Problem 5: When treated with Br2 and H2O allyl bromide chiefly the primary alcohol CH2BrCH2OH, in contrast to propylene which gives a secondary alcohol CH3CHOHCH2Br. Explain.
Solution: In the intermediate
* carbon can’t stabilize the positive charge as it is close to the electronegative Br atom, and hence attack of the OH– ion takes place on ** carbon.
OH– here goes to that carbon which stabilizes the positive charge easily.
Problem 6: An organic compound (A), C5H8O3 on heating with soda lime gives (B), which reacts with HCl to give (C). The compound (C) reacts with thionyl chloride to produce (D) which on reaction with KCN gives compound (E). Alkaline hydrolysis of (E) gives a salt (F) which on heating with sodalime produces n-butane. Careful oxidation of (A) with dichromate gives acetic acid and malonic acid. Gives structures of (A) to (F) with proper reasoning.
Solution: The statements given in problem suggest (A)to be
Problem 7: A 10 g mixture of isobutane and isobutene requires 20 g of Br2 (in CCl4) for complete addition. If 10g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at 127°C, which exclusive product and how much of it would be formed? (Atomic weight of bromine = 80).
Solution: Let isobutane by a g and isobutene be b g. Thus a + b = 10
Now \ 160g Br2 is needed for 56 g isobutene
\ 20g Br2 is needed for = 7g
Thus isobutene = 7g
Isobutane = 3 g
Now 7g isobutene is converted to isobutane by reduction to give g
= 7.25 g of isobutane
Now isobutane reacts with Br2/hn to give monobromo derivative
C4H10 + Br2 ¾® C4H9Br + HBr
Q 58 g C4H10 gives pOH = 137 g C4H9Br
\ (7.25 + 3)g C4H10 gives = 24.21g C4H9Br
Problem 8: When treated with C2H5 OK in C2H5OH diasteromer (I) and its enantiomer
gave cis-2 butene without loss of deuterium and trans-2-butene with loss of deuterium. Diasteromer (II) (and its enantiomer) gave trans-2-butene without loss of deuterium. Explain the formation of product. What is the stereochemistry of elimination in both the cases.
Solution: The stereochemistry of the E-2 elimination is anti coplanarity where the H (D) and Br should be anti-coplanar. The Fischer projection of (I) is converted to eclipsed saw horse conformation.
The C – C single bond is rotated till the H(D) and Br is in anti coplanar stereochemistry
Problem 6: Which one of the following does not of dissolve in conc. H2SO4?
(A) CH3 – C = C – CH3 (B) CH3 – CH2 – C º CH
(C) CH º CH (D) CH2 = CH2
Solution: If CH º CH were to dissolve in H2SO4 a bisulphite salt of vinyl carbocation H2C = C+H would be formed. The more s-character in the positively charged ‘C’ less stable is the carbocation and less likely to be formed.
Problem 7: Which one of the following compounds will give in the presence of peroxide a product different from that obtained in the absence of peroxide?
(A) 1-butane (B) 1-butene, HBr
(C) 2-butene, HCl (D) 2-butene, HBr
Solution: Peroixde effect is observed when unsymmetrical alkene is treated with HBr only (and not with HCl and HI).
Problem 8: Which of the following alkene on acid catalysed hydration form 2-methyl propan-2-ol.
(A) (CH3)2CH = CH2 (B) CH3 – CH = CH2
(C) CH3 – CH = CH – CH3 (D) CH3 – CH2 – CH = CH2
Solution: Addition of H2O occurs according to Markownikoff’s rule.
Problem 8: Which of the following compounds yields only one product on monobromination?
(A) Neopentane (B) Toluene
(C) Phenol (D) Aniline
Solution: CH3 – CH3 has twelve equivalent 1°H. Hence H forms only one product on monobromination.
Problem 9: Aqueous solution of the following compounds are electrolysed. Acetylene gas is obtained from.
(A) Sodium fumarate (B) Sopdium maleate
(C) Sodium succinate (D) Both (A) and (B)
Problem 10: Dehydration of butan-2-ol with conc. H2SO4 gives preferred product.
(A) but-1-ene (B) but-2-ene
(C) propene (D) ethane
Solution: CH3 – – – CH3 CH3 – CH = CH – CH3 (80%)
+ CH3 – CH2 – CH = CH2 (20%)
This is in accordance with saytzeff rule.
Problem 11: CH3 – C º C – CH3 ‘X’. What is X
(A) CH3CH2CH = CH2 (B) CH3CH2C º CH
(C) CH3 – CH = CH – CH3 (D) CH2 = C = CH – CH3
Solution: Isomerisation occurs, when 2-butyne is treated with NaNH2, it converts into terminal alkyne (1-butyne).
Problem 12: Identify the compound ‘Y’ in the following sequence of reactioin
HC º CH
Problem 13: Dehydration of 1-butanol gives 2-butene as a major product, by which of the following intermediate the compound 2-butene obtained
Problem 14: The principal organic compound formed in the reaction
CH2 = CH(CH2)COOH + HBr …………. is
|(A)||CH3 –– (CH3)8COOH||(B)||CH2 = CH(CH2)8COBr|
|(C)||–CH2(CH2)8COOH||(D)||CH2 = CH(CH3)7 – – COOH|
Solution: Follows the peroxide effect
\ CH2 = CH(CH2)8COOH – CH2(CH2)8COOH
Problem 15: The compound most likely to decolourise a solution of potassium permanganate is
Solution: It is a test for unsaturation. As benzene and naphthalene is also unsaturated, but they are stabilized due to resonance, and thus does not give Bayer’s test.
- Point out (A) in the given reaction sequence:
- End product of the following sequence is:
(A) Ethanol (B) Ethyl hydrogen sulphate
(C) Ethanal (D) Ethylene glycol
- 10 mL of a certain hydrocarbon require 25 mL of oxygen for complete combustion and the volume of CO2 product is 20 mL. What is the formula of hydrocarbon.
(A) C2H2 (B) C2H4
(C) CH4 (D) C2H6
- Hydrogenation of the compound:
In the presence of poisoned palladium catalyst gives:
(A) An optically active compound (B) An optically inactive compound
(C) A racemic mixture (D) A diastereomeric mixture
- CH3 –– CH = CH2 + HBr ¾® A (predominant), A is
|(C)||(D)||None is correct|
In the increasing order is
(A) I < III < IV < II (B) I < II < III < IV
(C) IV < III < II < I (D) II < III < IV = I
Which is true about this reaction?
(A) A is meso 1, 2-butan-di-ol formed by syn addition
(B) A is meso 1, 2-butan-di-ol fomred by anti addition
(C) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by anti addition
(D) A is a racemic mixture of d and l 1, 2-butan-di-ol formed by syn addition.
9 The treatment of C2H5MgI with water produces
(A) Methane (B) Ethane
(C) Ethanal (D) Ethanol
10. The order of reactivity of halogens towards halogenation of alkanes is
(A) F2 > Br2 > Cl2 (B) F2 > Cl2 > Br2
(C) Cl2 > F2 > Br2 (D) Cl2 > Br2 > F2
11. The chlorination of alkane involves
(A) Cl free radicals (B) Cl+ species
(C) Cl– species (D) CH3 free radicals
Which is true statement?
(A) A is formed by anti addition and is meso
(B) A is formed by syn addition and is meso
(C) A is formed by anti addition and is racemic
(D) A is formed by syn addition and is racemic
(A) cis (B) trans
(C) both (D) none
- (A) cis-2-butene
(B) trans-2-butene II
Correct statements are
(A) is racemic mixture by anti addition
(B) II is meso compound by anti addition
(C) I is meso compound by syn addition
(D) II is racemic compound by syn addition
A and B are
- C 2. C
- A 4. B
- B 6. C
- A 8. A
- B 10. B
- A 12. A
- A 14. C, D
- Three compounds A,B and C, all have the formula C6H10. All three compounds rapidly decolourise Br2 in CCl4. Compound A gives a precipitate when treated with silver nitrate in ammonia, but compound B and C do not. A and B both yield n-hexane when they are treated with excess hydrogen in presence when they are treated with excess hydrogen in presence of palladium catalyst. Under these conditions C absorbs only one molar equivalent of H2 and gives a product with formula C6H12. When A is oxidised with hot basic KMnO4 and the resulting solution acidified, the only organic product that can be isolated is n-pentanoic acid. Similar oxidation of B gives only propanoic acid and similar treatment of C gives only adipic acid. Find A,B and C and give the reactions involved.
- One of a graduate student had need of the hydroxy ester (CH3)2C(OH)CH2CO2C2H5. Turning to the Grignard reaction, he prepared methyl magnesium iodide and to it he added acetoacetic ester. Everything went well, indeed, even without the application of heat, the reaction mixture bubbled merrily. Working carefully and with great skill he isolated an excellent yield of the starting material, acetoacetic ester. What reaction had taken place? What was the bubbling due to? Explain.
- Identify organic products of the following reactions.
- Hydrocarbon (A) of molecular weight 68 reacts with an acid and excess of Br2 in CCl4 to give a compound (B), whose molecular weight is 470.6% more than that of (A). However, on catalytic hydrogenation with excess of hydrogen, (A) forms (C) whose molecular weight is only 5.8% more than that of (A). A reacts with CH3Br in the presence of NaNH2 to give another hydrocarbon (D). Which on ozonolysis yields ketone (E). E on oxidation gives ethanoic acid and isobutanoic acid. Give structures (A) to (E) with reactions.
- Hydrocarbon (A) of molecular weight 56 has the element % C (85.7%) and H(14.28) A is converted on reaction with sulphuric acid into a mixture of two alkenes B and C of molecular formula C8H16. Hydrogenation of either of these two alkenes produces the same substituted alkane of molecular formula (D) C8H18. D on reaction produces four different monochlorinated products E,F,G,H. Identify A to H.
- An unsaturated hydrocarbon (A) C7H10 readily gives (B) on treatment with NaNH2 in liq. NH3. When (B) is allowed to react with 1-bromopropane, a compound (C) is obtained. On partial hydrogenation in the presence of Lindlar catalyst (C) gives (D), C10H18, (C) on reaction with Na/LiqNH3 gives (E) (an isomer of D) vigorous oxidation of (C) or (D) gives cyclopentane carboxylic acid and Butanoic acid as the products. Identify (A) to (D).
- Deduce structure (C), (D) & (E). Where (C) and (D) are positional isomers having eight carbon atoms each, with one them is the major product and other one is the minor product of (A) and (E) reactants. (E) similar to (A) has 4 carbon atoms but they are not isomers to each others.
- A compound A (C8H16) undergoes hydrogenation to give n–octane, when treated with m – chloroperbenzoic acid A gives B (C8H16O) which when treated with an aqueous acids gives a compound C (C8H18O2) which can be resolved into enantiomers. When A is treated with OsO4 followed by treatment with NaHCO3, a compound D, an isomer of C, results. D can not be resolved into enantiomers. Treatment of B with LiAlH4 followed by hydrolysis gives compound E, which is also formed when A reacts with B2H6 followed by treatment with H2O2/OH–. Identify all the compounds including stereochemistry where appropriate.
- An organic compound (A), C5H9Br which readily decolourises bromine water and KMnO4 solution, gives (B), C5H11Br on treatment with Sn/HCl. The reaction of A with NaNH2 produces (C) with evolution of ammonia. (C) neither reacts with sodium nor forms any metal acetylide but reacts with Lindlar’s catalyst to give (D) and on reduction with Na/Liq. NH3 produces (E). Both the compounds (D) and (E) are isomeric. Give structures of (A) to (E) with proper reasoning.
- An unsaturated hydrocarbon A (C6H10) readily gives B on treatment with NaNH2 in liquid NH3. When B is allowed to react with 1-chloropropane a compound C is obtained. On partial hydrogenation in the presence of Lindlar catalyst, compound C gives D (C9H18). On ozonolysis D gives 2, 2-dimethyl propanal and 1-butanal. With proper reasoning give the structures of A, B, C and D.
- Compound A (C6H12) is treated with Br2 to form compound B (C6H12Br2). On treating B with alcoholic KOH followed by NaNH2 the compound C(C6H10) is formed. C on treatment with H2/Pt forms 2-methylpentane. The compound ‘C’ does not react with ammonical Cu2Cl2 or AgNO3. When A is treated with cold KMnO4 solution, a diol D is formed which gives two acids E and F when heated with KMnO4 solution. Compound E is found to be ethanoic acid. Deduce the structures from A to F.
- An alkene (A) on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid (B). When (B) is treated with bromine in presence of phosphorous it yields compound (C) which on hydrolysis gives a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrloysis. Identify the compounds A, B, C and D.
- 0.037 g of an alcohol, ROH was added to CH3MgI and the gas evolved measured 11.2Cm3 at STP. What is the molecular weight of ROH? On dehydration, ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Give structures of ROH and the acid with proper reasoning.
- Compound (A) C10H12O gives off hydrogen on treatment with sodium metal and also decolouriese Br2 in CCl4 to give (B), C10H12OBr2. (A) on treatment with I2 in NaOH gives iodoform and an acid (C) after acidification. Give structure of (A) to (C) and also of all the geometrical and optical isomers of (A).
- Two isomeric forms of an organic compound (A) C11H13OCl readily decolourise bromine water and give same compound (B) on catalytic hydrogenation. Both the isomeric forms on vigorous oxidation give (C) which on treatment with sodalime give 2-chloro ethoxy benzene. However, (C) on treatment with Ni/Al alloy in alkaline medium gives 3-ethoxy benzoic acid. Only one of the isomers of (A) give geometrical isomers (D) and (E). Identify (A) to (E) with proper reasoning.
|1. A =||CH3CH2CH2CH2CºCH|
- has an active methylene group with acidic hydrogens.
coming from CH3MgI can abstract this proton and form methane gas .
The conjugate base of aceto acetic ester on acidification once again gives aceto acetic ester.
the gas evolved was CH4
- A is alkyne having m.f. C5H8. Diketone E on oxidation yields isobutyric acid and ethanoic acid. Therefore E may be
Hydrocarbon D is
D is obtained from A on reaction with CH3Br in NaNH2. Therefore A is terminal alkyne with the structure
(C) Can also be represented in fischer projection as
- i) A decolorsize Br2/H2O and KMnO4 and thus it has unsaturation.
- iv) (C) being nonterminal alkyne
- v) (D) and (E) are geometrical isomers
- 11.2 ml 0.037 g ROH
22400 ml = = 74 g
Alcohol is C3H7CH2OH
Geometrical isomers of
(vi) Optical isomer