Hydrocarbons
Multiple Choice Questions (MCQs)
1. Arrange the following in decreasing order of their boiling points.
A.
B. 2-methylbutane
C.
D. 2, 2-dimethylpropane
(a)
(b)
(c)
(d)
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Answer
(d) As the number of carbon atom increases, boiling point increases.
Boiling point decreases with branching

(4 carbon atoms with no branching)
-
Option (a) is incorrect because
-butane has fewer carbon atoms and less branching compared to 2-methylbutane and -pentane, which would result in a lower boiling point than those compounds. -
Option (b) is incorrect because 2-methylbutane has more branching than
-pentane, which would result in a lower boiling point than -pentane. Additionally, -butane should have a lower boiling point than 2-methylbutane due to having fewer carbon atoms. -
Option (c) is incorrect because 2, 2-dimethylpropane has the most branching among the given compounds, which would result in the lowest boiling point, not the highest.
2. Arrange the halogens
(a)
(b)
(c)
(d)
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Answer
(a) Rate of reaction of alkanes with halogens is
Alkane react with
-
Option (b) is incorrect because it suggests that iodine (
) is the most reactive with alkanes, which contradicts the fact that iodine is the least reactive due to its lower electronegativity and bond dissociation energy compared to other halogens. -
Option (c) is incorrect because it suggests that fluorine (
) is the least reactive with alkanes, which contradicts the fact that fluorine is the most reactive due to its high electronegativity and low bond dissociation energy. -
Option (d) is incorrect because it places iodine (
) between bromine ( ) and chlorine ( ) in terms of reactivity with alkanes, which is incorrect. Iodine is the least reactive, and the correct order should be .
3. The increasing order of reduction of alkyl halides with zinc and dilute
(a)
(b)
(c)
(d)
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Answer
(b) The reactivity of halogens with alkane is
-
Option (a)
: This option is incorrect because it suggests that is less reactive than , which is less reactive than . However, the reactivity order should be based on the bond strength, where is the weakest and thus most reactive, followed by , and then . -
Option (c)
: This option is incorrect because it suggests that is the least reactive, followed by , and then . In reality, is the most reactive due to the weakest bond strength, followed by , and then . -
Option (d)
: This option is incorrect because it suggests that is less reactive than , which is less reactive than . The correct order should be being the most reactive, followed by , and then , based on the decreasing bond strength.
4. The correct IUPAC name of the following alkane is

(b) 5-isopropyl -3-ethyloctane
(c) 3-ethyl-5-isopropyloctane
(d) 3-isopropyl-6-ethyloctane
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Answer
(a) The correct IUPAC name is

Longest chain
Branch on 2, 3,6 follows lowest sum rule.
Branch of 2-C-methyl; 3, 6, C atom-ethyl.
Ethyl comes alphabetically before methyl.
Hence, 3,6-diethyl 2-methyl octane.
-
Option (b): 5-isopropyl-3-ethyloctane
- The longest chain is correctly identified as octane, but the numbering of the substituents does not follow the lowest sum rule. The correct numbering should give the lowest possible numbers to the substituents, which is not achieved here.
-
Option (c): 3-ethyl-5-isopropyloctane
- The substituents are not listed in alphabetical order. According to IUPAC rules, substituents should be listed alphabetically, and “ethyl” should come before “isopropyl.”
-
Option (d): 3-isopropyl-6-ethyloctane
- The substituents are not listed in alphabetical order. According to IUPAC rules, substituents should be listed alphabetically, and “ethyl” should come before “isopropyl.” Additionally, the numbering does not follow the lowest sum rule.
5. The addition of
The mixture consists of
(a)
(b)
(c)
(d)
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Answer
(a) The alkene is unsymmetrical, hence will follow Markownikoff’s rule to give major product.
Since, I contains, a chiral carbon, it exists in two enantiomers (


-
Option (b): This option is incorrect because it suggests that product B is the major product, while A and C are minor products. However, according to Markownikoff’s rule, the major product should be the one where the hydrogen atom from HBr adds to the carbon with the greater number of hydrogen atoms (leading to the formation of a more stable carbocation intermediate). This results in product A (and its enantiomer B) being the major products, not B alone.
-
Option (c): This option is incorrect because it suggests that product B is the minor product, while A and C are major products. However, product C is formed via anti-Markownikoff addition, which is less favored in the presence of HBr alone (without peroxides). Therefore, C should be the minor product, not a major one.
-
Option (d): This option is incorrect because it suggests that products A and B are minor products, while C is the major product. However, product C is formed via anti-Markownikoff addition, which is less favored in the presence of HBr alone (without peroxides). Therefore, A and B should be the major products, not minor ones.
6. Which of the following will not show geometrical isomerism?
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Thinking Process
This question is based upon geometrical isomerism. For geomterical isomerism, it is essential that each carbon atom of the double bond must have different substituents.
Answer
(d) In option (d), a carbon with double bond has two same functional groups
-
In option (a), the double bond is between two carbons, each of which has two different groups attached. This allows for the possibility of cis-trans isomerism, where the groups can be on the same side (cis) or opposite sides (trans) of the double bond.
-
In option (b), the double bond is between two carbons, each of which has two different groups attached. This also allows for the possibility of cis-trans isomerism.
-
In option (c), the double bond is between two carbons, each of which has two different groups attached. This configuration allows for the possibility of cis-trans isomerism.
7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
(a)
(b)
(c)
(d)
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Answer
(c) Bond energy of
-
Option (a) is incorrect because it suggests that HCl is more reactive than HBr and HI with propene. However, the bond energy of HCl is higher than that of HBr and HI, making it less reactive.
-
Option (b) is incorrect because it places HCl as the least reactive, which is correct, but it incorrectly suggests that HBr is more reactive than HI. The bond energy of HI is lower than that of HBr, making HI more reactive.
-
Option (d) is incorrect because it suggests that HCl is more reactive than HI and HBr. Given that HCl has the highest bond energy among the three, it is actually the least reactive.
8. Arrange the following carbanions in order of their decreasing stability.
A.
B.
C.
(a)
(b)
(c)
(d)
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Answer
(b)
Hence,
-
Option (a)
: This option is incorrect because it suggests that the carbanion with the methyl group ( ) is more stable than the carbanion without the methyl group ( ). However, the methyl group has a +I effect, which destabilizes the carbanion by intensifying the negative charge. Therefore, should be more stable than . -
Option (c)
: This option is incorrect because it suggests that the carbanion with the sp³ hybridized carbon ( ) is more stable than the sp hybridized carbanions ( and ). However, sp hybridized carbanions are more stable than sp³ hybridized carbanions due to the higher s-character, which allows for better stabilization of the negative charge. Therefore, should be the least stable. -
Option (d)
: This option is incorrect because it suggests that the sp³ hybridized carbanion ( ) is more stable than both sp hybridized carbanions ( and ). As mentioned earlier, sp hybridized carbanions are more stable than sp³ hybridized carbanions. Additionally, it incorrectly places as more stable than , despite the destabilizing +I effect of the methyl group in .
9. Arrange the following alkyl halides in decreasing order of the rate of

C.
(a)
(b)
(c)
(d)
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Answer
(d) Alkyl halides on heating with alcoholic potash eliminates one molecule of halogen acid to form alkene. Hydrogen is eliminated from
-
Option (a)
: This option is incorrect because it suggests that the primary alkyl halide (C) has a slower rate of -elimination than the secondary alkyl halide (B). However, the rate of -elimination follows the order , meaning tertiary alkyl halides react faster than secondary, which in turn react faster than primary. Therefore, (tertiary) should be faster than (secondary), and should be faster than (primary). -
Option (b)
: This option is incorrect because it suggests that the primary alkyl halide (C) has the fastest rate of -elimination, followed by the secondary (B), and then the tertiary (A). This contradicts the established order of reactivity for -elimination, which is . Therefore, should be the fastest, followed by , and then . -
Option (c)
: This option is incorrect because it suggests that the secondary alkyl halide (B) has a faster rate of -elimination than the primary (C) and tertiary (A). According to the reactivity order , the tertiary alkyl halide (A) should have the fastest rate, followed by the secondary (B), and then the primary (C). Therefore, should be faster than , and should be faster than .
10. Which of the following reactions of methane is incomplete combustion?
(a)
(b)
(c)
(d)
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Answer
(c) During incomplete combustion of alkanes with insufficient amount of air or dioxygen carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Thus,
-
(a) This reaction is not incomplete combustion because it results in the formation of methanol (
) rather than carbon black or carbon monoxide. It is a controlled oxidation reaction. -
(b) This reaction is not incomplete combustion because it produces formaldehyde (
) and water ( ) instead of carbon black or carbon monoxide. It is a partial oxidation reaction. -
(d) This reaction is not incomplete combustion because it results in the complete combustion of methane, producing carbon dioxide (
) and water ( ). Complete combustion occurs when there is a sufficient amount of oxygen.
Multiple Choice Questions (More Than One Options)
11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions?
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Answer
Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products.
-
Option (a) is incorrect because it represents the complete combustion of methane, which produces carbon dioxide and water. This is not a controlled oxidation reaction but rather a complete oxidation reaction.
-
Option (b) is incorrect because it represents the incomplete combustion of methane, which produces carbon monoxide and water. This is also not a controlled oxidation reaction but rather an incomplete oxidation reaction.
12. Which of the following alkenes on ozonolysis give a mixture of ketones only?
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Answer
Alkenes which have two substituents on each carbon atom of the double bond, give mixture of ketones on ozonolysis. Thus, option (c) and (d) give mixture of ketones.
-
Option (a): This alkene has one substituent on one carbon atom of the double bond and two substituents on the other carbon atom. Ozonolysis of such alkenes will produce a mixture of a ketone and an aldehyde, not just ketones.
-
Option (b): This alkene has one substituent on each carbon atom of the double bond. Ozonolysis of such alkenes will produce a mixture of two aldehydes, not ketones.
-
Option (e): This alkene has one substituent on one carbon atom of the double bond and two substituents on the other carbon atom. Ozonolysis of such alkenes will produce a mixture of a ketone and an aldehyde, not just ketones.
13. Which are the correct IUPAC names of the following compound?
(a) 5-Butyl-4-isopropyldecane
(b) 5-Ethyl-4-propyldecane
(c) 5-sec-Butyl -4-iso-propyldecane
(d) 4-(1-methylethyl) - 5 - (1-methylpropyl)-decane
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Answer

5- sec-Butyl -4 iso -propyldecane
4-(1-methylethyl)-5-(1-methylpropyl)- decane
Although IUPAC name for sec- butyl and isopropyl groups are 1methyl propyl and 1-methylethyl respectively yet both these names, are also recommended for IUPAC nomenclatiro
-
Option (a): 5-Butyl-4-isopropyldecane
- Incorrect because the substituents are not named according to the longest carbon chain rule. The correct substituents should be sec-butyl and isopropyl, not butyl and isopropyl.
-
Option (b): 5-Ethyl-4-propyldecane
- Incorrect because the substituents are not named correctly. The correct substituents should be sec-butyl and isopropyl, not ethyl and propyl.
14. Which are the correct IUPAC names of the following compound?
(a) 5-(2’, 2’ - Dimethylpropyl)-decane
(b) 4-Butyl-2,2-dimethylnonane
(c) 2,2- Dimethyl-4- pentyloctane
(d) 5-neo-Pentyldecane
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Answer

5- (2’,2’- Dimethylpropyl)- decane 5-neo- pentyldecane
The IUPAC name for neopentyl groups is 2, 2 dimethyl propyl.
-
Option (b): 4-Butyl-2,2-dimethylnonane
- Incorrect because the longest carbon chain should be identified first. The longest chain in this compound is decane (10 carbons), not nonane (9 carbons). The substituent should be named based on the longest chain, which is why “decane” is correct, not “nonane”.
-
Option (c): 2,2-Dimethyl-4-pentyloctane
- Incorrect because the longest carbon chain should be identified first. The longest chain in this compound is decane (10 carbons), not octane (8 carbons). The substituent should be named based on the longest chain, which is why “decane” is correct, not “octane”. Additionally, the substituent is not correctly named as “pentyloctane”; it should be “2,2-dimethylpropyl”.
15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring
(a) deactivates the ring by inductive effect
(b) deactivates the ring by reasonance
(c) increases the charge density at ortho and para position relative to meta position by resonance.
(d) directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position.
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Answer
(a, c)
For an electrophilic substitution reaction, the presence of halogen atom in the benzene ring deactivates the ring by inductive effect and increases the charge density at ortho and para position relative to meta position by resonance.
When chlorine is attached to benzene ring, chlorine being more electronegative pulls the electron i.e.,

The last structure contributes more to the orientation and hence halogen are
-
(b) deactivates the ring by resonance: This option is incorrect because halogens, despite being electron-withdrawing by inductive effect, have lone pairs that can participate in resonance with the benzene ring. This resonance effect actually increases the electron density at the ortho and para positions, not deactivating the ring by resonance.
-
(d) directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position: This option is incorrect because the resonance effect of halogens increases the electron density at the ortho and para positions, making these positions more reactive towards electrophilic substitution. Therefore, halogens are ortho and para directors, not meta directors.
16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group.
(a) deactivates the ring by inductive effect
(b) activates the ring by inductive effect
(c) decreases the charge density at ortho and para position of the ring relative to meta position by resonance
(d) increases the charge density at meta position relative to the ortho and para positions of the ring by resonance
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Answer
Nitro group by virtue of

In ortho, para-attack of electrophile on nitrobenzene, we are getting two structures (A) and
As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.
-
(b) activates the ring by inductive effect: This option is incorrect because the nitro group is an electron-withdrawing group due to its strong
(inductive) effect. This means it pulls electron density away from the benzene ring, thereby deactivating the ring towards electrophilic substitution reactions rather than activating it. -
(d) increases the charge density at meta position relative to the ortho and para positions of the ring by resonance: This option is incorrect because the nitro group, being an electron-withdrawing group, decreases the electron density at the ortho and para positions more significantly than at the meta position. This is due to the resonance effect, where the nitro group withdraws electron density through resonance, making the ortho and para positions less electron-rich compared to the meta position. However, the statement about increasing charge density at the meta position by resonance is misleading because the meta position is not directly affected by the resonance withdrawal of electrons by the nitro group.
17. Which of the following are correct?
(a)
(b)
(c)
(d)
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Answer
(i)
Thus,
(ii)
(iii)
(iv) In
-
Option (b):
is more stable than because the former is stabilized by the +I-effect of two groups. The presence of two methyl groups provides greater electron-donating inductive effects, which stabilize the carbocation more effectively than the single inductive effect from the ethyl group in . -
Option (d): In
, the positive charge is present on the more electronegative, sp-hybridized carbon, which makes it less stable. In contrast, in , the positive charge is on a less electronegative sp²-hybridized carbon, making it more stable. Therefore, is less stable than .
18. Four structures are given in options (a) to (d). Examine them and select the aromatic structures.
(a)

(b)

(c)

(d)

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Answer
(a, c) Aromaticity requires following condition
(i) planarity
(ii) complete delocalisation of
(iii) presence of
(a)
planar
(b)

(c)

(d)

-
(b): The structure does not satisfy the aromaticity criteria because it does not have a complete delocalization of π electrons in the ring. Additionally, it does not follow the
electron rule. -
(d): The structure does not satisfy the aromaticity criteria because the number of π electrons does not fit the
electron rule, where must be an integer.
19. The molecules having dipole moment are ……..
(a) 2,2-Dimethylpropane
(b) trans-Pent-2-ene
(c) cis-Hex-3-ene
(d) 2, 2, 3, 3 - Tetramethylbutane
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Answer
Thus, trans-pent-2-ene show net diple moment because different group attached and cis- Hex -3- ene show dipole moment because both groups
-
2,2-Dimethylpropane: This molecule is symmetrical, and the dipole moments of the individual C-H bonds cancel each other out, resulting in a net dipole moment of zero.
-
2, 2, 3, 3-Tetramethylbutane: This molecule is also symmetrical, and the dipole moments of the individual C-H bonds cancel each other out, resulting in a net dipole moment of zero.
Short Answer Type Questions
20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.
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Answer
Alkenes are rich source of loosely held pi

In arenes during electrophilic addition reactions, aromatic character of benzene ring is destroyed while during electrophilic substitution reaction it remains intact. Electrophilic substitution reactions of arenes are energetically more favourable than that of electrophilic addition reaction.
That’s why alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reaction.
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Thinking Process
In geometrical isomerism, when same groups are on the same side it is cis and if same groups are on the opposite side it is trans isomer.
Answer
Trans-2-butene formed by the reduction of 2-butyne is capable of showing geometrical isomerism.

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Thinking Process
The infinite number of momentary arrangements of the atoms in space which result through rotation about a single bond are called conformations.
In ethane, if one carbon atom is kept stationary and other rotated around
Answer
Alkanes can have infinite number of conformations by rotation around
Hence, minimum repulsive force. In eclipsed electron cloud of carbon-hydrogen become close resulting in increase in electron cloud repulsion. This repulsion affects stablity of a conformer.
In all the conformations of ethane the staggered form has least torsional strain and the eclipsed form has the maximum torsional strain. Hence, rotation around C-C bond in ethtane is not completely free.

Eclipsed

Staggered
Newman’s projection of ethane
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Answer
Staggered form of ethane is more stable than the eclipsed conformation, by about
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Answer
Addition of halogen acids to an alkene is an electrophilic addition reaction.

First step is slow so, it is rate determining step. The rate of this step depends on the availability of proton. This in turn depends upon the bond dissociation enthalpy of the
Lower the bond dissociation enthalpy of
Therefore, the reactivity of the halogen acids decreases from

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Answer

When Friedel-Craft alkylation is carried out with higher alkyl halide, e.g.,
(b) m-nitrobromobenzene
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Answer
Halogens attached to benzene ring is ortho and para directing where as nitro group is meta directing.
(a)



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Answer
The methoxy group

In case of alkyl halides, halogens are moderately deactivating because of their strong
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Answer
Halogens have

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Answer
The meta - directing substituents (like -

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Answer
Acetylene when passed through red hot iron tube at

Note In nitration of benzene ring conc
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Answer
In presence of organic peroxides, the addition of

However, in absence of peroxides, addition of

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
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Answer
Electrophiles are electron deficient species. They may be natural or positively charged e.g., (iii)
Nucleophiles are electron rich species. They may be neutral or negatively charged e.g.,
(i)
(ii)
(vi)
(vii)
(viii)
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Answer
The given organic compound is
This compound has 9 primary hydrogen, 2 secondary and one tertiary hydrogen atoms. The relative reactivity of
Total amount of mono chloro product
Percentage of
Percentage of
Percentage of
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Thinking Process
This question is based upon Wurtz reaction. Wurtz reaction represent that two alkyl groups can be coupled by reacting alkyl halide with
Answer
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Answer
2-methylpropane gives two types of radicals.
Radical (I) is more stable because it is
Radical (II) is less stable because it is
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Answer
From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl halide gives an alkane

Since, alkane

(i) Planar ring containing conjugated
(ii) Complete delocalisation of the
(iii) Presence of
Using this information classify the following compounds as aromatic/non-aromatic.

(A)

(B)

(C)

(D)

(E)

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Answer
Compound | Planar ring |
Complete delocalisation of |
Huckel rule |
Aromatic or non-aromatic |
|
---|---|---|---|---|---|
A. | P | Huckel rule obeyed |
Aromatic | ||
B. | í | Í Incomplete (sp hybrid carbon) |
Non-aromatic | ||
C. | P | P | pair verified |
Aromatic | |
D. | P | í | Anti-aromatic | ||
E. | P | P | Huckel rule obeyed | Aromatic | |
F. | P | P | verified |
Aromatic | |
G. | P | í | not verified |
Non-aromatic |
(A)
(B)
(C)
(D)
(E)
(F)
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Answer
A. The compound has
B. The compound is aromatic. It has
C. The compound contains
E. In this compound one six membered planar ring has
F. It has
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Answer
For preparation of ethyl hydrogensulphate
Step I Protonation of alcohol

Step II Attack of nucleophile

Temperature should not be allowed to rise above
Matching The Columns
40. Match the reagent from Column I which on reaction with
Column I | Column II | ||
---|---|---|---|
A. | 1. | Acetic acid and |
|
B. | 2. | Propan-1-ol | |
C. | 3. | Propan-2-ol | |
D. | 4. | Acetaldehyde and formaldehyde | |
E. | 5. | Propane-1, 2-diol |
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Answer
A.
B.
C.
D.
E.
Reagent | Recation with propene | ||
A. | ![]() |
||
B. | |||
C. | ![]() |
||
D. | |||
E. | ![]() |
Column I | Column II | ||
---|---|---|---|
A. | n-pentane | 1. | |
B. | iso-pentane | 2. | |
C. | neo-pentane | 3. |
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Thinking Process
To solve this question, it keep in mind that branching of hydrocarbons decreases boiling point of the compound
Answer
A.
B.
C.
Column I | Column II | ||
---|---|---|---|
A. | Benzene |
1. | Benzoic acid |
B. | Benzene |
2. | Methyl phenyl ketone |
C. | Benzene |
3. | Toluene |
D. | Toluene |
4. | Chlorobenzene |
5. | Benzene hexachloride |
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Answer
A.
B.
C.
D.
reactants | Products | ||
---|---|---|---|
A. | Benzene |
||
B. | Benzene |
||
C. | Benzene |
||
D. | Toluene |
||
Column I | Column II | ||
---|---|---|---|
A. | 1. | Hydrogenation | |
B. | 2. | Halogenation | |
C. | 3. | Polymerisation | |
4. | Hydration | ||
5. | Condensation |
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Answer
A.
B.
C.
D.

Assertion and Reason
In the following questions a statement of assertion (A) followed by a statement of reason

It is cyclic and has conjugated 8 8 -electron system but it is not an aromatic compound.
Reason (R)
(a) Both
(b) Both
(c) Both
(d)
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Answer
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
According to Huckel rule Aromaticity is shown by compounds possessing following characteristics
(i) Compound must be planar and cyclic
(ii) Complete delocalisation of
(iii) Presence of conjugated
Reason (R)
(a) Both
(b) Both
(c) Both
(d)
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Answer
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
Toluene has
In resonating structure of toluene, electronic density is more on ortho and para position. Hence, substitution takes place mainly at these positions.
Reason (R) The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile,
(a) Both
(b) Both
(c) Both
(d)
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Answer
(a) Both assertion and reason are correct and reason is the correct explanation of assertion. In nitration of benzene with nitric acid sulphuric acid acts as a calatyst. It helps in the formation of electrophile i.e., nitronium ion
Reason (R) Branching does not affect the boiling point.
(a) Both
(b) Both
(c) Both
(d)
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Answer
(c) Both assertion and reason are correct
Correct assertion Among isomeric pentanes, 2, 2 - dimethylpentane has the lowest boiling point.
Correct reason Branching decrease the boiling point.
Long Answer Type Questions
48. An alkyl halide
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Answer
The reaction scheme involved in the problem is
Hydrogenation of alkyne (D) gives straight chain alkane hence all the compounds (A), (B),

It is important point that alkyl halide (A) can not be 2-bromopentane because dehydrobromination of
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Answer
To determine the molecular mass of hydrocarbon (A)
Hence, molecular mass of
Element | % | Atomic mass | Relative ratio | Relative no. of atoms |
Simplest ratio |
---|---|---|---|---|---|
87.8 | 12 | 7.31 | 1 | 3 | |
12.19 | 1 | 12.19 | 1.66 |
Thus, Empirical formula of
Molecular mass is double of empirical formula mass.
To determine the structure of compounds
Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefore, hydrocarbon(A) is an alkyne having five carbon atoms in a staight chain and a methyl substituent at position 2 . Thus, the possible structures for the alkyne

Since, addition of
Now addition of
In contrast, addition of
Thus, hydrocabon
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Answer
The scheme of reaction is
Compound

Thus, structure of
The reactions involved in the question



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Answer
The mechanism of the reaction is

Step 1

Step II

Peroxide effect is effective only in the case of
(i)
(ii) lodine free radical