Hydrocarbons Questions and solutions

Solved SUbjectives

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

 

 

Solved Objectives

 

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.

                        \(C)

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).

                        \(B)

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.

                       

                        \ (A)

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.

                        \(A)

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)

 

Solution:

                        \ (D)

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.

\ (B)

Problem 11:     CH₃ – C º C – CH₃ ‘X’. What is X

                        (A)  CH­₃CH_­2CH = 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).

                        \ (B)

Problem 12:     Identify the compound ‘Y’ in the following sequence of reactioin

                        HC º CH

(A)			(B)
(C)			(D)	CH3COOH
Solution:

                        \ (A)

Problem 13:     Dehydration of 1-butanol gives 2-butene as a major product, by which of the following intermediate the compound 2-butene obtained

(A)			(B)
(C)			(D)
Solution:

                        \ (C)

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

                        \ (C)

Problem 15:     The compound most likely to decolourise a solution of potassium permanganate is

(A)		(B)
(C)		(D)

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.

                        \ (A)

 

Objectives

 

1. Point out (A) in the given reaction sequence:

 

(A)		(B)
(C)		(D)

2. End product of the following sequence is:

 

(A) Ethanol                                                (B) Ethyl hydrogen sulphate

(C) Ethanal                                                (D) Ethylene glycol

3. 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

4. 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

5.
(A)			(B)
(C)			(D)

6. CH₃ –– CH = CH₂ + HBr ¾® A (predominant), A is

(A)		(B)
(C)		(D)	None is correct

 

7.

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

8

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

12.

(A)

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

13.

(A) cis                                                        (B) trans

(C) both                                                     (D) none

14. (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

15.

A and B are

(A)		(B)
(C)		(D)

 

Answers

 

1. C 2.                                                         C
2. A                                                         4.         B
3. B                                                         6.         C
4. A                                                         8.         A
5. B                                                         10.       B
6. A                                                         12.       A
7. A                                                         14.       C, D                
8. D

 

 

 

Subjectives

 

1. 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.
2. 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.
3. Identify organic products of the following reactions.

	a)
	b)
	c)
d)

4. 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.
5. 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.
6. 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).
7. 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.

 

8. 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.
9. 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.
10. 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.
11. 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.
12. 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.
13. 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.
14. 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).
15. 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.

 

Answers

 

1. A =	CH3CH2CH2CH2CºCH
B =	CH3CH2CºCCH2CH3
C =
	CH3CH2CH2CH2CºCH
	CH3CH2CºCCH2CH3

 

2. 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

 

3.

 

 

 

4. 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

 

 

 

 

6.

 

7.

 

8.

(C) Can also be represented in fischer projection as

 

	(B)

 

9. i) A decolorsize Br₂/H₂O and KMnO₄ and thus it has unsaturation.

ii)

 

iii)

 

(A)

 

 

1. iv) (C) being  nonterminal alkyne
2. v) (D) and (E) are geometrical isomers

 

10.

 

 

11.
	(E)  CH3COOH

 

12.

 

13. 11.2 ml 0.037 g ROH

 

22400 ml =  = 74 g

Alcohol is C₃H₇CH₂OH

 

 

 

 

14.

 

Geometrical isomers of

 

 

(vi) Optical isomer

 

 

15.

 

i)

 

ii)