Practice Start Test

Hydrocarbons

NEET Chemistry · 53 questions · Page 1 of 6 · Click an option or "Show Solution" to reveal answer

Q1
The alkane that can be oxidized to the corresponding alcohol by KMnO4_4 as per the equation is, when :
A R1=H;R2=H;R3=H\mathrm{R}_1=\mathrm{H} ; \mathrm{R}_2=\mathrm{H} ; \mathrm{R}_3=\mathrm{H}
B R1=CH3;R2=CH3;R3=CH3\mathrm{R}_1=\mathrm{CH}_3 ; \mathrm{R}_2=\mathrm{CH}_3 ; \mathrm{R}_3=\mathrm{CH}_3
C R1=CH3;R2=H;R3=H\mathrm{R}_1=\mathrm{CH}_3 ; \mathrm{R}_2=\mathrm{H} ; \mathrm{R}_3=\mathrm{H}
D R1=CH3;R2=CH3;R3=H\mathrm{R}_1=\mathrm{CH}_3 ; \mathrm{R}_2=\mathrm{CH}_3 ; \mathrm{R}_3=\mathrm{H}
Correct Answer
Option B
Solution

Generally alkanes resist oxidation but alkane with tertiary H atom(s) can be oxidised to corresponding alcohols by KMnO

4_4
∴R1⇒CH3R2⇒CH3R3⇒CH3\therefore \mathrm{R}_1 \Rightarrow \mathrm{CH}_3 \quad \mathrm{R}_2 \Rightarrow \mathrm{CH}_3 \quad \mathrm{R}_3 \Rightarrow \mathrm{CH}_3
Q2
The incorrect method for the synthesis of alkenes is
A Treating vicinal dihalides with Zn metal
B Treating of alkynes with Na in liquid NH 3
C Heating alkyl halides with alcoholic KOH
D Treating alkyl halides in aqueous KOH solution
Correct Answer
Option D
Solution

Alkenes can be prepared

Q3
With respect to the conformers of ethane, which of the following statements is true?
A Bond angle changes but bond length remains same.
B Both bond angle and bond length change.
C Both bond angle and bond length remain same.
D Bond angle remains same but bond length changes.
Correct Answer
Option C
Solution

Conformers are the isomers which are formed by rotation about single bonds without any cleavage of any bond.

These conformers have same bond angle between them and have same bond length while their dihedral angle changes.

Q4
Which one is the correct order of acidity?
A CH≡ \equiv CH > CH 3 −-C≡ \equiv CH > CH 2 == CH 2 > CH 3 −-CH 3
B CH ≡ \equiv CH > CH 2 == CH 2 > CH 3 −-C ≡ \equiv CH > CH 3 −-CH 3
C CH 3 −-CH 3 > CH 2 == CH 2 > CH 3 −-C ≡ \equiv CH > CH≡ \equiv CH
D CH 2 ==CH 2 > CH 3 −- CH == CH 2 > CH 3 −- C≡ \equiv CH > CH ≡ \equiv CH
Correct Answer
Option A
Solution

Greater the character of C-atom in hydrocarbons, greater the electronegativity of that carbon and thus greater the acidic nature of the H attached to electronegative carbon.

Thus CH≡\equivCH > CH 3 −-C≡\equivCH > CH 2 == CH 2 > CH 3 −-CH 3

Q5
Which of the following can be used as the halide component for Friedel-Crafts reaction?
A Chlorobenzene
B Bromobenzene
C Chloroethene
D Isopropyl chloride
Correct Answer
Option D
Solution

Friedel–Crafts reaction : Chlorobenzene, bromobenzene and chloroethene are not suitable halide components as lone pair of electrons of halogen are delocalized with π\pi-bonds to attain double bond (C = X) character.

Q6
In pyrrole the electron density is maximum on
A 2 and 3
B 3 and 4
C 2 and 4
D 2 and 5
Correct Answer
Option D
Solution

Pyrrole has maximum electron density on 2 and 5.

It generally reacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbocation that is a hybrid of structures (I) and (II).

Attack at position 2 or 5 yields a carbocation that is a hybrid not only of structures (III) and (IV) (analogous to I and II) but also of structure (V).

The extra stabilization conferred by (V) makes this ion the more stable one.

Also, attack at position 2 or 5 is faster because the developing positive charge is accommodated by three atoms of the ring instead of by only two.

Q7
The compound that will react most readily with gaseous bromine has the formula
A C 3 H 6
B C 2 H 2
C C 4 H 10
D C 2 H 4
Correct Answer
Option A
Solution

Propene is most reactive towards Br 2 (gaseous) than CH 2 = CH 2 , HC ≡\equiv CH and butane due to most electron density.

Q8
For the following reactions : (A)  CH 3 CH 2 CH 2 Br + KOH → \to                  CH 3 CH == CH 2 + KBr + H 2 O Which of the following statements is correct?
A (A) is elimination, (B) and (C) are substitution reactions.
B (A) is substitution, (B) and (C) are addition reactions.
C (A) and (B) are elimination reactions and (C) is addition reaction.
D (A) is elimination, (B) is substitution and (C) is addition reaction.
Correct Answer
Option D
Solution

(A) Saturated compound is converted into unsaturated compound by removal of group of atoms hence, it is an elimination reaction.

(B) —Br group is replaced by —OH group hence, it is a substitution reaction.

(C) Addition of Br 2 converts an unsaturated compound into a saturated compound hence, it is an addition reaction.

Q9
Consider the nitration of benzene using mixed conc.H 2 SO 4 and HNO 3 . If a large amount of KHSO 4 is added to the mixture, the rate of nitration will be
A unchanged
B doubled
C faster
D slower
Correct Answer
Option D
Solution

H 2 SO 4 + HNO 3 →\to

NO2+NO_2^ +

+

HSO4−HSO_4^ -

+

H3O+{H_3}{O^ + }

KHSO 4 →\to K + +

HSO4−HSO_4^ -

Due to common ion effect backward reaction will take place so the formation

NO2+NO_2^ +

decrease so nitration process will become slower.

Q10
The pair of electrons in the given carbanion, CH 3 C ≡ \equiv C −- , is present in which of the following orbitals?
A sp 2
B sp
C 2p
D sp 3
Correct Answer
Option B
Solution

Thus, pair of electrons is present in sp-hybridised orbital.

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