Hydrocarbons Test

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Hydrocarbons Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

What is one of the products of the addition of $$HBr$$ to 2-Butene?

A
1-bromobutane

B
2-bromobutane

C
1, 2-dibromobutane

D
2, 3-dibromobutane

Solution

When $$\displaystyle HBr$$ adds to $$\displaystyle 2-butene$$, $$\displaystyle 2-bromobutane$$ is obtained.

$$\displaystyle CH_3-CH=CH-CH_3 +HBr \rightarrow CH_3-CH(Br)-CH_2-CH_3$$

A molecule of $$\displaystyle H-Br$$ adds to $$\displaystyle C=C$$ double bond.

Since, $$\displaystyle 2-butene$$ is symetrical, only one product is obtained.

Option B is correct.
Question 2
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The overlapping of orbitals in benzene is of the type:

A
$$sp - sp$$

B
$$p - p$$

C
$${sp}^{2} - {sp}^{2}$$

D
$${sp}^{3} - {sp}^{3}$$

Solution

In Benzene all the six carbon atoms undergo $$sp^2$$ (Double bonds between all six carbon atoms) hybridization.

So the overlapping of orbitals in Benzene is $$sp^2-sp^2$$.

Hence, option C is correct.
Question 3
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Which one of the following is not formed when a mixture of methyl bromide and bromobenzene is heated with sodium metal in the presence of dry ether?

A
Ethane

B
Diphenyl

C
Propane

D
Toluene

Solution

Propane is not formed when a mixture of methyl bromide and bromobenzene is heated with sodium metal in the presence of dry Ether.
Two moles of methyl bromide in presence of $$Na/ether$$ give ethane.
Two moles of bromobenzene in presence of $$Na/ether$$ give diphenyl.
One mole of methyl bromide and one mole of bromobenzene in presence of $$Na/ether$$ give toluene.
$$\displaystyle R-Br + R'-Br +2Na \rightarrow R-R' + 2 NaBr$$
Question 4
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When propyne is treated with dilute sulphuric acid in presence of mercury (II) sulphate, the major product is:

A
acetone

B
propene

C
propanol

D
propanal

Solution

When propyne is treated with dilute sulphuric acid in presence of mercury II sulphate, the major product is acetone. A molecule of water is added to $$\displaystyle C=C$$ double bond to from an enol which tautomerizes to keto form.
Question 5
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On reduction of glycolic acid with $$HI$$, the product formed is:

A
acetic acid

B
iodo aectic acid

C
formic acid

D
none of these

Solution

$$ \underset {Glycolic \quad acid}{HOOC-CH_2OH} \xrightarrow []{HI} \underset {Acetic \quad acid} {HOOC-CH_3} $$
$$HI$$ reduces $$CH_2OH$$ to $$CH_3$$ but $$HI + P$$ reduces both $$-COOH$$ and $$-CH_2OH$$ to $$-CH_3$$.
Question 6
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The product formed when toulene is heated in light with $${Cl}_{2}$$ and in absence of halogen carrier is:

A
Chlorobenzene

B
Gammexene

C
Benzotrichloride

D
DDT

Solution

answer is (c)
Question 7
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Which of the following reactions will not yield phenol?

A

B

C

D

Solution

The reaction of option (D) will not yield phenol. It is nucleophilic substitution of chlorobenzene which requires drastic conditions (high temperature of 613 K and high pressure of 300 atm).

Note:
The following reactions will yield phenol

(A) Dow process: Chlorbenzene is heated with excess aq NaOH at 613 K and 300 atm. This is followed by hydrolysis with dil HCl.

(B) From aniline diazotization: Diazotization of aniline (with nitrous acid at 0-5 deg C) followed by warming with water gives phenol.

(C) From benzene sulphonic acid: Benzene is sulphonated with oleum to form benzene sulphonic acid. It is then heated with NaOH and then followed with acid hydrolysis gives phenol.

Option D is correct.
Question 8
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The reactivity order of halogenation of alkanes is $$F_2>Cl_2>Br_2>I_2$$ . Choose the correct statements.
(I) Lower the activation energy for the chain initiation step, more reactive is the halogen.
(II) Lower the activation energy for the first chain-propagation step, more reactive is the halogen.
(III) Lower the activation energy for the second chain-propagation step, more reactive is the halogen.
(IV) More negative is the overall heat of the reaction ( $$\Delta H_r^o$$ of the halogenation of alkane) more reactive is the halogen.

A
I and II

B
I, II and III

C
II and IV

D
II, III and IV

Solution

There are two factors that determine the reactivity of the halogenation of alkanes.
(i) Lower the positive value of $$E_{act}$$ for the chain propagation step, more reactive is the halogen.
(ii) Higher the negative value of the overall heat of reaction ( $$\Delta H_r^o$$), more reactive is the halogen.
This explains high reactivity and explosive violence with which $$F_2$$ reacts with $$CH_4$$.
Question 9
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When $$n-$$hexane is heated with anhydrous $$AlCl_{3}$$ and $$HCl$$ gas, the major products obtained is:

A
$$1$$-chlorohexane

B
$$2$$-chlorohexane

C
$$3$$-chlorohexane

D
hex-$$3$$-ene

E
mixture of $$2$$-methylpentane and $$3$$-methylpentane

Solution

When $$n-hexane$$ is heated with anhydrous $$AlCl_{3}$$ and $$HCl$$ gas, at $$573\ K$$ under a pressure of about $$30-35$$ atmosphere, it isomerises to give a mixture of branched chain alkanes.
$$\underset{n-hexane}{CH_3(CH_2)_4CH_3}\xrightarrow[Heat]{AlCl_3.HCl}\underset{2-methyl\, pentane}{CH_3-\overset{CH_3}{\overset{|}{C}H}CH_2CH_2CH_3}+\underset{3-methyl \, pentane}{H_3CCH_2-\underset{CH_3}{\underset{|}{C}H}CH_2CH_3}$$
Question 10
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An alkane (molecular weight $$=86$$) on monochlorination gives only two monochloro derivatives. One isomer (X) of this alkane is obtained in catalytic reduction of alkene (Y) which gives four monochloro derivatives. Isomer (X) is?

A

B

C

D

Solution

Alkene $$\xrightarrow{\text{catalytic reduction}}\underset{isomers}{\underbrace{I+II}}$$

Mol. wt. $$=86\Rightarrow C_6H_{14}$$

$$Y(alkane)\xrightarrow [ reduction ]{ Catalytic } X(isomerofA)\xrightarrow{ Monochlorination }4$$ monochloro derivatives.

As $$X$$ is isomer of $$A$$ thus it should have molecular formula $$C_6H_{14}$$

This isomer of hexane gives four monochloro products on chlorination.