Hydrocarbons Test

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Hydrocarbons Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The major organic compound formed by the reaction of $$ 1, 1, 1-$$ trichloroethane with silver powder is:

A
$$2$$-butyne

B
$$2$$-butene

C
acetylene

D
ethene

Solution

$$H_3C-C\equiv C-CH_3\uparrow +6AgCl\downarrow$$
Product formed is 2-butyne
Question 2
1 / -0

The IUPAC name of the following compound is:

A
$$3$$-ethyl-$$4$$-methylhex-$$4$$-ene

B
$$4, 4$$-diethyl-$$3$$-methylbut-$$2$$-ene

C
$$4$$-Ethyl-$$3$$-methylhex-$$2$$-ene

D
$$4$$-methyl-$$3$$-ethylhex-$$4$$-ene

Solution

The IUPAC name of the following compound is 4−Ethyl−3-methylhex−2−ene.

Here the longest chain of carbon contains 6 carbon atoms and there is a double bond between carbon number 2 and 3. Thus parent chain is hexene.

Also, ethyl and methyl groups are attached to the carbon number 4 and 3 respectively.
Question 3
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The major product of the following addition reaction is :
$$H_3C - CH = CH_2 \xrightarrow{Cl_2/H_2O}$$

A
$$CH_3 - \underset{Cl}{\underset{|}{CH}} - \underset{OH}{\underset{|}{CH_2}}$$

B
$$H_3 C - \underset{OH}{\underset{|}{CH}} - \underset{Cl}{\underset{|}{CH_2}}$$

C

D

Solution

$$Cl^+$$ reacts as the electrophile so the C-O bond forms at the more stable cation center. In given structure, secondary carbocation (+ve charge at middle carbon) is more stable, hence$$-OH$$ group attached to 2nd carbon.
Question 4
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Directions For Questions

Treatment of benzene with $$CO/HCl$$ in the presence of anhydrous $$AlCl_3/CuCl$$ followed by reaction with $$Ac_2O/NaOAc$$ gives compound $$X$$ as the major product. Compound $$X$$ upon reaction with $$Br_2/Na_2CO_3$$, followed by heating at 473 K with moist $$KOH$$ furnishes $$Y$$ as the major product. The reaction of $$X$$ with $$H_2/PdC$$, followed by $$H_3PO_4$$ treatment givens $$Z$$ as the major product.

...view full instructions

The compound $$Y$$ is:

A

B

C

D

Solution

Hence, the answer is option $$C$$.
Question 5
1 / -0

Consider the given reaction.
What is the correct explanation for the reaction?

A
the product is formed due to nucleophilic substitution

B
the product is formed according to Saytzeff's rule

C
conjugated double bond product is formed due to higher stability because of resonance stabilisation

D
$${({CH}_{3})}_{3}{CO}^{-}$$ is better leaving group

Solution

$${({CH}_{3})}_{3}{CO}^{-}$$ is a better base than nucleophile. Hence elimination occurs. The product formed is resonance stabilised.
Hence the correct explanation is given by option C.
Question 6
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A fuel has the same knocking property as a mixture of 70% iso-octane (2, 2, 4-trimethylpentane) and 30% n-heptane by volume. The octane number of the fuel is:

A
100

B
70

C
50

D
40

Solution

Octane number is defined as the percentage of iso-octane (by volume) in a mixture of iso-octane and n-heptane which has the same anti-knocking properties as the fuel under consideration.
Thus, the octane number of the given fuel is 70 as it contains 70% iso-octane.
Question 7
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A compound X is obtained by the reaction of alkaline $$KMnO_3$$ with another compound Y followed by acidification. Compound X also reacts with compound Y in presence of few drops of $$H_2SO_4$$ to form a sweet smelling compound Z. The compound X, Y and Z are respectively.

A
Ethanol, Ethene, Ethanoic acid

B
Ethanoic acid, Ethanol, Ethylethanoate

C
Ethanoic Acid, Ethanal, Ethene

D
Ethanol, Ethanoic Acid, Sodium Ethanoate

Solution

Ethanoic acid is obtained by the reaction of alkaline KMnO4$$_4$$ with ethanol followed by acidification.

$$CH_3-CH_2-OH \xrightarrow [\text {ii. acidification}]{\text {i. alkaline KMnO}_4} CH_3-COOH$$

Ethanoic acid also reacts with ethanol in presence of few drops of H$$_2$$SO$$_4$$ to form a sweet smelling ethyl ethanoate.

$$CH_3-COOH+CH_3-CH_2-OH \xrightarrow {H_2SO_4} CH_3-COOCH_2-CH_3$$

The compound $$X$$, $$Y$$ and $$Z$$ are ethanoic acid, ethanol, ethylethanoate respectively.

Hence, option (B) is the correct answer.
Question 8
1 / -0

Bond length of C - C in benzene________.

A
$$1.34A^{0}$$

B
$$1.39A^{0}$$

C
$$1.54A^{0}$$

D
$$1.20A^{0}$$

Solution

In benzene all C-C bond are of equal length and bond length is $$1.39A^{0}$$ due to delocalisation of bonds.
Question 9
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The IUPAC name of $$H_{3}CH_{2}C-C(CH_{3})_{2}-C\equiv CH$$ is:

A
3,3 - dimethyl -1 pentyne

B
Propyne

C
Butyne

D
2-pentyne

Solution

The given compound is alkyne as it contains a triple bond. The parent chain contains 5 carbon.
As two methyl group are present as substituent group at carbon no.3 and a triple bond is present at carbon no. 1 so IUPAC name of the compound is 3,3 - dimethyl -1 pentyne.
Question 10
1 / -0

The name benzene was given by:

A
Kekule

B
Wohler

C
Mitscherlich

D
Davy

Solution

Eilhard Mitscherlich produced benzene by distillation of benzoic acid (from gum benzoin) and lime. He gave the compound the name benzene.