Hydrocarbons
Unit 13
Hydrocarbons
I. Multiple Choice Questions (Type-I)
1. Arrange the following in decreasing order of their boiling points.
(A) $n$-butane
(B) 2-methylbutane
(C) n-pentane
(D) 2,2-dimethylpropane
(i) $A>B>C>D$
(ii) $B>C>D>A$
(iii) $D>C>B>A$
(iv) $C>B>D>A$
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Answer: (iv)2. Arrange the halogens $F_2, Cl_2, Br_2, I_2$, in order of their increasing reactivity with alkanes.
(i) I $_2<Br_2<Cl_2<F_2$
(ii) $Br_2<Cl_2<F_2<I_2$
(iii) $F_2<Cl_2<Br_2<I_2$
(iv) $Br_2<I_2<Cl_2<F_2$
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Answer: (i)3. The increasing order of reduction of alkyl halides with zinc and dilute $HCl$ is
(i) $R-Cl<R-$ I $<R-Br$
(ii) $R-Cl<R-Br<R-$ I
(iii) $R-I<R-Br<R-Cl$
(iv) $R-Br<R-I<R-Cl$
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Answer: (ii)4. The correct IUPAC name of the following alkane is
(i) 3,6-Diethyl - 2 - methyloctane
(ii) 5 - Isopropyl - 3 - ethyloctane
(iii) 3-Ethyl - 5 - isopropyloctane
(iv) 3-Isopropyl - 6 - ethyloctane
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Answer: (i)5. The addition of $HBr$ to 1-butene gives a mixture of products $A, B$ and $C$
The mixture consists of
(i) A and B as major and C as minor products
(ii) B as major, A and C as minor products
(iii) B as minor, A and C as major products
(iv) A and B as minor and C as major products
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Answer: (i)6. Which of the following will not show geometrical isomerism?
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Answer: (iv)7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
(i) HCl > HBr > HI
(ii) HBr > HI > HCl
(iii) HI > HBr > HCl
(iv) HCl > HI > HBr
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Answer: (iii)8. Arrange the following carbanions in order of their decreasing stability.
(i) A > B > C
(ii) B > A > C
(iii) C > B > A
(iv) C > A > B
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Answer: (ii)9. Arrange the following alkyl halides in decreasing order of the rate of β– elimination reaction with alcoholic KOH.
(i) A > B > C
(ii) C > B > A
(iii) B > C > A
(iv) A > C > B
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Answer: (iv)10. Which of the following reactions of methane is incomplete combustion:
(i) $2 CH_4+O_2 \xrightarrow{Cu / 523 K / 100 atm} 2 CH_3 OH$
(ii) $CH_4+O_2 \xrightarrow{Mo_2 O_3} HCHO+H_2 O$
(iii) $CH_4+O_2 \longrightarrow C(s)+2 H_2 O$ (l)
(iv) $CH_4+2 O_2 \longrightarrow CO_2(g)+2 H_2 O(l)$
II. Multiple Choice Questions (Type-II)
In the following questions two or more options may be correct.
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Answer: (iii)
II. Multiple Choice Questions (Type-II)
11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions?
(i) $CH_4$ (g) $+2 O_2(g) \longrightarrow CO_2(g)+2 H_2 O$ (l)
(ii) $CH_4(g)+O_2(g) \longrightarrow C$ (s) $+2 H_2 O(l)$
(iii) $CH_4(g)+O_2(g) \xrightarrow{Mo_2 O_3} HCHO+H_2 O$ (iv) $2 CH_4(g)+O_2(g) \xrightarrow{Cu / 523 / 100 atm} 2 CH_3 OH$
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Answer: (iii), (iv)12. Which of the following alkenes on ozonolysis give a mixture of ketones only?
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Answer: (iii), (iv)13. Which are the correct IUPAC names of the following compound?
(i) 5- Butyl-4-isopropyldecane
(ii) 5- Ethyl - 4- propyldecane
(iii) 5- sec-Butyl - 4-iso-propyldecane
(iv) 4-(1-methylethyl)- 5 - (1-methylpropyl)-decane
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Answer: (iii), (iv)14. Which are the correct IUPAC names of the following compound?
(i) 5 - (2’, 2’-Dimethylpropyl)-decane
(ii) 4-Butyl-2,2-dimethylnonane
(iii) 2,2-Dimethyl-4- pentyloctane
(iv) 5 - neo-Pentyldecane
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Answer: (i), (iv)15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring ———.
(i) deactivates the ring by inductive effect
(ii) deactivates the ring by resonance
(iii) increases the charge density at ortho and para position relative to meta position by resonance
(iv) directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position.
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Answer: (i), (iii)16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group——–.
(i) deactivates the ring by inductive effect.
(ii) activates the ring by inductive effect.
(iii) decreases the charge density at ortho and para position of the ring relative to meta position by resonance.
(iv) increases the charge density at meta position relative to the ortho and para positions of the ring by resonance.
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Answer: (i), (iii)17.* Which of the following are correct?
(i) $CH_3-O-CH_2^{\oplus}$ is more stable than $CH_3-CH_2^{\oplus}$
(ii) $\quad(CH_3)_2 CH^{\oplus}$ is less stable than $CH_3-CH_2-CH_2^{\oplus}$
(iii) $CH_2=CH-CH_2^{\oplus}$ is more stable than $CH_3-CH_2-CH_2^{\oplus}$
(iv) $CH_2=CH^{\oplus}$ is more stable than $CH_3-CH_2^{\oplus}$
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Answer: (i), (iii)18. Four structures are given in options (i) to (iv). Examine them and select the aromatic structures.
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Answer: (i), (iii)19. The molecules having dipole moment are ——–.
(i) 2,2-Dimethylpropane
(ii) trans-Pent-2-ene
(iii) cis-Hex-3-ene
(iv) 2, 2, 3, 3 - Tetramethylbutane.
III. Short Answer Type
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Answer: (ii), (iii)
III. Short Answer Type
20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.
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Answer: Both alkenes and arenes are electron-rich. Therefore undergo electrophilic reactions. Olefins undergo addition reactions because addition of a reagent to an olefin gives a more stable product as $s p^{2}$ hybridisation changes to $s p^{3}$ hybridisation. Addition to the double bond of an arene would give a product with less or no resonance stability hence addition is difficult arenes. On the other hand in substitution reaction resonance stabilisation is retained therefore, arenes undergo substitution reaction.21. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
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Answer: 2-Butene is capable of showing geometrical isomerism.22. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
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Answer: The rotation about $C-C$ bond is restricted because of repulsion between electron cloud of $C-H$ bonds on either carbon atoms.23. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why?
24. The intermediate carbocation formed in the reactions of $HI, HBr$ and $HCl$ with propene is the same and the bond energy of $HCl, HBr$ and $HI$ is $430.5 kJ mol^{-1}$, $363.7 kJ mol^{-1}$ and $296.8 kJ mol^{-1}$ respectively. What will be the order of reactivity of these halogen acids?
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Answer: Bond dissociation energy is least for $HI$ and maximum for $HCl$ therefore, order of reactivity will be $HI>HBr>HCl$.25. What will be the product obtained as a result of the following reaction and why?
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Answer: Propyl chloride forms less stable $CH_3-CH_2-CH_2^{\oplus}$ carbocation with anhydrous $AlCl_3$ which rearranges to a more stable $CH_3-\stackrel{\oplus}{C} H-CH_3$ carbocation and gives isopropylbenzene as the product of the reaction.26. How will you convert benzene into (i) $p$ - nitrobromobenzene
(ii) $m$ - nitrobromobenzene
27. Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.
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Answer: The $+R$ effect of $-OCH_3>-Cl$ and $-NO_2$ has a $-R$ effect. Relative reactivity of the substituted benzene rings is as follows :
$C_6 H_5-OCH_3>C_6 H_5-Cl>C_6 H_5-NO_2$
28. Despite their - I effect, halogens are $o$ - and $p$-directing in haloarenes. Explain.
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Answer: Halogens attached to benzene rings exert $-I$ and $+R$ effect. $+R$ effect dominates $-I$ effect and increases the electron density at ortho and para positions of the benzene ring with respect to halogens.29. Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.
30. Suggest a route for the preparation of nitrobenzene starting from acetylene?
31. Predict the major product (s) of the following reactions and explain their formation.
32. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.
(i) $H_3 CO$
(ii)
(iii) $\dot{C}$
(iv) $Cl_2 C$ :
(v) $(H_3 C)_3 C^{+}$
(vi) $Br^{-}$
(vii) $H_3 COH$
(viii) R-NH-R
33. The relative reactivity of $1^{\circ}, 2^{\circ}, 3^{\circ}$ hydrogen’s towards chlorination is $1: 3.8: 5$. Calculate the percentages of all monochlorinated products obtained from 2 -methylbutane.
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Answer: 2-Methyl butane is . Possible compounds are A, B and $C$ given below :
Relative amounts of A , B and C compounds = number of hydrogen x relative reactivity
Total Amount of monohaloginated compounds $=9+7.6+5=21.6$
Percentage of $A=\frac{9}{21.6} \times 100=41.7 %$
Percentage of $B=\frac{7.6}{21.6} \times 100=35.2 %$
Percentage of $C \quad=\frac{5}{21.6} \times 100=23.1 %$
Relative amounts of A , B and C compounds = number of hydrogen x relative reactivity
Radical I is tertiary where as radical II is primary. Radical I is more stable due to hyperconjugation.
34. Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane.
35. Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.
36. An alkane $C_8 H _{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.
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Answer:
37. The ring systems having following characteristics are aromatic.
(i) Planar ring containing conjugated $\pi$ bonds .
(ii) Complete delocalisation of the $\pi$-electrons in ring system i.e. each atom in the ring has unhybridised $p$-orbital, and
(iii) Presence of $(4 n+2) \pi$-electrons in the ring where $n$ is an integer ( $n=0,1,2, \ldots . . . . .$. ) [Huckel rule].
Using this information classify the following compounds as aromatic/nonaromatic.
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Answer: $A=$ Planar ring, all atoms of the ring $s p^{2}$ hybridised, has six delocalised $\pi$ electrons, follows Huckel rule. It is aromatic.
$B=$ Has six $\pi$ electrons, but the delocalisation stops at $s p^{3}$ hybridised $CH_2$ - carbon. Hence, not aromatic.
$C=$ Six delocalised $\pi$-electrons ( $4 \pi$ electrons +2 unshared electrons on negatively charged carbon) in a planar ring, follows Huckel’s rule. It is aromatic.
$D=$ Has only four delocalised $\pi$-electrons. It is non aromatic.
$E=$ Six delocalised $\pi$-electrons follows Huckel’s rule. $\pi$ electrons are in $s p^{2}$ hybridised orbitals, conjugation all over the ring because of positively charged carbon. The ring is planar hence is aromatic.
$F=$ Follows Huckel’s rule, has $2 \pi$ electrons i.e. $(4 n+2) \pi$-electrons where $(n=0$ ), delocalised $\pi$-electrons. It is aromatic.
$G=8 \pi$ electrons, does not follow Huckel’s rule i.e., $(4 n+2) \pi$-electrons rule. It is not aromatic.
38. Which of the following compounds are aromatic according to Huckel’s rule?
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Answer: $A=$ Has $8 \pi$ electrons, does not follow Huckel rule. Orbitals of one carbon atom are not in conjugation. It is not aromatic.
$B=$ Has $6 \pi$ delocalised electrons. Hence, is aromatic.
$C=$ Has $6 \pi$ electrons in conjugation but not in the ring. Non aromatic.
$D=10 \pi$ electrons in planar rings, aromatic.
$E=$ Out of $8 \pi$ electrons it has delocalised $6 \pi$ electrons in one six membered planar ring, which follows Huckel’s rule due to which it will be aromatic.
$F=14 \pi$ electrons are in conjugation and are present in a ring. Huckel’s rule is being followed. Compound will be aromatic if ring is planar.
IV. Matching Type
39. Suggest a route to prepare ethyl hydrogensulphate $(CH_3-CH_2-OSO_2-OH)$ starting from ethanol $(C_2 H_5 OH)$.
IV. Matching Type
40. Match the reagent from Column I which on reaction with $CH_3-CH=CH_2$ gives some product given in Column II as per the codes given below :
| Column I | Column II |
|---|---|
| (i) $O_3 / Zn+H_2 O$ | (a) Acetic acid and $CO_2$ |
| (ii) $KMnO_4 / H^{+}$ | (b) Propan-1-ol |
| (iii) $KMnO_4 / OH^{-}$ | (c) Propan-2-ol |
| (iv) $H_2 O / H^{+}$ | (d) Acetaldehyde and formaldehyde |
| (v) $B_2 H_6 / NaOH$ and $H_2 O_2$ | (e) Propane-1,2-diol |
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Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (a) $\quad$ (iii) $\to$ (e) $\quad$ (iv) $\to$ (c) $\quad$ (v) $\to$ (b)41. Match the hydrocarbons in Column I with the boiling points given in Column II.
| Column I | Column II |
|---|---|
| (i) n-Pentane | (a) $282.5 K$ |
| (ii) iso-Pentane | (b) $309 K$ |
| (iii) neo-Pentane | (c) $301 K$ |
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Answer: (i) $\to$ (b) $\quad$ (ii) $\to$ (c) $\quad$ (iii) $\to$ (a)42. Match the following reactants in Column I with the corresponding reaction products in Column II.
| Column I | Column II |
|---|---|
| (i) Benzene $+Cl_2 \xrightarrow{AlCl_3}$ | (a) Benzoic acid |
| (ii) Benzene $+CH_3 Cl \xrightarrow{AlCl_3}$ | (b) Methyl phenyl ketone |
| (iii) Benzene $+CH_3 COCl \xrightarrow{AlCl_3}$ | (c) Toluene |
| (iv) Toluene $\xrightarrow{KMnO_4 / NaOH}$ | (d) Chlorobenzene |
| (e) Benzene hexachloride |
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Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (c) $\quad$ (iii) $\to$ (b) $\quad$ (iv) $\to$ (a)43. Match the reactions given in Column I with the reaction types in Column II.
| Column I | Column II |
|---|---|
| (i) $CH_2=CH_2+H_2 O \xrightarrow{H^{+}} CH_3 CH_2 OH$ | (a) Hydrogenation |
| (ii) $CH_2=CH_2+H_2 \xrightarrow{pd} CH_3 - CH_3$ | (b) Halogenation |
| (iii) $CH_2=CH_2+Cl_2 \longrightarrow Cl-CH_2-CH_2-Cl$ | (c) Polymerisation |
| (iv) $3 CH \equiv CH \xrightarrow[\text{ Heat }]{\text{ Cu tube }} C_6 H_6$ | (d) Hydration |
| (e) Condensation |
V. Assertion and Reason Type
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
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Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (a) $\quad$ (iii) $\to$ (b) $\quad$ (iv) $\to$ (c)
V. Assertion and Reason Type
44. Assertion (A) : The compound cyclooctane has the following structural formula :
It is cyclic and has conjugated $8 \pi$-electron system but it is not an aromatic compound.
Reason (R) : $\quad(4 n+2) \pi$ electrons rule does not hold good and ring is not planar.
(i) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.
(iii) Both $A$ and $R$ are not correct.
(iv) $A$ is not correct but $R$ is correct.
45. Assertion (A) : Toluene on Friedal Crafts methylation gives $o-$ and $p$-xylene.
Reason (R) : $\quad CH_3$-group bonded to benzene ring increases electron density at $o$ - and $p$ - position.
(i) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.
(iii) Both $A$ and $R$ are not correct.
(iv) $A$ is not correct but $R$ is correct.
46. Assertion (A) : Nitration of benzene with nitric acid requires the use of concentrated sulphuric acid.
Reason (R) : The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile, $NO_2^{+}$.
(i) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.
(iii) Both $A$ and $R$ are not correct.
(iv) $A$ is not correct but $R$ is correct.
47. Assertion (A) : Among isomeric pentanes, 2, 2-dimethylpentane has highest boiling point.
Reason (R) : Branching does not affect the boiling point.
(i) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.
(iii) Both $A$ and $R$ are not correct.
(iv) $A$ is not correct but $R$ is correct.
VI. Long Answer Type
48. An alkyl halide $C_5 H _{11} Br$ (A) reacts with ethanolic $KOH$ to give an alkene ’ $B$ ‘, which reacts with $Br_2$ to give a compound ’ $C$ ‘, which on dehydrobromination gives an alkyne ’ $D$ ‘. On treatment with sodium metal in liquid ammonia one mole of ’ $D$ ’ gives one mole of the sodium salt of ’ $D$ ’ and half a mole of hydrogen gas. Complete hydrogenation of ’ $D$ ’ yields a straight chain alkane. Identify A,B, C and D. Give the reactions invovled.
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Answer: 
(A)
(B)
(C)
The reactions suggest that (D) is a terminal alkyne. This means triple bond is at the end of the chain. It could be either (I) or (II).
Since alkyne ‘D’ on hydrogenation yields straight chain alkane, therefore structure I is the structure of alkyne (D).
Hence, the structures of A, B and C are as follows :
(A) $CH_3-CH_2-CH_2-CH_2-CH_2 Br$
(B) $CH_3-CH_2-CH_2-CH=CH_2$
(C) $CH_3-CH_2-CH_2-CH(Br)-CH_2 Br$
49. $896 mL$ vapour of a hydrocarbon ’ $A$ ’ having carbon $87.80 %$ and hydrogen $12.19 %$ weighs $3.28 g$ at STP. Hydrogenation of ‘A’ gives 2-methylpentane. Also ’ $A$ ’ on hydration in the presence of $H_2 SO_4$ and $HgSO_4$ gives a ketone ’ $B$ ’ having molecular formula $C_6 H _{12} O$. The ketone ’ $B$ ’ gives a positive iodoform test. Find the structure of ’ $A$ ’ and give the reactions involved.
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Answer: Step I
$896 mL$ vapour of $C_x H_y$ (A) weighs $3.28 g$
$22700 mL$ vapour of $C_x H_y(A)$ weighs $\frac{3.28 \times 22700}{896} g mol^{-1}=83.1 g mol^{-1}$
Step II
| Element | (%) | Atomic mass |
Relative ratio | Relative no. of atoms |
Simplest ratio |
|---|---|---|---|---|---|
| C | 87.8 | 12 | 7.31 | 1 | 3 |
| $H$ | 12.19 | 1 | 12.19 | 1.66 | $4.98 \approx 5$ |
Empirical formula of ’ $A$ ’ $C_3 H_5$
Empirical formula mass $=35+5=41 u$
$n=\frac{\text{ Molecular mass }}{\text{ Empirical formula mass }}=\frac{83.1}{41}=2.02 \approx 2$
$\Rightarrow$ Molecular mass is double of the empirical formula mass.
$\therefore$ Molecular Formula is $C_6 H _{10}$
Step III
$C_6 H _{10} \xrightarrow{2 H_2}$ 2-methylpentane
(A)
Structure of 2-methylpentane is
Hence, the molecule has a five carbon chain with a methyl group at the second carbon atom.
’ $A$ ’ adds a molecule of $H_2 O$ in the presence of $Hg^{2+}$ and $H^{+}$, it should be an alkyne. Two possible structures for ‘A’ are :
II
Since the ketone (B) gives a positive iodoform test, it should contain a $-COCH_3$ group. Hence the structure of ketone is as follows :
Therefore structure of alkyne is II.
50. An unsaturated hydrocarbon ’ $A$ ’ adds two molecules of $H_2$ and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.
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Answer: Two molecules of hydrogen add on ’ $A$ ’ this shows that ’ $A$ ’ is either an alkadiene or an alkyne.
On reductive ozonolysis ’ $A$ ’ gives three fragments, one of which is dialdehyde. Hence, the molecule has broken down at two sites. Therefore, ’ $A$ ’ has two double bonds. It gives the following three fragments:
$OHC-CH_2-CH_2-CHO, CH_3 CHO$ and $CH_3-CO-CH_3$
Hence, its structure as deduced from the three fragments must be
$CH_3-CH=CH-CH_2-CH_2-CH=\underset{ _C^{+} CH_3}{C}-CH_3$
(A)
Reactions
51. In the presence of peroxide addition of $HBr$ to propene takes place according to anti Markovnikov’s rule but peroxide effect is not seen in the case of $HCl$ and HI. Explain.
