6. INDUCCION PRE-LABORAL PARA PERSONAL INGRESANTE
7.3 Inducción de Seguridad Personal Contratista
EXERCISE -I
Q.1 The uptake of 2 H atoms shows the presence of one >C = C< along with C H — and —COOH, which6 5 accounts for the 6° unsaturation. Furthermore H and I are monosubstituted benzene derivatives. H is
*
H3C.C HC6H5COOH
C H – C(COOH) = CH giving6 5 2 with one asymmetric carbon atom. I is C H CH = CH.COOH, giving C H CH CH COOH with no asymmetric carbon.6 5 6 5 2 2
CH2CH2COOH
H2 CH CH.COOH Pd
(I)
Q.2 (A) = C H CH OH, (B) = C H CH Cl, (C) = C H CH CH CH OH,15 31 2
(D) = C H CH CH COOH.
15 31 2 15 31 2 2 2 15 31 2 2
(Y) is an ester because it is hydrolysed to acid and alcohol. Since the alcohol is not oxidized by acidified KMnO and gives cloudiness at once with Lucas reagent, hence it is a t-alcohol.
Q.3
4
CH3
O CH3 O
HOHH C C OH H3C C OH
H C3 C O C CH3 3
(A) CH3
(B)
CH3
(Y)
Q.4 This is because C H OH undergoes dehydration to form C H at 170°C in presence of excess of conc.2 5 2 4 H SO .2 4
H2SO4(conc.)
CH3— CH2OH 170C H2O CH2 CH2
Q.5 In the presence of strong acids, the H is captured by the carboxylic acid and the following equilibrium is established:
O OH
H2SO4–
H2SO4 R C
R C
OH OH
O
H3C CH2CH2C CH3 Q.6
Q.7 It is because the carboxylate group (—COO–) of the branched acid is more shielded from the solvent molecules, therefore, it cannot be stabilized effectively by salvation.
Q.8 It is because carboxylic group does not have a true carbonyl group due to resonance.
O O
R C O H R C OH
Due to resonance >C = O bond of —COOH develops partial double bond character and cannot show reactions with hydroxylamine, phenyl hydrazine, etc.
It is because formic acid combines the properties of both an aldehyde and an acid.
Q.9
Aldehyde
H C OH
Hence it has reducing character of aldehydes.
HCOOH 2[Ag(NH3)2] OH HCOONH4 3NH3 H2O 2Ag or HCOOH Ag2O CO2 H2O 2Ag
Q.10 Both these unsaturated acids have two ionisable hydogens. After the release of first hydrogen, second hydrogen of maleiate ion is involved in H-bonding, whereas no H-bonding is possible in fumarate ion.
–
O H
O C
O
H C C
O H
H-bond C OH
H C C OH O
(Maleiate ion)
O
(Fumarate ion) (H-bond not possible)
Due to the formation of H-bond in maleiate ion more enegy is required to remove H from it than from fumarate ion, in which H release is easy comparatively. Thus, K2for fumaric acid is more than maleic acid.
O
CH3–CH2–MgBr H2C – CH2
Q.11
O
CH3–CH2–CH2CH2OMgBrH3
(X)
CH3–CH2–CH2CH2–OH (Y)
CH3CH2–CH2COOH (Z) Butanoic acid
KMnO4
O
Q.12 HCCHNaNH2NaCCNa2 CH3
CH3 – C C – CH3
H2SO4CH3– C = C H – C H3
(X) HgSO4 OH|
O
Tautomerises
CH3–|| –CH2–CH3
C
(Y)
Na+O¯ –O–CH
NaOH/ Br
||
Haloform reaction CH
C 2 3
H3O CH CH COOH3 2
(Z)
O---H – O
C – C H3
Q.13 CH3 – C (Dimmer)
O – H ---O
Dimerization of acetic acid occur in benzene via intermolecular H–bonding. Hydrogen bond is a special type of dipole –dipole attraction.
Sp hybridized carbon of – C C – of acid () and SP2 hybridized carbon of – C = C – of acid () attract the bonded electron more than do the SP3 – hybridzed carbon atoms. Consequently –CC– and –C=C– are acid strengthening E WG’s [Electron withdrawing group, stabilizes anion, thus stengthen’s acid] This makes CH3CH2COOH weakest ofallthese three acids since –CC– is more acid strengthining group than –C=C– group. This make acid () stranger than acid ()
Q.14
Q.15
The electron charge in carboxylate ion is more dispersed in comparision to phenoxide ion, since there are two electro negative oxygen carboxylate ion as compared to oxygen atom in phenoxide ion.
CH3COCl will after least stearic hinderance hence it hydrolysis will be more vigrous.
Amide = CH3CONH2 Therefore acid is CH3COOH
Acid (Y) obtained after decarboxylation must be mono carboxylic acid thus molecular weight = Equiva
lent weight
The acid must be (COOH 45g / mol) Given mass = 60g
= 60 – 45 = 15g/mol
Which is definetely due to –CH3 Hence Y is CH3COOH
Carboxylic acid (X) has second COOH replacing H of CH3COOH
COOH
Q.16 Q.17 Q.18
So(X) is malonic acidH2C of molecular mass 60 + 44 = 104
COOH
Since it has two – COOH group so its equivalent mass = 104/2 = 52 g/eq.
Q.19 Dehydration occur with all the three reagent
C6H5CH2CONH2P2O5C H CH CN + H O6 5 2 2
C H CH CONH6 5 2 2 SOCl2 C6H5CH2CN + 2HCl + SO2
POCl3 Or
C6H5CH2CONH2 PCl5 C H CH CN+H O6 5 2 2
Q.20 The mechanism of esterification
OH|
R 'OHR – C – OH
OH|
H +
R – C – OH2
| O R'
Slow
O
||
R – C – OR'
O
||
R – C – OR'
H
H2O
As the size of the substituent on – carbon increases, the tetrahedral bonded intermediate become more crowded. The greater the crowding the slower is the r reaction
(A) (CH CO) O (Acetic anhydride)
Q.21 3 2
(B) CH COOH (Ethanoic acid)3 (C) CH COOC H (Ethyl ethanoate)3 2 5 (D) C H OH (Ethanol)3 5
(E) CH COCH3 3
Q.22 (A) CH CH COOC H (Ethyl propanonate)3 2 2 5
(B) CH3.CH2.CO.CH.COOC2H5 (Ethyl (2-methyl, 3-ketopentanoate) CH3
(C) CH3.CH2.CO.CH.COOC2H5 (2-methyl, 3-ketopentanoic acid) CH3
Q.23 (A) CH .CH .CH .CH COOCH .CH or CH COOCH CH CH CH3 2 2 2
(B) C H OH
2 3 3 2 2 2 3 2 5
(C) CH .CH .CH3 2 2 2.CH OH (D) CH CHO3
(E) CH CH==CH.CHO3 (F) CH COOH3
CH3–C–COOH
Q.24 (A) CH3–C–COOH
CH3–C–H Cis
H–C–CH3
Trans
H
(B) CH3.CH2.C*–COOH (2-methylbutanoic acid) CH3
Q.26 A B C D E
=
=
=
=
=
CH OH (Methanol)
(Methyl ethanoate) (Methanal)
(Methanoic acid)
(Formamide or methanamide)
3
CH COCH3 HCHO HCOOH HCONH
3
2
Q.27
OH
Q.28 A = HO OH B =H2C O
CH3 CH3 O
D =H2C N
C =H2C Cl
CH3 O O
E =H3C OH F =H3C CH3
O O O
G =H3C CH3
Q.29
H3C
H3C
O O
Q.30 X =H3C CH Y =
3
HO propionic acid O
propyl propionate
H3C H3C
Z = D = CH
OH 2
propan-1-ol H3C
EXERCISE - II
Q.1 (a) The isomers have 1° of unsaturation that must be due to —COOH, since CO is evolved on2 adding NaHCO . The remaining oxygen may be present as —OH or —OR.3
COOH CH2COOH COOH COOH
H3C CH
(b) LiAlH converts —COOH to —CH OH. Only (D) is reduced to an achiral product.4 2
Not chiral
HCl C C COOMgCl
H3C C
SOCl2 NaN3 D HYdrolysis
Q.5 Q.6
RCO2H RCOCl RCON3 RNCO RNH2
bond is very stable due to large Hf of CO; so the decomposition reaction C O
O
H C Cl C O HCl is favoured. Formly chloride is not stable above –60°C.
Q.7 An extremely mild but selective oxidizing agent for aldehydes is silver oxide suspended in aqueous base.
An unsaturated acid is obtained with this reagent because the >C=C< remains untaouched by this reagent.
CHO COOH
H H
C C
H2CH3C C C
CH3 CH3
H2CH3C
Q.8 C H O 5 6 4 C H O 4 6 2 H2 CH CH CH COOH
(B) 3 2 2
– CO2
COOH
CH CH — CH COOH H2/ Ni CH CH CH COOH
H2C CHHC – CO2 2 2 3 2 2
COOH ( B)
Q.9 (A) (CH CO) O3 2 (B) CH COOH3 (C) CH COOC H3 2 5 (D) C H OH2 5
(E) CH COCH3 3
Q.10 (A) C H COOC H2 5 2 5 (B) C H CO—CH(CH )COOC H2 5 3 2 5 (C) C H COCH(CH )COOH2 5 3
Q.11 An alkali salt of palmitic acid is known as soap. The general formula of palmitic acid C15H31COOH.
Which on hydrolysis in presence of alkali give soap (C15H31 COONa) and glycerol as by product.
Acids do not reacts with NaHSO3 though they have >C = O group because of resonance stabilization.
Q.12
The resonance take place as follows.
Q.13 CH3COOH (X)
Acetic acid
CH3
(CH3COO)2 Ca (Y) Cal. acetate
Ca(OH)2
dry
distillati on
C = O Conc. H SO2 4 CH3
Acetone
CH3
CH3
(Z) Mesitylene
CH3
[o]
Q.14 CH3CH2COOH SeO2 CH CO COOH+H O3 2 Propionic acid Pyruvic acid
Q.15 Acid are directly reduced to the corresponding primary alcohol with powerful reactant like LiAlH4. It attack only on the carbonyl group of a fatty acid.
O
Ethanoic acid Methane CH3CN H3O CH3COOH NH Ammonium ethanoate Ethanamide
Br2/ KOH CH NH
Methyl chloride Acetyl chloride CH3Cl MgCH3MgCl
Ethyl alcohol Acetamide
[O]
(c) CH3COOH
Urea (E) Diethyl oxalate
NH– C = O nPen tan oic acid
CH3 CH3 CH3
CH3 CH3 CH3 CH3 H3C C OH SOCl2 H C3 C Cl EtherMg H C3 C MgCl (ii ) H O(i ) CO2H3C C COOH
3
CH3 CH3 CH3 CH3
H H
HOOCCH2 HOOCCH2 HOOCCH2
HOOCCH2
Q.22 Br 2H CH
C Ni2 2
Br
H
F Meso compound
H
E Meso compound
CH3 C H3
and
CH3 C OOH
C = C C = C
Q.23 H COOH H CH3
( A) Geo metrical isomers
CH3 CH3
C2H5—C—COOH or HOOC—C—C2H5
H H B
Q.24 (A) CH CHOHCH COOH3 2 (B) CH CH=CHCOOH3
(D) CH —CH = HCON(CH ) (C) CH CH = CHCOCl3
(E) CH COCH COOH
3
(F) CH COCH
3 2
3 2 3 3
(G) CH CH CH3 2 3
Q.25 is a saturated monoester with M.W = 186
CH3 O CH3 CH3 O
CH3 H
H3C O
F = G = H3C O
CH3 CH3 CH3
OH
CH3
H =H3C I =
CH3
OSO2C6H5 Br
CH3 CH3
H3C H3C
J = K =
CH3
CH3 O
Q.26 A = H3C B = O
H3C OH
CH3 CH3
O O
H3C H H3C H
C = D =
O O O
CH3 CH3 CH3
Cl N
H3C
E = H3C F =
O CH3
O
Q.27
CH3
CH3
Q.28 A C = CH—C—OCH2CH2CH2CH3 or C = CH—C—O—CH—CH2CH3;
O H3C
CH3 O
CH3
CH3 H3C
CH3CHCH2CH3 OH
CH3CH2CH2CH2OH ; Dehydration
B C = CHCO2H ; C
2 - Butene CH3
H3C
D O=CH—CO2H
MF : C2H2O3 C—CH—CO2H ; E
OH OH M.F C5H10O4
Q.29 The given reactions are as follows.
The compound E must be ketonic compound as it does not give Tollens test and does not rce
Fehling’s solution but forms a 2, 4-dinitrophenyl-hydrazone. Therefore, its structure would be CH COCH
3 3
(acetone).
Since E is obtained by heating B with Ca(OH) , the compound B must be CH COOH (acetic acid).2 3 Since B is obtained by oxidation of D with KMnO , the compound D must be an alcohol with molecular4 formula CH CH OH (ethanol).3 2
Since B and D are obtained by acid hydrolysis of C, the compound C must be an ester CH COOC H3 2 5 (ethyl acetate).
Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treatingAwith ethanol, the compound Amust be an anhydride (CH CO) O (acetic anhydride).
3 2
The given reactions are
(C H3CO)2O C2H5OH ` CH3C OOH + CH3COOC2H5
Q.30 Acetoacetic ester shows tautomerism and the two forms are called as keto and enol forms.
OH O
Keto form Enol form
OH
)gives blue-violet colour with FeCl3solution. When Br2is added, of the enol form.
The enol (H3C C CH
it reacts at once with
Br
As soon as enol form is consumed, its colouration with FeCl disappears and excess of bromine gives3 brown colour.As keto and enol forms are in equilibrium, when enol form is used, the equilibrium shifts to right hand side to give more enol form which discharges the colour of excess of Br and gives blue violet2 colour with excess of FeCl present in the reaction mixture.3