propene into propyne?
[1]
Ans3. CH3-CH=CH2
+Br2
CCl4
CH3-CH-CH2 + 2KOH (alc)  CH3-CCH + 2KBr+ 2H2O
( Propene)
| |
( Propyne )
Br Br
Q4a. Markovnikov’s rule ?
Ans4.When addition across an unsymmetrical alkene takes place, the -ve part of the
addendum goes to that carbon atom of the double bond which holds the lesser
number of hydrogen atoms. It takes place in the presence of mercuric acetate
Hg(CH3COO)2.
Or
absence
of
peroxides
Br

CH3-CH=CH2 + HBr
MARKONIKOV’S RULE
CH3-CH-CH3
Propene (Unsymmetrical alkene)
2-Bromopropane
Q4b. Anti Markonikov rule or peroxide effect or kharash effect.
Ans4b. During addition of HBr (only) to unsymmetical alkenes in the presence of
organic peroxide such as benzoyl peroxicle, (C6H5CO)2O2 –ve part
of
addendum goes
to that C
atom of double bond which has more no. of H
atoms around double bond. It takes place contrary to the Markovnikov's rule. Also
known as Peroxide effect or Kharasch effect.
CH3-CH=CH2 + HBr
ANTI-MARKONIKOV RULE (Presence of peroxides)
CH3-CH2-CH2Br
Q4c. Saytzeff rule in dehydrohalogention reaction?
[3]
ANS4C DEHYDROHALOGENATION : (1,2 ELIMINATION, β ELIMINATION)
Saytzeff rule: It states that “In dehydrohalogenation reactions, the preferred product is that alkene
which has lesser no. of H atom on doubly bonded carbon atoms or preferred product is that alkene
which is more substituted.
CH3-CH2-CH2-CH2Br + KOH(alc) Δ CH3-CH=CH-CH3 + CH3-CH2-CH=CH2 + KBr + H2O
(more substituted) (Less substituted)
2-Butene (80%)
1-Butene (20%)
4
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Q5. Conversions: (a) Ethyl bromide into Propanoic acid?
Ans5(a) C2H5Br+KCN(alc)
(nculeophylic substitution ) C2H5CN
C2H5CN+2H2O H+ acidic hydrolysis  CH3COOH +NH3
[1]
+KBr
(b) How will you convert propene to 1-bromopropane?
Ans5(b) CH3-CH=CH2 + HBr ANTI-MARKONIKOV’S RULE  CH3-CH2-CH2Br
[1]
(c) Convert bromo methane to acetic acid?
Ans5(c) C2H5Br+KOH(aq) hydrolysis CH3CH2OH +KBr
CH3CH2OH
+
[O]
KMnO4/H2SO4

CH3CHO
+
[O]
KMnO4/H2SO4

[1]
CH3CHO
CH3COOH
+H2O
+H2O
(d) Name the reagent used to convert 1-chroropropane to 1-nitropropane?
Ans5(d) AgNO2
(e) Convert bromo ethane to butane?
Ans5(e) C2H5Br
+2Na +C2H5Br dry ether Wurtz reaction C2H5C2H5
Bromo ethane
Butane
[1]
[1]
+ 2NaBr
(f)1-Propanol to 2-bromopropane?
Ans5f CH3CH2CH2-OH + Conc. H2SO4 443K  CH3CH=CH2
CH3CH=CH2 + HBr MK RULE  CH3CH(Br)CH2 2-bromopropane
(g) CH3CH2Br + AgCN 
?
+?
ANS5(g) CH3CH2Br + AgCN  CH3CH2NC + AgBr
(h) C2H5Br+ KOH (alc) 
?
+?
ANS5(H) C2H5Br+ KOH (alc) DEHYDROHALOGENATION  C2H4 + KBr + HOH
[1]
(i) Convert ethyl chloride to ethyl alcohol?
ANS(I) C2H5Br + KOH (aq) HYDROLYSIS  C2H5OH + KBr
[1]
(j) Convert silver acetate to methyl bromide?
Ans (j) CH3COOAg +Br2 CCl4 REFLUX BORODINE HUNSDIECKER  CH3Br + CO2 + AgBr
[1]
[1]
[1]
Q6. Alkyl halides though polar, are immiscible with water, why?
[2]
ANS6 Although alkyl halides are polar molecules, neither they form H bonds with
water nor can they break H- bonds already existing between water molecules. As a
result, the solubility of haloalkane in water is very low.
Q7. Why is thionyl chloride method preferred for preparing alkyl chlorides from alcohol?
[1]
ANS7.Reaction: C2H5OH + SOCl2 C2H5Cl + SO2↑ + HCl ↑
Thionyl chloride (SOCl2) method is preferred to hydrogen halide or PCl5, PCl3
method as by products obtained are SO2 and HCl which are in gaseous state and
which can escape leaving behind the chloroalkane in almost pure state.
Q8. Treatment of alkyl halides with alc. AgNO2 give mainly nitroalkanes while that with aq. NaNO2 give mainly
alkyl nitrites. Explain.
[2]
+
ANS8. Because sodium nitrite ( NaNO2 ) is an ionic compounds and hence have –ve
charge on one of the oxygen atoms. Nucleophile attack through this –ve charged
Oxygen atom on the alkyl halide mainly gives the alkyl nitrite.
R+ X+ +Na-NO2-  RONO (alkyl nitrite) + NaX
RX
+ Ag-NO2
C2H5OH/ H2O  RNO2
+ AgX
Alkyl halide
silver nitrite
Nitro alkane
In contrast silver nitrate (AgNO2) is a covalent compound and hence do not have a
–ve charge on the O atom. Instead both the O atoms and the N atoms carry lone
pair of electrons. Since N is less electronegative than oxygen, therefore lone pair of
electrons on the nitrogen atom is more easily available for bond formation.
In other hand, nucleophillic attack occurs through N and hence silver nitrite
predominantly gives nitro compounds.
Q9. Explain, why the treatment of alkyl halides with aqueous KOH leads to the formation of alcohols, but in the
presence of alcoholic KOH. Alkenes are major products.
[2]
ANS9.SUSTITUTION
OR
ELIMINATION of alkyl halides depends upon nature of alkyl
halide, strength and size of base or nucleophile and reaction conditions.
RBr + KOH (aq)  ROH + KBr
RBr + KOH (alc)  alkene + KBr + H2O it undergoes β dehydrohalogenation
If the base is weak (KOH) or nucleophile is less concnentrated it results
into
substitution. Moreover in polar solvent, OH- ions are highly hydrated (solvated)
hydration reduces basic character and therefore fails to abstract H from  Carbon of
alkyl halide to form alkene.
If base is bulkier (stronger base) elimination predominates over substitution.
5
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
On the other hand alc KOH , contains alkoxide ion (RO-) which being stronger base than OH- preferentially
eliminates a molecule of HCl from an alkyl halide to form alkene.
2-Bromobutane + (CH3)2COelimination

CH2=CH-CH3
+ HBr
2-Bromobutane + OHsubstitution

CH2-CH(OH)-CH3
+ BrQ10. What are ambident nucleophiles ? Explain with an example.
[2]
Ans10. Nucleophiles which have more than one site through which the reaction can
occurs are called ambient nucleophiles. E.g. CN-,NO2-
Q11.Which compound in each case reacts faster inSN2 reactions with -OH.
(i) CH3Br or CH3I
(ii) (CH3)3CCl or CH3Cl
Ans11.CH3I (less bond enthalpy) (ii) CH3Cl (less stearic hinderance)
Q12. Complete (CH3)3CBr+KOH ethanol/heat?
Ans12. (CH3)3CBr+KOH Ethanol/heat  (CH3)2C=CH2+KBr
[1]
+H2O
Q13.Why Grignard reagent should be prepared in anhydrous condition?
[1]
Ans13. Grignard reagent are very reactive. They react with moisture present in the apparatus or the starting
materials RX or Mg.
RMgX +H2O RH +MgX(OH)
Therefore they must be prepared in anhydrous conditions.
Q14.Acetylene into propyne?
Ans14. CHCH +Na liquid NH3  CHC-Na
CHC-Na + CH3I  NaI
+ CHC-CH3
[1]
Q15.Propyne into 2-butyne?
Ans15. CHC-CH3
+
Na-CC-CH3 +
[1]
Q16.Ethane into ethylene glycol?
Ans16. C2H6
+
C2H5Cl
+
CH2=CH2
+[O]
H2O
Q17. Ethane into ethylene ?
Ans17. C2H6 +Cl2
hv
C2H5Cl
+
Na liquid NH3
CH3I
 Na-CC-CH3
 CH3-CC-CH3
+ NaI
+ NaI
[1]
Cl2 hv 573K
 C2H5Cl
+ HCl
KOH (alc)
CH2=CH2
+KCl +H2O
alkaline KMnO4

CH2-CH2
(Baeyer ‘s reagent)
|
|
OH OH
[1]
C2H5Cl + HCl
KOH (alc) dehydrohalogenation CH2=CH2 + KCl + H2O
Q18.Methyl bromide to ethylamine?
C2H5Br
+
KCN(alc)
 C2H5CN
+KBr
C2H5CN
+
[H] Na/alcohol 
C2H5CH2NH2
Mendius-reduction
Q19.Propane into propene?
Ans
C 3H 8
+ Cl2 hv
 C3H7Cl
+
HCl
C3H7Cl
+ KOH(alc)
CH3-CH=CH2 +
KCl
[1]
[1]
+H2O
Q20.Difference between SN1 and SN2 mechanism of nucleophillic substitution reactions?
[2]
SNI
SNII
First order reaction
Second order reaction
Rate =k [RX]
Rate =k [RX] [Nu]
One step reaction
Two step reaction
Reaction rate influence by electronic Reaction rate is determined by steric
factors
factors
Racemic mixture: d+ l forms
Inversion of configuration (Walden
inversion) d to l form(vice-versa)
Order: CH3X <10<20<30 < Allyl < benzyl
Order : CH3X > 10 > 20 > 30
Nu can attack from back side as well as well The nucleophile(Nu-) can attack from
from front side,however former predominates back side
Rearrangement is possible.
Rearrangement is not possible.
30 RX are common substrate i.e. presence of 10 RX are common substrates i.e.
bulky group facilitates SN1
presence of simple and fewer R groups
facilitates SN2 mechanism
Reaction favoured by mild nucleophiles e.g. Favoured by strong nucleophile e.g.
alcohol, water etc
alkoxide ion(30alcohol)
Favoured
by
low
concentration
of Favoured by high concentration of
nucleophiles.
nucleophiles
Favoured by polar solvents(water).
Favoured by non polar solvents or low
polarity solvents(alcohol).
Favoured by the presence of electrophilic Favoured by the absence of electrophilic
catalysts
catalyst.
--xxx—
6
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Max. Marks: 40
Q1.What is meant by hydroboration reaction? Give example?
Max. Time: 40min
[1]
Refer Reaction:34 of named reaction notes
CH3-CH=CH2 + H-BH2 THB  CH3-CH-CH2
CH3-CH=CH2  (CH3CH2CH2)2B-H
Propene
Borane


H BH2
CH3-CH=CH2  (CH3CH2CH2)3B
Tri-n-propylborane
(CH3CH2CH2)3B + 3H2O2 OH ,H2O 3CH3CH2CH2OH+H3BO3
Diborane (B2H6) is an electron deficient molecule. Therefore, it acts as an electrophile
and reacts with alkene to form trialkylboranes which upon subsequent oxidation with
alkaline H2O2 to give alcohols. Thus, B2H6  2BH3
CH3-CH=CH2 + H-BH2 THB CH3-CH-CH2
CH3-CH=CH2  (CH3CH2CH2)2B-H
Propene
Borane


H
BH2
+ CH3-CH=CH2  (CH3CH2CH2)3B
tri-n-propylborane
(CH3CH2CH2)3B
+ 3H2O2 OH- , H2O  3CH2CH2CH2OH + H3BO3
tri-n-propylborane
Propan-1-ol
Boric acid
The addition of borane to the double bond takes place in such a manner that the boron
atom gets attached to that carbon atom of the double bond which has greater number of
hydrogen atoms.
This two step process is called hydroboration oxidation and gives alcohols which seem
to have been formed by Anti-Markovnikov’s addition of water to alkene.
Q2. Complete the reaction: CH3CH2CH2OCH3 + HBr  ?
+
CH3CH2CH2OCH3
+ HBr  CH3CH2CH2OH
+ CH3Br
?
[1]
Q3. Write Williamson’s synthesis reaction?
[1]
Reaction:13 of named reaction notes
C2H5-Br
+ NaOC2H5
Williamson syn C2H5-O-C2H5 + NaBr
Ethyl bromide
Sodium Ethoxide
Diethyl ether
Q4. Write the reaction: (a) Propene to propan-1-ol ?
Reaction:34 hydroboration oxidation
CH3-CH=CH2 +
H2O H-BH2 THB  3CH3CH2CH2OH
[1]
+
H3BO3
Q5.Write IUPAC name for the following?
CH3CH(OH)CH(CH3)CH2CH3
Ans5. 3-Methylpentan-2-ol
[1]
Q6. Explain: Why 10 alcohols are more acidic than 30 alcohol?
[2]
Due to +I effect electron density increases thus acidic character decreases, tendency to loose –OH
increases with increase of +I effect. Release of H+ ions is easy in case of 10alcohol as compared to
30alcohol.
Q7. Write reactions of acidic dehydration of ethanol to give ethene?
CH3CH2OH
Conc. H2SO4 443K  CH2=CH2 + H2O
[1]
Q8.Which of two C6H5OH or C2H5OH is more acidic and Why?
C6H5OH
 C6H5O- +
H+
[2]
resonance
C2H5OH

stabilised
C2H5O- +
not
resonance
H+
stabilised
+
Phenoxide ion formed after release of H ion is resonance stabilised as compared to alkoxide ion formed
after release of H+ ion.
Details:
Why alcohols are weaker acids than water? (Marks 2)
Ans. ACIDIC NATURE OF ALCOHOL:
a. H2O > Primary alcohol > Secondary alcohol > Tertiary alcohol
The acidic character of alcohols is due to the polar nature of O–H bond
7
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Alkyl group is an electron-releasing group (–CH3, –C2H5) or it has electron releasing
inductive effect (+I effect). Due to +I effect of alkyl groups, the electron density on
oxygen increases. This decreases the polarity of O-H bond. And hence the acid
strength decreases.
V imp Q13. why phenols are more acidic than alcohols?
ACIDIC NATURE OF PHENOL:
a. Phenol > H2O > Primary alcohol > Secondary alcohol > Tertiary alcohol
The acidic character of alcohols is due to the polar nature of O–H bond
Alkyl group is an electron-releasing group (–CH3, –C2H5) or it has electron releasing
inductive effect (+I effect). Due to +I effect of alkyl groups, the electron density on
oxygen increases. This decreases the polarity of O-H bond. And hence the acid
strength decreases.
b. Phenol is more acidic than alcohol because:
In phenol, the hydroxyl group is directly attached to the sp2 hybridised carbon of
benzene ring which acts as an electron withdrawing group. Whereas in alcohols, the
hydroxyl group is attached to the alkyl group which have electron releasing
inductive effect.
In phenol, the hydroxyl group is directly attached to the sp2 hybridised carbon of
benzene ring. Whereas in alcohols, the hydroxyl group is attached to the sp3
hybridised carbon of the alkyl group. The sp2 hybridised carbon has higher
electronegativity than sp3 hybridised carbon. Thus, the polarity of O–H bond of
phenols is higher than those of alcohols. Hence, the ionisation of phenols id higher
than that of alcohols.
The ionisation of an alcohol and a phenol takes place as follows:
In alkoxide ion, the negative charge is localised on oxygen while in phenoxide ion,
the charge is delocalised.
The delocalisation of negative charge makes phenoxide ion more stable and favours
the ionisation of phenol. Although there is also charge delocalisation in phenol, its
resonance structures have charge separation due to which the phenol molecule is
less stable than phenoxide ion.
c. In substituted phenols, the presence of electron withdrawing groups such as nitro
group enhances the acidic strength of phenol. On the other hand, electron releasing
groups, such as alkyl groups, in general, decreases the acid strength. It is because
8
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
electron withdrawing groups lead to effective delocalisation of negative charge in
phenoxide ion.
(b) Phenols are less acidic than p-nitrophenol. Explain. (Marks 1,2)
Ans. In –NO2 substituted phenols (p-nitrophenol), the presence of electron withdrawing groups –NO2
nitro group enhances the acidic strength of phenol. It is because electron withdrawing groups lead to
effective delocalisation of negative charge in phenoxide ion.
Q9. Write : (i) Lucas reagent (ii) Baeyer ‘s Reagent (iii) Jone’s reagent
Anhyd. ZnCl2 +HCl
Alkaline
KMnO4
Jones reagent: chromic acid in aqueous acetone for controlled oxidation
PCC: Pyridinium chloro chromate forcontrolled oxidation
Refer: reaction:50
(v) Hybridisation of C bonded to –OH
sp3
(iv) PCC
Q10. Write conversions
.
(i) Ethane to ethyl chloride to butane to chlorobutane to butane to butanol?
C2H6 +
Cl2
hv,573 C2H5Cl
+HCl
Wurtz reaction C2H5Cl
+ C2H5Cl
Na, Dry Ether C2H5-C2H5 +NaCl
C4H10 +
Cl2
hv,573 C4H9Cl
+
HCl
C4H9Cl +
KOH (aq)
C4H9OH
+
KCl
[5]
[3]
(ii) Ethane to ethyl chloride to butane to chlorobutane to butanol?
Same as above
(iii) Convert Propanal to 2-butanol?
CH3CHO
+ CH3MgBr
dry ether  CH3C(OMgBr)HCH3
+
CH3C(OMgBr)HCH3 + H2O / H
acidic hydrolysis  CH3CH(OH)CH2CH3
Q11.Give chemical test to distinguish between?
(i) Methanol and ethanol
Iodoform
test:ethanol
 Yellowppt.
CH3CH2OH + 3NaOI (I2+ NaOH)  CHI3 (yellow + HCOONa
Write reactions: ref reaction No.51(details)
I2 + 2NaOH  NaI +
NaOI
+ H 2O
CH3CH2-OH + NaOI  CH3CHO + NaI + H2O
10 alcohol
aldehyde
3NaOI + CH3CHO
 CI3.CHO + 3NaOH
NaOH + CI3CHO
 CHl3  Iodoform (yellow) + HCOONa
2-butanol
[3]
CH3COOH  no yellow ppt.
(ii) 1-propanol & 2-propanol
Iodoform
test:2-propanol  Yellow ppt.
Write reactions: ref reactionNo.51
CH3COCH3 + 3NaOI  CI3COCH3 + 3NaOH
Acetone
Tri-Iodoacetone
NaOH + CI3COCH3  CHI3  (yellow) Iodoform + CH3COONa
(iii) n-propyl chloride & iso-propylchloride
iso-propylchloride
+KOH (aq)
 2-propanol +KCl
Iodoform
test:2-propanol
 yellow ppt.
Write reactions: ref reactionNo.51
2-propanol + NaOI  CH3COCH3 + NaI
CH3COCH3 + 3NaOI  CI3COCH3 + 3NaOH
Acetone
Tri-Iodoacetone
NaOH + CI3COCH3  CHI3  (yellow) Iodoform + CH3COONa
Q12.Complete the following: (i) C3H7ONa + C2H5Cl

C3H7ONa
+ C2H5Cl
Williamson synthesis C3H7OC2H5
[1]
+NaCl
Q13. Convert ethanol to propanoic acid?
Ethnaol + SOCl2 pyridine  C2H5Cl + SO2+ HCl
C2H5Cl + KCN (alc)  C2H5CN + KCl
C2H5CN + 2H2O
H+  C2H5COOH + NH3
[1]
Q14. Convert Propan-1-ol to propan-2-ol?
[1]
9
SACRED HEART SCHOOL
Moga, Punjab
Test XII-Chemistry [tests-organic-chemistry]
Propan-1-ol + conc H2SO4 443K  propene + water
propene + H2O dil H2SO4 MK rule  propan-2-ol
Q15. Discuss the oxidation of 10, 20 and 30 alcohols by KMnO4 and by Cu-573K by taking suitable
example?
[6]
Refer reaction : 50 , 52
CH3CH2OH + [O] KMnO4  CH3CHO + [O]  CH3COOH Ethanoic acid
Ethanol
-H2O
Ethanal
OH
O

║
CH-CH-CH3 + [O] KMnO4  CH3-C-CH3 propanone + [O]  CH3COOH + H2O + CO2 or
Propan-2-ol
-H2O
 CH3COOH + HCOOH
OH
O

║
CH3-C-CH3 + [O] KMnO4  CH3-C-CH3+CO2+ 2H2O CH3COOH + H2O + CO2

CH3
2-Methyl propan-2-ol
Propanone ( Lesser no. of Carbon atoms)
Q16. A organic compound A (C2H6O) reacts with sodium to form a compound B with the evolution of H2
and gives a yellow compound C when treated with iodine and NaOH. When heated with conc. H2SO4 at
413K, it gives a compound D (C4H10O) which on treatment with conc. HI at 373K gives E. D is also
obtained when B is heated with E. identify A, B, C, D and E. write equations for the reactants and
products involved?
[5]
Ans16.
A : 2CH3CH2OH + Na  C2H5ONa (B) + H2
A: CH3CH2OH + NaOI (I2+ NaOH)  CHI3 (C) + HCOONa
A : 2 CH3CH2OH + conc H2SO4 413K  C2H5OC2H5 (D)
D
C2H5OC2H5 + HI, 373K  C2H5I (E) + C2H5OH
C2H5ONa (B) + C2H5I (E)  C2H5OC2H5 (D) + NaI
Q17. Write CH3MgBr reactions with
(a) Ethylene oxide (oxirane) (b) CO2 (c) Propanone (d) Ethyl acetate (e) Ethyl alcohol
Products: propanol-1-ol
(b) CH3COOH
(c) 2-methyl propanol (e) C2H6
Refer reaction no. 33
--xxx--
[5]
10