Topic: Hydrocarbons (Test 1)

Topic: Hydrocarbons
Q.1
In Friedel-Crafts alkylation reaction, when benzene is treated with an alkyl halide in the presence of a catalyst, alkylbenene is formed. The catalyst used is
A. anhydrous aluminium chloride
B.

C.
Silver
D.

Explaination / Solution:

the acid chloride or anhydride reacts with anhydrous aluminium chloride to form the acylium ion which serves as an electrophile which then attacks the benzene ring to give alkylbenzene.

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Q.2
Some of the polynuclear hydrocarbons having fused benzene ring system have the property of being
A. malleable
B. nonflammable
C. conductive
D. carcinogenic
Explaination / Solution:

Polynuclear hydrocarbons are carcinogenic,i.e. cancer producing

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Q.3
Arrange the following in decreasing order of their boiling points. (A) n–butane (B) 2–methylbutane (C) n-pentane (D) 2,2–dimethylpropane
A. A > B > C > D
B. D > C > B > A
C. C > B > D > A
D. B > C > D > A
Explaination / Solution:

Boiling point depends on molecular mass and surface area.As the number of carbon atoms increase,boiling point increases ,hence n-butane has the minimum boiling point.Among isomeric alkanes,boiling point decreases with branching.Hence the order.

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Q.4

Arrange the halogens in order of their increasing reactivity with alkanes.

A. Br2< Cl2< F2< I2
B. Br2< I2< Cl2< F2
C. I
2
< Br2< Cl2< F2
D. F2< Cl2< Br2< I2
Explaination / Solution:

The reaction follows free radical mechanism in which the first step is generation of halogen free radical which is the fastest for fluorine.Hence the order.

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Q.5
The increasing order of reduction of alkyl halides with zinc and dilute HCl is
A. R–Cl < R–I < R–Br
B. R–Br < R–I < R–Cl
C. R–Cl < R–Br < R–I
D. R–I < R–Br < R–Cl
Explaination / Solution:

As we move from I to Br to Cl,atomic size decreases,it becomes difficult to release the halogen as the bond length decreases.Hence I- is the best leaving group among halogens and reacts the fastest.

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Q.6
The correct IUPAC name of the following alkane is

A. 3,6 – Diethyl – 2 – methyloctane
B. 5 – Isopropyl – 3 – ethyloctane
C. 3 – Isopropyl – 6 – ethyloctane
D. 3 – Ethyl – 5 – isopropyloctane
Explaination / Solution:

Following the rules of nomenclature,the IUPAC name of the given compound is 3,6-Diethyl-2-methyloctane

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Q.7
Which of the following will not show geometrical isomerism?

A. (ii)
B. (iii)
C. (iv)
D. (i)
Explaination / Solution:

This is because geometrical isomerism is not possible if three groups are same.

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Q.8
Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
A. HBr > HI > HCl
B. HI > HBr > HCl
C. HCl > HBr > HI
D. HCl > HI > HBr
Explaination / Solution:

As we move from Cl to I,the atomic size increases,bond length increases,bond strength decreases,hence it becomes easier to release halogen and hence the order.I- is the a better leaving group.

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Q.9
Arrange the following carbanions in order of their decreasing stability.(A) H3C – C $\equiv$C (B) H – C ≡ C (C)
A. C > A > B
B. A > B > C
C. C > B > A
D. B > A > C
Explaination / Solution:

In B,the '-' charge is on sp hybridised C atom so the negative charge is stabilised.In A,there is an alkyl group attached to sp hybridised carbon which destabilises the negative charge. In C,the '-' charge is on a sp3 hybridised carbon and is hence the least stable.

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Q.10
Arrange the following alkyl halides in decreasing order of the rate of  elimination reaction with alcoholic KOH

A. C > B > A
B. A > C > B
C. B > C > A
D. A > B > C